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3. (6 points) A construction company was building a tunnel. When $\frac{1}{3}$ of the tunnel was completed at the original speed, they started using new equipment, which increased the construction speed by $20 \%$ and reduced the working hours to $80 \%$ of the original. As a result, it took a total of 185 days to complete the tunnel. If they had not used the new equipment and continued at the original speed, it would have taken $\qquad$ days to complete the tunnel. | 3. (6 points) A construction company builds a tunnel. When $\frac{1}{3}$ of the tunnel is completed at the original speed, new equipment is used, increasing the construction speed by $20 \%$ and reducing the daily working hours to $80 \%$ of the original. As a result, the tunnel is completed in 185 days. If the new equipment was not used and the original speed was maintained, it would take $\qquad$ 180 days.
【Solution】Solution: $\left(1-\frac{1}{3}\right) \div[(1+20 \%) \times 80 \%]$
$$
\begin{array}{l}
=\frac{2}{3} \div[120 \% \times 80 \%], \\
=\frac{2}{3} \div \frac{24}{25} \\
=\frac{25}{36} ; \\
185 \div\left(\frac{1}{3}+\frac{25}{36}\right) \\
=185 \div \frac{37}{36}, \\
=180 \text { (days). }
\end{array}
$$
Answer: If the original speed was maintained, it would take 180 days.
Therefore, the answer is: 180. | 180 | Algebra | math-word-problem | olympiads | 1d57f4b3-c516-5522-91cc-a60dd91f6f66 | 2 |
Prove that number $1$ can be represented as a sum of a finite number $n$ of real numbers, less than $1,$ not necessarily distinct, which contain in their decimal representation only the digits $0$ and/or $7.$ Which is the least possible number $n$? | 1. **Restate the problem in a more manageable form:**
We need to prove that the number \(1\) can be represented as a sum of a finite number \(n\) of real numbers, each less than \(1\), and each containing only the digits \(0\) and \(7\) in their decimal representation. We also need to find the least possible value of \(n\).
2. **Reformulate the problem using a known rational number:**
Consider the rational number \(\frac{1}{7} = 0.\overline{142857}\). We need to express \(\frac{1}{7}\) as a sum of \(n\) real numbers, each less than \(\frac{1}{7}\) and containing only the digits \(0\) and \(1\) in their decimal representation.
3. **Assume \(n < 10\) and find the least value of \(n\):**
We assume that all the real numbers are positive and that \(n < 10\). The least value of \(n\) is found to be \(8\), for which we have the identity:
\[
\frac{1}{7} = 0.\overline{1} + 0.\overline{011111} + 0.\overline{010111} + 0.\overline{010111} + 0.\overline{000111} + 0.\overline{000101} + 0.\overline{000101} + 0.\overline{000100}.
\]
4. **Consider the case with both positive and negative real numbers:**
Suppose now that \(\frac{1}{7}\) is the sum of \(i > 0\) positive real numbers and \(j > 0\) negative real numbers, with \(i + j \leq 7\). Denote by \(u = 0.u_1u_2\ldots\) (respectively by \(-v = -0.v_1v_2\ldots\)) the sum of the positive (respectively negative) real numbers, so that \(\frac{1}{7} + v = u\). Note that \(u\) and \(v\) are necessarily strictly less than \(1\).
5. **Analyze the constraints on \(u\) and \(v\):**
Since \(i, j < 7\), it follows that \(v_6 \geq 3\), so that \(j \geq 3\) and \(i \leq 4\). Then, we must have \(v_5 \geq 4\), so that \(j \geq 4\) and \(i \leq 3\). In this case, we find \(u_3 \geq 3\) and thus \(i = 3\). This finally implies that \(v_2 \geq 6\), which is impossible since \(j = 4\).
6. **Conclude the least value of \(n\):**
The least value of \(n\) is therefore \(8\).
The final answer is \( \boxed{ 8 } \). | 8 | Number Theory | math-word-problem | aops_forum | 1cd82e3d-2e7a-5eca-9571-62f5669824c4 | 1 |
A right-angled triangle has side lengths that are integers. What could be the last digit of the area's measure, if the length of the hypotenuse is not divisible by 5? | Let the lengths of the legs be $a$ and $b$, and the length of the hypotenuse be $c$. According to the Pythagorean theorem, we then have
$$
a^{2}+b^{2}=c^{2}
$$
A square number can give a remainder of 0, 1, or 4 when divided by 5. If neither $a$ nor $b$ is divisible by 5, then the left side of (1) can give a remainder of 2 (in the form $1+1$), 0 (in the form $1+4$), or 3 (in the form $4+4$) when divided by 5. Since 2 and 3 cannot be the remainder of a square number $-c^{2}$ when divided by 5, and 0 is excluded by the condition, if the length of the hypotenuse is not divisible by 5, then the length of one of the legs will be divisible by 5.
If both legs are even, then their product is divisible by 4. We claim that this is also true if there is an odd leg.
First, note that both legs cannot be odd. In this case, in (1), $a^{2}+b^{2}$ would be even, and thus $c^{2}$, as an even square number, would be divisible by 4, while $a^{2}$ and $b^{2}$, when divided by 4, would each give a remainder of 1, making their sum 2. In this case, however, equality could not hold in (1).
If now one of the legs, say $a$, is odd, then $b$ is even, and $c$ is odd. From (1),
$$
b^{2}=(c+a)(c-a)
$$
Here, both factors on the right side are even. If neither is divisible by 4, then they each give a remainder of 2 when divided by 4, so their sum, $2c$, is divisible by 4. This would, however, make $c$ even, which is not true.
Thus, in (2), the right side is also divisible by $2^{3}=8$. We know that a square number, in this case $b^{2}$, has every prime factor in its prime factorization with an even exponent. This means that $b^{2}$ is divisible by $2^{4}$, and thus $b$ is divisible by 4.
This proves that if (1) holds, then $ab$ is divisible by 4. The area of the triangle, $ab/2$, is thus an even number. Since there is a leg divisible by 5, the area is also divisible by $2 \cdot 5=10$, so the last digit of its measure is 0.
Remark. The statement - that on the one hand, in every Pythagorean triple there is a number divisible by 5, and on the other hand, the product of the legs is divisible by 4 - can also be easily shown based on the known characterization of Pythagorean triples. According to this characterization, Pythagorean triples can be written in the form $a=k\left(u^{2}-v^{2}\right)$, $b=2 k \cdot u v$, and $c=k\left(u^{2}+v^{2}\right)$, where $k$ is any natural number, and $u$ and $v$ are relatively prime numbers of different parities, with $u>v$. See, for example, Rademacher and Toeplitz's book *Numbers and Figures*. | 0 | Geometry | math-word-problem | olympiads | 79e837be-2801-5a6a-bc7a-0acee3cdd5fd | 2 |
8.59 For each pair of real numbers $x, y$, the function $f$ satisfies the functional equation
$$
f(x)+f(y)=f(x+y)-x y-1 .
$$
If $f(1)=1$, then the number of integers $n$ (where $n \neq 1$) that satisfy $f(n)=n$ is
(A) 0.
(B) 1.
(C) 2.
(D) 3.
(E) infinitely many.
(30th American High School Mathematics Examination, 1979) | [Solution] Substituting $x=1$ into the functional equation, we get
$$
f(y+1)=f(y)+y+2 \text {. }
$$
Since $f(1)=1$, substituting $y=2,3,4, \cdots$ consecutively, we can see that for $y$ being a positive integer, $f(y)>0$. Therefore, for $y$ being a positive integer,
$$
f(y+1)>y+2>y+1 ;
$$
Thus, for integers $n>1$, $f(n)=n$ has no solution. Solving the equation for $f(y)$:
From $\circledast$ we have
$$
f(y)=f(y+1)-(y+2),
$$
Substituting $y=0,-1,-2, \cdots$ consecutively into this equation, we get
$$
\begin{array}{l}
f(0)=-1, \quad f(-1)=-2, \quad f(-2)=-2, \quad f(-3)=-1, \\
f(-4)=1 .
\end{array}
$$
Since $f(-4)>0$ and for $y0$.
Thus, for $y0$.
Therefore, for $n<-4, f(n) \neq n$.
Hence, $f(n)=n$ has only the solutions $n=1,-2$.
