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Example 2 Let $n \in \mathbf{N}^{*}$, consider the region in the coordinate plane bounded by the inequalities $y>0, 0<x \leqslant n, y \leqslant$ $\sqrt{x}$. Try to express the number of integer points within this region using $n$ and $a(=[\sqrt{n}])$. | From property 6, we know that the number of integer points in the region sought in the problem is
$$T=\sum_{k=1}^{n}[\sqrt{k}]$$
We use the following method to calculate $T$.
For a proposition $P$, we introduce the notation $[P]$: if $P$ holds, then $[P]=1$; if $P$ does not hold, then $[P]=0$. Thus, we have
$$\begin{aligned}
\sum_{k=1}^{n}[\sqrt{k}] & =\sum_{k=1}^{n} \sum_{j=1}^{a}\left[j^{2} \leqslant k\right] \\
& =\sum_{j=1}^{a} \sum_{k=1}^{n}\left[j^{2} \leqslant k\right] \\
& =\sum_{j=1}^{a}\left(n-j^{2}+1\right) \\
& =(n+1) a-\frac{1}{6} a(a+1)(2 a+1)
\end{aligned}$$ | (n+1) a-\frac{1}{6} a(a+1)(2 a+1) | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
Example 4 Find all pairs of positive real numbers $(a, b)$ such that for any $n \in \mathbf{N}^{*}$, we have
$$[a[b n]]=n-1$$ | From (5), we know that for any $n \in \mathbf{N}^{*}$, we have
$$\begin{aligned}
n-1 & =[a[b n]] \leqslant a[b n] \leqslant a b n1$, then for the irrational number $b$, there exists $n \in \mathbf{N}^{*}$ such that $\{n b\}>\frac{1}{a}$, which contradicts (6). Therefore, $a<1$.
In summary, the positive real numbers $(a, b)$ should satisfy: $a b=1,0<a<1$, and both $a, b$ are irrational numbers.
Finally, let $(a, b)$ be a pair of positive real numbers satisfying the above conditions. Then, since $b$ is irrational, for any $n \in \mathbf{N}^{*}$, we have $0<\{b n\}<1$, thus,
$$-1<-a<-a\{b n\}<0$$
This shows that (6) holds. Therefore, $(a, b)$ satisfies (5).
Hence, the set of $(a, b)$ that satisfy the conditions is $\{(a, b) \mid a b=1,0<a<1, a, b$ are irrational numbers\}. | (a, b) \mid a b=1,0<a<1, a, b \text{ are irrational numbers} | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 7 Find the largest real number $c$ such that for any $n \in \mathbf{N}^{*}$, we have $\{\sqrt{2} n\}>\frac{c}{n}$.
| Notice that, for any $n \in \mathbf{N}^{*}$, let $k=[\sqrt{2} n]$, then
$$\begin{aligned}
\{\sqrt{2} n\} & =\sqrt{2} n-[\sqrt{2} n]=\sqrt{2} n-k=\frac{2 n^{2}-k^{2}}{\sqrt{2} n+k} \\
& \geqslant \frac{1}{\sqrt{2} n+k}>\frac{1}{\sqrt{2} n+\sqrt{2} n}=\frac{1}{2 \sqrt{2} n}
\end{aligned}$$
Here we use the fact that $\sqrt{2}$ is irrational, so $\{\sqrt{2} n\}>0$. Thus, $2 n^{2}-k^{2}>0$ (of course, $2 n^{2}-k^{2} \geqslant 1$), and $k<\sqrt{2} n$. Therefore, $c \geqslant \frac{1}{2 \sqrt{2}}$.
On the other hand, for all positive integer solutions $\left(k_{m}, n_{m}\right)$ of the Pell equation $x^{2}-2 y^{2}=-1$, we have
$$n_{m}\left\{\sqrt{2} n_{m}\right\}=\frac{n_{m}}{\sqrt{2} n_{m}+k_{m}}=\frac{n_{m}}{\sqrt{2} n_{m}+\sqrt{1+2 n_{m}^{2}}} \rightarrow \frac{2}{2 \sqrt{2}}$$
Here the fundamental solution is $\left(k_{1}, n_{1}\right)=(1,1)$, and
$$k_{m}+n_{m} \sqrt{2}=(1+\sqrt{2})^{2 m+1}$$
The above limit holds as $m \rightarrow+\infty$, so $c \leqslant \frac{1}{2 \sqrt{2}}$.
In conclusion, the maximum real number $c=\frac{1}{2 \sqrt{2}}$. | \frac{1}{2 \sqrt{2}} | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 10 Find all pairs of positive integers $(m, n)$ such that
$$[n \sqrt{2}]=[2+m \sqrt{2}] .$$ | Let $(m, n)$ be a pair of positive integers satisfying (14), then $n>m$.
Notice
$$\begin{aligned}
{[(m+3) \sqrt{2}] } & =[m \sqrt{2}+3 \sqrt{2}] \geqslant[m \sqrt{2}]+[3 \sqrt{2}] \\
& =[m \sqrt{2}+4]
\end{aligned}$$
Therefore, $n=m+1$ or $n=m+2$.
If $n=m+1$, since $\frac{1}{\sqrt{2}}+\frac{1}{2+\sqrt{2}}=1$, by the Beatty-Rayleigh theorem, we know that $\{[\sqrt{2} m]\}$ and $\{[(2+\sqrt{2}) h]\}$ are complementary sequences, and at this time
$$[n \sqrt{2}]=[(m+1) \sqrt{2}]=[2+m \sqrt{2}]=2+[m \sqrt{2}],$$
Thus, there is exactly one integer $[(2+\sqrt{2}) h]$ between $[m \sqrt{2}]$ and $[(m+1) \sqrt{2}]$, i.e.,
$$m \sqrt{2}<(2+\sqrt{2}) h<(m+1) \sqrt{2}$$
Hence, $m<(\sqrt{2}+1) h<m+1$, i.e.,
$$m=[(\sqrt{2}+1) h]$$
Reversing the above process, we know that
$$(m, n)=([(\sqrt{2}+1) h],[(\sqrt{2}+1) h]+1)$$
is a solution to (14). Therefore, the positive integer solutions to (14) in this case are
If $n=m+2$, similarly, we know $[(m+2) \sqrt{2}]=[2+m \sqrt{2}]=2+[m \sqrt{2}]$, so $[m \sqrt{2}],[(m+1) \sqrt{2}]$, and $[(m+2) \sqrt{2}]$ are three consecutive positive integers. Therefore, there exists $h \in \mathbf{N}^{*}$, such that
$$[(2+\sqrt{2})(h+1)]-[(2+\sqrt{2}) h] \geqslant 4$$
However,
$$[(2+\sqrt{2})(h+1)]-[(2+\sqrt{2}) h] \leqslant 1+[2+\sqrt{2}]=4$$
Thus, there exists $h \in \mathbf{N}^{*}$, such that
$$\left\{\begin{array}{l}
{[(m-1) \sqrt{2}]<[(2+\sqrt{2}) h]<[m \sqrt{2}]} \\
{[(2+\sqrt{2})(h+1)]=[(2+\sqrt{2}) h]+4}
\end{array}\right.$$
From (15), we know $m=[(1+\sqrt{2}) h]+1$, and (16) is $[\sqrt{2}(h+1)]=[\sqrt{2} h+2]$. Therefore, using the discussion of the first case ($h$ is equivalent to $m$), we know there exists $k \in \mathbf{N}^{*}$, such that $h=[(1+\sqrt{2}) k]$. Reversing the above process, we know that the positive integer solutions to (14) in this case are
$$\left\{\begin{array}{l}
m=[(1+\sqrt{2}) h]+1=3 k+2[\sqrt{2} k], \\
n=3 k+2+2[\sqrt{2} k]
\end{array}\right.$$
In the last step, we use
$$\begin{aligned}
m & =[(1+\sqrt{2})[(1+\sqrt{2}) k]]+1 \\
& =[(3+2 \sqrt{2}) k+1-(1+\sqrt{2}) \alpha] \\
& =[3 k+2[\sqrt{2} k]+1-(\sqrt{2}-1) \alpha] \\
& =3 k+2[\sqrt{2} k]
\end{aligned}$$
Here $\alpha=\{(1+\sqrt{2}) k\}=\{\sqrt{2} k\}$. | (m, n)=([(\sqrt{2}+1) h],[(\sqrt{2}+1) h]+1) \text{ or } \left\{\begin{array}{l} m=3 k+2[\sqrt{2} k], \\ n=3 k+2+2[\sqrt{2} k] \end{array}\right.} | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 5 Find all positive integers $n$ such that the indeterminate equation
$$n=x_{1}^{2}+x_{2}^{2}+\cdots+x_{5}^{2}$$
has a unique solution in integers satisfying $0 \leqslant x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{5}$. | Using Lagrange's four-square theorem, we know that when $n \geqslant 17$, there exists the following representation:
$$\begin{aligned}
n & =x_{0}^{2}+y_{0}^{2}+z_{0}^{2}+w_{0}^{2}, \\
n-1^{2} & =x_{1}^{2}+y_{1}^{2}+z_{1}^{2}+w_{1}^{2}, \\
n-2^{2} & =x_{2}^{2}+y_{2}^{2}+z_{2}^{2}+w_{2}^{2}, \\
n-3^{2} & =x_{3}^{2}+y_{3}^{2}+z_{3}^{2}+w_{3}^{2}, \\
n-4^{2} & =x_{4}^{2}+y_{4}^{2}+z_{4}^{2}+w_{4}^{2},
\end{aligned}$$
where $0 \leqslant x_{i} \leqslant y_{i} \leqslant z_{i} \leqslant w_{i}$ are all integers, $i=1,2,3,4$.
Now, if $n \neq 1^{2}+2^{2}+3^{2}+4^{2}$, then
$$\{1,2,3,4\} \neq\left\{x_{0}, y_{0}, z_{0}, w_{0}\right\}$$
Let $k \in\{1,2,3,4\} \backslash\left\{x_{0}, y_{0}, z_{0}, w_{0}\right\}$, then by the previous conclusion, we know that $n$ has the following two different representations:
$$n=x_{0}^{2}+y_{0}^{2}+z_{0}^{2}+w_{0}^{2}=k^{2}+x_{k}^{2}+y_{k}^{2}+z_{k}^{2}+w_{k}^{2}$$
Since $30=1^{2}+2^{2}+3^{2}+4^{2}=1^{2}+2^{2}+5^{2}$, it follows that the required $n \leqslant 16$.
By verifying $n=1,2, \cdots, 16$ one by one, we find that $1,2,3,6,7$ and 15 meet the requirements, with their unique representations as follows:
$$\begin{array}{l}
1=1^{2}, 2=1^{2}+1^{2}, 3=1^{2}+1^{2}+1^{2}, 6=1^{2}+1^{2}+2^{2} \\
7=1^{2}+1^{2}+1^{2}+2^{2}, 15=1^{2}+1^{2}+2^{2}+3^{2}
\end{array}$$
While the other numbers have at least two representations:
$$\begin{aligned}
4 & =2^{2}=1^{2}+1^{2}+1^{2}+1^{2} \\
5 & =1^{2}+2^{2}=1^{2}+1^{2}+1^{2}+1^{2}+1^{2} \\
8 & =2^{2}+2^{2}=1^{2}+1^{2}+1^{2}+1^{2}+2^{2}, \\
9 & =3^{2}=1^{2}+2^{2}+2^{2} \\
10 & =1^{2}+3^{2}=1^{2}+1^{2}+2^{2}+2^{2} \\
11 & =1^{2}+1^{2}+3^{2}=1^{2}+1^{2}+1^{2}+2^{2}+2^{2} \\
12 & =1^{2}+1^{2}+1^{2}+3^{2}=2^{2}+2^{2}+2^{2} \\
13 & =1^{2}+1^{2}+1^{2}+1^{2}+3^{2}=2^{2}+3^{2} \\
14 & =1^{2}+2^{2}+3^{2}=1^{2}+1^{2}+2^{2}+2^{2}+2^{2} \\
16 & =4^{2}=2^{2}+2^{2}+2^{2}+2^{2}
\end{aligned}$$
In summary, the numbers that meet the conditions are $n=1,2,3,6,7$ or 15. | 1,2,3,6,7,15 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
4. (1) Find all positive integers $n$, such that there exist $a, b \in \mathbf{N}^{*}$, satisfying: $[a, b]=n!$, $(a, b)=1998 ;$
(2) Under the condition that (1) holds, to ensure that the number of pairs of positive integers $(a, b)$ with $a \leqslant b$ does not exceed 1998, what condition should $n$ satisfy? | 4. (1) From $(a, b)=2 \times 3^{3} \times 37$, we know $37 \mid[a, b]$, which means $37 \mid n$, so $n \geqslant 37$. When $n \geqslant 37$, $1998 \mid n!$, at this time, taking $a=1998, b=n!$ will suffice. Therefore, the positive integers $n$ that satisfy the condition are all positive integers not less than 37.
(2) Let $a=1998 x, b=1998 y$, then $(x, y)=1$. When $n \geqslant 37$, we have $x y=2^{33} \times 3^{14} \times 5^{8} \times 7^{5} \times 11^{3} \times 13^{2} \times 17^{2} \times 19 \times 23 \times 29 \times 31 \times\left(\frac{n!}{37!}\right)$.
Thus, when $37 \leqslant n \leqslant 40$, the number of pairs $(x, y)$ is $\frac{1}{2} \times 2^{11}=1024$ pairs, and when $n \geqslant 41$, there are at least $\frac{1}{2} \times 2^{12}=2048$ pairs, hence $37 \leqslant n \leqslant 40$. | 37 \leqslant n \leqslant 40 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 5 Let the set of all integer points (points with integer coordinates) in the plane be denoted as $S$. It is known that for any $n$ points $A_{1}, A_{2}, \cdots, A_{n}$ in $S$, there exists another point $P$ in $S$ such that the segments $A_{i} P(i=1,2, \cdots, n)$ do not contain any points from $S$ internally. Find the maximum possible value of $n$.
| The maximum value of the required $n$ is 3.
On one hand, when $n \geqslant 4$, take 4 points $A(0,0), B(1,0), C(0,1)$, $D(1,1)$ in $S$ (the remaining $n-4$ points are chosen arbitrarily), then for any other point $P$ in $S$, analyzing the parity of the coordinates of $P$, there are only 4 cases: (odd, odd), (odd, even), (even, odd), and (even, even). Therefore, among $A, B, C, D$, there is one point such that the midpoint of the line segment connecting it to $P$ is a point in $S$, hence $n \leqslant 3$.
On the other hand, we prove: for any three points $A, B, C$ in $S$, there exists another point $P$ in $S$ such that the interiors of $A P, B P, C P$ do not contain any integer points.
Note that for integer points $X\left(x_{1}, y_{1}\right)$ and $Y\left(x_{2}, y_{2}\right)$ on the plane, the necessary and sufficient condition for the line segment $X Y$ to have no integer points in its interior is $\left(x_{1}-x_{2}, y_{1}-y_{2}\right)=1$. Therefore, we can assume $A\left(a_{1}, a_{2}\right), B\left(b_{1}, b_{2}\right), C(0,0)$.
To find a point $P(x, y)$ such that $A P, B P, C P$ have no integer points in their interiors, $x, y$ need to satisfy the following conditions:
$$\left\{\begin{array}{l}
x \equiv 1(\bmod y) \\
x-a_{1} \equiv 1\left(\bmod y-a_{2}\right) \\
x-b_{1} \equiv 1\left(\bmod y-b_{2}\right)
\end{array}\right.$$
Using the Chinese Remainder Theorem, if there exists $y \in \mathbf{Z}$ such that $y, y-a_{2}, y-b_{2}$ are pairwise coprime, then an $x$ satisfying (2) exists, thus finding a point $P$ that meets the requirements.
Take $y=m\left[a_{2}, b_{2}\right]+1, m \in \mathbf{Z}, m$ to be determined, then $\left(y, y-a_{2}\right)=\left(y, a_{2}\right)=\left(1, a_{2}\right)=1$, similarly $\left(y, y-b_{2}\right)=1$. Therefore, it is only necessary to take $m$ such that
$$\left(y-a_{2}, y-b_{2}\right)=1$$
which is equivalent to
$$\left(y-a_{2}, a_{2}-b_{2}\right)=1$$
To make (3) hold, set $a_{2}=a_{2}^{\prime} d, b_{2}=b_{2}^{\prime} d$, where $a_{2}^{\prime}, b_{2}^{\prime}, d \in \mathbf{Z}$, and $\left(a_{2}^{\prime}, b_{2}^{\prime}\right)=1$, then $y=m d a_{2}^{\prime} b_{2}^{\prime}+1$, and thus $y-a_{2}=d a_{2}^{\prime}\left(m b_{2}^{\prime}-1\right)+1$. Noting that when $\left(a_{2}^{\prime}, b_{2}^{\prime}\right)=1$, $\left(b_{2}^{\prime}, a_{2}^{\prime}-b_{2}^{\prime}\right)=1$, so there exists $m \in \mathbf{Z}$ such that $m b_{2}^{\prime} \equiv 1\left(\bmod a_{2}^{\prime}-b_{2}^{\prime}\right)$, such a determined $m$ satisfies: $d\left(a_{2}^{\prime}-b_{2}^{\prime}\right) \mid d a_{2}^{\prime}\left(m b_{2}^{\prime}-1\right)$, at this time
$$y-a_{2} \equiv 1\left(\bmod a_{2}-b_{2}\right),$$
Therefore, (3) holds.
In summary, the maximum value of $n$ that satisfies the conditions is 3. | 3 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
2. In decimal notation, how many $m \in\{1,2, \cdots, 2009\}$ are there such that there exists $n \in \mathbf{N}^{*}$, satisfying: $S\left(n^{2}\right)=m$? Here $S(x)$ denotes the sum of the digits of the positive integer $x$. | 2. Notice that, perfect squares $\equiv 0,1,4,7(\bmod 9)$, and for $k \in \mathbf{N}^{*}$, let $n=$ $\underbrace{11 \cdots 1} \underbrace{22 \cdots 25}$, then $S(n)=3 k+4$, and
$k-1$ 个 $\underbrace{2+0}_{k \text { 个 }}$
$$\begin{aligned}
n & =\frac{10^{k-1}-1}{9} \times 10^{k+1}+\frac{2 \times\left(10^{k}-1\right)}{9} \times 10+5 \\
& =\frac{1}{9}\left(10^{2 k}-10^{k+1}+2 \times 10^{k+1}-20+45\right) \\
& =\frac{1}{9}\left(10^{2 k}+10 \times 10^{k}+25\right) \\
& =\frac{1}{9}\left(10^{k}+5\right)^{2}=(\underbrace{33 \cdots 35}_{k-1 \uparrow})^{2}
\end{aligned}$$
That is, $n$ is a perfect square.
By setting $k=3 t+2,3 t, 3 t+1, t \in \mathbf{N}$, and combining $S\left(1^{2}\right)=1$, we can see that for any $m \in\{0,1,2, \cdots, 2009\}$, if $m \equiv 1,4$ or $7(\bmod 9)$, then there exists $n \in \mathbf{N}^{*}$, such that $S\left(n^{2}\right)=m$.
Furthermore, for $k \in \mathbf{N}^{*}, n=\underbrace{99 \cdots 9}_{k \text { 个 }}$, we have
$$\begin{aligned}
n^{2} & =\left(10^{k}-1\right)^{2}=10^{2 k}-2 \times 10^{k}+1 \\
& =\underbrace{99 \cdots 9}_{k-1 \uparrow} \underbrace{00 \cdots 01}_{k-1 \uparrow},
\end{aligned}$$
Thus, $S\left(n^{2}\right)=9 k$. Therefore, when $m \equiv 0(\bmod 9)$, there also exists $n \in \mathbf{N}^{*}$, such that $S\left(n^{2}\right)=$ $m$.
The above discussion shows that the number of numbers that meet the condition is $\frac{4}{9} \times 2007+1=893$. | 893 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
6. Let \( x=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{100000}} \). Find the value of \([x]\). | 6. Notice that, when $k \geqslant 2$, we have
$$\begin{array}{l}
\frac{1}{\sqrt{k}}=\frac{2}{2 \sqrt{k}}\frac{2}{\sqrt{k}+\sqrt{k+1}}=2(\sqrt{k+1}-\sqrt{k}) .
\end{array}$$
Therefore,
$$\begin{aligned}
x & =1+\sum_{k=2}^{10^{6}} \frac{1}{\sqrt{k}}2 \sum_{k=1}^{10^{6}}(\sqrt{k+1}-\sqrt{k}) \\
& =2\left(\sqrt{10^{6}+1}-1\right)>1998 .
\end{aligned}$$
From this, we know that $[x]=1998$. | 1998 | Calculus | math-word-problem | Yes | Yes | number_theory | false |
9. Find all positive integers $a, b$, such that
$$\left[\frac{a^{2}}{b}\right]+\left[\frac{b^{2}}{a}\right]=\left[\frac{a^{2}+b^{2}}{a b}\right]+a b$$ | 9. When $a=b$, the left side $=a+b$, and the right side $=ab+2$. Also, $ab+2-a-b=(a-1)(b-1)+1>0$, so the left and right sides are not equal, hence $a \neq b$.
Assume without loss of generality that $ab>a$, thus $a^2-b2a$, which is a contradiction. Therefore, $b<a^2+2$.
In summary, $b=a^2+1$. Direct verification shows that $b=a^2+1$ meets the requirements. Therefore, the solutions $(a, b)=\left(m, m^2+1\right)$ or $\left(m^2+1, m\right), m \in \mathbf{N}^{*}$. | (a, b)=\left(m, m^2+1\right) \text{ or } \left(m^2+1, m\right), m \in \mathbf{N}^{*} | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
6 . Find all positive integers $a, b$ such that
$$(a, b)+9[a, b]+9(a+b)=7 a b .$$ | 6. Let $(a, b)=d, a=d x, b=d y$, then $(x, y)=1$. Substituting into the condition, we get
$$1+9 x y+9(x+y)=7 d x y$$
Thus, $\square$
$$9<7 d=9+9\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{1}{x y} \leqslant 9+9 \times 2+1=28 \text {. }$$
From this, we know that $d=2,3$ or 4. Substituting these values into the indeterminate equation for $x, y$, we can find that $(a, b)=(4,4),(4,38)$ or $(38,4)$. | (a, b)=(4,4),(4,38),(38,4) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
39. The sequence $\left\{a_{n}\right\}$ is defined as follows:
$$a_{1}=0, a_{n}=a\left[\frac{n}{2}\right]+(-1)^{\frac{n(n+1)}{2}}, n=2,3, \cdots$$
For each $k \in \mathbf{N}$, find the number of indices $n$ that satisfy $2^{k} \leqslant n<2^{k+1}$ and $a_{n}=0$. | For each $n \in \mathbf{N}^{*}$, let $x_{n}$ be the number of times the adjacent pairs 00 and 11 appear in the binary representation of $n$, and $y_{n}$ be the number of times the adjacent pairs 01 and 10 appear. We prove:
$$a_{n}=x_{n}-y_{n}$$
In fact, when $n=1$, $x_{1}=y_{1}=0$, so (7) holds for $n=1$.
Now assume (7) holds for indices $1,2, \cdots, n-1(n \geqslant 2)$, and consider the case for $n$.
If the last two digits of the binary representation of $n$ are 00 or 11, then $n \equiv 0$ or $3(\bmod 4)$, in this case, $a_{n}=a\left[\frac{n}{2}\right]+1$, and
$$x_{n}=x\left[\frac{n}{2}\right]+1, y_{n}=y\left[\frac{n}{2}\right]$$
Therefore, (7) holds for $n$.
If the last two digits of the binary representation of $n$ are 01 or 10, then $n \equiv 1$ or $2(\bmod 4)$, in this case, $a_{n}=a\left[\frac{n}{2}\right]-1$, and
$$x_{n}=x\left[\frac{n}{2}\right], y_{n}=y\left[\frac{n}{2}\right]+1,$$
Therefore, (7) holds for $n$.
In summary, (7) holds for all $n \in \mathbf{N}^{*}$.
Now we need to calculate the number of $n$ such that $2^{k} \leqslant n<2^{k+1}$ and $x_{n}$ equals $y_{n}$ in the binary representation.
In this case, $n$ in binary representation is a $(k+1)$-bit number, denoted as $B_{n}$. When $k \geqslant 1$, subtract the next bit from each bit of $B_{n}$ from left to right, and take the absolute value, to obtain a $k$-element 0,1 array $C_{n}$ (for example: if $B_{n}=(1101)_{2}$, then $\left.C_{n}=(011)\right)$. Note that, each adjacent pair 00 and 11 in $B_{n}$ becomes a 0 in $C_{n}$, and 01 and 10 become a 1 in $C_{n}$. Therefore, if $x_{n}=y_{n}$, then the number of 1s in $C_{n}$ is the same as the number of 0s. Conversely, for a $k$-element 0,1 array $C_{n}=\left(c_{1} c_{2} \cdots c_{k}\right)$, then in $\bmod 2$ sense, compute the following sum:
$$b_{1}=1+c_{1}, b_{2}=b_{1}+c_{2}, \cdots b_{k}=b_{k-1}+b_{k}$$
Here $b_{0}=1$, then $B_{n}=\left(b_{0} b_{1} \cdots b_{k}\right)_{2}$ is a binary representation of a number $n$ satisfying $2^{k} \leqslant n<2^{k+1}$. This indicates: $B_{n}$ and $C_{n}$ form a one-to-one correspondence.
Therefore, the answer to the original problem is equal to the number of $k$-element 0,1 arrays where the number of 0s and 1s are the same. Thus, when $k$ is odd, the answer is 0; when $k$ is even, the answer is $\mathrm{C}_{k}^{\frac{k}{2}}$ (note that $\mathrm{C}_{0}^{0}=1$). | \mathrm{C}_{k}^{\frac{k}{2}} | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
41. Let $a \bmod n$ denote the remainder of $a$ divided by $n$ (i.e., $\left.a-\left[\frac{a}{n}\right] \cdot n\right)$. For given $m, n \in \mathbf{N}^{*}$, find $\max _{0 \leqslant a_{1}, a_{2}, \cdots, a_{m}<n} \min _{0 \leqslant k<n} \sum_{j=1}^{m}\left(\left(a_{j}+k\right) \bmod n\right)$, where $a_{1}$, $a_{2}, \cdots, a_{m} \in \mathbf{Z}$ | 41. The answer is: $\frac{1}{2}((m-1)(n-1)+(m, n)-1)$.
Because
$$a_{j}+k(\bmod n)=\left\{\begin{array}{l}
a_{j}+k, a_{j}<n-k \\
a_{j}+k-n, a_{j} \geqslant n-k
\end{array}\right.$$
Hence
$$\begin{aligned}
S & =\sum_{j=1}^{m}\left(\left(a_{j}+k\right) \bmod n\right) \\
& =\sum_{j=1}^{m} a_{j}+m k-\left(\mu_{n-1}+\mu_{n-2}+\cdots+\mu_{n-k}\right) \cdot n
\end{aligned}$$
where $\mu_{i}$ is the number of times $i$ appears in $a_{1}, a_{2}, \cdots, a_{n}$.
From (9), the minimum value of $S$ is achieved when $k=0$ if and only if
$$\mu_{n-1}+\mu_{n-2}+\cdots+\mu_{n-k} \leqslant\left[\frac{m k}{n}\right], 1 \leqslant k \leqslant n-1$$
Notice that, if the minimum value of $S$ is achieved at $0 \leqslant l \leqslant n-1$, let $b_{j} \equiv a_{i}+l(\bmod n)$, then the minimum value of $S$ for $b_{1}, b_{2}, \cdots, b_{n}$ (all $\in\{0,1, \cdots, n-1\}$) is also achieved at $k=0$. Therefore, we can assume that the minimum value of $S$ is achieved at $k=0$. In this case, (10) holds, and $S_{\min }=\sum_{j=1}^{m} a_{j}$.
Next, we find the maximum value of $\sum_{j=1}^{m} a_{j}$ under condition (10). Since
$$\sum_{j=1}^{m} a_{j}=\sum_{i=1}^{n-1}(n-i) \mu_{n-i}$$
it is equivalent to finding the maximum value of $T=\sum_{i=1}^{n-1}(n-i) \mu_{n-i}$ under conditions (10) and $\sum_{i=1}^{n} \mu_{n-i}=m$.
For $1 \leqslant k \leqslant n$, let
$$\bar{\mu}_{n-k}=\left[\frac{m k}{n}\right]-\left[\frac{m(k-1)}{n}\right]$$
Then $\sum_{i=1}^{n} \bar{\mu}_{n-i}=m$, and the sequence $\left(\bar{\mu}_{0}, \bar{\mu}_{1}, \cdots, \bar{\mu}_{n-1}\right)$ satisfies (10), and
$$\sum_{i=1}^{k} \mu_{n-i} \leqslant\left[\frac{m k}{n}\right]=\sum_{i=1}^{k} \bar{\mu}_{n-i}$$
Thus
$$\begin{aligned}
T & =\sum_{i=1}^{n-1}(n-i) \mu_{n-i}=\sum_{k=1}^{n-1} \sum_{i=1}^{k} \mu_{n-k} \leqslant \sum_{k=1}^{n-1} \sum_{i=1}^{k} \bar{\mu}_{n-i} \\
& =\sum_{i=1}^{n-1}(n-i) \bar{\mu}_{n-i}
\end{aligned}$$
Therefore
$$T_{\max }=\sum_{i=1}^{n-1}(n-i) \bar{\mu}_{n-i}=\sum_{k=1}^{n-1}\left[\frac{m k}{n}\right]=\frac{1}{2}((m-1)(n-1)+(m, n)-1)$$
The last equation uses the result from Example 1 in Section 5.2. | \frac{1}{2}((m-1)(n-1)+(m, n)-1) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
45. Find the largest real number $\alpha$, such that there exists an infinite sequence of positive integers $\left\{a_{n}\right\}_{n=1}^{+\infty}$, satisfying:
(1) For any $n \in \mathbf{N}^{*}$, we have $a_{n}>2008^{n}$;
(2) For any $n \in \mathbf{N}^{*}, n \geqslant 2$, the number $a_{n}^{\alpha}$ does not exceed the greatest common divisor of all numbers in the set $\left\{a_{i}+a_{j} \mid i+j=n, i, j \in \mathbf{N}^{*}\right\}$. | 45. First, prove $\alpha \leqslant \frac{1}{2}$.
In fact, let $\left\{a_{n}\right\}$ be a sequence of positive integers satisfying conditions (1) and (2). We have: for any $0 < \varepsilon < 1$, there are infinitely many positive integers $n$ such that
$$a_{n}^{2-\varepsilon} \leqslant a_{2 n}.$$
If (11) does not hold, then there exists a positive integer $n$ such that
$$a_{n}^{2-\varepsilon} > a_{2 n}.$$
Since $a_{n} \geqslant 2008^n$, we have
$$a_{2 n} \geqslant 2008^{2 n}.$$
Thus,
$$a_{n}^{2-\varepsilon} > 2008^{2 n},$$
which implies
$$a_{n} > 2008^{\frac{2 n}{2-\varepsilon}}.$$
Since $a_{n} \geqslant 2008^n$, we have
$$2008^n > 2008^{\frac{2 n}{2-\varepsilon}},$$
which simplifies to
$$n > \frac{2 n}{2-\varepsilon},$$
or
$$2-\varepsilon > 2,$$
which is a contradiction. Therefore, (11) holds.
Now, let $n$ be a positive integer satisfying (11). Then,
$$a_{n}^{2-\varepsilon} \leqslant a_{2 n},$$
so
$$a_{n}^{(2-\varepsilon) \alpha} \leqslant a_{2 n}^{\alpha} \leqslant \operatorname{gcd}\left\{a_{i}+a_{j} \mid i+j=2 n, i, j \in \mathbf{N}^{*}\right\} \leqslant 2 a_{n},$$
which implies
$$a_{n}^{(2-\varepsilon) \alpha - 1} \leqslant 2.$$
Thus,
$$2008^{n((2-\varepsilon) \alpha - 1)} \leqslant 2.$$
Since there are infinitely many such $n$, we have
$$(2-\varepsilon) \alpha - 1 \leqslant 0,$$
or
$$\alpha \leqslant \frac{1}{2-\varepsilon}.$$
By the arbitrariness of $\varepsilon$, we conclude that
$$\alpha \leqslant \frac{1}{2}.$$
Next, prove that $\alpha = \frac{1}{2}$ is achievable by constructing a sequence $\left\{a_{n}\right\}$ that satisfies the conditions. Use the Fibonacci sequence $\left\{F_{n}\right\}$, which satisfies: $F_{1} = F_{2} = 1$, $F_{n+2} = F_{n+1} + F_{n}$, $n = 0, 1, 2, \cdots$. Then,
$$F_{n} = \frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^{n} - \left(\frac{1-\sqrt{5}}{2}\right)^{n}\right),$$
and by its recurrence relation, we have: when $m > n$,
$$F_{m} = F_{m-n} F_{n+1} + F_{m-n-1} F_{n}.$$
Notice that $\frac{1+\sqrt{5}}{2} > 1$ and $-1 < \frac{1-\sqrt{5}}{2} < 0$. Therefore, for sufficiently large $n$,
$$F_{n} > 2008^n.$$
Now define the sequence $\left\{a_{n}\right\}$ as $a_{n} = 3 F_{2m}$, $n = 1, 2, \cdots$. Condition (1) is clearly satisfied. For condition (2), we have
$$\begin{array}{l}
F_{2i} = F_{t(i+j)} F_{t(i-j)+1} + F_{t(i+j)-1} F_{t(i-j)}, \\
F_{2tj} = F_{t(i+j)} F_{t(j-i)-1} + F_{t(i+j)+1} F_{t(j-i)},
\end{array}$$
where we extend $\left\{F_{n}\right\}_{n=1}^{+\infty}$ to negative indices using the recurrence relation $F_{n+2} = F_{n+1} + F_{n}$. This gives us $F_{-2m} = -F_{2m}$ and $F_{-(2m-1)} = F_{2m-1}$, $m = 1, 2, \cdots$. Since $t$ is even, we have
$$\begin{aligned}
F_{2t} + F_{2tj} & = 2 F_{t(i+j)} F_{t(i-j)+1} + (F_{t(i+j)+1} - F_{t(i+j)-1}) F_{t(j-i)} \\
& = F_{t(i+j)} (2 F_{t(i-j)+1} + F_{t(j-i)}).
\end{aligned}$$
Thus, $F_{t(i+j)} \mid (F_{2i} + F_{2tj})$, and therefore $F_{nt} \mid (F_{2i} + F_{2tj})$ when $i + j = n$. This implies
$$\operatorname{gcd}\left\{a_{i} + a_{j} \mid i + j = n\right\} \geqslant 3 F_{m}.$$
Now,
$$a_{n} = 3 F_{20} = 3 F_{on} (F_{or+1} + F_{m-1}) \leqslant 9 F_{m}^2 \leqslant \left(\operatorname{gcd}\left\{a_{i} + a_{j} \mid i + j = n\right\}\right)^2,$$
so
$$a_{n}^{\frac{1}{2}} \leqslant \operatorname{gcd}\left\{a_{i} + a_{j} \mid i + j = n\right\}.$$
In conclusion, the maximum real number $\alpha = \frac{1}{2}$. | \frac{1}{2} | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
47. Find all positive integers $p(\geqslant 2)$, such that there exists $k \in \mathbf{N}^{*}$, satisfying: the number $4 k^{2}$ in base $p$ representation consists only of the digit 1. | 47. It can be proven that the positive integers $p \in \{3,7\}$ satisfy the condition.
When $p=3$, taking $k=1$ is sufficient. When $p=7$, taking $k=10$, we have $4 k^{2}=400=(1111)_7$. Therefore, when $p \in \{3,7\}$, the condition is satisfied.
On the other hand, let $p$ be a positive integer that satisfies the condition. The problem is equivalent to finding all $p \in \mathbf{N}^{*}$ such that the equation
$$4 k^{2}=(\underbrace{11 \cdots 1}_{n \uparrow})_{p}=1+p+\cdots+p^{n-1}=\frac{p^{n}-1}{p-1}$$
has positive integer solutions $(k, n)$.
From (12), we know that $p$ is odd (otherwise, $1+p+\cdots+p^{n-1}$ is odd, and (12) cannot hold). Furthermore, $1+p+\cdots+p^{n-1}$ is the sum of $n$ odd numbers, so $n$ is even. Let $n=2^{a} \cdot q, \alpha \in \mathbf{N}^{*}, q$ be an odd number, then from (12) we have
$$\begin{aligned}
4 k^{2} & =\frac{p^{n}-1}{p-1} \\
& =\left(\frac{p^{q}-1}{p-1}\right)\left(p^{q}+1\right)\left(p^{2 q}+1\right) \cdot \cdots \cdot\left(p^{2^{\alpha-1} q}+1\right)
\end{aligned}$$
Notice that $\frac{p^{q}-1}{p-1}=1+p+\cdots+p^{q-1}$ is the sum of $q$ odd numbers, so $\frac{p^{q}-1}{p-1}$ is odd. For any $0 \leqslant \beta \leqslant \alpha-1$, we have
$$\left(p^{q}-1, p^{2^{\beta} q}+1\right)=\left(p^{q}-1,\left(p^{q}\right)^{2^{\beta}}-1+2\right)=\left(p^{q}-1,2\right)=2,$$
hence $\left(\frac{p^{q}-1}{p-1}, p^{2^{\beta} q}+1\right)=1$.
Furthermore, for $1 \leqslant \beta \leqslant \alpha-1$ (if such $\beta$ exists), we have
$$\left(p^{q}+1, p^{2^{\beta} q}+1\right)=\left(p^{q}+1,(-1)^{2^{\beta}}+1\right)=\left(p^{q}+1,2\right)=2 .$$
Using these conclusions and (13), we know there exists $a \in \mathbf{N}^{*},(a \mid k)$, such that $p^{q}+1=4 a^{2}$ or $p^{q}+1=2 a^{2}$.
If the former holds, then $p^{q}=4 a^{2}-1=(2 a-1)(2 a+1)$, and $(2 a-1,2 a+1)=1$, so $2 a-1$ and $2 a+1$ are both $q$-th powers. Let $2 a-1=x^{q}, 2 a+1=y^{q}, x y=p$, $(x, y)=1$, then $x<1$. Considering $p^{2 q}+1$, note that $p^{q}+1$, $p^{2 q}+1, \cdots, p^{2^{\alpha-1} q}+1$ have pairwise greatest common divisors of 2. Combining (13) and $p^{2 q}+1 \equiv 2(\bmod 4)$, we know there exists $b \in \mathbf{N}^{*}$ such that $p^{2 q}+1=2 b^{2}$. Combining $p^{q}+1=2 a^{2}$, we get
$$2 a^{4}-2 a^{2}+1=b^{2}$$
We now solve (14) for positive integer solutions.
$$(14) \Leftrightarrow\left(a^{2}-1\right)^{2}+\left(a^{2}\right)^{2}=b^{2}$$
Since $\left(a^{2}-1, a^{2}\right)=1$, by the structure of primitive Pythagorean triples, we have
$$\text { (I) }\left\{\begin{array} { l }
{ a ^ { 2 } - 1 = c ^ { 2 } - d ^ { 2 } , } \\
{ a ^ { 2 } = 2 c d , } \\
{ b = c ^ { 2 } + d ^ { 2 } , }
\end{array} \text { or (II) } \left\{\begin{array}{l}
a^{2}-1=2 c d, \\
a^{2}=c^{2}-d^{2}, \\
b=c^{2}+d^{2},
\end{array}\right.\right.$$
where $c, d \in \mathbf{N}^{*}, c, d$ are of different parities, and $(c, d)=1$.
For case (I), $a$ is even. Combining $a^{2}-1=c^{2}-d^{2}$, taking $\bmod 4$ on both sides, we find $c$ is even and $d$ is odd. From $a^{2}=2 c d$ and $(c, d)=1$, we know there exist $x, y \in \mathbf{N}^{*}$ such that $(c, d, a)=\left(2 x^{2}, y^{2}, 2 x y\right)$, where $(x, y)=1$ and $y$ is odd. Substituting back into $a^{2}-1=c^{2}-d^{2}$, we get
$$4 x^{2} y^{2}-1=4 x^{4}-y^{4} \Leftrightarrow\left(y^{2}+2 x^{2}\right)^{2}-8 x^{4}=1$$
The positive integer solutions to (15) are only $\left(y^{2}+2 x^{2}, x\right)=(3,1)$ (see Exercise 4, Question 25) $\Rightarrow$ $x=y=1 \Rightarrow a=2, b=5 \Rightarrow p=7, q=1$.
For case (II), $a$ is odd. Using $a^{2}=c^{2}-d^{2}$, i.e., $a^{2}+d^{2}=c^{2}$ and $(c, d)=1$, we know there exist $x, y \in \mathbf{N}^{*}, (x, y)=1$, $x, y$ of different parities, such that $(a, d, c)=\left(x^{2}-y^{2}, 2 x y, x^{2}+y^{2}\right)$. Substituting into $a^{2}-1=2 c d$, we get
$$\begin{array}{c}
x^{4}-2 x^{2} y^{2}+y^{4}-1=4 x y\left(x^{2}+y^{2}\right), \\
(x-y)^{4}-8(x y)^{2}=1,
\end{array}$$
where $x>y$, so we need (16) to have positive integer solutions. However, (16) has no positive integer solutions (similar to the proof in Exercise 4, Question 25), so there are no positive integers $a, b$ that satisfy (II). | p \in \{3,7\} | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
16. Given $a, b \in \mathbf{N}^{*}$, and in decimal notation, the number $a^{a} \cdot b^{b}$ ends with exactly 98 zeros. Find all pairs $(a, b)$ that minimize $ab$.
untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The last sentence is a note and not part of the problem statement, so it is not translated. | 16. Let the powers of 2 and 5 in the prime factorizations of $a$ and $b$ be $\alpha_{1}, \beta_{1}$ and $\alpha_{2}, \beta_{2}$, respectively. Then, by the given condition, one of the equations $a \cdot \alpha_{1} + b \cdot \alpha_{2} = 98$ or $a \cdot \beta_{1} + b \cdot \beta_{2} = 98$ must hold. If $a \cdot \beta_{1} + b \cdot \beta_{2} = 98$, then when $\beta_{1}, \beta_{2}$ are both positive integers, the left side of the equation is a multiple of 5, while the right side is not; when exactly one of $\beta_{1}, \beta_{2}$ is zero, taking the equation modulo 5 still leads to a contradiction. Therefore, it must be that $a \cdot \alpha_{1} + b \cdot \alpha_{2} = 98$. In this case, if one of $\alpha_{1}, \alpha_{2}$ is zero, without loss of generality, let $\alpha_{2} = 0$, then $\alpha_{1} > 0$. If $\alpha_{1} \geqslant 2$, then $4 \mid a \cdot \alpha_{1}$. Since $4 \times 98$, this leads to a contradiction. Hence, $\alpha_{1} = 1$, which implies $a = 98$. By the problem's condition, in this case, $a \cdot \beta_{1} + b \cdot \beta_{2} \geqslant 98$, and since $a = 98$, it follows that $\beta_{1} = 0$. Thus, $b \cdot \beta_{2} \geqslant 98$, and the smallest possible value for $b$ is $3 \times 5^{2} = 75$.