Therefore, the answer is $(B)$. | 1 | Algebra | MCQ | olympiads | 4cb423e0-0c1b-5c53-943d-58dadc0e8955 | 1 |
9.6. Find the minimum value of the expression $(\sqrt{2(1+\cos 2 x)}-\sqrt{36-4 \sqrt{5}} \sin x+2) \cdot(3+2 \sqrt{10-\sqrt{5}} \cos y-\cos 2 y) \cdot$ If the answer is not an integer, round it to the nearest integer. | Answer: -27 . Instructions. Exact answer: $4 \sqrt{5}-36$. | -27 | Algebra | math-word-problem | olympiads | 9224d27e-798f-569e-a808-122e2d76c789 | 2 |
2. (9th Canadian Mathematical Competition) $N$ is an integer, its representation in base $b$ is 777. Find the smallest positive integer $b$ such that $N$ is a fourth power of an integer in decimal notation. | 2. This problem is equivalent to finding the smallest positive integer $b$, such that the equation $7 b^{2}+7 b+7=x^{4}$, (1) has an integer solution for $x$. Since 7 is a prime number, it follows from equation (1) that 7 is a divisor of $x$. Therefore, let $x=7 k$, then equation (1) becomes $b^{2}+b+1=7^{3} k^{4}$. The smallest $b$ occurs when $k$ is at its minimum. Taking $k=1$, we then have $b^{2}+b+1=343, b^{2}+b-342=0$, which is $(b-18)(b+19)=0$, yielding the positive integer solution $b=18$. Thus, we have $(777)_{18}=\left(7^{4}\right)_{10}$. | 18 | Number Theory | math-word-problem | olympiads | 39a6c760-dfa3-5201-aa87-105127951e16 | 2 |
Problem 6. (8 points) In the plane, there is a non-closed, non-self-intersecting broken line consisting of 31 segments (adjacent segments do not lie on the same straight line). For each segment, the line defined by it is constructed. It is possible for some of the 31 constructed lines to coincide. What is the minimum number of different lines that can be obtained?
Answer. 9. | Solution. Evaluation. Besides the ends, the broken line has 30 vertices, each of which is the intersection point of two lines. If the lines are no more than 8, then the intersection points are no more
than 7.8/2 = $28<30$, a contradiction. An example is shown in the drawing. | 9 | Geometry | math-word-problem | olympiads | bdcc2598-b04d-5cb8-8634-d35d03fc6bf2 | 2 |
7.1. Solve the equation $\frac{n!}{2}=k!+l!$ in natural numbers, where $n!=1 \cdot 2 \cdot \ldots n$. If there are no solutions, write 0; if there is one solution, write $n$; if there are multiple solutions, write the sum of the values of $n$ for all solutions. Recall that a solution is a triplet $(n, k, l)$; if solutions differ in at least one component, they are considered different. | Answer: 10 (all triples of solutions $(n, k, l):(3,1,2),(3,2,1),(4,3,3)$).
Solution. Note that if $k4$, then $n!>4 \cdot(n-1)!\geqslant 2 \cdot(k!+l!)$, so there are no such solutions. If $n=2$, we get $1=k!+l!$ - no solutions; if $n=3$, the equation $3=k!+l!$ has two solutions: $k=1, l=2$ and $k=2, l=1$; if $n=4$, the equation $12=k!+l!$ gives one more solution $k=l=3$. | 10 | Number Theory | math-word-problem | olympiads | 09902996-6e6f-5507-94c7-f8b8045a702b | 2 |
Example 1 (Question from the 13th "Hope Cup" Invitational Competition) The real roots of the equations $x^{5}+x+1=0$ and $x+\sqrt[5]{x}+1=0$ are $\alpha, \beta$ respectively, then $\alpha+\beta$ equals ( ).
A. -1
B. $-\frac{1}{2}$
C. $\frac{1}{2}$
D. 1 | Solution: Choose A. Reason: Consider the function $f(x)=x^{5}+x+1$, which is an increasing function on $\mathbf{R}$,
when $x^{5}+x+1=0$,
we get $x=\sqrt[5]{-x-1}$, substituting $-x-1$ for $x$ yields $-x-1=\sqrt[5]{x}$.
From (2), we have $(-x-1)^{5}=x$, which is also $(-x-1)^{5}+(-x-1)+1=0$.
Let $\alpha$ be a root of equation (1), i.e., $f(\alpha)=0$. Let $\beta$ be a root of equation (2), i.e., $-1-\beta=\sqrt[5]{\beta}$, thus from (3) we know that $f(-1-\beta)=0$.
Therefore, we have $f(\alpha)=f(-1-\beta)$, which means $\alpha+\beta=-1$. | -1 | Algebra | MCQ | olympiads | 1eb53f35-0255-526f-b139-9f8efdda61ec | 1 |
Example 5 Given that $x_{1}, x_{2}, \cdots, x_{10}$ are all positive integers, and $x_{1}+x_{2}+\cdots+x_{10}=2005$, find the maximum and minimum values of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}$. | Solve: The number of ways to write 2005 as the sum of 10 positive integers is finite, so there must be a way that maximizes or minimizes the sum of their squares.
If the positive integers $x_{1}, x_{2}, \cdots, x_{10}$ satisfy $x_{1}+x_{2}+\cdots+x_{10}=2005$, and $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}$ reaches its maximum.
Assume without loss of generality that $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{10}$. If $x_{1}>1$, then $x_{1}+x_{2}=\left(x_{1}-1\right)+\left(x_{2}+1\right)$. $\left[\left(x_{1}-1\right)^{2}+\left(x_{2}+1\right)^{2}\right]-\left(x_{1}^{2}+x_{2}^{2}\right)=-2 x_{1}+2 x_{2}+2=2\left(x_{2}-x_{1}\right)+2 \geqslant 2$. Therefore, replacing $x_{1}, x_{2}$ with $x_{1}-1, x_{2}+1$ keeps their sum unchanged but increases the sum of their squares. This contradicts the assumption that $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}$ is maximized, so $x_{1}=1$.
Similarly, $x_{2}=x_{3}=\cdots=x_{9}=1, x_{10}=1996$.
Thus, the maximum value of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}$ is: $9+1996^{2}=3984025$.
Now assume $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{10}$, satisfying $x_{1}+x_{2}+\cdots+x_{10}=2005$, and $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}$ reaches its minimum value. If there exist $x_{i}, x_{j}$ and $x_{j}-x_{i} \geqslant 2(j>i)$, then $\left[\left(x_{j}-1\right)^{2}+\left(x_{i}+1\right)^{2}\right]-\left(x_{j}^{2}+x_{i}^{2}\right)=-2 x_{j}+2 x_{i}+2=2\left(x_{i}-x_{j}\right)+2 \leqslant-2$. Therefore, replacing $x_{j}, x_{i}$ with $x_{j}-1, x_{i}+1$ keeps their sum unchanged but decreases the sum of their squares, contradicting the assumption that $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}$ is minimized. Thus, the absolute difference between any two of $x_{1}, x_{2}, \cdots, x_{10}$ is at most 1. Therefore, these 10 numbers can only be 5 instances of 200 and 5 instances of 201. Thus, the minimum value is $5 \times 200^{2}+5 \times 201^{2}=402005$. | 402005 | Algebra | math-word-problem | olympiads | 09750c65-bcba-59f8-bcbd-58bba44f0c4f | 1 |
Four, (50 points) In an $n \times n$ grid, fill each cell with one of the numbers 1 to $n^{2}$. If no matter how you fill it, there must be two adjacent cells where the difference between the two numbers is at least 1011, find the minimum value of $n$.
---
The translation preserves the original text's formatting and structure. | Four, take the row $l$ where the number 1 is located and the column $t$ where the number $n^{2}$ is located, and approach $n^{2}$ sequentially along row $l$ and column $t$ from 1. Assume the numbers filled in the small grids passed through are $a_{1}, a_{2}, \cdots, a_{k}$.
From Figure 5, we know $k \leqslant 2 n-3$.
Then make the sum
$$
\begin{array}{l}
\left|1-a_{1}\right|+\left|a_{1}-a_{2}\right|+\left|a_{2}-a_{3}\right|+\cdots+ \\
\left|a_{k-1}-a_{k}\right|+\left|a_{k}-n^{2}\right| \\
\geqslant \mid\left(1-a_{1}\right)+\left(a_{1}-a_{2}\right)+\left(a_{2}-a_{3}\right)+\cdots+ \\
\quad\left(a_{k-1}-a_{k}\right)+\left(a_{k}-n^{2}\right) \mid \\
=n^{2}-1 .
\end{array}
$$
By the pigeonhole principle, we know that among
$$
\left|a_{i}-a_{i+1}\right|\left(a_{0}=1, a_{k+1}=n^{2}, i=0,1, \cdots, k\right)
$$
the $k+1$ positive integers, at least one number is not less than
$$
\left[\frac{n^{2}-1}{k+1}\right]+1 \text {, and } k+1 \leqslant 2 n-2 \text {. }
$$
According to the problem, we have
$$
\begin{array}{l}
\frac{n^{2}-1}{k+1} \geqslant \frac{n^{2}-1}{2 n-2}>1011-1 \\
\Rightarrow n>2019 .
\end{array}
$$
Therefore, the minimum value of $n$ is 2020.