In the equation $a \cdot \alpha_{1} + b \cdot \alpha_{2} = 98$, if $\alpha_{1}, \alpha_{2}$ are both positive integers, without loss of generality, let $\alpha_{1} \geqslant \alpha_{2} \geqslant 1$. If $\alpha_{2} \geqslant 2$, then $4 \mid (a \cdot \alpha_{1} + b \cdot \alpha_{2})$, leading to a contradiction. Therefore, $\alpha_{2} = 1$. From this, we can also deduce that $\alpha_{1} > 1$. We set $a = 2^{a_{1}} \cdot 5^{\beta_{1}} \cdot x$ and $b = 2 \cdot 5^{\beta_{2}} \cdot y$, then we should have
$$2^{a_{1}-1} \cdot 5^{\beta_{1}} \cdot x \cdot \alpha_{1} + 5^{\beta_{2}} \cdot y = 49$$
It follows that one of $\beta_{1}, \beta_{2}$ must be zero. Combining this with $a \cdot \beta_{1} + b \cdot \beta_{2} > 98$, if $\beta_{2} = 0$, then the number of trailing zeros in $a^{a}$ is greater than 98, leading to a contradiction. If $\beta_{1} = 0$, then the number of trailing zeros in $b^{b}$ is greater than 98, also leading to a contradiction.
Therefore, the pairs $(a, b)$ that satisfy the conditions and minimize $a b$ are $(98, 75)$ or $(75, 98)$. | (98, 75) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
25. Find the largest positive integer $m$, such that for $k \in \mathbf{N}^{*}$, if $1<k<m$ and $(k, m)=1$, then $k$ is a power of some prime.
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The translation maintains the original format and line breaks as requested. | 25. The required maximum positive integer $m=60$.
On the one hand, it can be directly verified that $m=60$ meets the requirement (it is only necessary to note that the product of the smallest two primes coprime with 60, 7 and 11, is greater than 60).
On the other hand, let $p_{1}, p_{2}, \cdots$ represent the sequence of primes in ascending order. Suppose $m$ is a positive integer that meets the requirement. We first prove: if $p_{n} p_{n+1} \leqslant n$, then
$$p_{1} p_{2} \cdots p_{n} \leqslant m$$
In fact, if there are two different numbers $p_{u}, p_{v}$ in $p_{1}, p_{2}, \cdots, p_{n+1}$ that are not divisors of $m$, then $\left(p_{u} p_{v}, m\right)=1, p_{u} p_{v} \leqslant p_{n} p_{n+1} \leqslant m$, which contradicts the property of $m$. Therefore, at most one of $p_{1}, p_{2}, \cdots, p_{n+1}$ is not a divisor of $m$, so (5) holds.
Now, if there exists $m(>60)$ that meets the requirement, then when $m \geqslant 77=p_{4} p_{5}$, let $n$ be the largest positive integer such that $p_{1} p_{2} \cdots p_{n} \leqslant m$. Then, by (5), we know $n \geqslant 4$. Combining the conclusion from the previous problem, we have $p_{n+1} p_{n+2}5 \times 7$, so $2,3,5,7$ have at most one that is not a divisor of $m$ (see the proof of (5)), thus $m$ is a multiple of one of the numbers 105, 70, 42, 30, hence $m=70$. But in this case, $(33, m)=1$, which contradicts the property of $m$. | 60 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
26. (1) Find all prime sequences $p_{1}<p_{2}<\cdots<p_{n}$, such that
$$\left(1+\frac{1}{p_{1}}\right)\left(1+\frac{1}{p_{2}}\right) \cdot \cdots \cdot\left(1+\frac{1}{p_{n}}\right)$$
is an integer; $\square$
(2) Does there exist $n$ distinct positive integers $a_{1}, a_{2}, \cdots, a_{n}, n \in \mathbf{N}^{\vee}$, greater than 1, such that
$$\left(1+\frac{1}{a_{1}^{2}}\right)\left(1+\frac{1}{a_{2}^{2}}\right) \cdot \cdots \cdot\left(1+\frac{1}{a_{n}^{2}}\right)$$
is an integer? | 26. (1) When $n \geqslant 3$, since $p_{n}>p_{i}+1, 1 \leqslant i \leqslant n-1$, thus $p_{n}$ is coprime with the number $\prod_{i=1}^{n-1}\left(1+p_{i}\right)$, hence it requires $p_{n} \mid\left(1+p_{n}\right)$, leading to $p_{n}=1$, which is a contradiction. When $n=1$, $1+\frac{1}{p_{1}} \notin \mathbf{N}^{*}$. When $n=2$, it should have $p_{2} \mid\left(p_{1}+1\right)$, thus $p_{2}=p_{1}+1$, which can only be $p_{1}=2, p_{2}=3$.
In summary, the required sequence is only: $\left(p_{1}, p_{2}\right)=(2,3)$.
(2) The conclusion is: it does not exist.
In fact, when $1<a_{1}<a_{2}<\cdots<a_{n}$, let $a_{n} \leqslant m$, then
$$\begin{aligned}
1<A & =\prod_{i=1}^{n}\left(1+\frac{1}{a_{i}^{2}}\right) \leqslant \prod_{i=2}^{m}\left(1+\frac{1}{i^{2}}\right)=\prod_{i=2}^{m} \frac{i^{2}+1}{i^{2}}<\prod_{i=2}^{m} \frac{i^{2}}{i^{2}-1} \\
& =\prod_{i=2}^{m} \frac{i^{2}}{(i-1)(i+1)}=\frac{2 m}{m+1}<2
\end{aligned}$$
Thus, $A \notin \mathbf{N}^{*}$. | (2,3) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
28. Let $n \in \mathbf{N}^{\cdot}, I$ be an open interval on the number line with length $\frac{1}{n}$. Find the maximum number of simplest fractions $\frac{a}{b}$ such that $1 \leqslant b \leqslant n$ and $\frac{a}{b} \in I$. | 28. Let $\frac{a_{1}}{b_{1}}, \frac{a_{2}}{b_{2}}, \cdots, \frac{a_{r}}{b_{r}} \in I$ be all the simplest fractions in $I^{\prime}$ satisfying: $1 \leqslant b_{1} \leqslant b_{2} \leqslant \cdots \leqslant b_{r} \leqslant n$. Then for $1 \leqslant i\left|\frac{a_{j}}{b_{j}}-\frac{a_{i}}{b_{i}}\right|=\frac{M}{\left[b_{i}, b_{j}\right]}$
Here $M$ is a positive integer. Therefore, $\left[b_{i}, b_{j}\right]>n$. That is, the least common multiple of any two numbers in $b_{1}, b_{2}, \cdots, b_{r}$ is greater than $n$. We will prove: $r \leqslant\left[\frac{n+1}{2}\right]$.
In fact, let the odd numbers from $1,2, \cdots, n$ be $x_{1}, x_{2}, \cdots, x_{m}$ in ascending order, then $m \leqslant\left[\frac{n+1}{2}\right]$. Let $x_{j}=\left\{2^{a} \cdot x_{j} \mid \alpha \in \mathbf{N}\right\}$, and $T_{j}=x_{j} \cap\{1,2, \cdots, n\}$, then $T_{1}$, $T_{2}, \cdots, T_{m}$ form a partition of $\{1,2, \cdots, n\}$, and in $T_{j}$, the larger number is a multiple of the smaller number. If $r>\left[\frac{n+1}{2}\right]$, then there must be two numbers in $b_{1}, b_{2}, \cdots, b_{r}$ that belong to the same $T_{j}$, and the least common multiple of these two numbers is no greater than $n$, which is a contradiction. Therefore, $r \leqslant\left[\frac{n+1}{2}\right]$.
On the other hand, when $n=2 p+1$, for any $j \in\{1,2, \cdots, p+1\}$, we have $\frac{1}{p+j} \in \left(\frac{1}{n}-\varepsilon, \frac{2}{n}-\varepsilon\right)$, where $\varepsilon=\frac{1}{2(2 p+1)(p+1)}$; and when $n=2 p$, for any $j \in \{1,2, \cdots, p\}$, we have $\frac{1}{p+j} \in\left(\frac{1}{n}-\varepsilon, \frac{2}{n}-\varepsilon\right)$, where $\varepsilon=\frac{1}{2 p(p+1)}$. Therefore, some $I$ can contain $\left[\frac{n+1}{2}\right]$ numbers.
In summary, the maximum value sought is $\left[\frac{n+1}{2}\right]$. | \left[\frac{n+1}{2}\right] | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
29. Find all prime numbers $p$ such that there exist $m, n \in \mathbf{N}^{*}$, satisfying: $p=m^{2}+n^{2}$, and $p \mid\left(m^{3}+n^{3}-4\right)$. | 29. From the given conditions, we know that \( p \mid (2m^3 + 2n^3 - 8) \), and \( p = m^2 + n^2 \), thus,
\[ p \mid (2m^3 + 2n^3 - 8 - 3(m^2 + n^2)(m + n)) \]
which means \( p \mid (-(m+n)^3 - 8) \). Therefore,
\[ p \mid (m+n+2)((m+n)^2 - 2(m+n) + 4) \]
Since \( p \) is a prime, we have \( p \mid (m+n+2) \) or \( p \mid ((m+n)^2 - 2(m+n) + 4) \).
If \( p \mid (m+n+2) \), then \( m^2 + n^2 \leq m + n + 2 \). Without loss of generality, assume \( m \leq n \). If \( m \geq 2 \), then \( m^2 - m \geq 2 \) and \( n^2 - n \geq 2 \), leading to \( m^2 + n^2 \geq m + n + 4 \), which is a contradiction. Therefore, \( m = 1 \), and thus \( n^2 \leq n + 2 \), giving \( n = 1 \) or 2. Hence, \( p = 2 \) or 5, and the corresponding pairs \((m, n) = (1,1)\) or \((m, n) = (1,2), (2,1)\) satisfy the conditions.
If \( p \mid ((m+n)^2 - 2(m+n) + 4) \), without loss of generality, assume \( p > 2 \) (otherwise, it falls into the previous discussion). In this case, \( p \mid (m^2 + n^2 + 2(mn - m - n + 2)) \). Combining \( p = m^2 + n^2 \) and \( p > 2 \), we get \( p \mid (mn - m - n + 2) \). However, at this point,
\[ 0 < (m-1)(n-1) + 1 = mn - m - n + 2 < mn \leq \frac{m^2 + n^2}{2} < p \]
which is a contradiction.
In summary, the required \( p = 2 \) or 5. | p = 2 \text{ or } 5 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
30. Find all integers $n > 1$ such that any of its divisors greater than 1 have the form $a^{r}+1$, where $a \in \mathbf{N}^{*}, r \geqslant 2, r \in \mathbf{N}^{*}$. | 30. Let all numbers that meet the conditions form a set $S$. For $x \in S(x>2)$, it is known that there exist $a, r \in \mathbf{N}^{\cdot}$, such that $x=a^{r}+1$, where $a, r \geqslant 2$. For this $x$, we set $a$ to be the smallest positive integer in this representation of $x$, then $r$ must be even (otherwise $(a+1) \mid x$, thus there exist $b, t \in \mathbf{N}^{*}, t \geqslant 2$, such that $a+1=b^{t}+1$, i.e., $a=b^{t}$, which leads to $x=b^{*}+1$ contradicting the minimality of $a$). Therefore, every number in $S$ can be expressed in the form $m^{2}+1$ (note that $2 \in S$, and also $2=1^{2}+1$), where $m \in \mathbf{N}^{*}$.
Below, we use the above property to determine the elements $p$ in $S$.
If $p$ is a prime, then $p$ is a prime of the form $m^{2}+1\left(m \in \mathbf{N}^{*}\right)$.
If $p$ is an odd composite number, then there exist odd primes $u \leqslant v$, such that $u, v, u v \in S$, thus there exist $a, b, c \in \mathbf{N}^{*}$, such that $a \leqslant b<c$, and
$$u=4 a^{2}+1, v=4 b^{2}+1, u v=4 c^{2}+1$$
At this time, $v(u-1)=4\left(c^{2}-b^{2}\right)$, hence $v \mid(c-b)$ or $v \mid(c+b)$, both leading to $v<2 c, u v \leqslant v^{2}<4 c^{2}$, a contradiction. Therefore, $p$ is not an odd composite number.
If $p$ is an even composite number, note that $4 \notin S$, and odd composites $\notin S$, so we can set $p=2 q$, where $q$ is an odd prime. In this case, there exist $a, b \in \mathbf{N}^{*}$, such that
$$q=4 a^{2}+1,2 q=b^{2}+1$$
Thus, $q=b^{2}-4 a^{2}=(b-2 a)(b+2 a)$, so
$$(b-2 a, b+2 a)=(1, q)$$
This gives $q=4 a+1$, so $4 a^{2}+1=4 a+1$, yielding $a=1, q=5$, and thus $p=10$.
In summary, $S=\left\{10\right.$ or primes of the form $a^{2}+1$, where $\left.a \in \mathbf{N}^{*}\right\}$ (note that the numbers described in the set clearly meet the requirements). | S=\left\{10\right. \text{ or primes of the form } a^{2}+1, \text{ where } a \in \mathbf{N}^{*}\} | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 1 Find the number of positive integers $a$ that satisfy the following condition: there exist non-negative integers $x_{0}$, $x_{1}, \cdots, x_{2009}$, such that
$$a^{x_{0}}=a^{x_{1}}+a^{x_{2}}+\cdots+a^{x_{2009}} .$$ | Solution: Clearly, $a$ that satisfies the condition is greater than 1. Taking both sides of (1) modulo $(a-1)$, we get a necessary condition as
$$1^{x_{0}} \equiv 1^{x_{1}}+1^{x_{2}}+\cdots+1^{x_{2009}}(\bmod a-1)$$
This indicates: $(a-1) \mid 2008$, the number of such $a$ is $=d(2008)=8$.
On the other hand, let $a \in \mathbf{N}^{*}$, satisfying $(a-1) \mid 2008$, set $2008=(a-1) \cdot k$, in $x_{1}, x_{2}, \cdots, x_{2009}$ take $a$ to be $0, a-1$ to be $1, \cdots, a-1$ to be $k-1$, and let $x_{0}=k$, then we know that (1) holds.
Therefore, there are 8 positive integers $a$ that satisfy the condition. | 8 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 2 Find all sequences of positive integers $\left\{a_{n}\right\}$ that satisfy the following conditions:
(1) The sequence $\left\{a_{n}\right\}$ is bounded above, i.e., there exists $M>0$, such that for any $n \in \mathbf{N}^{\cdot}$, we have $a_{n} \leqslant M ;$
(2) For any $n \in \mathbf{N}^{*}$, we have $a_{n+2}=\frac{a_{n+1}+a_{n}}{\left(a_{n}, a_{n+1}\right)}$. | Let $g_{n}=\left(a_{n}, a_{n+1}\right)$, then $g_{n} a_{n+2}=a_{n+1}+a_{n}$. Combining this with $g_{n+1}\left|a_{n+1}, g_{n+1}\right| a_{n+2}$, we know that $g_{n+1} \mid a_{n}$, which indicates that $g_{n+1}$ is a common divisor of $a_{n}$ and $a_{n+1}$, hence $g_{n+1} \leqslant g_{n}$. This means that the sequence $\left\{g_{n}\right\}_{n=1}^{+\infty}$ is a non-increasing sequence of positive integers, so there exists $N$ such that for any $n \geqslant N$, we have $g_{n}=g_{n+1}=\cdots$, denoted by $g$. Now we discuss the value of $g$.
$1^{\circ}$ If $g=1$, then for any $n \geqslant N$, we have $a_{n+2}>a_{n+1}$, i.e., $a_{n+2} \geqslant a_{n+1}+1$, thus for any $m \in \mathbf{N}^{*}$, we have
$$a_{N+m} \geqslant a_{N+1}+m-1$$
When $m$ is sufficiently large, it will contradict condition (1).
$2^{\circ}$ If $g>2$, i.e., $g \geqslant 3$, then for any $n \geqslant N$, we have $a_{n+2}<\max \left\{a_{n+1}, a_{n}\right\}$, thus for any $m \in \mathbf{N}^{*}$, we have
$$a_{N+m+2} \leqslant \max \left\{a_{N+1}, a_{N}\right\}-m$$
When $m$ is sufficiently large, the sequence $\left\{a_{n}\right\}$ will contain negative integers, which is also a contradiction.
Therefore, $g=2$. In this case, when $n \geqslant N$, we have $a_{n}=a_{n+1}$ (if not, suppose there exists $n \geqslant N$ such that $a_{n} \neq a_{n+1}$, then $\min \left\{a_{n}, a_{n+1}\right\}<a_{n+2}<\max \left\{a_{n}, a_{n+1}\right\}$, from which we can deduce that for any $m \in \mathbf{N}^{*}$, we have $\min \left\{a_{n}, a_{n+1}\right\}<a_{n+m+1}<\max \left\{a_{n}, a_{n+1}\right\}$, which is impossible), further, combining $g=2$, we know that for any $n \geqslant N$, we have $a_{n}=2$, from $a_{N+1}=a_{N}=2, g_{N}=2$ we can deduce backwards that $a_{N-1}=2, g_{N-1}=2$, and so on, we can deduce that for any $n \in \mathbf{N}^{*}$, we have $a_{n}=2$.
In summary, the sequence $\left\{a_{n}\right\}$ that satisfies the conditions is unique, and its general term is $a_{n}=2$. | a_{n}=2 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 3 Find all pairs of positive integers $(m, n), m, n \geqslant 2$, such that for any $a \in\{1$, $2, \cdots, n\}$, we have $a^{n} \equiv 1(\bmod m)$. | When $m$ is an odd prime and $n=m-1$, by Fermat's Little Theorem, the pair $(m, n)$ satisfies the condition. Conversely, must $(m, n)=(p, p-1)$, where $p$ is an odd prime?
Now, let $(m, n)$ be a pair of positive integers that satisfy the condition, and let $p$ be a prime factor of $m$. Then $p>n$ (otherwise, if $p \leqslant n$, take $a=p$, then $m \mid\left(p^{n}-1\right)$, but $p \nmid\left(p^{n}-1\right)$, a contradiction).
Consider the polynomial
$$f(x)=(x-1)(x-2) \cdot \cdots \cdot(x-n)-\left(x^{n}-1\right)$$
Given $p>n$, and for $a \in\{1,2, \cdots, n\}$, we have $m \mid\left(a^{n}-1\right)$ (and thus $p \mid\left(a^{n}-1\right)$), it follows that in modulo $p$, $f(x) \equiv 0(\bmod p)$ has $n$ distinct roots. Since $f(x)$ is an $n-1$ degree polynomial, by Lagrange's Theorem, $f(x)$ is the zero polynomial modulo $p$, meaning every coefficient of $f(x)$ is a multiple of $p$. In particular, the coefficient of $x$ in $f(x)$ is $\equiv 0(\bmod p)$, i.e.,
$$p \mid(1+2+\cdots+n)$$
Thus, $p \mid n(n+1)$. Given $p>n$, it must be that $p=n+1$.
The above discussion shows that $(m, n)=\left(p^{\alpha}, p-1\right)$, where $p$ is an odd prime, and $\alpha \in \mathbf{N}^{*}$.
If $\alpha>1$, by the condition, we have $p^{\alpha} \mid\left((p-1)^{p-1}-1\right)$, hence
$$(p-1)^{p-1} \equiv 1\left(\bmod p^{2}\right)$$
Using the binomial theorem, we get
$$\begin{aligned}
1 & \equiv(p-1)^{p-1} \equiv \mathrm{C}_{p-1}^{1} \cdot p \cdot(-1)^{p-2}+1 \\
& \equiv 1-p(p-1) \equiv 1+p\left(\bmod p^{2}\right)
\end{aligned}$$
This is a contradiction, so $\alpha=1$.
In summary, the pairs $(m, n)=(p, p-1)$, where $p$ is any odd prime, satisfy the condition. | (m, n)=(p, p-1) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 5 By Fermat's Little Theorem, for any odd prime $p$, we have $2^{p-1} \equiv 1(\bmod p)$. Question: Does there exist a composite number $n$ such that $2^{n-1} \equiv 1(\bmod n)$ holds? | $$\begin{array}{l}
\text { Hence } \\
\text { Therefore }
\end{array}$$
$$\begin{array}{l}
2^{10}-1=1023=341 \times 3 \\
2^{10} \equiv 1(\bmod 341) \\
2^{340} \equiv 1^{34} \equiv 1(\bmod 341)
\end{array}$$
Hence 341 meets the requirement.
Furthermore, let $a$ be an odd composite number that meets the requirement, then $2^{a}-1$ is also an odd composite number (this can be known through factorization). Let $2^{a-1}-1=a \times q, q$ be an odd positive integer, then
$$\begin{aligned}
2^{2^{a}-1-1}-1 & =2^{2\left(2^{a-1}-1\right)}-1 \\
& =2^{2 a q}-1 \\
& =\left(2^{a}\right)^{2 q}-1 \\
& \equiv 1^{2 q}-1 \\
& \equiv 0\left(\bmod 2^{a}-1\right)
\end{aligned}$$
Therefore $2^{a}-1$ is also a number that meets the requirement. By this recursion (combined with 341 meeting the requirement), it is known that there are infinitely many composite numbers that meet the condition. | 341 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 6 Find all prime numbers $p$ such that $\frac{2^{p-1}-1}{p}$ is a perfect square.
Find all prime numbers $p$ such that $\frac{2^{p-1}-1}{p}$ is a perfect square. | Let $p$ be a prime number that satisfies the condition, then obviously $p$ is an odd prime. By Fermat's Little Theorem, we have
Thus,
$$\begin{array}{c}
p \mid 2^{p-1}-1 \\
2^{p-1}-1=\left(2^{\frac{p-1}{2}}-1\right)\left(2^{\frac{p-1}{2}}+1\right) \\
p \left\lvert\, 2^{\frac{p-1}{2}}-1\right. \text { or } p \left\lvert\, 2^{\frac{p-1}{2}}+1\right.
\end{array}$$
Since $\left(2^{\frac{p-1}{2}}-1,2^{\frac{p-1}{2}}+1\right)=\left(2^{\frac{p-1}{2}}-1,2\right)=1$, one of $p \left\lvert\, 2^{\frac{p-1}{2}}-1\right.$ and $p \left\lvert\, 2^{\frac{p-1}{2}}+1\right.$ must hold.
If $p \left\lvert\, 2^{\frac{p-1}{2}}-1\right.$, then by the condition and $\left(2^{\frac{p-1}{2}}-1,2^{\frac{p-1}{2}}+1\right)=1$, there exists a positive integer $x$ such that
At this point,
$$\begin{array}{c}
2^{\frac{p-1}{2}}+1=x^{2} \\
(x-1)(x+1)=2^{\frac{p-1}{2}}
\end{array}$$
This indicates that $x-1$ and $x+1$ are both powers of 2, and $x$ is odd, so $x-1$ and $x+1$ are two consecutive even numbers. Therefore, it can only be
$$x-1=2, x+1=4$$
Thus,
$$\begin{array}{l}
x=3 \\
p=7
\end{array}$$
If $p \left\lvert\, 2^{\frac{p-1}{2}}+1\right.$, then similarly, there exists a positive integer $x$ such that
$$2^{\frac{p-1}{2}}-1=x^{2},$$
When $p>3$, this leads to
$$x^{2}=2^{\frac{p-1}{2}}-1 \equiv-1(\bmod 4)$$
which is a contradiction, so $p=3$.
On the other hand, when $p=3$ and 7, $\frac{2^{p-1}-1}{p}$ are 1 and 9, respectively, both of which are perfect squares. In summary, $p=3$ or 7. | p=3 \text{ or } 7 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 1 Given that $p$ is a prime number, find all integer pairs $(x, y)$ such that $|x+y|+(x-y)^{2}=p$. | Notice that, $x+y$ and $x-y$ are either both odd or both even, so $|x+y|+(x-y)^{2}$ is even, which implies $p=2$. This indicates that $|x+y|+(x-y)^{2}=2$.
Since $|x+y|$ and $(x-y)^{2}$ have the same parity, and $(x-y)^{2}$ is a perfect square, we have $\left(|x+y|,(x-y)^{2}\right)=(2,0),(1,1)$. Solving these respectively, we find
$$(x, y)=(1,1),(-1,-1),(0,1),(0,-1),(1,0),(-1,0) .$$ | (x, y)=(1,1),(-1,-1),(0,1),(0,-1),(1,0),(-1,0) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 5 It is known that among 4 coins, there may be counterfeit coins, where genuine coins each weigh 10 grams, and counterfeit coins each weigh 9 grams. Now there is a balance scale, which can weigh the total weight of the objects on the tray. Question: What is the minimum number of weighings needed to ensure that the authenticity of each coin can be identified? | At least 3 weighings can achieve this.
In fact, let the 4 coins be $a, b, c, d$. Weigh $a+b+c, a+b+d$, and $a+c+d$ three times. The sum of these three weights is $3a + 2(b+c+d)$, so if the sum of these three weights is odd, then $a$ is the counterfeit coin; otherwise, $a$ is genuine. Once $a$ is determined, solving the system of three linear equations for $b, c, d$ can determine the authenticity of $b, c, d$. Therefore, 3 weighings are sufficient.
Next, we prove that two weighings cannot guarantee the determination of the authenticity of each coin.
Notice that if two coins, for example, $a, b$, either both appear or both do not appear in each weighing, then when $a, b$ are one genuine and one counterfeit, swapping the authenticity of $a, b$ does not affect the weighing results, so their authenticity cannot be determined.
If there is a weighing in which at most two coins, for example, $a, b$, appear, then in the other weighing, $c, d$ can only have exactly one on the scale (otherwise, swapping the odd/even nature of $c, d$ does not affect the result), at this point, one coin does not appear in both weighings, and changing its authenticity does not affect the weighing results, thus its authenticity cannot be determined. Therefore, at least 3 coins must be on the scale in each weighing, which means there must be two coins that appear in both weighings, leading to a contradiction.
In summary, at least 3 weighings are required. | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | number_theory | false |
Example 6 A cube with a side length of 3 is divided into 27 unit cubes. The numbers $1, 2, \cdots$, 27 are randomly placed into the unit cubes, one number in each. Calculate the sum of the 3 numbers in each row (horizontal, vertical, and column), resulting in 27 sum numbers. Question: What is the maximum number of odd numbers among these 27 sum numbers? | Solve: To calculate the sum $S$ of these 27 sums, since each number appears in exactly 3 rows, we have
$$S=3 \times(1+2+\cdots+27)=3 \times 27 \times 14$$
Thus, $S$ is an even number, so the number of odd numbers among these 27 sums must be even.
If 26 of these 27 sums are odd, let's assume that the even number is the sum of the 3 numbers in the first row of Figure 1, i.e., $a_{1}+a_{2}+a_{3}$ is even, while the sums of the other 5 rows in Figure 1 are all odd. In this case, summing the numbers in Figure 1 by rows and columns, we get
$$\begin{aligned}
& \left(a_{1}+a_{2}+a_{3}\right)+\left(a_{4}+a_{5}+a_{6}\right)+\left(a_{7}+a_{8}+a_{9}\right) \\
= & \left(a_{1}+a_{4}+a_{7}\right)+\left(a_{2}+a_{5}+a_{8}\right)+\left(a_{3}+a_{6}+a_{9}\right)
\end{aligned}$$
However, the left side of this equation is the sum of two odd numbers and one even number, while the right side is the sum of 3 odd numbers, leading to a contradiction where the left side is even and the right side is odd.
Therefore, among these 27 sums, there can be at most 24 odd numbers.
The example below (as shown in Figure 2) demonstrates that there exists a way to fill the numbers such that 24 of the 27 sums can be odd. In the tables of Figure 2, 0 represents an even number, and 1 represents an odd number, from left to right, representing the top, middle, and bottom layers of the unit cubes.
Thus, among these 27 sums, the maximum number of odd numbers is 24. | 24 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
Example 2 Consider the following sequence:
$$101,10101,1010101, \cdots$$
Question: How many prime numbers are there in this sequence? | It is easy to know that 101 is a prime number. Next, we prove that this is the only prime number in the sequence.
Let $a_{n}=\underbrace{10101 \cdots 01}_{n \uparrow 01}$, then when $n \geqslant 2$, we have
$$\begin{aligned}
a_{n} & =10^{2 n}+10^{2(n-1)}+\cdots+1 \\
& =\frac{10^{2(n+1)}-1}{10^{2}-1} \\
& =\frac{\left(10^{n+1}-1\right)\left(10^{n+1}+1\right)}{99} .
\end{aligned}$$
Notice that, $99<10^{n+1}-1,99<10^{n+1}+1$, and $a_{n}$ is a positive integer, so $a_{n}$ is a composite number (because the terms $10^{n+1}-1$ and $10^{n+1}+1$ in the numerator cannot be reduced to 1 by 99). | 1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 4 Find all prime numbers $p$ such that $p^{3}-4 p+9$ is a perfect square.
Find all prime numbers $p$ such that $p^{3}-4 p+9$ is a perfect square. | Let $p^{3}-4 p+9=x^{2}, x$ be a non-negative integer, then $p \mid x^{2}-9$, i.e., $p \mid (x-3)(x+3)$. Combining with $p$ being a prime, we can set $x=k p \pm 3, k$ as a non-negative integer. Thus,
$$p^{3}-4 p=x^{2}-9=k^{2} p^{2} \pm 6 k p$$
We get $p^{2}-4=k^{2} p \pm 6 k$, which indicates: $p \mid 6 k \pm 4$.
When $p>2$, $p$ is an odd prime, it can be known that $p \mid 3 k \pm 2$, so $p \leqslant 3 k+2$, which indicates: $\frac{1}{3}\left(p^{2}-2 p-9\right) \leqslant p k-3 \leqslant x$
If $x \leqslant \frac{p^{2}}{4}$, then $\frac{1}{3}\left(p^{2}-2 p-9\right) \leqslant \frac{p^{2}}{4}$, we get $p \leqslant 8+\frac{36}{p}$, which means $p \leqslant 11$; if $x>\frac{p^{2}}{4}$, then $p^{3}-4 p+9=x^{2}>\frac{p^{4}}{16}$, we get $p<16-\frac{16(4 p-9)}{p^{3}}$, which means $p \leqslant 13$.
In summary, $p \leqslant 13$, direct enumeration yields $(p, x)=(2,3),(7,18),(11,36)$. Therefore, $p=2,7$ or 11. | p=2,7,11 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 7 Find all pairs of positive integers $(a, b)$ such that
$$a^{3}+6 a b+1, b^{3}+6 a b+1,$$
are both perfect cubes. | Let's assume $a \leqslant b$, then
$$b^{3}<b^{3}+6 a b+1 \leqslant b^{3}+6 b^{2}+1<(b+2)^{3}$$
Since $b^{3}+6 a b+1$ is a perfect cube, we have
$$b^{3}+6 a b+1=(b+1)^{3},$$
which implies $\square$
$$\begin{aligned}
6 a b & =3 b^{2}+3 b \\
b & =2 a-1
\end{aligned}$$
Thus, $\square$
$$a^{3}+6 a b+1=a^{3}+12 a^{2}-6 a+1$$
Notice that $\quad(a+1)^{3} \leqslant a^{3}+12 a^{2}-6 a+1<(a+4)^{3}$, therefore, since $a^{3}+12 a^{2}-6 a+1$ is a perfect cube, it can only be
$$a^{3}+12 a^{2}-6 a+1=(a+1)^{3},(a+2)^{3},(a+3)^{3}$$
Solving each case, we find that the only solution is $a=1$.
Therefore, the pair $(a, b)=(1,1)$ satisfies the condition. | (1,1) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 8 Find the smallest positive integer $n$, such that there exist integers $x_{1}, x_{2}, \cdots, x_{n}$, satisfying
$$x_{1}^{4}+x_{2}^{4}+\cdots+x_{n}^{4}=1599$$ | From property 1, for any integer $a$, we know that
$$a^{2} \equiv 0(\bmod 4) \text { or } a^{2} \equiv 1(\bmod 8),$$
From this, we can deduce that
$$a^{4} \equiv 0 \text { or } 1(\bmod 16).$$
Using this conclusion, if $n<15$, let
$$x_{1}^{4}+x_{2}^{4}+\cdots+x_{n}^{4} \equiv m(\bmod 16),$$
then
and
$$\begin{array}{c}
m \leqslant n<15 \\
1599 \equiv 15(\bmod 16)
\end{array}$$
This is a contradiction, so
$$n \geqslant 15.$$
Furthermore, when $n=15$, it is required that
$$x_{1}^{4} \equiv x_{2}^{4} \equiv \cdots \equiv x_{n}^{4} \equiv 1(\bmod 16)$$
This means that $x_{1}, x_{2}, \cdots, x_{n}$ must all be odd numbers, which points us in the right direction to find suitable numbers. In fact, among $x_{1}, x_{2}, \cdots, x_{15}$, if one number is 5, twelve are 3, and the other two are 1, then
$$\begin{aligned}
& x_{1}^{4}+x_{2}^{4}+\cdots+x_{15}^{4} \\
= & 5^{4}+12 \times 3^{4}+2 \\
= & 625+972+2 \\
= & 1599
\end{aligned}$$
Therefore, the minimum value of $n$ is 15. | 15 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
1 Let $p$ and $q$ both be prime numbers, and $7p + q$, $pq + 11$ are also prime numbers. Find the value of $\left(p^{2} + q^{p}\right)\left(q^{2} + p^{q}\right)$. | 1. If $p$ and $q$ are both odd, then $7p + q$ is even, and it is not a prime number. Therefore, one of $p$ and $q$ must be even.
Case one: Suppose $p$ is even, then $p = 2$. In this case, since $7p + q$ is a prime number, $q$ must be an odd prime. If $q \neq 3$, then $q \equiv 1$ or $2 \pmod{3}$.
If $q \equiv 1 \pmod{3}$, then
$$7p + q = 14 + q \equiv 0 \pmod{3}$$
This is a contradiction;
If $q \equiv 2 \pmod{3}$, then
$$pq + 11 = 2q + 11 \equiv 4 + 11 \equiv 0 \pmod{3}$$
This is also a contradiction. Therefore, $q = 3$. In this case,
$$7p + q = 17, \quad pq + 11 = 17$$
Both are prime numbers, so
$$\left(p^2 + q^p\right)\left(q^2 + p^q\right) = \left(2^2 + 3^2\right)\left(3^2 + 2^3\right) = 221$$
Case two: Suppose $q$ is even, then $q = 2$. Similarly, we find that $p = 3$. In this case,
$$\left(p^2 + q^p\right)\left(q^2 + p^q\right) = \left(3^2 + 2^3\right)\left(2^2 + 3^2\right) = 221$$
In conclusion, the desired value is 221. | 221 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
2 Let $p_{1}<p_{2}<p_{3}<p_{4}<p_{5}$ be 5 prime numbers, and $p_{1}, p_{2}, p_{3}, p_{4}, p_{5}$ form an arithmetic sequence. Find the minimum value of $p_{5}$. | 2. Let $d$ be the common difference, then $p_{1}, p_{1}+d, p_{1}+2 d, p_{1}+3 d, p_{1}+4 d$ are all primes. If $2 \nmid d$, i.e., $d$ is odd, then one of $p_{1}+d$ and $p_{1}+2 d$ is even, and it is not a prime. If $3 \nmid d$, then one of $p_{1}+d, p_{1}+2 d, p_{1}+3 d$ is a multiple of 3 (they form a complete residue system modulo 3), which is a contradiction.
If $5 \nmid d$, then one of $p_{1}, p_{1}+d, \cdots, p_{1}+4 d$ is a multiple of 5, which can only be $p_{1}=$ 5, in this case the common difference $d$ is a multiple of 6.
And $5,11,17,23,29$ is a sequence of 5 primes in arithmetic progression, so, $p_{5}$ is at least 29. | 29 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 3 Find all positive integers $n$, such that $\frac{n(n+1)}{2}-1$ is a prime number.
Find all positive integers $n$, such that $\frac{n(n+1)}{2}-1$ is a prime number. | Let $a_{n}=\frac{n(n+1)}{2}-1$, then $a_{1}=0$ is not a prime number, so we only need to discuss the case when $n>1$. We will use the fact that $n$ can only be of the form $4k$, $4k+1$, $4k+2$, or $4k+3$ and discuss each case separately.
When $n$ is of the form $4k+2$ or $4k+1$, $a_{n}$ is always even. For $a_{n}$ to be a prime number, it can only be
$$\frac{n(n+1)}{2}-1=2$$
Solving this, we get
$$n=2$$
When $n=4k$, we have
$$\begin{aligned}
a_{n} & =2k(4k+1)-1 \\
& =8k^2+2k-1 \\
& =(4k-1)(2k+1)
\end{aligned}$$
This is a composite number.
When $n=4k+3$, we have
$$\begin{aligned}
a_{n} & =2(k+1)(4k+3)-1 \\
& =8k^2+14k+5 \\
& =(4k+5)(2k+1)
\end{aligned}$$
Only when $k=0$, i.e., $n=3$, is $a_{n}$ a prime number.
Therefore, the values of $n$ that satisfy the condition are $n=2$ or $3$.
Explanation: Classifying $n$ is necessary both for eliminating the denominator and for preparing to factorize. | n=2 \text{ or } 3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
4 Find the largest positive integer $k$, such that there exists a positive integer $n$, satisfying $2^{k} \mid 3^{n}+1$.
| 4. Notice that, when $n$ is even, let $n=2 m$, we have
$$3^{n}=9^{m} \equiv 1(\bmod 8)$$
When $n=2 m+1$,
$$3^{n}=9^{m} \times 3 \equiv 3(\bmod 8)$$
Therefore, for any positive integer $n$, we have
$$3^{n}+1 \equiv 2 \text { or } 4(\bmod 8),$$
so $k \leqslant 2$. Also, $2^{2} \mid 3^{1}+1$, hence, the maximum value of $k$ is 2. | 2 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
14 Labeled as $1,2, \cdots, 100$, there are some matches in the matchboxes. If each question allows asking about the parity of the sum of matches in any 15 boxes, then to determine the parity of the number of matches in box 1, at least how many questions are needed? | 14. At least 3 questions are needed.
First, prove that "3 questions are sufficient." For example:
The first question is: $a_{1}, a_{2}, \cdots, a_{15}$;
The second question is: $a_{1}, a_{2}, \cdots, a_{8}, a_{16}, a_{17}, \cdots, a_{22}$;
The third question is: $a_{1}, a_{9}, a_{10}, \cdots, a_{22}$.
Here, $a_{i}$ represents the number of matches in the $i$-th box. In this way, the parity (odd or even) of the sum of the three answers is the same as the parity of $a_{1}$ (the other boxes each appear exactly twice in the three questions). Therefore, after 3 questions, the parity of $a_{1}$ can be determined.
Next, prove that "at least 3 questions are needed." If there are only two questions, and $a_{1}$ appears in both questions, then there must be $a_{i}$ and $a_{j}$ in the two questions such that $a_{i}$ appears only in the first question, and $a_{j}$ appears only in the second question. By changing the parity of $a_{1}$, $a_{i}$, and $a_{j}$ simultaneously, the answers to each question remain the same, thus the parity of $a_{1}$ cannot be determined. If $a_{1}$ does not appear in both questions, when $a_{1}$ does not appear at all, change the parity of $a_{1}$; when $a_{1}$ appears only once, change the parity of $a_{1}$ and $a_{i}$ (where $a_{i}$ is a box that appears with $a_{1}$ in the question), then the answers to the two questions remain the same, and the parity of $a_{1}$ cannot be determined.
In summary, at least 3 questions are needed. | 3 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
15 Find all positive integers $n$, such that it is possible to write +1 or -1 in each cell of an $n \times n$ grid, satisfying: each cell labeled +1 has exactly one adjacent cell labeled -1, and each cell labeled -1 has exactly one adjacent cell labeled +1. | 15. Let $a_{ij}$ denote the number filled in the square at the $i$-th row and $j$-th column. If there exists a valid filling method, we can assume $a_{11}=1$ (otherwise, change the signs of all numbers in the table and then discuss), at this point, one of $a_{21}$ and $a_{12}$ must be -1, let's assume $a_{21}=-1$ (otherwise, interchange the 2nd row and the 2nd column and then discuss), then $a_{12}=1$, further discussion shows that $a_{22}=-1, a_{13}=1, \cdots$, it follows that all numbers in the 1st row are 1, all numbers in the 2nd row are -1, and so on, all numbers in the 3rd row are -1, all numbers in the 4th row are 1, and so forth, it is known that when and only when $i \equiv 1(\bmod 3)$, all numbers in the $i$-th row are 1, and all numbers in the other rows are -1. If $n \equiv 0(\bmod 3)$, then the numbers in the $n$-th row are -1, and each square in this row has adjacent squares with numbers -1, which does not meet the requirements. Direct verification shows that the other cases all meet the requirements.