(Xie Wenxiao, Huanggang High School, Hubei Province, 438000) | 2020 | Combinatorics | math-word-problem | olympiads | 0c6f58a5-af16-5fc6-b982-cdbffc29350d | 1 |
Augusto has a wire that is $10 \mathrm{~m}$ long. He makes a cut at a point on the wire, obtaining two pieces. One piece has a length of $x$ and the other has a length of $10-x$ as shown in the figure below:
Augusto uses the two pieces of wire to make two squares.
a) What is the length of the side of each of the squares? What is the area of each?
b) What is the length of each of the two pieces of wire so that the sum of the areas of the squares obtained is minimized?
c) Suppose Augusto cuts the wire into ten pieces and uses each one to make a square. What should be the size of each piece so that the sum of the areas of the squares obtained is minimized? | Solution
In this problem, all lengths are given in meters and areas in square meters.
a) A piece of rope has length $x$ and another piece of rope has length $10-x$. Since a square has four sides of equal length, one square will have a side length of $\frac{x}{4}$ and the other square will have a side length of $\frac{10-x}{4}$.
The area of a square with side length $\ell$ is $\ell^{2}$. Therefore, one square will have an area of $\left(\frac{x}{4}\right)^{2}=\frac{x^{2}}{16}$, while the other square will have an area of $\left(\frac{10-x}{4}\right)^{2}=\frac{100-20 x+x^{2}}{16}$.
b) Let $S(x)$ be the sum of the areas of the two squares. From the previous part, we have
$$
S(x)=\frac{x^{2}}{16}+\frac{100-20 x+x^{2}}{16}=\frac{100-20 x+2 x^{2}}{16}=\frac{1}{8} x^{2}-\frac{5}{4} x+\frac{25}{4}
$$
which is a quadratic function. The minimum of a function of the form
$$
f(x)=a x^{2}+b x+c
$$
with $a>0$ is achieved at $x=\frac{-b}{2 a}$. Thus, the minimum area will be achieved if
$$
x=-\frac{\left(-\frac{5}{4}\right)}{2 \frac{1}{8}}=5
$$
In other words, if the rope is cut exactly in the middle!
c) From the previous part, we know that to minimize the sum of the areas, it is necessary to cut the rope exactly in the middle. Well, we claim that to minimize the area with nine cuts (i.e., creating ten squares), it is necessary that all pieces of rope be equal. To show this, consider the following argument: if two of the ten pieces of rope were different, it would be possible to reduce the area by cutting the pieces of rope so that these two were equal (we are using the previous part). Therefore, any two pieces of rope must be equal. Hence, all must be equal! | 5 | Algebra | math-word-problem | olympiads | e8e60a66-4791-5625-bbfd-31ac1c2af0d6 | 2 |
29. Choose any three numbers from $a, b, c, d, e$ and find their sum, exactly obtaining the ten different numbers $7,11,13,14,19,21,22,25,26,28$. Then $a+b+c+d+e=$ ( i). MATHEMATICE YoUTH CLUE A. 25
B. 31
C. 37
D. 43 | Reference answer: B | 31 | Number Theory | MCQ | olympiads | 900794e8-c2d7-5919-88da-32579cf20635 | 1 |
Problem 3. In the school, there are 50 teachers, of whom 29 drink coffee, 28 drink tea, and 16 do not drink either coffee or tea. How many teachers drink only coffee, and how many drink only tea? | Solution. Since 16 teachers do not drink either coffee or tea, we get that $50-16=34$ teachers drink either coffee or tea. However, 28 teachers drink tea, and 29 drink coffee, so $(28+29)-34=57-34=23$ teachers drink
both coffee and tea. Only coffee is drunk by $29-23=6$ teachers, and only tea is drunk by $28-23=5$ teachers. | 6 | Combinatorics | math-word-problem | olympiads | ae9d787b-d6ca-55d6-867f-283c9cdfc754 | 1 |
83. Fill in the following squares with $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ respectively, so that the sum of the two five-digit numbers is 99999. Then the number of different addition equations is $\qquad$. $(a+b$ and $b+a$ are considered the same equation) | Answer: 1536 | 1536 | Combinatorics | math-word-problem | olympiads | 5b87c250-7b32-5287-8810-f83a8c6a5b3c | 2 |
On a board, the numbers from 1 to 2009 are written. A couple of them are erased and instead of them, on the board is written the remainder of the sum of the erased numbers divided by 13. After a couple of repetition of this erasing, only 3 numbers are left, of which two are 9 and 999. Find the third number. | 1. **Define the invariant:**
Let \( I \) be the remainder when the sum of all the integers on the board is divided by 13. Initially, the numbers from 1 to 2009 are written on the board. The sum of these numbers is:
\[
S = \sum_{k=1}^{2009} k = \frac{2009 \times 2010}{2} = 2009 \times 1005
\]
We need to find \( S \mod 13 \).
2. **Calculate \( 2009 \mod 13 \) and \( 1005 \mod 13 \):**
\[
2009 \div 13 = 154 \quad \text{remainder} \quad 7 \quad \Rightarrow \quad 2009 \equiv 7 \pmod{13}
\]
\[
1005 \div 13 = 77 \quad \text{remainder} \quad 4 \quad \Rightarrow \quad 1005 \equiv 4 \pmod{13}
\]
3. **Calculate \( S \mod 13 \):**
\[
S = 2009 \times 1005 \equiv 7 \times 4 = 28 \equiv 2 \pmod{13}
\]
Therefore, \( I = 2 \).
4. **Determine the third number \( a \):**
After several operations, only three numbers are left on the board: 9, 999, and \( a \). The sum of these three numbers modulo 13 must be equal to the invariant \( I \):
\[
a + 9 + 999 \equiv 2 \pmod{13}
\]
5. **Calculate \( 999 \mod 13 \):**
\[
999 \div 13 = 76 \quad \text{remainder} \quad 11 \quad \Rightarrow \quad 999 \equiv 11 \pmod{13}
\]
6. **Set up the equation:**
\[
a + 9 + 11 \equiv 2 \pmod{13}
\]
Simplify the equation:
\[
a + 20 \equiv 2 \pmod{13}
\]
Subtract 20 from both sides:
\[
a \equiv 2 - 20 \pmod{13}
\]
\[
a \equiv -18 \pmod{13}
\]
Since \(-18 \equiv -18 + 26 \equiv 8 \pmod{13}\), we have:
\[
a \equiv 8 \pmod{13}
\]
7. **Find the possible values of \( a \) between 1 and 2009:**
The possible values of \( a \) are of the form:
\[
a = 8 + 13k \quad \text{for integer} \quad k
\]
We need \( 1 \leq a \leq 2009 \):
\[
1 \leq 8 + 13k \leq 2009
\]
\[
-7 \leq 13k \leq 2001
\]
\[
0 \leq k \leq 154
\]
Therefore, the possible values of \( a \) are:
\[
a = 8, 21, 34, \ldots, 2002
\]
The final answer is \( \boxed{8} \). | 8 | Number Theory | math-word-problem | aops_forum | 26ac9e7a-f8f3-5ef9-b051-f84dfb5129ce | 2 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{(n+1)^{4}-(n-1)^{4}}{(n+1)^{3}+(n-1)^{3}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(n+1)^{4}-(n-1)^{4}}{(n+1)^{3}+(n-1)^{3}}=\lim _{n \rightarrow \infty} \frac{\left((n+1)^{2}-(n-1)^{2}\right) \cdot\left((n+1)^{2}+(n-1)^{2}\right)}{(n+1)^{3}+(n-1)^{3}}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(n^{2}+2 n+1-n^{2}+2 n-1\right) \cdot\left(n^{2}+2 n+1+n^{2}-2 n+1\right)}{(n+1)^{3}+(n-1)^{3}}= \\
& =\lim _{n \rightarrow \infty} \frac{4 n\left(2 n^{2}+2\right)}{(n+1)^{3}+(n-1)^{3}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{3}} 8 n\left(n^{2}+1\right)}{n^{3}\left((n+1)^{3}+(n-1)^{3}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{8\left(1+\frac{1}{n^{2}}\right)}{\left(1+\frac{1}{n}\right)^{3}+\left(1-\frac{1}{n}\right)^{3}}=\frac{8 \cdot 1}{1^{3}+1^{3}}=4
\end{aligned}
$$
## Problem Kuznetsov Limits 3-24 | 4 | Algebra | math-word-problem | olympiads | 59b0b709-19a9-5073-aedc-14576d104c24 | 2 |
6. Let $[x]$ denote the greatest integer not exceeding the real number $x$,
$$
\begin{array}{c}
S=\left[\frac{1}{1}\right]+\left[\frac{2}{1}\right]+\left[\frac{1}{2}\right]+\left[\frac{2}{2}\right]+\left[\frac{3}{2}\right]+ \\
{\left[\frac{4}{2}\right]+\left[\frac{1}{3}\right]+\left[\frac{2}{3}\right]+\left[\frac{3}{3}\right]+\left[\frac{4}{3}\right]+} \\
{\left[\frac{5}{3}\right]+\left[\frac{6}{3}\right]+\cdots}
\end{array}
$$
up to 2016 terms, where, for a segment with denominator $k$, there are $2 k$ terms $\left[\frac{1}{k}\right],\left[\frac{2}{k}\right], \cdots,\left[\frac{2 k}{k}\right]$, and only the last segment may have fewer than $2 k$ terms. Then the value of $S$ is | 6.1078.
$$
2+4+\cdots+2 \times 44=1980 \text {. }
$$
For any integer \( k \) satisfying \( 1 \leqslant k \leqslant 44 \), the sum includes \( 2k \) terms with the denominator \( k \): \(\left[\frac{1}{k}\right],\left[\frac{2}{k}\right], \cdots,\left[\frac{2 k}{k}\right]\), whose sum is \( k+2 \).