Therefore, a valid filling method exists if and only if $3 \nmid n, n>1$. | 3 \nmid n, n>1 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
17 Find the number of all positive integers $a$ that satisfy the following condition: there exist non-negative integers $x_{0}, x_{1}, x_{2}, \cdots$, $x_{2001}$, such that $a^{x_{0}}=a^{x_{1}}+a^{x_{2}}+\cdots+a^{x_{2001}}$. | 17. If $a$ is a number that satisfies the condition, then $a^{x_{0}}>1$, so $a>1$. At this point, taking
both sides modulo $a-1$, we know
so $\square$
$$\begin{array}{c}
a^{x_{0}}=a^{x_{1}}+a^{x_{2}}+\cdots+a^{x_{2001}} \\
1 \equiv \underbrace{1+\cdots+1}_{2001 \uparrow}(\bmod a-1), \\
a-1 \mid 2000 .
\end{array}$$
On the other hand, if $a>1$ and $a-1 \mid 2000$, then we can take $x_{1}, x_{2}, \cdots, x_{2001}$ to be $a$ numbers as $0$, $a-1$ numbers as $1$, $a-1$ numbers as $2$, ..., $a-1$ numbers as $k-1$, where $k=\frac{2000}{a-1}$, and take $x_{0}=k$, then we have $a^{x_{0}}=a^{x_{1}}+a^{x_{2}}+\cdots+a^{x_{2001}}$.
Therefore, $a$ is a number that satisfies the condition if and only if $a>1$ and $a-1 \mid 2000$, and there are 20 such $a$. | 20 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
20 Find the smallest positive integer $a$, such that for any integer $x$, we have $65 \mid\left(5 x^{13}+13 x^{5}+9 a x\right)$. | 20. From the condition, we know $65 \mid (18 + 9a)$ (take $x=1$), and since $(9,65)=1$, it follows that $65 \mid a+2$, hence $a \geq 63$.
When $a=63$, using Fermat's Little Theorem, we know that for any integer $x$, we have
$$\begin{aligned}
& 5 x^{13} + 13 x^{5} + 9 a x \\
\equiv & 13 x + 9 a x \\
\equiv & (3 + (-1) \times 3) x \\
\equiv & 0 \pmod{5}
\end{aligned}$$
$$\begin{aligned}
& 5 x^{13} + 13 x^{5} + 9 a x \\
\equiv & 5 x + 9 a x \\
\equiv & (5 + 9 \times (-2)) x \\
\equiv & 0 \pmod{13}
\end{aligned}$$
Therefore,
$$65 \mid 5 x^{13} + 13 x^{5} + 9 a x$$
In summary, the smallest positive integer $a=63$. | 63 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
22 Find all positive integer tuples $(x, y, z, w)$, such that $x!+y!+z!=w!$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | $$
\begin{array}{l}
22. \text{Let } x \leqslant y \leqslant z, \text{ and } w! = z! + y! + x! \\
z \leqslant 2
\end{array}
$$
If $z=1$, then $x=y=z=1$, at this time $w!=3$, there is no such $w$, so $z=2$. At this time $w \geqslant 3$, hence
$$w!\equiv 0(\bmod 3)$$
Therefore,
$$x!+y!\equiv 1(\bmod 3)$$
And
thus it can only be
$$\begin{array}{l}
x \leqslant y \leqslant 2 \\
x=y=2
\end{array}$$
At this time
hence
$$(x, y, z, w)=(2,2,2,3)$$ | (2,2,2,3) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
23 Find the number of integer pairs $(a, b)$ that satisfy the following conditions: $0 \leqslant a, b \leqslant 36$, and $a^{2}+b^{2}=$ $0(\bmod 37)$. | 23. Notice that, $a^{2}+b^{2} \equiv a^{2}-36 b^{2}(\bmod 37)$, so from the condition we have
$$37 \mid a^{2}-36 b^{2},$$
which means
$$37 \mid(a-6 b)(a+6 b),$$
thus
$37 \mid a-6 b$ or $37 \mid a+6 b$.
Therefore, for each $1 \leqslant b \leqslant 36$, there are exactly two $a(a \equiv \pm 6 b(\bmod 37))$ that satisfy the condition, and when $b=0$, from $a^{2}+b^{2} \equiv 0(\bmod 37)$ we know $a=0$. Hence, the number of pairs $(a, b)$ that satisfy the condition is $2 \times 36+1=73$ (pairs). | 73 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
26 Find all positive integers $n$, such that the cube root of $n$ equals the positive integer obtained by removing the last three digits of $n$.
Find all positive integers $n$, such that the cube root of $n$ equals the positive integer obtained by removing the last three digits of $n$. | 26. Let $n=1000 x+y$, where $x$ is a positive integer, $y$ is an integer, and $0 \leqslant y \leqslant 999$. According to the problem,
$$x^{3}=1000 x+y$$
From $0 \leqslant y \leqslant 999$, we know
$$1000 x \leqslant x^{3}<1000 x+1000=1000(x+1)$$
Thus,
$$x^{2} \geqslant 1000, x^{3}+1 \leqslant 1000(x+1)$$
Therefore,
$$x^{2} \geqslant 1000, x^{2}-x+1 \leqslant 1000$$
So,
$$32 \leqslant x<33$$
Hence,
$$x=32,$$
Then,
$$y=768$$
Therefore,
$$n=32768$$ | 32768 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
28 Find the smallest positive integer $n$, such that in decimal notation $n^{3}$ ends with the digits 888. | 28. From the condition, we know $n^{3} \equiv 888(\bmod 1000)$, hence
$$n^{3} \equiv 888(\bmod 8), n^{3} \equiv 888(\bmod 125)$$
From the former, we know $n$ is even, let $n=2 m$, then
$$m^{3} \equiv 111(\bmod 125)$$
Therefore
$$m^{3} \equiv 111 \equiv 1(\bmod 5)$$
Noting that when $m=0,1,2,3,4(\bmod 5)$, correspondingly
$$m^{3} \equiv 0,1,3,2,4(\bmod 5)$$
So, from $m^{3} \equiv 1(\bmod 5)$ we know $m \equiv 1(\bmod 5)$, we can set $m=5 k+1$, at this time
$$m^{3}=(5 k+1)^{3}=125 k^{3}+75 k^{2}+15 k+1 \equiv 111(\bmod 125)$$
Hence
Thus
That is, $\square$
Therefore
This requires
Hence
$$\begin{array}{c}
75 k^{2}+15 k \equiv 110(\bmod 125) \\
15 k^{2}+3 k \equiv 22(\bmod 25) \\
15 k^{2}+3 k+3 \equiv 0(\bmod 25) \\
5 k^{2}+k+1 \equiv 0(\bmod 25) \\
5 k^{2}+k+1 \equiv 0(\bmod 5) \\
5 \mid k+1
\end{array}$$
We can set $k+1=5 l$, then
$$\begin{aligned}
5 k^{2}+k+1 & =5 \times(5 l-1)^{2}+5 l \\
& =125 l^{2}-50 l+5(l+1) \\
& \equiv 0(\bmod 25) \\
& 5 \mid l+1
\end{aligned}$$
We can set $l+1=5 r$, therefore
$$\begin{aligned}
n & =2 m=10 k+2=10(5 l-1)+2 \\
& =50 l-8=50(5 r-1)-8 \\
& =250 r-58
\end{aligned}$$
Combining that $n$ is a positive integer, we know
$$n \geqslant 250-58=192$$
Also, $192^{3}=7077888$ meets the requirement, hence the smallest positive integer satisfying the condition is 192. | 192 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
33 Find the largest positive integer that cannot be expressed as the sum of a positive multiple of 42 and a composite number.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 33. For any positive integer $n$ that cannot be expressed as a positive multiple of 42 and a composite number, consider the remainder $r$ when $n$ is divided by 42. If $r=0$ or $r$ is a composite number, then $n \leqslant 42$.
Now consider the case where $r=1$ or $r$ is a prime number.
If $r \equiv 1(\bmod 5)$, then
$$84+r \equiv 0(\bmod 5)$$
In this case,
$$n<3 \times 42=126$$
If $r \equiv 2(\bmod 5)$, then
$$4 \times 42+r \equiv 0(\bmod 5)$$
In this case,
$$n<5 \times 42=210$$
If $r \equiv 3(\bmod 5)$, then
$$42+r \equiv 0(\bmod 5)$$
In this case,
$$n<2 \times 42=84$$
If $r \equiv 4(\bmod 5)$, then
$$3 \times 42+r \equiv 0(\bmod 5)$$
In this case,
$$n<4 \times 42=168$$
If $r \equiv 0(\bmod 5)$, then
$$r=5,$$
In this case, since 5, 47, 89, 131, and 173 are all prime numbers, $n$ is at most 215.
In summary, the largest positive integer sought is 215. | 215 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
34 Find a positive integer $n$, such that each number $n, n+1, \cdots, n+20$ is not coprime with 30030. | 34. Since $30030=2 \times 3 \times 5 \times 7 \times 11 \times 13$, if we take $N=210 k$, then $N$ and $N \pm r$ are not coprime with 30030, where $r$ is a number in $2,3, \cdots, 10$. Now consider the numbers $N \pm 1$. We choose $k$ such that
$$210 k \equiv 1(\bmod 11) \text { and } 210 k \equiv-1(\bmod 13),$$
The first condition requires $k \equiv 1(\bmod 11)$, set $k=11 m+1$,
The second condition requires
$$210(11 m+1) \equiv-1(\bmod 13)$$
Solving this, we get
$$m \equiv 4(\bmod 13)$$
Therefore, let $k=45$, then the 21 numbers $9440,9441, \cdots, 9460$ are not coprime with 30030, so taking $n=9440$ is sufficient. | 9440 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
37 points are arranged on a circle, and one of the points is labeled with the number 1. Moving clockwise, label the next point with the number 2 after counting two points, then label the next point with the number 3 after counting three points, and so on, labeling the points with the numbers $1,2, \cdots, 2000$. In this way, some points are not labeled, and some points are labeled with more than one number. Question: What is the smallest number among the numbers labeled on the point that is labeled with 2000? | 37. Equivalent to finding the smallest positive integer $n$, such that
$$1+2+\cdots+n \equiv 1+2+\cdots+2000(\bmod 2000)$$
i.e. $\square$
$$\frac{n(n+1)}{2} \equiv 1000(\bmod 2000),$$
which is equivalent to
$$n(n+1) \equiv 2000(\bmod 4000),$$
This requires
$$2000 \mid n(n+1)$$
Notice that
$$(n, n+1)=1$$
and
$$2000=2^{4} \times 5^{3},$$
so $2^{4}\left|n, 5^{3}\right| n+1$; or $5^{3}\left|n, 2^{4}\right| n+1$; or one of $n$ and $n+1$ is a multiple of 2000. Solving these, the smallest $n$ is found to be $624, 1375, 1999$, with the smallest number satisfying (1) being 624. Therefore, the smallest number marked on the point labeled 2000 is 624. | 624 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
38 There are 800 points on a circle, labeled $1,2, \cdots, 800$ in a clockwise direction, dividing the circle into 800 gaps. Now, choose one of these points and color it red, then perform the following operation: if the $k$-th point is colored red, then move $k$ gaps in a clockwise direction and color the point reached red. Question: According to this rule, what is the maximum number of points that can be colored red on the circle? Prove your conclusion. | 38. This is equivalent to finding the maximum number of distinct numbers in the sequence $a, 2a, 2^2a, 2^3a, \cdots$ under modulo 800, where $a$ takes values in $1, 2, \cdots, 800$.
Notice that when $2^n \not\equiv 2^m \pmod{800}$, it is not necessarily true that $2^n a \not\equiv 2^m a \pmod{800}$. Conversely, when $2^n a \not\equiv 2^m a \pmod{800}$ holds, $2^n \equiv 2^m \pmod{800}$ must hold. Therefore, the maximum number of distinct elements in the sequence $a, 2a, 2^2a, \cdots$ under modulo 800 can be achieved when $a=1$. Thus, we only need to find the number of distinct elements in the sequence $1, 2, 2^2, \cdots$ under modulo 800.
Since $800 = 2^5 \times 5^2$, and for $n \geq 5$ we have
$$2^n \equiv 0 \pmod{2^5}$$
Additionally, $\{2^n \pmod{25}\}$ is: $2, 4, 8, 16, 7, 14, 3, 6, 12, -1, -2, -4, -8, -16, -7, -14, -3, -6, -12, 1, \cdots$, so $\{2^n \pmod{25}\}$ contains exactly 20 distinct elements. Combining this with $\{2^n \pmod{2^5}\}$ being $2, 4, 8, 16, 0, 0, \cdots$, we get that $\{2^n \pmod{800}\}$ contains $20 + 4 = 24$ (distinct) numbers.
Therefore, there are at most 24 points on the circle that are colored red. | 24 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 1 Find the number of positive integer solutions to the indeterminate equation
$$7 x+19 y=2012$$ | Solve: First, find a particular solution of (1).
$$x=\frac{1}{7}(2012-19 y)=287-3 y+\frac{1}{7}(3+2 y) .$$
Therefore, $\frac{1}{7}(3+2 y)$ must be an integer. Taking $y_{0}=2$, then $x_{0}=282$.
Using the conclusion of Theorem 2, the general solution of equation (1) is
$$\left\{\begin{array}{l}
x=282-19 t, \\
y=2+7 t .
\end{array} \text { ( } t \text { is an integer }\right)$$
Combining $x>0, y>0$ and $t$ being an integer, we can solve to get $0 \leqslant t \leqslant 14$.
Therefore, equation (1) has 15 sets of positive integer solutions. | 15 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 4 Find the number of positive integer solutions to the indeterminate equation
$$x+2 y+3 z=2012$$ | Let $(x, y, z)$ be a positive integer solution to (1), then $3 z \leqslant 2009$, i.e., $1 \leqslant z \leqslant 669$, which respectively yield
$$x+2 y=2009,2006, \cdots, 5$$
Correspondingly, the range of values for $y$ are
$$\begin{array}{l}
1 \leqslant y \leqslant 1004,1 \leqslant y \leqslant 1002,1 \leqslant y \leqslant 1001 \\
1 \leqslant y \leqslant 999, \cdots, 1 \leqslant y \leqslant 2
\end{array}$$
Since when $y$ and $z$ are determined, the value of $x$ is uniquely determined, the number of positive integer solutions to (1) is
$$\begin{aligned}
& (1004+1002)+(1001+999)+\cdots+(5+3)+2 \\
= & 2006+2000+\cdots+8+2 \\
= & \frac{1}{2}(2006+2) \times 335=336340
\end{aligned}$$
In summary, there are 336340 sets of positive integer solutions. | 336340 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
Example 5 Find all positive integer arrays $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$, such that
$$\left\{\begin{array}{l}
a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}, \\
a_{1}+a_{2}+\cdots+a_{n}=26, \\
a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=62, \\
a_{1}^{3}+a_{2}^{3}+\cdots+a_{n}^{3}=164 .
\end{array}\right.$$ | Given that $a_{i}$ are all positive integers, and $6^{3}=216>164$, we know that each $a_{i}$ is no greater than 5. Let $x_{1}, x_{2}, \cdots, x_{5}$ be the number of 1, 2, 3, 4, 5 in $a_{1}, a_{2}, \cdots, a_{n}$, respectively. Then,
$$\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=n \\
x_{1}+2 x_{2}+3 x_{3}+4 x_{4}+5 x_{5}=26 \\
x_{1}+4 x_{2}+9 x_{3}+16 x_{4}+25 x_{5}=62 \\
x_{1}+8 x_{2}+27 x_{3}+64 x_{4}+125 x_{5}=164
\end{array}\right.$$
where $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ are all non-negative integers.
First, determine the values of $x_{1}, x_{2}, \cdots, x_{5}$, which is equivalent to finding the non-negative integer solutions of an indeterminate system of equations.
From (4), we know $x_{5} \leqslant 1$. If $x_{5}=1$, then (3) and (4) become
$$\left\{\begin{array}{l}
x_{1}+4 x_{2}+9 x_{3}+16 x_{4}=37 \\
x_{1}+8 x_{2}+27 x_{3}+64 x_{4}=39
\end{array}\right.$$
Subtracting (5) from (6) gives
$$4 x_{2}+18 x_{3}+48 x_{4}=2$$
This cannot hold when $x_{2}, x_{3}, x_{4}$ are non-negative integers, so
$$x_{5}=0$$
Subtracting (2) from (3) gives
$$2 x_{2}+6 x_{3}+12 x_{4}=36$$
which simplifies to
$$x_{2}+3 x_{3}+6 x_{4}=18$$
Subtracting (2) from (4) gives
$$x_{2}+4 x_{3}+10 x_{4}=23$$
Subtracting (7) from (8) gives
$$x_{3}+4 x_{4}=5$$
Thus,
$$\left(x_{3}, x_{4}\right)=(5,0),(1,1)$$
Consequently, we have $\quad\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=(5,3,5,0),(1,9,1,1)$.
Using the above conclusions, we get
$$\begin{aligned}
& \left(a_{1}, a_{2}, \cdots, a_{n}\right) \\
= & (1,1,1,1,1,2,2,2,3,3,3,3,3) \\
& (1,2,2,2,2,2,2,2,2,2,3,4)
\end{aligned}$$ | (1,1,1,1,1,2,2,2,3,3,3,3,3) \text{ and } (1,2,2,2,2,2,2,2,2,2,3,4) | Algebra | math-word-problem | Yes | Yes | number_theory | false |
Example 6 Arrange all the simplest fractions with denominators no greater than 99 in ascending order, find the two numbers adjacent to $\frac{17}{76}$. | Let $\frac{q}{p}$ and $\frac{n}{m}$ be the two numbers adjacent to $\frac{17}{76}$, and
$$\frac{q}{p}<\frac{17}{76}<\frac{n}{m}$$
with
$$\begin{array}{l}
17 p-76 q>0 \\
17 p-76 q \geqslant 1
\end{array}$$
First, consider the equation $17 p-76 q=1$ and find the largest positive integer solution for $p \leqslant 99$. To do this, we need to find a particular solution. Using
$$p=\frac{1}{17}(76 q+1)=4 q+\frac{1}{17}(8 q+1)$$
we can find a particular solution $(p, q)=(9,2)$. Thus, the general solution to this indeterminate equation is
$$(p, q)=(9+76 t, 2+17 t)$$
where $t$ is an integer. Under the condition $p \leqslant 99$, the maximum value of $p$ is 85, at which $q=19$.
On the other hand, from
$$\frac{17}{76}-\frac{q}{p}=\frac{17 p-76 q}{76 p}$$
we know that if $17 p-76 q \geqslant 2$, then
$$\frac{17}{76}-\frac{q}{p} \geqslant \frac{2}{76 p}=\frac{1}{38 p} \geqslant \frac{1}{38 \times 99}>\frac{1}{76 \times 85}=\frac{17}{76}-\frac{19}{85}$$
Therefore, under the given conditions, the number smaller than and closest to $\frac{17}{76}$ is $\frac{19}{85}$.
Similarly, we can determine that $\frac{n}{m}=\frac{15}{67}$.
In summary, the two numbers adjacent to $\frac{17}{76}$ in the sequence of these fractions are $\frac{19}{85}$ and $\frac{15}{67}$. | \frac{19}{85} \text{ and } \frac{15}{67} | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 1 Find the integer solutions of the equation
$$x y-10(x+y)=1$$ | Solving by using the cross-multiplication method, we can transform (1) into
$$(x-10)(y-10)=101$$
Since 101 is a prime number, we have
$$\begin{aligned}
& (x-10, y-10) \\
= & (1,101),(101,1),(-1,-101),(-101,-1) .
\end{aligned}$$
Solving each case, we get the integer solutions of the equation as
$$(x, y)=(11,111),(111,11),(9,-91),(-91,9)$$ | (x, y)=(11,111),(111,11),(9,-91),(-91,9) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 3 Find all pairs of positive integers $(m, n)$ such that
$$n^{5}+n^{4}=7^{m}-1$$ | Solving, by rearranging and factoring (1), we get
$$\begin{aligned}
7^{m} & =n^{5}+n^{4}+1=n^{5}+n^{4}+n^{3}-\left(n^{3}-1\right) \\
& =n^{3}\left(n^{2}+n+1\right)-(n-1)\left(n^{2}+n+1\right) \\
& =\left(n^{3}-n+1\right)\left(n^{2}+n+1\right) .
\end{aligned}$$
From (1), we know $n>1$, and when $n=2$, we get $m=2$.
Next, consider the case where $n>2$. We first look at the greatest common divisor of the two expressions on the right side of (2).
$$\begin{aligned}
& \left(n^{3}-n+1, n^{2}+n+1\right)=\left(n^{3}-n+1-\left(n^{2}+n+1\right)(n-1), n^{2}+n+1\right) \\
= & \left(-n+2, n^{2}+n+1\right)=\left(-n+2, n^{2}+n+1+(-n+2)(n+3)\right) \\
= & (-n+2,7) .
\end{aligned}$$
Thus, $\left(n^{3}-n+1, n^{2}+n+1\right) \mid 7$.
Combining this with (2), we know that $n^{3}-n+1$ and $n^{2}+n+1$ are both powers of 7, and when $n \geqslant 3$, they are both greater than 7, which implies $7^{2} \mid\left(n^{3}-n+1, n^{2}+n+1\right)$, leading to a contradiction with the previous result.