Also, \( 2016-1980=36 \), so the sum includes 36 terms with the denominator 45: \(\left[\frac{1}{45}\right],\left[\frac{2}{45}\right], \cdots,\left[\frac{36}{45}\right]\), whose sum is zero.
$$
\text { Therefore, } S=\sum_{k=1}^{44}(k+2)=1078 \text {. }
$$ | 1078 | Number Theory | math-word-problem | cn_contest | 98326df5-29ae-59b9-8286-d2ddeb417dd3 | 2 |
40. The sum of three consecutive odd numbers is equal to the fourth power of a single-digit number. Find all such triples of numbers. | 40. The sum of three consecutive odd numbers is an odd number, and if the fourth power is an odd number, then the base is also odd. Therefore, single-digit numbers should be sought among the numbers $3,5,7,9$.
1) $(2 x-1)+(2 x+1)+(2 x+3)=3^{4}, 6 x+3=3^{4}$, from which $2 x+1=3^{3}$ and $x=$ $=13 ; 2 x-1=25 ; 2 x+1=27 ; 2 x+3=29 ; 25+27+29=81=3^{4}$;
2) $6 x+3=5^{4}$ (no integer solutions);
3) $6 x+3=7^{4}=2401$ (no integer solutions);
4) $6 x+3=9^{4} ; 2 x+1=2187 ; 2 x-1=2185 ; 2 x+3=2189$.
Thus, the problem has two solutions:
a) $25,27,29$; b) $2185,2187,2189$. | 25,27,292185,2187,2189 | Number Theory | math-word-problem | olympiads | 6196c962-baa9-559f-9221-352113314f5b | 2 |
Two is $10 \%$ of $x$ and $20 \%$ of $y$. What is $x - y$?
$(\mathrm {A}) \ 1 \qquad (\mathrm {B}) \ 2 \qquad (\mathrm {C})\ 5 \qquad (\mathrm {D}) \ 10 \qquad (\mathrm {E})\ 20$ | $2 = \frac {1}{10}x \Longrightarrow x = 20,\quad 2 = \frac{1}{5}y \Longrightarrow y = 10,\quad x-y = 20 - 10=10 \mathrm{(D)}$. | 10 | Algebra | MCQ | amc_aime | 5fafd2ee-f7d4-5e50-8a7e-7fe29cc17f17 | 2 |
1. (6 points) Today is January 31, 2015, welcome to the 2015 "Spring Cup" Final. The calculation result of the expression $\frac{\frac{2015}{1}+\frac{2015}{0.31}}{1+0.31}$ is | 【Solution】Solve: $\frac{\frac{2015}{1}+\frac{2015}{0.31}}{1+0.31}$
$$
\begin{array}{l}
=\frac{\left(\frac{2015}{1}+\frac{2015}{0.31}\right) \times 0.31}{(1+0.31) \times 0.31} \\
=\frac{(1+0.31) \times 2015}{(1+0.31) \times 0.31} \\
=6500 ;
\end{array}
$$
Therefore, the answer is: 6500. | 6500 | Algebra | math-word-problem | olympiads | 46eb6c3b-bc5d-5fc0-8913-2c4a06359b71 | 2 |
30. Find the remainder when the 2018-digit number $\underbrace{\overline{55 \cdots}}_{2018 \text { 555 }}$ is divided by 13. | Reference answer: 3 | 3 | Number Theory | math-word-problem | olympiads | 12b14f92-3f01-5f14-9fb9-abf68de16852 | 2 |
7.242. $\left(16 \cdot 5^{2 x-1}-2 \cdot 5^{x-1}-0.048\right) \lg \left(x^{3}+2 x+1\right)=0$. | Solution.
Domain: $x^{3}+2 x+1>0$.
From the condition $16 \cdot 5^{2 x-1}-2^{x-1}-0.048=0$ or $\lg \left(x^{3}+2 x+1\right)=0$. Rewrite the first equation as
$\frac{16}{5} \cdot 5^{2 x}-\frac{2}{5} \cdot 5^{x}-0.048=0 \Leftrightarrow 16 \cdot 5^{2 x}-2 \cdot 5^{x}-0.24=0$.
Solving this equation as a quadratic in terms of $5^{x}$, we get $5^{x}=-\frac{3}{40}$ (no solutions), or $5^{x}=5^{-1} \Leftrightarrow x_{1}=-1$ (does not satisfy the domain). From the second equation, we have
$x^{3}+2 x+1=1 \Leftrightarrow x^{3}+2 x=0 \Leftrightarrow x\left(x^{2}+2\right)=0, x_{3}=0, x^{2}+2 \neq 0$.
Answer: 0. | 0 | Algebra | math-word-problem | olympiads | 2bfb42a5-11c9-5aee-b6df-edb9b8efee0e | 2 |
Example 1 In $\triangle ABC$, it is known that $x \sin A + y \sin B + z \sin C = 0$. Find the value of $(y + z \cos A)(z + x \cos B)(x + y \cos C) + (y \cos A + z)(z \cos B + x)(x \cos C + y)$. | In $\triangle A B C$, $\sin C=\sin (A+B)=\sin A \cos B+\cos A \sin B$. Substituting the known conditions, we get $x \sin A+y \sin B+z(\sin A \cos B+\cos A \sin B)=0$, which simplifies to $\sin A(x+z \cos B)=-\sin B(y+z \cos A)$.
Similarly, $\sin B(y+x \cos C)=-\sin C(z+x \cos B)$, $\sin C(z+y \cos A)=-\sin A(x+y \cos C)$,
Multiplying the three equations, we have:
$(x+z \cos B)(y+x \cos C)(z+y \cos A)=-(y+z \cos A)(z+x \cos B)(x+y \cos C)$, which simplifies to $(y+z \cos A)(z+x \cos B)(x+y \cos C)+(y \cos A+z)(z \cos B+x)(x \cos C+y)=0$. | 0 | Algebra | math-word-problem | olympiads | 80c5345a-b5f8-5bef-8c9f-f624dc51008a | 2 |
In order for Mateen to walk a kilometer (1000m) in his rectangular backyard, he must walk the length 25 times or walk its perimeter 10 times. What is the area of Mateen's backyard in square meters?
$\text{(A)}\ 40 \qquad \text{(B)}\ 200 \qquad \text{(C)}\ 400 \qquad \text{(D)}\ 500 \qquad \text{(E)}\ 1000$ | The length $L$ of the rectangle is $\frac{1000}{25}=40$ meters. The perimeter $P$ is $\frac{1000}{10}=100$ meters. Since $P_{rect} = 2L + 2W$, we plug values in to get:
$100 = 2\cdot 40 + 2W$
$100 = 80 + 2W$
$2W = 20$
$W = 10$ meters
Since $A_{rect} = LW$, the area is $40\cdot 10=400$ square meters or $\boxed{C}$. | 400 | Geometry | MCQ | amc_aime | b1d3189a-8bdf-5597-81b5-0e8ef947e53f | 2 |
A father wants to divide his property among his children: first, he gives 1000 yuan and one-tenth of the remaining property to the eldest child, then 2000 yuan and one-tenth of the remaining property to the second child, then 3000 yuan and one-tenth of the remaining property to the third child, and so on. It turns out that each child receives the same amount of property. The father has $\qquad$ children. | (Method 1)
Let the father's property be $x$ yuan;
Since the eldest and the second son receive the same amount of property, we have
$$
1000+(x-1000) \times \frac{1}{10}=2000+\left[(x-1000) \times\left(1-\frac{1}{10}\right)-2000\right] \times \frac{1}{10} ;
$$
Solving this, we get $x=81000$;
The eldest son receives $1000+(81000-1000) \times \frac{1}{10}=9000$ yuan;
That is, each child receives 9000 yuan;
This father has a total of $81000 \div 9000=9$ children.