In conclusion, the only solution is $(m, n)=(2,2)$. | (m, n)=(2,2) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 4 Find all integer solutions of the indefinite equation $3 x^{2}-4 x y+3 y^{2}=35$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solve: Multiply both sides of the equation by 3, and complete the square to get
$$(3 x-2 y)^{2}+5 y^{2}=105$$
From equation (1), we get
$$5 y^{2} \leqslant 105$$
Therefore,
$$|y| \leqslant 4$$
When $|y|=4$, $|3 x-2 y|=5$, at this time the solutions to the original equation are
$$(x, y)=(1,4),(-1,-4)$$
When $|y|=1$, $|3 x-2 y|=10$, at this time the solutions to the original equation are
$$(x, y)=(4,1),(-4,-1)$$
When $|y|=0,2,3$, $(3 x-2 y)^{2}$ are $105,85,60$ respectively. In these cases, the resulting equations clearly have no integer solutions.
The above discussion shows that the original equation has 4 solutions:
$$(x, y)=(4,1),(1,4),(-4,-1),(-1,-4)$$ | (x, y)=(4,1),(1,4),(-4,-1),(-1,-4) | Algebra | math-word-problem | Yes | Yes | number_theory | false |
Example 5 Find the integer solutions of the equation $x^{2}+x=y^{4}+y^{3}+y^{2}+y$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solving as in the previous example, multiply both sides of the equation by 4 and complete the square on the left side, we get
$$(2 x+1)^{2}=4\left(y^{4}+y^{3}+y^{2}+y\right)+1 .$$
Next, we estimate the right side of equation (1). Since
$$\begin{aligned}
& 4\left(y^{4}+y^{3}+y^{2}+y\right)+1 \\
= & \left(2 y^{2}+y+1\right)^{2}-y^{2}+2 y \\
= & \left(2 y^{2}+y\right)^{2}+3 y^{2}+4 y+1,
\end{aligned}$$
thus, when $y>2$ or $y<-1$, we have
$$\left(2 y^{2}+y\right)^{2}<(2 x+1)^{2}<\left(2 y^{2}+y+1\right)^{2} .$$
Since $2 y^{2}+y$ and $2 y^{2}+y+1$ are two consecutive integers, there cannot be a perfect square between their squares, so the above inequality does not hold.
Therefore, we only need to consider the solutions of the equation when $-1 \leqslant y \leqslant 2$, which is trivial. It is easy to find that all integer solutions of the original equation are
$$(x, y)=(0,-1),(-1,-1),(0,0),(-1,0),(-6,2),(5,2)$$ | (x, y)=(0,-1),(-1,-1),(0,0),(-1,0),(-6,2),(5,2) | Algebra | math-word-problem | Yes | Yes | number_theory | false |
Example 6 Find all positive integers $n \geqslant 2$, such that the system of equations
$$\left\{\begin{array}{c}
x_{1}^{2}+x_{2}^{2}+50=16 x_{1}+12 x_{2} \\
x_{2}^{2}+x_{3}^{2}+50=16 x_{2}+12 x_{3} \\
\cdots \\
x_{n-1}^{2}+x_{n}^{2}+50=16 x_{n-1}+12 x_{n} \\
x_{n}^{2}+x_{1}^{2}+50=16 x_{n}+12 x_{1}
\end{array}\right.$$
has integer solutions. | Solving by rearranging and completing the square, the system of equations transforms to
$$\left\{\begin{array}{c}
\left(x_{1}-8\right)^{2}+\left(x_{2}-6\right)^{2}=50 \\
\left(x_{2}-8\right)^{2}+\left(x_{3}-6\right)^{2}=50 \\
\cdots \\
\left(x_{n-1}-8\right)^{2}+\left(x_{n}-6\right)^{2}=50 \\
\left(x_{n}-8\right)^{2}+\left(x_{1}-6\right)^{2}=50
\end{array}\right.$$
Since 50 can be expressed as the sum of squares of two positive integers in only two ways: \(50=1^{2}+7^{2}=5^{2}+5^{2}\), it follows from (1) that \(\left|x_{2}-6\right|=1, 5\) or 7, and from (2) that \(\left|x_{2}-8\right|=1, 5\) or 7, thus \(x_{2}=1, 7\) or 13.
Furthermore, for each \(1 \leqslant i \leqslant n\), we have \(x_{i}=1, 7\) or 13. Depending on \(x_{1}=1, 7, 13\), we consider three cases.
If \(x_{1}=1\), then from (1) we know \(x_{2}=7\), and from (2) we know \(x_{3}=13\), and so on. Thus, when \(k \equiv 1(\bmod 3)\), \(x_{k}=1\); when \(k \equiv 2(\bmod 3)\), \(x_{k}=7\); and when \(k \equiv 0(\bmod 3)\), \(x_{k}=13\). Therefore, by the second equation, we know that the original system of equations has integer solutions if and only if \(n+1 \equiv 1(\bmod 3)\), i.e., if and only if \(3 \mid n\), \(n\) meets the requirement.
For the other two cases \(x_{1}=7\) and \(x_{1}=13\), similar discussions yield the same condition.
In summary, the \(n\) that satisfies the condition are all multiples of 3. | n = 3k | Algebra | math-word-problem | Yes | Yes | number_theory | false |
Example 7 Find the positive integer solutions of the indeterminate equation $x^{3}-y^{3}=x y+61$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Let $(x, y)$ be a positive integer solution of the equation, then $x>y$. Let $x=y+d$, then $d$ is a positive integer, and
$$\begin{aligned}
(y+d) y+61 & =(y+d)^{3}-y^{3} \\
& =3 d y^{2}+3 y d^{2}+d^{3}
\end{aligned}$$
Thus we have
$$(3 d-1) y^{2}+d(3 d-1) y+d^{3}=61$$
Therefore,
$$\begin{array}{l}
d^{3}<61 \\
d \leqslant 3
\end{array}$$
Substituting $d=1, 2, 3$ respectively, we get
$$\begin{array}{l}
2 y^{2}+2 y+1=61 \\
5 y^{2}+10 y+8=61 \\
8 y^{2}+24 y+27=61
\end{array}$$
Only the first equation has integer solutions, and since $y$ is a positive integer, $y=5$, and thus $x=6$. Therefore, the original equation has only one positive integer solution $(x, y)=(6,5)$. | (x, y)=(6,5) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 8 Find all positive integers $a, b$ such that
$$4^{a}+4 a^{2}+4=b^{2}$$ | If $(a, b)$ is a pair of positive integers satisfying (1), then $b^{2}$ is even, and $b^{2}>4^{a}$, so $b$ is even, and $b>2^{a}$, hence $b \geqslant 2^{a}+2$. Therefore,
$$4^{a}+4 a^{2}+4=b^{2} \geqslant\left(2^{a}+2\right)^{2}=4^{a}+4 \cdot 2^{a}+4$$
we have $a^{2} \geqslant 2^{a}$, which implies $a \leqslant 4$ (by induction on $a$, we can prove that when $a \geqslant 5$, $a^{2}<2^{a}$).
Substituting $a=1,2,3,4$ into equation (1), we find that all positive integer solutions of the equation are $(a, b)=$ $(2,6)$ or $(4,18)$. | (2,6) \text{ or } (4,18) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 9 Find all positive integer tuples $(a, b, c, x, y, z)$ such that
$$\left\{\begin{array}{l}
a+b+c=x y z, \\
x+y+z=a b c,
\end{array}\right.$$
where $a \geqslant b \geqslant c, x \geqslant y \geqslant z$. | By symmetry, we only need to consider the case $x \geqslant a$. At this time,
$$x y z=a+b+c \leqslant 3 a \leqslant 3 x$$
Thus,
$$y z \leqslant 3$$
Therefore,
$$(y, z)=(1,1),(2,1),(3,1)$$
When $(y, z)=(1,1)$, $a+b+c=x$ and $x+2=a b c$, thus
$$a b c=a+b+c+2$$
If $c \geqslant 2$, then
$$a+b+c+2 \leqslant 3 a+2 \leqslant 4 a \leqslant a b c$$
Equality holds if and only if $a=b=c=2$.
If $c=1$, then
$$a b=a+b+3$$
$$(a-1)(b-1)=4$$
$$(a, b)=(5,2),(3,3)$$
When $(y, z)=(2,1)$, $2 a b c=2 x+6=a+b+c+6$, similar discussion shows that $c=1$, thus
$$(2 a-1)(2 b-1)=15$$
We get
$$(a, b)=(3,2)$$
When $(y, z)=(3,1)$, $3 a b c=3 x+12=a+b+c+12$, similarly, there is no solution in this case.
In summary, we have
$$\begin{aligned}
& (a, b, c, x, y, z) \\
= & (2,2,2,6,1,1),(5,2,1,8,1,1),(3,3,1,7,1,1) \\
& (3,2,1,3,2,1),(6,1,1,2,2,2),(8,1,1,5,2,1) \\
& (7,1,1,3,3,1)
\end{aligned}$$ | (2,2,2,6,1,1),(5,2,1,8,1,1),(3,3,1,7,1,1),(3,2,1,3,2,1),(6,1,1,2,2,2),(8,1,1,5,2,1),(7,1,1,3,3,1) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 11 Find all non-negative integers $x, y, z$, such that
$$2^{x}+3^{y}=z^{2}$$ | Solve (1) When $y=0$, we have
$$2^{x}=z^{2}-1=(z-1)(z+1)$$
Thus, we can set
$$z-1=2^{\alpha}, z+1=2^{\beta}, 0 \leqslant \alpha \leqslant \beta$$
Therefore,
$$2^{\beta}-2^{\alpha}=2$$
At this point, if $\alpha \geqslant 2$, then $4 \mid 2^{\beta}-2^{\alpha}$, which contradicts $4 \nmid 2$, so $\alpha \leqslant 1$. And $\alpha=0$ leads to $2^{\beta}=$
3, which is a contradiction, hence
So
$$\begin{array}{l}
\alpha=1, \beta=2 \\
z=3, x=3
\end{array}$$
We get
$$(x, y, z)=(3,0,3)$$
(2) When $y>0$, since $3 \nmid 2^{x}+3^{y}$, it follows that $3 \nmid z$, so
$$z^{2} \equiv 1(\bmod 3)$$
Taking (1) modulo 3, we know
$$(-1)^{x} \equiv 1(\bmod 3)$$
Thus, $x$ is even. Now set $x=2 m$, then
$$\left(z-2^{m}\right)\left(z+2^{m}\right)=3^{y}$$
So we can set
$$z-2^{m}=3^{\alpha}, z+2^{m}=3^{\beta}, 0 \leqslant \alpha \leqslant \beta, \alpha+\beta=y$$
Thus,
$$3^{\beta}-3^{\alpha}=2^{m+1},$$
If $\alpha \geqslant 1$, then $3 \mid 3^{\beta}-3^{\alpha}$, but $3 \nmid 2^{m+1}$, which is a contradiction, so $\alpha=0$, hence
$$3^{\beta}-1=2^{m+1}$$
When $m=0$, $\beta=1$, we get
$$(x, y, z)=(0,1,2)$$
When $m>0$, $2^{m+1} \equiv 0(\bmod 4)$, so
$$3^{\beta} \equiv 1(\bmod 4)$$
This requires $\beta$ to be even. Set $\beta=2 n$, then
$$2^{m+1}=3^{2 n}-1=\left(3^{n}-1\right)\left(3^{n}+1\right)$$
As in the case when $y=0$, we know
$$3^{n}-1=2$$
Thus $n=1$, and $m=2$, we get
$$(x, y, z)=(4,2,5)$$
Therefore, $(x, y, z)=(3,0,3),(0,1,2),(4,2,5)$. | (x, y, z)=(3,0,3),(0,1,2),(4,2,5) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 12 Let $m, n$ be positive integers, and $n>1$. Find the minimum value of $\left|2^{m}-5^{n}\right|$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
(Note: The note is for you, the assistant, and should not be included in the output.)
Example 12 Let $m, n$ be positive integers, and $n>1$. Find the minimum value of $\left|2^{m}-5^{n}\right|$. | Since $\left|2^{m}-5^{n}\right|$ is an odd number, and when $m=7, n=3$, $\left|2^{m}-5^{n}\right|=3$, if we can prove that when $n>1$, $\left|2^{m}-5^{n}\right| \neq 1$, then the minimum value sought is 3.
If there exist positive integers $m, n$, such that $n>1$, and $\left|2^{m}-5^{n}\right|=1$, then
$$2^{m}-5^{n}=1 \text{ or } 2^{m}-5^{n}=-1 \text{.}$$
If $2^{m}-5^{n}=1$, then $m \geqslant 3$. Taking both sides modulo 8, we require
$$5^{n} \equiv 7(\bmod 8)$$
But for any positive integer $n$, $5^{n} \equiv 1$ or $5(\bmod 8)$, which is a contradiction. Therefore, $2^{m}-5^{n}=1$ does not hold.
If $2^{m}-5^{n}=-1$, then by $n>1$, we know $m \geqslant 3$. Taking both sides modulo 8, we get
$$5^{n} \equiv 1(\bmod 8)$$
This implies that $n$ is even. Let $n=2x$, where $x$ is a positive integer, then
$$2^{m}=\left(5^{x}-1\right)\left(5^{x}+1\right)$$
Since $5^{x}-1$ and $5^{x}+1$ are two consecutive even numbers, this requires
$$5^{x}-1=2,5^{x}+1=4$$
which is impossible.
Therefore, the minimum value of $\left|2^{m}-5^{n}\right|$ is 3. | 3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
1 Given that the ages of A, B, and C are all positive integers, A's age does not exceed twice B's age, B is 7 years younger than C, the sum of the three people's ages is a prime number less than 70, and the sum of the digits of this prime number is 13. Question: What is the maximum age of A? | 1. Let Jia's age be $x$ years, and Yi's age be $y$ years, then Bing is $y+7$ years old, and $x \leqslant 2 y$.
Since the positive integers less than 70 and with a digit sum of 13 are only $49$, $58$, and $67$, the sum of the three people's ages (which is a prime number) can only be 67 years, i.e.,
$$x+y+(y+7)=67$$
This gives
$$x+2 y=60 .$$
Combining with
$$x \leqslant 2 y$$
We get
$$4 y \geqslant 60 \text{, i.e., } y \geqslant 15,$$
Thus,
$$x=60-2 y \leqslant 60-30=30,$$
Therefore, Jia is at most 30 years old (Note: Jia, Yi, and Bing being 30 years, 15 years, and 22 years old, respectively, meet the requirements). | 30 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
2 Given 12 bamboo poles, each 13 units long, divide them into smaller segments of lengths 3, 4, or 5, and then assemble them into 13 triangles with side lengths of 3, 4, or 5. How should they be divided? Please explain your reasoning. | 2. First, find the non-negative integer solutions of $3 a+4 b+5 c=13$. It is known that the lengths obtained after each bamboo stick is divided can only be $(3,3,3,4),(3,5,5),(4,4,5)$, a total of 3 cases. Then, let the number of times each case occurs be $x, y, z$, respectively, so we need to solve
$$\left\{\begin{array}{l}
3 x+y=13 \\
x+2 z=13 \\
2 y+z=13
\end{array}\right.$$
Solving this, we get
$$(x, y, z)=(3,4,5)$$
This means there are 3 bamboo sticks divided into $(3,3,3,4)$; 4 bamboo sticks divided into $(3,5,5)$; and another 5 bamboo sticks divided into $(4,4,5)$. | (x, y, z)=(3,4,5) | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
3i Let $n$ be a positive integer, and the indeterminate equation $x+2 y+2 z=n$ has exactly 28 sets of positive integer solutions. Find the value of $n$.
| 3. When $n$ is even, let $n=2 m$, then $x$ is even. For $x=2 k, 1 \leqslant k \leqslant m-1$, we have
$$y+z=m-k$$
In this case, there are $m-k-1$ solutions, meaning $x+2 y+2 z=2 m$ has a total of $(m-2)+(m-3)+\cdots+$
$$1+0=\frac{1}{2}(m-1)(m-2)$$
sets of positive integer solutions. Therefore,
$$\frac{1}{2}(m-1)(m-2)=28$$
we get
$$\begin{array}{l}
m=9 \\
n=18
\end{array}$$
Similarly, discussing the case $n=2 m+1$, we find $n=17$.
In summary, $n=17$ or 18. | n=17 \text{ or } 18 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
4. Positive integers $a, b, c, d$ satisfy: $1<a<b<c<d<1000$, and $a+d=b+c$, $bc-ad=2004$. Find the number of all such positive integer tuples $(a, b, c, d)$. | 4. Let $b=a+x, c=a+y$, then $x<y$, and $d=a+x+y$ (this is obtained from $a+d=b+c$), thus
$$b c-a d=(a+x)(a+y)-a(a+x+y)=x y$$
That is
$$x y=2004$$
Combining $a+x+y<1000$ and $2004=2^{2} \times 3 \times 167$, we know that $(x, y)=(3,668),(4,501),(6,334),(12,167)$.
Accordingly, $1<a<329,1<a<495,1<a<660,1<a<821$. Thus, the number of qualifying arrays is $327+493+658+819=2297$ (groups). | 2297 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
5 Find the smallest positive integer $c$, such that the indeterminate equation $x y^{2}-y^{2}-x+y=c$ has exactly three sets of positive integer solutions.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 5. Factorizing the left side of the equation, we get
$$(y-1)(x y+x-y)=c$$
Notice that, for any positive integer $c$, there is a solution
$$(x, y)=(1, c+1)$$
When $c$ is a prime number, there is at most one additional set of positive integer solutions. Therefore, to make the equation have exactly 3 sets of positive integer solutions, $c$ should be a composite number.
By direct calculation, it is known that the smallest $c$ that has exactly 3 sets of positive integer solutions is 10, and they are
$$(x, y)=(4,2),(2,3),(1,11)$$
The smallest positive integer
$$c=10$$ | 10 | Algebra | math-word-problem | Yes | Yes | number_theory | false |
7 rectangles can be divided into $n$ identical squares, and it can also be divided into $n+76$ identical squares, find the value of the positive integer $n$.
---
Note: The translation keeps the original format and line breaks as requested. However, the phrase "7-个长方形" is translated as "7 rectangles" for clarity in English, assuming it refers to 7 rectangles. If it refers to a single rectangle, the correct translation would be "A rectangle can be divided into $n$ identical squares...". Please clarify if needed. | 7. Let the side length of the squares when divided into $n$ squares be $x$, and the side length when divided into $n+76$ squares be $y$, then
$$n x^{2}=(n+76) y^{2}$$
Since both divisions are performed on the same rectangle, $\frac{x}{y}$ is a rational number (this can be seen by considering one side of the rectangle), thus $\frac{n+76}{n}=\left(\frac{x}{y}\right)^{2}$ is the square of a rational number. Therefore, we can set
$$n=k a^{2}, n+76=k b^{2}$$
where $k$, $a$, and $b$ are all positive integers. Consequently,
that is
$$\begin{array}{c}
k\left(b^{2}-a^{2}\right)=76 \\
k(b-a)(b+a)=76=2^{2} \times 19
\end{array}$$
Noting that $b-a$ and $b+a$ have the same parity, we have
$$(k, b-a, b+a)=(1,2,38),(4,1,19)$$
which gives
$$(k, a, b)=(1,18,20),(4,9,10)$$
The resulting $n$ is 324 in both cases. Therefore, $n=324$. | 324 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
8 Positive integers $x, y, z$ satisfy $\left\{\begin{array}{l}7 x^{2}-3 y^{2}+4 z^{2}=8, \\ 16 x^{2}-7 y^{2}+9 z^{2}=-3 .\end{array}\right.$ Find the value of $x^{2}+y^{2}+z^{2}$. | 8. Let $x, y, z$ satisfy
$$\left\{\begin{array}{l}
7 x^{2}-3 y^{2}+4 z^{2}=8 \\
16 x^{2}-7 y^{2}+9 z^{2}=-3
\end{array}\right.$$
Multiplying (1) by 7 and (2) by 3, we get
Substituting back into (1) yields
$$\begin{array}{c}
x^{2}+z^{2}=65 \\
3 x^{2}-3 y^{2}=8-260
\end{array}$$
Thus,
$$\begin{aligned}
y^{2}-x^{2} & =84 \\
(y-x)(y+x) & =2^{2} \times 3 \times 7
\end{aligned}$$
Since $y-x$ and $y+x$ have the same parity, we have
$$(y-x, y+x)=(2,42),(6,14)$$
Solving these, we get
$$(x, y)=(20,22),(4,10)$$
However,
$$x^{2}+z^{2}=65$$
Thus, it can only be
$$(x, y)=(4,10)$$
In this case,
$$z=7$$
Therefore,
$$x^{2}+y^{2}+z^{2}=165$$ | 165 | Algebra | math-word-problem | Yes | Yes | number_theory | false |
9 Find the integer solutions of the indeterminate equation $\left(x^{2}-y^{2}\right)^{2}=1+16 y$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 9. From the equation, we know $y \geqslant 0$, and $x \neq y$. Without loss of generality, let $x \geqslant 0$.
If $x>y$, then
we get
$$\begin{aligned}
1+16 y & =\left(x^{2}-y^{2}\right)^{2} \\
& \geqslant\left[(y+1)^{2}-y^{2}\right]^{2} \\
& =(2 y+1)^{2} \\
& =4 y^{2}+4 y+1 \\
0 & \leqslant y \leqslant 3 .