(Method 2)
Since each child receives the same amount of property;
The eldest and the second son receive the same amount of property;
The eldest son receives 1000 yuan plus $\frac{1}{10}$ of the remaining property (let's denote this remaining as $A$); the second son receives 2000 yuan plus $\frac{1}{10}$ of the remaining property (let's denote this remaining as $B$); thus, $\frac{1}{10}$ of $A$ is 1000 yuan more than $\frac{1}{10}$ of $B$;
$A$ is $1000 \div \frac{1}{10}=10000$ yuan more than $B$;
From $A$ to $B$, it decreases by $\frac{1}{10}$ of $A$ and 2000 yuan;
Therefore, $\frac{1}{10}$ of $A$ is $10000-2000=8000$ yuan;
$A$ is $8000 \div \frac{1}{10}=80000$ yuan;
This father's property is $1000+80000=81000$ yuan;
The eldest son receives $1000+(81000-1000) \times \frac{1}{10}=9000$ yuan;
That is, each child receives 9000 yuan;
This father has a total of $81000 \div 9000=9$ children.
(Method 3)
Let this father have $n$ children;
Then the second-to-last child receives $1000(n-1)$ yuan plus $\frac{1}{10}$ of the remainder;
The last child receives $1000n$ yuan;
These $1000n$ yuan are $\frac{9}{10}$ of what remains after the second-to-last child receives $1000(n-1)$; each child receives the same amount of property;
From the equality of the property received by the second-to-last and the last child, we have
$$
1000(n-1)+1000n \times \frac{1}{9}=1000n \text {; }
$$
Solving this, we get $n=9$;
This father has a total of 9 children. | 9 | Algebra | math-word-problem | olympiads | 16dbe1c7-62c1-5abf-88b4-a4d83a5c8276 | 2 |
## Zadatak B-1.2.
Na slici su prikazani pravilni peterokut. $A B C D E$ i kvadrat $A B F G$. Odredite mjeru kuta $F A D$.
| ## Rješenje.
Zbroj mjera unutrašnjih kutova pravilnog peterokuta iznosi $(5-2) 180^{\circ}=540^{\circ}$ pa je mjera unutrašnjeg kuta pravilnog peterokuta jednaka $\frac{540^{\circ}}{5}=108^{\circ}$.
Budući da je trokut $A E D$ jednakokračan, $\varangle D A E=\frac{180^{\circ}-108^{\circ}}{2}=36^{\circ}$
Kut $\varangle B A F=45^{\circ}$ jer je to kut između dijagonale i stranice kvadrata.
Tada je $\varangle F A D=\varangle B A E-\varangle B A F-\varangle D A E=108^{\circ}-45^{\circ}-36^{\circ}=27^{\circ}$.
| 27 | Geometry | math-word-problem | olympiads | 95bcc8e7-a809-533d-8cbd-45e190531b31 | 2 |
The equation of line $\ell_1$ is $24x-7y = 319$ and the equation of line $\ell_2$ is $12x-5y = 125$. Let $a$ be the number of positive integer values $n$ less than $2023$ such that for both $\ell_1$ and $\ell_2$ there exists a lattice point on that line that is a distance of $n$ from the point $(20,23)$. Determine $a$.
[i]Proposed by Christopher Cheng[/i]
[hide=Solution][i]Solution. [/i] $\boxed{6}$
Note that $(20,23)$ is the intersection of the lines $\ell_1$ and $\ell_2$. Thus, we only care about lattice points on the the two lines that are an integer distance away from $(20,23)$. Notice that $7$ and $24$ are part of the Pythagorean triple $(7,24,25)$ and $5$ and $12$ are part of the Pythagorean triple $(5,12,13)$. Thus, points on $\ell_1$ only satisfy the conditions when $n$ is divisible by $25$ and points on $\ell_2$ only satisfy the conditions when $n$ is divisible by $13$. Therefore, $a$ is just the number of positive integers less than $2023$ that are divisible by both $25$ and $13$. The LCM of $25$ and $13$ is $325$, so the answer is $\boxed{6}$.[/hide] | 1. **Intersection of Lines**:
We start by finding the intersection of the lines $\ell_1$ and $\ell_2$. The equations are:
\[
24x - 7y = 319
\]
\[
12x - 5y = 125
\]
To find the intersection, we solve this system of linear equations. We can use the method of elimination or substitution. Here, we will use elimination.
2. **Elimination Method**:
Multiply the second equation by 2 to align the coefficients of $x$:
\[
24x - 10y = 250
\]
Now subtract the second equation from the first:
\[
(24x - 7y) - (24x - 10y) = 319 - 250
\]
Simplifying, we get:
\[
3y = 69 \implies y = 23
\]
Substitute $y = 23$ back into the first equation:
\[
24x - 7(23) = 319 \implies 24x - 161 = 319 \implies 24x = 480 \implies x = 20
\]
Therefore, the intersection point is $(20, 23)$.
3. **Distance from $(20, 23)$**:
We need to find lattice points on $\ell_1$ and $\ell_2$ that are an integer distance $n$ from $(20, 23)$. The distance formula is:
\[
d = \sqrt{(x - 20)^2 + (y - 23)^2}
\]
For $d$ to be an integer, $(x - 20)^2 + (y - 23)^2$ must be a perfect square, say $k^2$. Thus:
\[
(x - 20)^2 + (y - 23)^2 = n^2
\]
4. **Lattice Points on $\ell_1$**:
The line $\ell_1$ can be rewritten in slope-intercept form:
\[
y = \frac{24}{7}x - \frac{319}{7}
\]
Notice that $24$ and $7$ are part of the Pythagorean triple $(7, 24, 25)$. Therefore, the distance $n$ must be a multiple of $25$ for there to be lattice points on $\ell_1$.
5. **Lattice Points on $\ell_2$**:
The line $\ell_2$ can be rewritten in slope-intercept form:
\[
y = \frac{12}{5}x - \frac{125}{5}
\]
Notice that $12$ and $5$ are part of the Pythagorean triple $(5, 12, 13)$. Therefore, the distance $n$ must be a multiple of $13$ for there to be lattice points on $\ell_2$.
6. **Common Multiples**:
We need $n$ to be a multiple of both $25$ and $13$. The least common multiple (LCM) of $25$ and $13$ is:
\[
\text{LCM}(25, 13) = 325
\]
We need to count the multiples of $325$ that are less than $2023$:
\[
\left\lfloor \frac{2023}{325} \right\rfloor = 6
\]
Conclusion:
The number of positive integer values $n$ less than $2023$ such that there exists a lattice point on both $\ell_1$ and $\ell_2$ at a distance $n$ from $(20, 23)$ is $\boxed{6}$. | 6 | Number Theory | math-word-problem | aops_forum | b325fc9f-7ff8-5556-976b-f72348f99405 | 2 |
2. How many integers $b$ exist such that the equation $x^{2}+b x-9600=0$ has an integer solution that is a multiple of both 10 and 12? Specify the largest possible $b$. | 2. Solution. Since the desired integer solution $x$ is divisible by 10 and 12, it is divisible by 60, hence it can be written in the form $x=60 t, t \in Z$. Substitute $x=60 t$ into the original equation:
$3600 t^{2}+60 b t-9600=0$. Express $b: b=\frac{160}{t}-60 t$. For $b$ to be an integer, $t$ must be a divisor of the number 160. $160=2^{5} \cdot 5$. The number $160=2^{5} \cdot 5$ has 12 positive divisors: $1,2,2^{2}, 2^{3}, 2^{4}, 2^{5}, 5,2 \cdot 5,2^{2} \cdot 5,2^{3} \cdot 5,2^{4} \cdot 5,2^{5} \cdot 5$. Taking the sign into account, we get 24 divisors. Each of them corresponds to a solution $x=60 t$ and an integer $b=\frac{160}{t}-60 t$. Therefore, the number of integers $b$ that satisfy the condition of the problem is 24. Since $b$ decreases as $t$ increases, the maximum $b$ corresponds to the smallest $t=-160$, so $b_{\max }=-1+60 \cdot 160=9599$. | 9599 | Algebra | math-word-problem | olympiads | 622e4608-b0d7-5071-90bd-fa39367cb1ac | 1 |
4.206 There are two forces $f_{1}$ and $f_{2}$ acting on the origin $O$ of the coordinate axis,
$$\begin{array}{l}
\vec{f}_{1}=\overrightarrow{O A}=\sqrt{2}\left(\cos 45^{\circ}+i \sin 45^{\circ}\right) \\
\vec{f}_{2}=\overrightarrow{O B}=2\left[\cos \left(-30^{\circ}\right)+i \sin \left(-30^{\circ}\right)\right]
\end{array}$$
(1) Find the magnitude and direction of their resultant force;
(2) Find the distance between points $A$ and $B$ (accurate to 0.1). | [Solution]
$$\text { (1) Resultant force } \begin{aligned}
\vec{f}= & \overrightarrow{f_{1}}+\overrightarrow{f_{2}} \\
= & \sqrt{2}\left(\cos 45^{\circ}+i \sin 45^{\circ}\right) \\
& +2\left[\cos \left(-30^{\circ}\right)+i \sin \left(-30^{\circ}\right)\right] \\
= & 1+i+\sqrt{3}-i \\
= & \sqrt{3}+1 \\
= & (\sqrt{3}+1)\left(\cos 0^{\circ}+i \sin 0^{\circ}\right) .
\end{aligned}$$
So the magnitude of the resultant force is $\sqrt{3}+1$, and its direction is the same as the $x$-axis.
(2) Since $\overrightarrow{O A}=1+i$, the Cartesian coordinates of point $A$ are $(1,1)$, and $\overrightarrow{O B}=\sqrt{3}-i$, so the Cartesian coordinates of point $B$ are $(\sqrt{3},-1)$.
$$\text { Therefore } \begin{aligned}
|A B| & =\sqrt{(\sqrt{3}-1)^{2}+(-1-1)^{2}} \\
& =\sqrt{8-2 \sqrt{3}} \approx 2.1
\end{aligned}$$ | 2.1 | Algebra | math-word-problem | inequalities | e93a8cbd-336b-5aa7-9e39-b81d5037510d | 2 |
6. Given that $\overrightarrow{O A} \perp \overrightarrow{O B}$, and $|\overrightarrow{O A}|=|\overrightarrow{O B}|=24$. If $t \in[0,1]$, then
$$
|t \overrightarrow{A B}-\overrightarrow{A O}|+\left|\frac{5}{12} \overrightarrow{B O}-(1-t) \overrightarrow{B A}\right|
$$
the minimum value is ( ).
(A) $2 \sqrt{193}$
(B) 26
(C) $24 \sqrt{2}$
(D) 24 | 6. B.
Construct a square $O A C B$, connect the diagonal $A B$, and let $D$ and $E$ be points on the diagonal $A B$ and the side $O B$ respectively, such that
$$
\begin{array}{l}
t \overrightarrow{A B}-\overrightarrow{A O}=\overrightarrow{O D}, \\
\frac{5}{12} \overrightarrow{B O}-(1-t) \overrightarrow{B A}=\overrightarrow{E D}, \\
E B=10, O D=D C . \\
\text { Therefore }|t \overrightarrow{A B}-\overrightarrow{A O}|+\left|\frac{5}{12} \overrightarrow{B O}-(1-t) \overrightarrow{B A}\right| \\
=|\overrightarrow{E D}|+|\overrightarrow{D C}| \leqslant|\overrightarrow{E C}|=26 .
\end{array}
$$ | 26 | Geometry | MCQ | olympiads | 09c374b3-283a-5d5c-b647-4b3336197400 | 1 |
All three vertices of $\bigtriangleup ABC$ lie on the parabola defined by $y=x^2$, with $A$ at the origin and $\overline{BC}$ parallel to the $x$-axis. The area of the triangle is $64$. What is the length of $BC$?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 16$ | By: Albert471
Plotting points $B$ and $C$ on the graph shows that they are at $\left( -x,x^2\right)$ and $\left( x,x^2\right)$, which is isosceles. By setting up the triangle area formula you get: $64=\frac{1}{2}*2x*x^2 = 64=x^3$ Making x=4, and the length of $BC$ is $2x$, so the answer is $\boxed{\textbf{(C)}\ 8}$. | 8 | Geometry | MCQ | amc_aime | 00d33721-4fbb-510f-b4d7-2a941104e29b | 2 |
1B. If for the non-zero real numbers $a, b$ and $c$ the equalities $a^{2}+a=b^{2}, b^{2}+b=c^{2}$ and $c^{2}+c=a^{2}$ hold, determine the value of the expression $(a-b)(b-c)(c-a)$. | Solution. By adding the three equations, we obtain
$$
a^{2}+b^{2}+c^{2}+a+b+c=a^{2}+b^{2}+c^{2}, \text { i.e., } a+b+c=0
$$
Accordingly, $a+b=-c, b+c=-a, c+a=-b$. From the equation $a^{2}+a=b^{2}$, it follows that $(a-b)(a+b)=-a$, and since $a+b=-c \neq 0$, we get $a-b=-\frac{a}{a+b}$. Similarly, $b-c=-\frac{b}{b+c}$ and $c-a=-\frac{c}{c+a}$. Therefore,
$$
(a-b)(b-c)(c-a)=\frac{-a}{a+b} \cdot \frac{-b}{b+c} \cdot \frac{-c}{c+a}=\frac{-a}{-c} \cdot \frac{-b}{-a} \cdot \frac{-c}{-b}=1
$$
2AB. Prove that for any real numbers $a, b$, and $c$, the equation
$$
3(a+b+c) x^{2}+4(a b+b c+c a) x+4 a b c=0
$$
has real solutions. Under what conditions are the solutions of the equation equal to each other? | 1 | Algebra | math-word-problem | olympiads | 9941a501-7cbb-5f6c-9005-e629b9d55555 | 2 |
1. Given $a, b>0, a \neq 1$, and $a^{b}=\log _{a} b$, then the value of $a^{a^{b}}-\log _{a} \log _{a} b^{a}$ is | 1. -1 Explanation: From $a^{b}=\log _{a} b$ we know: $b=a^{a^{b}}, b=\log _{a} \log _{a} b$,
$$
a^{a^{b}}-\log _{a} \log _{a} b^{a}=b-\log _{a}\left(a \log _{a} b\right)=b-\log _{a} a-\log _{a} \log _{a} b=b-1-b=-1
$$ | -1 | Algebra | math-word-problem | olympiads | c8a6178c-e74a-5a3d-a899-590ea913cc0e | 2 |
4. Let the sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=0, a_{2}=1$, and for all $n \geqslant 3, a_{n}$ is the smallest positive integer greater than $a_{n-1}$ such that there is no subsequence of $a_{1}, a_{2}, \cdots, a_{n}$ that forms an arithmetic sequence. Find $a_{2014}$. | 4. First, prove a lemma using mathematical induction.
Lemma A non-negative integer appears in the sequence if and only if its ternary expansion contains only 0 and 1.
Proof It is obvious that the proposition holds for 0.
Assume the proposition holds for all non-negative integers less than \( N \), and consider \( N \).
If the ternary expansion of \( N \) contains the digit 2, replace all occurrences of 2 with 0 to get the number \( N_{0} \); replace all occurrences of 2 with 1 to get the number \( N_{1} \). Thus, the ternary expansions of \( N_{0} \) and \( N_{1} \) do not contain the digit 2.
By the induction hypothesis, the numbers \( N_{0} \) and \( N_{1} \) are in the sequence.
Since \( N_{0} \), \( N_{1} \), and \( N \) form an arithmetic sequence, \( N \) does not appear in the sequence. If the ternary expansion of \( N \) does not contain the digit 2, then we need to prove that \( N \) must appear in the sequence. If not, then there exist terms \( N_{0} \) and \( N_{1} \) in the sequence such that \( N_{0} \), \( N_{1} \), and \( N \) form an arithmetic sequence. Let the common difference be \( d \), and \( 3^{k} \| d \). Then the ternary expansions of these three numbers have the same lowest \( k-1 \) digits, and the \( k \)-th digit is different for each pair, so one of the numbers must have a 2 in the \( k \)-th digit, which contradicts the assumption.
Returning to the original problem.
We only need to find the 2014th non-negative integer whose ternary expansion does not contain the digit 2.
Notice that under this restriction, the ternary carry-over method is equivalent to binary, so we just need to write 2013 (note \( a_{1}=0 \)) in binary \((11111011101)_{2}\) and then convert it to ternary \( a_{2014}=88327 \). | 88327 | Combinatorics | math-word-problem | cn_contest | 27ef0a0a-88ec-50b0-a32a-0ae07e230bed | 3 |
18. (3 points) Li Shuang rides a bike at a speed of 320 meters per minute from location $A$ to location $B$. On the way, due to a bicycle malfunction, he pushes the bike and walks for 5 minutes to a place 1800 meters from $B$ to repair the bike. After 15 minutes, he continues towards $B$ at 1.5 times his original riding speed, and arrives at $B$ 17 minutes later than the expected time. What is Li Shuang's walking speed in meters per minute? | 【Solution】Solve: $1800 \div 320-1800 \div(320 \times 1.5)$
$$
\begin{array}{l}
=5.625-3.75 \\
=1.875 \text { (minutes) } \\
320 \times[5-(17-15+1.875)] \div 5 \\
=320 \times[5-3.875] \div 5 \\
=320 \times 1.125 \div 5 \\
=360 \div 5 \\
=72 \text { (meters/minute) }
\end{array}
$$
Answer: Li Shuang's speed of pushing the cart while walking is 72 meters/minute.