\end{aligned}$$
By
$$y=0,1,2,3,$$
we obtain the integer solutions of the original equation as
$$(x, y)=(1,0),(4,3)$$
If $x<y$, then
we get
$$\begin{aligned}
1+16 y & =\left(y^{2}-x^{2}\right)^{2} \\
& \geqslant\left[y^{2}-(y-1)^{2}\right]^{2} \\
& =(2 y-1)^{2} \\
& =4 y^{2}-4 y+1, \\
0 & \leqslant y \leqslant 5
\end{aligned}$$
By
$$y=0,1,2,3,4,5,$$
we obtain the integer solutions of the original equation as
$$(x, y)=(4,5)$$
When $x \leqslant 0$, we get
$$(x, y)=(-1,0),(-4,3),(-4,5)$$
In summary, the solutions of the equation are $(x, y)=( \pm 1,0),( \pm 4,3),( \pm 4,5)$. | (x, y)=( \pm 1,0),( \pm 4,3),( \pm 4,5) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
10 Find all integers $x, y$ such that $x^{2} + xy + y^{2} = 1$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 10. Multiply both sides by 4, then complete the square, to get
$$(2 x+y)^{2}+3 y^{2}=4$$
Thus, $4-3 y^{2}$ is a perfect square, which requires $y^{2}=0$ or 1, corresponding to $(2 x+y)^{2}=4,1$. Solving these respectively yields
$$(x, y)=( \pm 1,0),(0, \pm 1),(1,-1),(-1,1)$$ | (x, y)=( \pm 1,0),(0, \pm 1),(1,-1),(-1,1) | Algebra | math-word-problem | Yes | Yes | number_theory | false |
111 Determine all positive integers $x, y$ that satisfy $x+y^{2}+z^{3}=x y z$, where $z$ is the greatest common divisor of $x$ and $y$. | 11. From the conditions, we can set \(x = za, y = zb\), where \(a, b\) are positive integers, and \((a, b) = 1\). Substituting into the equation, we get
$$a + z b^{2} + z^{2} = z^{2} a b$$
Thus, \(z \mid a\). Let \(a = zm\), then the equation becomes
$$m + b^{2} + z = z^{2} m b$$
Therefore, \(b \mid m + z\). Let \(m + z = bk\), then
Hence
$$\begin{aligned}
b^{2} + b k & = z^{2} m b \\
b + k & = z^{2} m
\end{aligned}$$
Notice that \(b k - (b + k) + 1 = (b - 1)(k - 1) \geqslant 0\),
Thus
$$b + k \leqslant b k + 1$$
Therefore
$$z^{2} m = b + k \leqslant b k + 1 = z + m + 1$$
Thus
$$\begin{array}{c}
m(z^{2} - 1) \leqslant z + 1 \\
m(z - 1) \leqslant 1 \\
z \leqslant 2
\end{array}$$
If \(z = 1\), then from (1) we have
Thus
$$\begin{array}{c}
m + b^{2} + 1 = m b \\
b^{2} - m b + (m + 1) = 0
\end{array}$$
This requires \(\Delta = m^{2} - 4(m + 1)\) to be a perfect square. Let \(m^{2} - 4(m + 1) = n^{2}\), where \(n\) is a non-negative integer, then
$$(m - n - 2)(m + n - 2) = 8$$
Solving this, we get
$$(m, n) = (5, 1)$$
Thus
$$b = 2, 3,$$
Therefore
$$(x, y, z) = (5, 2, 1), (5, 3, 1)$$
Thus
$$\begin{array}{c}
m(z - 1) \leqslant 1 \\
m = 1
\end{array}$$
Therefore, from (1) we get
Thus
$$\begin{array}{c}
b^{2} - 4 b + 3 = 0 \\
b = 1, 3
\end{array}$$
Therefore
$$(x, y, z) = (4, 2, 2), (4, 6, 2)$$
In summary, the solutions \((x, y)\) that satisfy the conditions are \((5, 2), (5, 3), (4, 2), (4, 6)\). | (5, 2), (5, 3), (4, 2), (4, 6) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
12 Find all integers $m, n$ such that $m^{4}+(m+1)^{4}=n^{2}+(n+1)^{2}$. | 12. After expanding and rearranging, we get
$$m^{4}+2 m^{3}+3 m^{2}+2 m=n^{2}+n$$
Adding 1 to both sides and completing the square, we get
$$\left(m^{2}+m+1\right)^{2}=n^{2}+n+1$$
When $n>0$,
$$n^{2}<n^{2}+n+1<(n+1)^{2}$$
When $n<-1$, $\quad(n+1)^{2}<n^{2}+n+1<n^{2}$,
the number $n^{2}+n+1$ is never a perfect square. Therefore, for (1) to hold, we need $n=-1$ or 0, which leads to
$$(m, n)=(-1,0),(0,0),(-1,-1),(0,-1)$$ | (m, n)=(-1,0),(0,0),(-1,-1),(0,-1) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
14 Let $x, y$ be real numbers greater than 1, and let $a=\sqrt{x-1}+\sqrt{y-1}, b=\sqrt{x+1}+$ $\sqrt{y+1}$, where $a, b$ are two non-consecutive positive integers. Find the values of $x, y$. | 14. From the conditions, we know
$$\begin{aligned}
b-a & =(\sqrt{x+1}-\sqrt{x-1})+(\sqrt{y+1}-\sqrt{y-1}) \\
& =\frac{2}{\sqrt{x+1}+\sqrt{x-1}}+\frac{2}{\sqrt{y+1}+\sqrt{y-1}} \\
& 2-\frac{2}{\sqrt{2}}=2-\sqrt{2}
\end{aligned}$$
Therefore,
$$\sqrt{x+1}+\sqrt{x-1}<\frac{2}{2-\sqrt{2}}=2+\sqrt{2}$$
Similarly,
$$\sqrt{y+1}+\sqrt{y-1}<2+\sqrt{2}$$
This indicates
$$a+b=(\sqrt{x+1}+\sqrt{x-1})+(\sqrt{y+1}+\sqrt{y-1})<4+2 \sqrt{2},$$
Thus,
$$a+b \leqslant 6 \text {. }$$
Combining $b-a=2$ and the fact that $b-a$ and $b+a$ have the same parity, we have
$$\begin{array}{c}
(b-a, b+a)=(2,4),(2,6) \\
(a, b)=(1,3),(2,4)
\end{array}$$
Solving each case, we find that only $(a, b)=(1,3)$ has a solution, which is
$$x=y=\frac{5}{4} .$$ | x=y=\frac{5}{4} | Algebra | math-word-problem | Yes | Yes | number_theory | false |
15 Positive integers $a, b, c$ satisfy: $[a, b]=1000,[b, c]=2000,[c, a]=2000$. Find the number of such ordered positive integer triples $(a, b, c)$. | 15. From the conditions, we can set
$$a=2^{\alpha_{1}} \cdot 5^{\beta_{1}}, b=2^{\alpha_{2}} \cdot 5^{\beta_{2}}, c=2^{\alpha_{3}} \cdot 5^{\beta_{3}}$$
Then
$$\begin{array}{l}
\max \left\{\alpha_{1}, \alpha_{2}\right\}=3, \max \left\{\alpha_{2}, \alpha_{3}\right\}=4, \max \left\{\alpha_{3}, \alpha_{1}\right\}=4 \\
\max \left\{\beta_{1}, \beta_{2}\right\}=\max \left\{\beta_{2}, \beta_{3}\right\}=\max \left\{\beta_{3}, \beta_{1}\right\}=3
\end{array}$$
Note that, the array $(a, b, c)$ is determined only when the non-negative integer arrays $\left(\alpha_{1}, \alpha_{2}, \alpha_{3}\right)$ and $\left(\beta_{1}, \beta_{2}, \beta_{3}\right)$ are both determined. From the conditions above, we know that $\alpha_{3}=4$, and at least one of $\alpha_{1}$ or $\alpha_{2}$ is 3, indicating that $\left(\alpha_{1}, \alpha_{2}, \alpha_{3}\right)$ has 7 possible combinations; at least two of $\beta_{1}$, $\beta_{2}$, and $\beta_{3}$ are equal to 3, so $\left(\beta_{1}, \beta_{2}, \beta_{3}\right)$ has 10 possible combinations.
In summary, the number of ordered arrays $(a, b, c)$ that satisfy the conditions is $7 \times 10 = 70$ (groups). | 70 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
18 Let $n$ be a given positive integer. Find all positive integers $m$, such that there exist positive integers $x_{1}<x_{2}<\cdots<x_{n}$, satisfying: $\frac{1}{x_{1}}+\frac{2}{x_{2}}+\cdots+\frac{n}{x_{n}}=m$. | 18. Let $m$ be a positive integer that meets the requirement, i.e., there exist positive integers $x_{1}<x_{2}<\cdots<x_{n}$, such that
$$\frac{1}{x_{1}}+\frac{2}{x_{2}}+\cdots+\frac{n}{x_{n}}=m$$
Then
$$x_{i} \geqslant i(1 \leqslant i \leqslant n)$$
Hence
$$\frac{i}{x_{i}} \leqslant 1,1 \leqslant i \leqslant n$$
Thus $\square$
$$m \leqslant n$$
Next, we prove: For any positive integer $m$, if $1 \leqslant m \leqslant n$, then there exist $x_{1}, x_{2}, \cdots, x_{n}$ that satisfy (1). When $m=n$, take $x_{i}=i$;
When $m=1$, take $x_{i}=n \cdot i$;
When $1<m<n$, take $x_{1}=1, x_{2}=2, \cdots, x_{n-1}=m-1, x_{m}=(n-m+1) m, x_{m+1}=(n-m+1)(m+1), \cdots, x_{n}=(n-m+1) \cdot n$.
Therefore, the $m$ that satisfies the condition is $1,2, \cdots, n$. | 1,2, \cdots, n | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
20 Find all integer pairs $(x, y)$, such that $x^{2}+3 y^{2}=1998 x$. | 20. Let $(x, y)$ be the integer pairs that satisfy the equation, it is known that $3 \mid x^{2}$, hence $3 \mid x$, set
then
$$\begin{array}{l}
x=3 x_{1} \\
3 x_{1}^{2}+y^{2}=1998 x_{1}
\end{array}$$
Thus,
$$3 \mid y^{2},$$
hence
$$3 \mid y \text {. }$$
Set again
$$y=3 y_{1},$$
then
$$x_{1}^{2}+3 y_{1}^{2}=666 x_{1}$$
Proceeding similarly, we can set
$$x=27 m, y=27 n$$
we get
$$m^{2}+3 n^{2}=74 m$$
Rearranging and completing the square, we get
$$(m-37)^{2}+3 n^{2}=37^{2}$$
Analyzing the parity of both sides of this equation, we find that $m$ and $n$ are both even, thus
$$3 n^{2}=37^{2}-(m-37)^{2} \equiv 0(\bmod 8)$$
Hence
set
then
thus
hence
$$\begin{array}{c}
n=4 r \\
48 r^{2} \leqslant 37^{2} \\
r^{2} \leqslant 28 \\
|r| \leqslant 5
\end{array}$$
For each
$$|r|=0,1, \cdots, 5$$
calculate the value of $37^{2}-48 r^{2}$, we find that only when $|r|=0,5$, $37^{2}-48 r^{2}$ is a perfect square, so $(x, y)=(0,0),(1998,0),(1350, \pm 540),(648, \pm 540)$ | (x, y)=(0,0),(1998,0),(1350, \pm 540),(648, \pm 540) | Algebra | math-word-problem | Yes | Yes | number_theory | false |
21 Find all positive integer solutions to the indeterminate equation $7^{x}-3 \cdot 2^{y}=1$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 21. When $x=1$, it is known that $y=1$. Now consider the case $x \geqslant 2$, at this time $y \geqslant 4$, taking both sides modulo 8, we should have
$$(-1)^{x} \equiv 1(\bmod 8)$$
Thus $x$ is even, let $x=2 m$, then
$$\left(7^{m}-1\right)\left(7^{m}+1\right)=3 \cdot 2^{y}$$
Since $7^{m}-1$ and $7^{m}+1$ differ by 2, exactly one of them is a multiple of 3, there are two scenarios: $\square$
We get
Thus
$$\begin{array}{c}
7^{m}-1=3 \cdot 2^{u} \\
u=1 \\
y=u+v=4
\end{array}$$
If $m>1$, then
$$7^{m}+1=8 \cdot\left(7^{m-1}-7^{m-2}+\cdots-7+1\right)$$
Here, $7^{m-1}-7^{m-2}+\cdots-7+1$ is the sum of an odd number of odd numbers, which contradicts $7^{m}+1=2^{u}$.
In summary, the pairs $(x, y)$ that satisfy the conditions are $(1,1),(2,4)$. | (1,1),(2,4) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
23 Let $k$ and $m$ be positive integers. Find the minimum possible value of $\left|36^{k}-5^{m}\right|$. | 23. Notice that
$$36^{k}-5^{m} \equiv \pm 1(\bmod 6)$$
and
$$36^{k}-5^{m} \equiv 1(\bmod 5)$$
Therefore, the positive integers that $36^{k}-5^{m}$ can take, in ascending order, are $1, 11, \cdots$.
If $36^{k}-5^{m}=1$, then $k>1$, so taking both sides modulo 8, we require
$$5^{m} \equiv-1(\bmod 8)$$
However, $5^{m} \equiv 1$ or $5(\bmod 8)$, which is a contradiction.
Additionally, when $k=1, m=2$, we have $36^{k}-5^{m}=11$. Therefore, the smallest positive integer that $36^{k}-5^{m}$ can take is 11.
Now consider the smallest positive integer that $5^{m}-36^{k}$ can take. From
$$\begin{array}{l}
5^{m}-36^{k} \equiv 4(\bmod 5) \\
5^{m}-36^{k} \equiv \pm 1(\bmod 6)
\end{array}$$
we know that
$$5^{m}-36^{k} \geqslant 19$$
In summary, the smallest possible value of $\left|36^{k}-5^{m}\right|$ is 11. | 11 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
24 Find all positive integers $x, y, z$, such that $3^{x}+4^{y}=5^{z}$. | 24. Taking both sides of the equation modulo 3, we can see that $z$ is even, so $3^{x}+2^{2 y}$ is a perfect square. Thus, using the conclusion from Example 11 in Section 3.2, we can determine that $x=y=z=2$. | x=y=z=2 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
25 Find all Pythagorean triples $(x, y, z)$ such that $x<y<z$, and $x, y, z$ form an arithmetic sequence. | 25. From the conditions, we know $x^{2}+y^{2}=z^{2}$ and $x+z=2 y$, then
$$y^{2}=(z-x)(z+x)=2 y(z-x)$$
Thus,
$$y=2(z-x)$$
Therefore, $y$ is even. Let $y=2 m$, then
we get
$$\begin{array}{c}
x+z=4 m, z-x=m \\
x=\frac{3 m}{2}, z=\frac{5 m}{2}
\end{array}$$
Hence, $m$ is even. Therefore,
$$(x, y, z)=(3 n, 4 n, 5 n)$$
where $n=\frac{m}{2}$ is a positive integer. | (x, y, z) = (3n, 4n, 5n) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
30 Given that the three sides of $\triangle A B C$ are all integers, $\angle A=2 \angle B, \angle C>90^{\circ}$. Find the minimum perimeter of $\triangle A B C$. | 30. Let the corresponding side lengths of $\triangle ABC$ be $a, b, c$. Draw the angle bisector of $\angle A$ intersecting $BC$ at point $D$, then
$$C D=\frac{a b}{b+c},$$
Using $\triangle A C D \backsim \triangle B C A$, we know
$$\begin{array}{c}
\frac{C D}{b}=\frac{b}{a} \\
a^{2}=b(b+c)
\end{array}$$
That is $\square$
And
Therefore
By
Let
$$\begin{array}{c}
\angle C>90^{\circ} \\
c^{2}>a^{2}+b^{2} \\
a^{2}=b(b+c) \\
(b, b+c)=d
\end{array}$$
Then $(b, c)=d$, and $d^{2} \mid a^{2}$, hence $d \mid a$.
To find the minimum value of $a+b+c$, we can set $d=1$, in which case both $b$ and $b+c$ are perfect squares. Let
$$b=m^{2}, b+c=n^{2}, m, n \in \mathbf{N}^{*}$$
Then $a=m n$. Using $a+b>c$ and $c^{2}>a^{2}+b^{2}$, we know
And $\square$
$$\begin{array}{c}
m n+m^{2}>n^{2}-m^{2} \\
\left(n^{2}-m^{2}\right)^{2}>(m n)^{2}+m^{4}
\end{array}$$
Thus
$$m>n-m$$
That is $\square$
$$n3 m^{2} n^{2},$$
That is $\square$
$$n^{2}>3 m^{2},$$
So $\square$
$$3 m^{2}<n^{2}<4 m^{2}$$
Thus, there is a perfect square between $3 m^{2}$ and $4 m^{2}$, which requires $m \geqslant 4$, at this time $n \geqslant 7$, hence
$$a+b+c \geqslant 4 \times 7+7^{2}=77$$
Clearly, $(a, b, c)=(28,16,33)$ satisfies the conditions, so the minimum perimeter of $\triangle A B C$ is 77. | 77 | Geometry | math-word-problem | Yes | Yes | number_theory | false |
33 Let $n$ be a positive integer, and let $d(n)$ denote the number of positive divisors of $n$, and $\varphi(n)$ denote the number of integers in $1,2, \cdots, n$ that are coprime to $n$. Find all $n$ such that $d(n)+\varphi(n)=n$. | 33. Let $n$ be a number that satisfies the condition, then $n>1$, and at this time, only one number (i.e., 1) is both a divisor of $n$ and coprime with $n$. Therefore, among the numbers $1,2, \cdots, n$, there is exactly one number that is neither a divisor of $n$ nor coprime with $n$.
Since $\varphi(n)$ is even when $n>1$, if $n$ is even, then for $n \geqslant 10$, the numbers $n-2$ and $n-4$ are neither coprime with $n$ nor divisors of $n$, so $n \geqslant 10$ and $n$ being even does not satisfy the condition; if $n$ is odd, then $d(n)$ is odd, in which case $n$ is a perfect square. Let $n=(2 m+1)^{2}$, if $m>1$, then the numbers $n-(2 m+1)$ and $n-2(2 m+1)$ are neither divisors of $n$ nor coprime with $n$, at this time, $n$ does not satisfy the condition.
In summary, we only need to verify for $n=2,4,6,8,9$, and it can be known that the numbers satisfying the condition are $n=6,8,9$. | n=6,8,9 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
38 Find all integers $a$ such that the equation $x^{2}+a x y+y^{2}=1$ has infinitely many integer solutions.
Find all integers $a$ such that the equation $x^{2}+a x y+y^{2}=1$ has infinitely many integer solutions. | 38. Discuss the number of integer solutions of the equation
$$x^{2}+a x y+y^{2}=1$$
for different values of $a$.
When $a=0$, the equation becomes $x^{2}+y^{2}=1$, which has only 4 sets of integer solutions.
If $a \neq 0$, then $(x, y)$ is a solution to equation (1) if and only if: $(x, -y)$ is a solution to the equation $x^{2}-a x y+y^{2}=1$. Therefore, we only need to discuss $ay$, then $(x, -a x+y)$ is also a solution to (1) (this solution can be obtained using Vieta's formulas when viewing (1) as a quadratic equation in $y$), and of course $( -a x+y, x)$ (satisfying $-a x+y>x>y$) is also a positive integer solution to (1), and by this recursion, we know that (1) has infinitely many positive integer solutions.
In summary, when $|a|>1$, equation (1) has infinitely many integer solutions, and when $|a| \leqslant 1$, (1) has only finitely many integer solutions. | |a|>1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
39 Find all positive integers $x, k, n (n \geqslant 2)$, such that
$$3^{k}-1=x^{n}$$ | 39. From the condition, we know $x \geqslant 2$.
If $n$ is even, then by the property of perfect squares $\equiv 0$ or $1(\bmod 3)$, we have $x^{n} \equiv 0$ or $1(\bmod 3)$, but $3^{k}-1 \equiv 2(\bmod 3)$, so there is no solution in this case.
If $n$ is odd, then
$$3^{k}=x^{n}-1=(x+1)\left(x^{n-1}-x^{n-2}+\cdots-x+1\right) .$$
Thus, both $x+1$ and $A=x^{n-1}-x^{n-2}+\cdots-x+1$ are powers of 3.
Notice that,
$$A \equiv(-1)^{n-1}-(-1)^{n-2}+\cdots-(-1)+1=n(\bmod (x+1))$$
Therefore, $n$ is a multiple of 3. Let $n=3 m$, and set $y=x^{m}$, then
$$3^{k}=y^{3}+1=(y+1)\left(y^{2}-y+1\right)$$
Thus, both $y+1$ and $y^{2}-y+1$ are powers of 3.
Let $y+1=3^{t}$, then from (1) we have
$$3^{k}=\left(3^{t}-1\right)^{3}+1=3^{3 t}-3^{2 t+1}+3^{t+1} .$$
If $t>1$, then $3 t>2 t+1>t+1$, in this case $3^{t+1} \| 3^{3 t}-3^{2 t+1}+3^{t+1}$, comparing both sides of (2) we require $k=t+1$, but in this case the left side of (2) is less than the right side. Therefore, $t=1$, hence $y=2, k=2$. Thus, $(x, k, n)=(2,2,3)$.