Therefore, the answer is: 72. | 72 | Algebra | math-word-problem | olympiads | e32aa5a1-3bc5-5982-9571-0440038ed832 | 2 |
53. How many four-digit numbers contain at least one even digit? | 53. From four-digit numbers, we need to discard all those numbers that do not have a single even digit. We will get: $9 \cdot 10 \cdot 10 \cdot 10-5 \cdot 5 \cdot 5 \cdot 5 \cdot 5=8375$ numbers. | 8375 | Combinatorics | math-word-problem | olympiads | 9981f4e1-f944-5276-b5c5-9399972a55c5 | 2 |
There is a regular $17$-gon $\mathcal{P}$ and its circumcircle $\mathcal{Y}$ on the plane.
The vertices of $\mathcal{P}$ are coloured in such a way that $A,B \in \mathcal{P}$ are of diff erent colour, if the shorter arc connecting $A$ and $B$ on $\mathcal{Y}$ has $2^k+1$ vertices, for some $k \in \mathbb{N},$ including $A$ and $B.$
What is the least number of colours which suffices? | 1. **Identify the relevant values of \(2^k + 1\):**
Since the minor arc \(AB\) has at most 9 vertices on it, the relevant values of \(2^k + 1\) are \(3, 5,\) and \(9\). This is because:
\[
2^1 + 1 = 3, \quad 2^2 + 1 = 5, \quad 2^3 + 1 = 9
\]
2. **Enumerate the vertices:**
Enumerate the vertices of the 17-gon as \(V_0, V_1, \ldots, V_{16}\).
3. **Determine the conditions for different colors:**
We have that \(V_i\) and \(V_j\) are of different colors if \(i - j \equiv \pm 2, \pm 4, \pm 8 \pmod{17}\). This is derived from the relevant values of \(2^k + 1\).
4. **Establish a lower bound for the number of colors:**
Since \(V_0, V_2, V_4\) must have pairwise distinct colors, we have \(n \geq 3\).
5. **Check if 4 colors suffice:**
If there are 4 colors \(A, B, C,\) and \(D\), we can try to color the vertices such that the condition is satisfied. One possible coloring is:
\[
\{ V_0, V_1, \cdots, V_{16} \} = \{ A, D, B, B, C, C, D, D, B, B, C, A, D, D, B, B, C \}
\]
This coloring ensures that no two vertices \(V_i\) and \(V_j\) with \(i - j \equiv \pm 2, \pm 4, \pm 8 \pmod{17}\) have the same color. Therefore, \(n \leq 4\).
6. **Prove that 3 colors are insufficient:**
Assume there exists a solution with \(n = 3\). Without loss of generality, let \(V_0 = A\) and \(V_2 = B\). Then:
\[
V_4 = C, \quad V_6 = A, \quad V_8 = B, \quad V_{10} = C, \quad V_{12} = A, \quad V_{14} = B, \quad V_{16} = C
\]
Continuing this pattern, we find that \(V_{13} = A\), which contradicts the assumption that \(n = 3\) is a solution because \(V_0\) and \(V_{13}\) would be the same color, violating the condition for different colors.
Therefore, the least number of colors required is \(4\).
The final answer is \(\boxed{4}\). | 4 | Combinatorics | math-word-problem | aops_forum | ec0462f1-31cb-5753-b048-823b60e8ac48 | 1 |
8. Find the last four digits of $7^{7^{-7}}$ (100 sevens).
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 8. Solution: $\because 7 \equiv-1(\bmod 4), \therefore 7^{7^{7}} \equiv$ $-1(\bmod 4)(98$ sevens$)$, let $7^{\gamma^{\prime 7}}=4 k+3(98$ sevens$), k \in \mathbf{Z}^{+}$, then $7^{7^{\prime 7}}=7^{4 k+3}$ (99 sevens), $\because 7^{4} \equiv$ $1(\bmod 100), \therefore 7^{4 k} \equiv 1(\bmod 100), 7^{7^{\prime 2}} \equiv 7^{4 k+3}$ $\equiv 7^{3} \equiv 43(\bmod 100)$. Let $7^{\gamma^{7}}=7^{100 m+43}$ (100 sevens), $m \in \mathbf{Z}^{+}$.
$$
\because 7^{4} \equiv 2401(\bmod 10000), \therefore 7^{8} \equiv 4801
$$
$(\bmod 10000), 7^{16} \equiv 9601(\bmod 10000)$,
$$
7^{32} \equiv 9201(\bmod 10000), \quad 7^{64} \equiv 8401
$$
$(\bmod 10000), 7^{100} \equiv 7^{4} \times 7^{32} \times 7^{64} \equiv 2401 \times 9201$ $\times 8401 \equiv 1(\bmod 10000), 7^{100 m} \equiv 1(\bmod 10000)$, $7^{100 m+43} \equiv 7^{43} \equiv 7^{3} \times 7^{8} \times 7^{32} \equiv 2343$ $(\bmod 10000), \therefore 7^{\gamma^{7}}(100$ sevens$) \equiv 7^{100 m+43} \equiv$ $2343(\bmod 10000), 7^{7^{77}}(100$ sevens$)$ has the last four digits as 2343. | 2343 | Number Theory | math-word-problem | olympiads | 5c348b9a-119c-5e81-8cef-b2905b202dbd | 1 |
Let $a_1, a_2, a_3, a_4$ be integers with distinct absolute values. In the coordinate plane, let $A_1=(a_1,a_1^2)$, $A_2=(a_2,a_2^2)$, $A_3=(a_3,a_3^2)$ and $A_4=(a_4,a_4^2)$. Assume that lines $A_1A_2$ and $A_3A_4$ intersect on the $y$-axis at an acute angle of $\theta$. The maximum possible value for $\tan \theta$ can be expressed in the form $\dfrac mn$ for relatively prime positive integers $m$ and $n$. Find $100m+n$.
[i]Proposed by James Lin[/i] | 1. **Given Information and Setup:**
- We have integers \(a_1, a_2, a_3, a_4\) with distinct absolute values.
- Points in the coordinate plane are \(A_1 = (a_1, a_1^2)\), \(A_2 = (a_2, a_2^2)\), \(A_3 = (a_3, a_3^2)\), and \(A_4 = (a_4, a_4^2)\).
- Lines \(A_1A_2\) and \(A_3A_4\) intersect on the \(y\)-axis at an acute angle \(\theta\).
- We need to find the maximum possible value for \(\tan \theta\) expressed as \(\frac{m}{n}\) for relatively prime positive integers \(m\) and \(n\), and then find \(100m + n\).
2. **Slope Calculation:**
- The slope of line \(A_1A_2\) is:
\[
m_1 = \frac{a_2^2 - a_1^2}{a_2 - a_1} = a_2 + a_1
\]
- The slope of line \(A_3A_4\) is:
\[
m_2 = \frac{a_4^2 - a_3^2}{a_4 - a_3} = a_4 + a_3
\]
3. **Intersection on the \(y\)-axis:**
- Since the lines intersect on the \(y\)-axis, their \(x\)-coordinates at the intersection point are zero. Thus, the \(y\)-intercepts of the lines are:
\[
c_1 = a_1^2 \quad \text{and} \quad c_2 = a_3^2
\]
4. **Angle Between Lines:**
- The tangent of the angle \(\theta\) between the two lines is given by:
\[
\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{(a_1 + a_2) - (a_3 + a_4)}{1 + (a_1 + a_2)(a_3 + a_4)} \right|
\]
5. **Maximizing \(\tan \theta\):**
- Let \(x = a_1 + a_2\) and \(y = a_3 + a_4\). We need to maximize:
\[
f(x, y) = \left| \frac{x - y}{1 + xy} \right|
\]
6. **Case Analysis:**
- **Case 1: \(x, y > 0\)**
- Here, \(x, y \geq 1\) since they are integers. In this case, \(|f(x, y)| < 1\), which cannot exceed our current maximum.
- **Case 2: \(x > 0, y < 0\)**
- The numerator is positive, and the denominator is negative, so \(|f(x, y)|\) cannot be our maximum.
- **Case 3: \(x < 0, y > 0\)**
- Let \(x = -u\) and \(y = v\). We want to maximize:
\[
\frac{u + v}{uv - 1}
\]
- We need to check when:
\[
\frac{u + v}{uv - 1} > \frac{3}{2} \implies 3(uv - 1) < 2(u + v) \implies 3uv - 3 < 2u + 2v \implies 3uv - 2u - 2v < 3
\]
- If \(u, v > 1\), then it's impossible because \(3uv - 2u - 2v \geq 3\) is false. Thus, we can let \(v = 1\) and need \(u < 5\).
7. **Checking Specific Values:**
- For \(u = 4\) and \(v = 1\):
- We have \(a_1 = -6, a_2 = 2, a_3 = 4, a_4 = -3\).