In summary, the positive integers $(x, k, n)=(2,2,3)$ satisfy the conditions. | (x, k, n)=(2,2,3) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
40 Find the smallest positive integer $n$ such that the indeterminate equation
$$n=x^{3}-x^{2} y+y^{2}+x-y$$
has no positive integer solutions. | 40. Let $F(x, y)=x^{3}-x^{2} y+y^{2}+x-y$, then $F(1,1)=1, F(1,2)=2$.
Therefore, when $n=1,2$, the equation has positive integer solutions.
Next, we prove that $F(x, y)=3$ has no positive integer solutions.
Consider the equation $F(x, y)=3$ as a quadratic equation in $y$
$$y^{2}-\left(x^{2}+1\right) y+x^{3}+x-3=0$$
If there exist positive integer solutions, then
$$\Delta=\left(x^{2}+1\right)^{2}-4\left(x^{3}+x-3\right)=x^{4}-4 x^{3}+2 x^{2}-4 x+13$$
must be a perfect square.
Notice that, when $x \geqslant 2$, we have $\Delta < (x^{2}-2 x-2)^{2}$, so when $x \geqslant 6$, $\Delta$ is not a perfect square. And when $x=1,2,3,4,5$, the corresponding $\Delta=8,-3,-8,29,168$ are not perfect squares, hence there are no positive integer solutions when $n=3$.
In summary, the smallest positive integer $n=3$. | 3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 3 Find all positive integers $a, b (a \leqslant b)$, such that
$$a b=300+7[a, b]+5(a, b) .$$ | Let $[a, b]=x,(a, b)=y$, by property 8 we know $a b=x y$, thus, (1) becomes
$$x y=300+7 x+5 y$$
which is $(x-5)(y-7)=5 \times 67$.
Since $[a, b] \geqslant(a, b)$, hence $x \geqslant y$, thus $x-5>y-7$, there are only the following two cases.
Case one: $x-5=67$ and $y-7=5$; in this case, $x=72, y=12$, thus, we can set $a=12 n, b=12 m,(m, n)=1$, and have $(12 n)(12 m)=a b=x y=12 \times 72$, combining with $a \leqslant b$, it can only be $(m, n)=(1,6)$ or $(2,3)$, corresponding to $(a, b)=(12,72)$ or $(24, 36)$.
Case two: $x-5=335$ and $y-7=1$; correspondingly, $x=340, y=8$, but $y=(a, b)$ is a factor of $x=[a, b]$, and $8 \nmid 340$, so there is no solution in this case.
In summary, the pairs $(a, b)$ that meet the conditions are $(12,72)$ or $(24,36)$. | (12,72) \text{ or } (24,36) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 4 Find all positive integers $a, b$ such that
$$(a, b)+9[a, b]+9(a+b)=7 a b .$$ | Let $(a, b)=d$, set $a=d x, b=d y$, then $(x, y)=1$ (by property 11), $[a, b]=d x y$ (by property 8), thus substituting into (1) we get
$$\begin{array}{c}
1+9 x y+9(x+y)=7 d x y \\
7 d=9+9\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{1}{x y}
\end{array}$$
Therefore,
$$9<7 d \leqslant 9+9\left(\frac{1}{1}+\frac{1}{1}\right)+\frac{1}{1 \times 1}=28$$
Hence,
$$2 \leqslant d \leqslant 4$$
When $d=2$, from (2) we get
$$5 x y-9(x+y)=1$$
Multiplying both sides by 5 and factoring the left side, we get
$$(5 x-9)(5 y-9)=86=2 \times 43$$
Thus, $(5 x-9,5 y-9)=(1,86), (86,1), (2,43), (43,2)$. Solving each case, we find that only $(x, y)=(2,19), (19,2)$, corresponding to $(a, b)=(4,38), (38,4)$.
Similarly discussing for $d=3,4$, we get $(a, b)=(4,4)$.
Therefore, the pairs $(a, b)$ that satisfy the condition are $(4,38), (38,4), (4,4)$. | (4,38), (38,4), (4,4) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 1 There are 2012 lamps, numbered $1, 2, \cdots, 2012$, arranged in a row in a corridor, and initially, each lamp is on. A mischievous student performed the following 2012 operations: for $1 \leqslant k \leqslant 2012$, during the $k$-th operation, the student toggled the switch of all lamps whose numbers are multiples of $k$. Question: How many lamps are still on at the end? | Let $1 \leqslant n \leqslant 2012$, we examine the state of the $n$-th lamp. According to the problem, the switch of this lamp is pulled $d(n)$ times. An even number of pulls does not change the initial state of the lamp, while an odd number of pulls changes the state of the lamp from its initial state.
Using the properties of $d(n)$ and the previous discussion, since there are exactly 44 perfect squares among the numbers $1,2, \cdots, 2012$, it follows that finally, $2012-44=1968$ lamps are still on. | 1968 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 2 Find all positive integers $n$ such that $n=d(n)^{2}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | When $n=1$, the condition is satisfied. Below, we consider the case where $n>1$.
From the condition, we know that $n$ is a perfect square, so $d(n)$ is odd. Let $d(n)=2k+1$. For any positive integer $d$, when $d \mid n$, we have $\left.\frac{n}{d} \right\rvert\, n$. Therefore, by pairing $d$ with $\frac{n}{d}$, we know that $d(n)$ equals twice the number of divisors of $n$ among the numbers $1, 2, \cdots, 2k-1$ plus 1. Since the even numbers in $1, 2, \cdots, 2k-1$ are not divisors of $n\left(=(2k+1)^{2}\right)$, combining this with $d(n)=2k+1$, we know that every odd number in $1, 2, \cdots, 2k-1$ is a divisor of $n$.
Notice that, when $k>1$, $(2k-1, 2k+1) = (2k-1, 2) = 1$, so $2k-1 \nmid (2k+1)^{2}$. Therefore, when $k>1$, $n=(2k+1)^{2}$ does not meet the requirement, so $k=1$, and $n$ can only be 9.
Direct verification shows that 1 and 9 satisfy the condition, so $n=1$ or 9. | n=1 \text{ or } 9 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 4 Let $m$ and $n$ be positive integers, and the product of all positive divisors of $m$ equals the product of all positive divisors of $n$. Ask: Must $m$ and $n$ be equal? | Solve $m$ and $n$ must be equal.
In fact, by pairing the positive divisor $d$ of $m$ with $\frac{m}{d}$, we know that the product of all positive divisors of $m$ is $m^{\frac{d(m)}{2}}$, thus, the condition is equivalent to
$$m^{d(m)}=n^{d(n)},$$
This equation indicates that $m$ and $n$ have the same prime factors. We can assume
$$m=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{k}^{\alpha_{k}}, n=p_{1}^{\beta_{1}} p_{2}^{\beta_{2}} \cdots p_{k}^{\beta_{k}},$$
where $p_{1}d(n)$, then for $1 \leqslant i \leqslant k$, we have $\alpha_{i}<\beta_{i}$, thus, $\alpha_{i}+1<\beta_{i}+1$, hence $\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \cdots\left(\alpha_{k}+1\right)<\left(\beta_{1}+1\right)\left(\beta_{2}+1\right) \cdots\left(\beta_{k}+1\right)$, which leads to $d(m)<d(n)$, a contradiction. Similarly, from $d(m)<d(n)$, using (2) can also lead to a contradiction. Therefore, $d(m)=d(n)$, and by (1) we get $m=n$. | m=n | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 5 Find all positive integers $x, y$ such that
$$y^{x}=x^{50}$$ | Let $x, y$ be positive integers satisfying the conditions, and let $x=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{k}^{a_{k}}$ be the prime factorization of $x$. Then, since $y$ is a positive integer, for $1 \leqslant i \leqslant k$, we have $x \mid 50 \alpha_{i}$. Now, let's first discuss the prime factors of $x$.
If $x$ has a prime factor $p$ different from 2 and 5, and let $p^{\alpha} \| x$, then from the previous result, we know $x \mid 50 \alpha$, which implies $p^{\alpha} \mid 50 \alpha$. Since $p \neq 2, 5$, it follows that $p^{\alpha} \mid \alpha$. However, for any prime $p$ and positive integer $\alpha$, we have $p^{\alpha} > \alpha$, so $p^{\alpha} \mid \alpha$ cannot hold. This indicates that the prime factors of $x$ can only be 2 or 5.
Thus, we can set $x=2^{\alpha} \cdot 5^{\beta}$ (where $\alpha, \beta$ are non-negative integers). In this case, $x \mid 50 \alpha, x \mid 50 \beta$, so $2^{\alpha} \mid 50 \alpha, 5^{\beta} \mid 50 \beta$. The former requires $2^{\alpha-1} \mid \alpha$, and the latter requires $5^{\beta-2} \mid \beta$. Note that when $\alpha \geqslant 3$, $2^{\alpha-1} > \alpha$, and when $\beta \geqslant 3$, $5^{\beta-2} > \beta$. Therefore, $0 \leqslant \alpha \leqslant 2, 0 \leqslant \beta \leqslant 2$. This indicates that $x$ can only take the values $1, 2, 2^{2}, 5, 5^{2}, 2 \times 5, 2^{2} \times 5, 2 \times 5^{2}, 2^{2} \times 5^{2}$.
Substituting the above values of $x$ into equation (1), we can obtain all solutions as $(x, y) = (1,1)$, $\left(2,2^{25}\right), \left(2^{2}, 2^{25}\right), \left(5,5^{10}\right), \left(5^{2}, 5^{4}\right), \left(10,10^{5}\right), (50,50), (100,10)$, a total of 8 solutions. | (x, y) = (1,1), \left(2,2^{25}\right), \left(2^{2}, 2^{25}\right), \left(5,5^{10}\right), \left(5^{2}, 5^{4}\right), \left(10,10^{5}\right), (50,50), (100,10) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
2 Find all integer $x$ such that $\left|4 x^{2}-12 x-27\right|$ is a prime number.
Find all integer $x$ such that $\left|4 x^{2}-12 x-27\right|$ is a prime number. | 2. From $\left|4 x^{2}-12 x-27\right|=|(2 x+3)(2 x-9)|$, we know that only when $|2 x+3|=1$ or $|2 x-9|=1$, the number $\left|4 x^{2}-12 x-27\right|$ can possibly be a prime number. According to this, the required $x=-2,-1,4$ or 5, and the corresponding $\left|4 x^{2}-12 x-27\right|$ are 13, 11, 11 or 13, all of which are prime numbers. | x=-2, -1, 4, 5 | Algebra | math-word-problem | Yes | Yes | number_theory | false |
Find all primes of the form $n^{n}+1$ that do not exceed $10^{19}$, where $n$ is a positive integer. | 7. When $n=1$, $n^{n}+1=2$ satisfies the condition. When $n>1$, let $n=2^{k} q$, where $q$ is an odd number. If $q>1$, as shown in the previous problem, $n^{n}+1$ is not a prime number, so $n=2^{k}$, where $k$ is a positive integer. At this point,
$$n^{n}+1=2^{k \cdot 2^{k}}+1=\left(2^{2^{k}}\right)^{k}+1$$
Further analysis shows that there exists a non-negative integer $m$ such that $k=2^{m}$, hence
$$n^{n}+1=2^{2^{2^{m}+m}}+1$$
When $m \geqslant 2$, $2^{m}+m \geqslant 6$, so $2^{2^{m}+m} \geqslant 2^{6}$, thus
$$\begin{aligned}
n^{n}+1 & \geqslant 2^{2^{6}}+1=2^{64}+1 \\
& =16 \times(1024)^{6}+1 \\
& >16 \times\left(10^{3}\right)^{6}+1 \\
& >10^{19}
\end{aligned}$$
Therefore, by $n^{n}+1 \leqslant 10^{19}$, we know $m \leqslant 1$. By setting $m=0,1$, we find $n^{n}+1=5,257$, both of which are prime numbers.
In summary, the required prime numbers are 2, 5, and 257. | 2, 5, 257 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
11 Given a positive integer $n$, among its positive divisors, there is at least one positive integer ending in each of the digits $0,1,2, \cdots, 9$. Find the smallest $n$ that satisfies this condition. | 11. The smallest $n$ that satisfies the condition is $270$.
In fact, from the condition, we know that $10 \mid n$, and we start the discussion from the factor of the last digit of $n$ being 9. If $9 \mid n$, then $90 \mid n$, and it can be directly verified that 90 and 180 are not multiples of any number ending in 7; if $19 \mid n$, then $190 \mid n$, and $n=190$ is not a multiple of any number ending in 7. However, 270 is a multiple of $10, 1, 2, 3, 54, 5, 6, 27, 18, 9$, which meets the conditions. Therefore, the smallest $n$ is 270. | 270 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
12 Find a 9-digit number $M$, such that the digits of $M$ are all different and non-zero, and for $m=2,3, \cdots$, 9, the left $m$ digits of $M$ are multiples of $m$.
Find a 9-digit number $M$, such that the digits of $M$ are all different and non-zero, and for $m=2,3, \cdots$, 9, the number formed by the left $m$ digits of $M$ is a multiple of $m$. | 12. Let $M=\overline{a_{1} a_{2} \cdots a_{9}}$ be a number that satisfies the given conditions. From the conditions, we know that $a_{5}=5$, and $a_{2}$, $a_{4}$, $a_{6}$, $a_{8}$ are a permutation of $2$, $4$, $6$, $8$. Consequently, $a_{1}$, $a_{3}$, $a_{7}$, $a_{9}$ are a permutation of $1$, $3$, $7$, $9$. Therefore,
$$a_{4}=2 \text { or } 6\left(\text { because } 4 \mid \overline{a_{3} a_{4}}\right),$$
Furthermore,
$$8 \mid \overline{a_{7} a_{8}}$$
Thus,
$$a_{8}=2,6$$
Hence,
$$\left(a_{4}, a_{8}\right)=(2,6),(6,2)$$
By analyzing these two cases further, we can find a number $M=381654729$ that satisfies the conditions. | 381654729 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
17 Let $a, b, c, d$ all be prime numbers, and $a>3b>6c>12d, a^{2}-b^{2}+c^{2}-d^{2}=1749$. Find all possible values of $a^{2}+b^{2}+c^{2}+d^{2}$. | 17. From $a^{2}-b^{2}+c^{2}-d^{2}=1749$ being odd, we know that one of $a, b, c, d$ must be even, indicating that $d=2$. Then,
from
$$\begin{array}{l}
a^{2}-b^{2}+c^{2}=1753 \\
a>3 b>6 c>12 d
\end{array}$$
we know $c \geqslant 5, b \geqslant 2 c+1, a \geqslant 3 b+1$, so
$$\begin{aligned}
a^{2}-b^{2}+c^{2} & \geqslant(3 b+1)^{2}-b^{2}+c^{2} \\
& =8 b^{2}+6 b+c^{2}+1 \\
& \geqslant 8(2 c+1)^{2}+6(2 c+1)+c^{2}+1 \\
& =33 c^{2}+44 c+15
\end{aligned}$$
Thus,
$$33 c^{2}+44 c+15 \leqslant 1753$$
Therefore, $c<7$, and combining $c \geqslant 5$ and $c$ being a prime number, we have $c=5$, leading to
$$a^{2}-b^{2}=1728=2^{6} \times 3^{3}$$
Using
$$b \geqslant 2 c+1=11, a \geqslant 3 b+1$$
we know
$$a-b \geqslant 2 b+1 \geqslant 23, a+b \geqslant 4 b+1 \geqslant 45$$
From $(a-b)(a+b)=2^{6} \times 3^{3}$ and $a, b$ both being odd primes, we have
$$(a-b, a+b)=(32,54)$$
Thus,
$$\begin{array}{c}
(a, b)=(43,11) \\
a^{2}+b^{2}+c^{2}+d^{2}=1749+2 \times\left(11^{2}+2^{2}\right)=1999
\end{array}$$ | 1999 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
19 The sequence of positive integers $\left\{a_{n}\right\}$ satisfies: for any positive integers $m, n$, if $m \mid n, m<n$, then $a_{m} \mid a_{n}$, and $a_{m}<a_{n}$. Find the minimum possible value of $a_{2000}$. | 19. From the conditions, when $m \mid n$ and $m < n$, we have $a_{n} \geqslant 2 a_{m}$. Therefore, $a_{1} \geqslant 1, a_{2} \geqslant 2, a_{4} \geqslant 2 a_{2} \geqslant 2^{2}$, similarly, $a_{8} \geqslant 2^{3}, a_{16} \geqslant 2^{4}, a_{30} \geqslant 2^{5}, a_{400} \geqslant 2^{6}, a_{2000} \geqslant 2^{7}$, which means $a_{2000} \geqslant 128$.
On the other hand, for any positive integer $n$, let the prime factorization of $n$ be
$$n=p_{1}^{q_{1}} p_{2}^{q_{2}} \cdots p_{k^{\prime}}^{q_{k}}$$
where $p_{1}<p_{2}<\cdots<p_{k}$ are prime numbers, and $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{k}$ are positive integers. Define
$$a_{n}=2^{a_{1}+a_{2}+\cdots+a_{k}}$$
Then the sequence $\left\{a_{n}\right\}$ satisfies the requirements of the problem, and
$$a_{2000}=2^{4+3} \leqslant 2^{7}$$
Therefore, the minimum value of $a_{2000}$ is 128. | 128 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
32 Given the pair of positive integers $(a, b)$ satisfies: the number $a^{a} \cdot b^{b}$ in decimal notation ends with exactly 98 zeros. Find the minimum value of $a b$.
| 32. Let the prime factorization of $a$ and $b$ have the powers of $2$ and $5$ as $\alpha_{1}$, $\beta_{1}$ and $\alpha_{2}$, $\beta_{2}$, respectively, then
$$\left\{\begin{array}{l}
a \cdot \alpha_{1} + b \cdot \alpha_{2} \geqslant 98 \\
a \cdot \beta_{1} + b \cdot \beta_{2} \geqslant 98
\end{array}\right.$$
and one of (1) and (2) must be an equality.
If (2) is an equality, i.e., $a \cdot \beta_{1} + b \cdot \beta_{2} = 98$, then when $\beta_{1}$ and $\beta_{2}$ are both positive integers, the left side is a multiple of 5. When $\beta_{1}$ or $\beta_{2}$ is zero, the other must be greater than zero, in which case the left side is still a multiple of 5, leading to a contradiction. Therefore, (1) must be an equality.
From $a \cdot \alpha_{1} + b \cdot \alpha_{2} = 98$, if $\alpha_{1}$ or $\alpha_{2}$ is zero, without loss of generality, let $\alpha_{2} = 0$, then $\alpha_{1} > 0$. In this case, $a \cdot \alpha_{1} = 98$. If $\alpha_{1} \geqslant 2$, then $4 \mid a$, which is a contradiction. Hence, $\alpha_{1} = 1$, and thus $a = 98$. Substituting $a = 98$ into (2), we know $\beta_{1} = 0$, so $b \cdot \beta_{2} > 98$. Combining $\alpha_{2} = 0$, we find that the minimum value of $b$ is 75.
If $\alpha_{1}$ and $\alpha_{2}$ are both positive integers, without loss of generality, let $\alpha_{1} \geqslant \alpha_{2}$. If $\alpha_{2} \geqslant 2$, then $4 \mid a$ and $4 \mid b$, leading to $4 \mid 98$, which is a contradiction. Hence, $\alpha_{2} = 1$. Further, if $\alpha_{1} = 1$, then $a + b = 98$, but $\frac{a}{2}$ and $\frac{b}{2}$ are both odd, so $\frac{a}{2} + \frac{b}{2}$ is even, which is a contradiction. Therefore, $\alpha_{1} > 1$. In this case, if $\beta_{1}$ and $\beta_{2}$ are both positive integers, then $5 \mid a$ and $5 \mid b$, which contradicts $a \cdot \alpha_{1} + b \cdot \alpha_{2} = 98$. Hence, one of $\beta_{1}$ and $\beta_{2}$ must be zero. If $\beta_{1} = 0$, then from (2) we know $b \cdot \beta_{2} > 98$, in which case the number of trailing zeros in $b^b$ is greater than 98 (since, in this case, $10 \mid b$. When $\beta_{2} = 1$, $b \geqslant 100$, so $10^{100} \mid b^b$. When $\beta_{2} \geqslant 2$, $50 \mid b$, if $b > 50$, then $10^{100} \mid b^b$; if $b = 50$, then $a \cdot \alpha_{1} = 48$, in which case when $\alpha_{1} \geqslant 4$, $2^5 \mid a \cdot \alpha_{1}$, and when $\alpha_{1} \leqslant 3$, $2^4 \nmid a \cdot \alpha_{1}$, both leading to a contradiction, so the number of trailing zeros in $b^b$ is greater than 98).
Similarly, if $\beta_{2} = 0$, then $a \cdot \beta_{1} > 98$, and similarly, the number of trailing zeros in $a^a$ is greater than 98, which is a contradiction. In summary, the minimum value of $ab$ is 7350 (when $(a, b) = (98, 75)$ or $(75, 98)$). | 7350 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
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