- This gives:
\[
\tan \theta = \left| \frac{(-6 + 2) - (4 - 3)}{1 + (-6 + 2)(4 - 3)} \right| = \left| \frac{-4 - 1}{1 - 4} \right| = \left| \frac{-5}{-3} \right| = \frac{5}{3}
\]
The final answer is \(\boxed{503}\). | 503 | Geometry | math-word-problem | aops_forum | 079495cf-64e8-565d-ab82-523255814ae6 | 2 |
15. As shown in the figure, the area of square $\mathrm{ABCD}$ is 196 square centimeters, and it contains two partially overlapping smaller squares. The larger of the two smaller squares has an area that is 4 times the area of the smaller one, and the overlapping area of the two squares is 1 square centimeter. Therefore, the area of the shaded part is $\qquad$ square centimeters. | 【Answer】 72
【Analysis】Key point: Skillful area calculation
The area of the square is 196, so the side length is 14. The overlapping area is 1, so the side length is 1; the area of the larger square is 4 times that of the smaller square, so the side length of the larger square is twice that of the smaller square, and the sum of the side lengths of the larger and smaller squares is $14+1=15$.
Therefore, the side length of the smaller square is $15 \div 3=5$, and the side length of the larger square is $5 \times 2=10$.
The area of the smaller rectangle is $(5-1) \times(10-1)=36$, so the area of the two smaller rectangles is $36 \times 2=72\left(\mathrm{~cm}^{2}\right)$. | 72 | Geometry | math-word-problem | olympiads | a62b9e76-5993-59bd-81e5-342b61459c26 | 1 |
## Problem 1
Perform the calculations:
a) $7 \cdot 147 - 7 \cdot 47$ (1p)
b) $(2+4+6+8+\cdots+50)-(1+3+5+7+\cdots+49)$ (2p)
c) $10 \cdot 9^{2} : 3^{2} - 3^{4} \quad(2 \text{p})$
d) $(\overline{a b} + \overline{b c} + \overline{c a}) : (a + b + c) \quad$ (2p) | ## Problem 1
a) 7400 ..................................................................................................................... 1p
c) \(10 \cdot(9: 3)^{2}-3^{4}=\)..................................................................................... 1p
= 11 ...................................................................................................................... 1p | 700 | Algebra | math-word-problem | olympiads | e255d78d-a4f6-556e-b5ac-4db7f09aa347 | 2 |
1. The curve $(x+2 y+a)\left(x^{2}-y^{2}\right)=0$ represents three straight lines intersecting at one point on the plane if and only if
A. $a=0$
B. $a=1$
C. $a=-1$
D. $a \in \mathbf{R}$ | Obviously, the three lines are $x+2y+a=0$, $x+y=0$, and $x-y=0$. Since $x+y=0$ and $x-y=0$ intersect at point $(0,0)$, substituting into $x+2y+a=0 \Rightarrow a=0$, so the answer is $A$.
| 0 | Algebra | MCQ | olympiads | 03f68a8f-fc1e-5d88-af8f-b36e21348a84 | 2 |
Of the 36 students in Richelle's class, 12 prefer chocolate pie, 8 prefer apple, and 6 prefer blueberry. Half of the remaining students prefer cherry pie and half prefer lemon. For Richelle's pie graph showing this data, how many degrees should she use for cherry pie?
$ \text{(A)}\ 10\qquad\text{(B)}\ 20\qquad\text{(C)}\ 30\qquad\text{(D)}\ 50\qquad\text{(E)}\ 72 $ | 1. First, we need to determine the number of students who prefer each type of pie. We know:
- 12 students prefer chocolate pie.
- 8 students prefer apple pie.
- 6 students prefer blueberry pie.
2. Calculate the total number of students who prefer chocolate, apple, or blueberry pie:
\[
12 + 8 + 6 = 26
\]
3. Subtract this number from the total number of students in the class to find the number of students who prefer either cherry or lemon pie:
\[
36 - 26 = 10
\]
4. Since half of the remaining students prefer cherry pie and the other half prefer lemon pie, we divide the remaining students by 2:
\[
\frac{10}{2} = 5
\]
Therefore, 5 students prefer cherry pie.
5. To find the number of degrees Richelle should use for cherry pie in her pie graph, we need to calculate the proportion of students who prefer cherry pie and then convert this proportion to degrees. The proportion of students who prefer cherry pie is:
\[
\frac{5}{36}
\]
6. Multiply this proportion by the total number of degrees in a circle (360 degrees):
\[
\frac{5}{36} \times 360 = 50
\]
7. Therefore, Richelle should use 50 degrees for cherry pie in her pie graph.
The final answer is $\boxed{50}$. | 50 | Combinatorics | MCQ | aops_forum | 4f89da07-3f71-5edc-85a6-7460e959c360 | 2 |
Father played chess with uncle. For a won game, the winner received 8 crowns from the opponent, and for a draw, nobody got anything. Uncle won four times, there were five draws, and in the end, father earned 24 crowns.
How many games did father play with uncle?
(M. Volfová) | Father lost four times, so he had to pay his uncle $4 \cdot 8=32$ crowns.
However, Father won so many times that even after paying these 32 crowns, he still gained 24 crowns. His total winnings were $32+24=56$ crowns, so he won $56: 8=7$ games.
Father won seven times, lost four times, and drew five times, so he played a total of $7+4+5=16$ games with his uncle.
Suggested scoring. 2 points for determining Father's total winnings; 2 points for the number of games Father won; 2 points for the total number of games played. | 16 | Logic and Puzzles | math-word-problem | olympiads | 467078db-4e34-536b-a2e4-adb36eeb4b23 | 2 |
## Problem 4
Given the numbers $1,2,3, \ldots, 1000$. Find the largest number $m$ with the property that by removing any $m$ numbers from these 1000 numbers, among the $1000-m$ remaining numbers, there exist two such that one divides the other.
Selected problems by Prof. Cicortas Marius
Note: a) The actual working time is 3 hours.
b) All problems are mandatory.
c) Each problem is graded from 0 to 7.
## NATIONAL MATHEMATICS OLYMPIAD
Local stage - 15.02.2014
## Grade IX
## Grading Rubric | ## Problem 4
If $m \geq 500$ then, for example, by removing the numbers from 1 to 500, among the remaining numbers, from 501 to 1000, there is no pair of numbers that can be divisible by each other, because their ratio is $<2$. 2 points We will show that $m=499$ has the required property. 1 point We show that among any 501 numbers from 1 to 1000, there is one that divides another. Each of the 501 numbers is of the form $2^{k}(2 t+1), 0 \leq k \leq 9, t$ is a natural number; to each such number, we associate $2 t+1$. There are 500 odd numbers less than 1000, so two of the 501 numbers will correspond to the same odd number. Among these two numbers, one will be obtained from the other by multiplying by a power of 2. 4 points | 499 | Number Theory | math-word-problem | olympiads | 3761a0f2-35cd-5bdc-973a-1d63f6ae5131 | 2 |
4. Let $A$ and $B$ be $n$-digit numbers, where $n$ is odd, which give the same remainder $r \neq 0$ when divided by $k$. Find at least one number $k$, which does not depend on $n$, such that the number $C$, obtained by appending the digits of $A$ and $B$, is divisible by $k$. | Solution. Let $A=k a+r, B=k b+r, 0<r<k$; then we have:
$$
C=10^{n} A+B=10^{n}(k a+r)+(k b+r)=k\left(10^{n} a+b\right)+\left(10^{n}+1\right) r .
$$
The number $C$ is divisible by $k$ if and only if the number $10^{n}+1$ is divisible by $k$. Since $n=2 m+1$, we have
$$
10^{n}+1=10^{2 m+1}+1=(10+1)\left(10^{2 m}-10^{m-1}+\ldots+1\right)=11 D
$$
which means that a number $k$ that satisfies the condition of the problem is the number 11.
## 4th year | 11 | Number Theory | math-word-problem | olympiads | fa88ab0b-4061-5ebc-8c4f-5dc03c241d9e | 2 |
5. Given positive real numbers $a$ and $b$ satisfy $a+b=1$, then $M=$ $\sqrt{1+a^{2}}+\sqrt{1+2 b}$ the integer part is | 5. 2
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 2 | Algebra | math-word-problem | olympiads | 30f87501-618e-5c88-94ef-7c3ded50115e | 2 |
1. Arrange the consecutive natural numbers from 1 to 99 in sequence to form a large number:
1234567891011…979899,
By extracting four consecutive digits, you can get a four-digit number, such as 5678, 1011, etc. Among the four-digit numbers obtained by this method, the largest is . $\qquad$ | $9909$ | 9909 | Number Theory | math-word-problem | olympiads | cc8b9ff4-2e54-5def-ae83-efd1defacfa5 | 2 |
8. (15 points) A thin ray of light falls on a thin converging lens at a distance of $x=10$ cm from its optical center. The angle between the incident ray and the plane of the lens $\alpha=45^{\circ}$, and the angle between the refracted ray and the plane of the lens $\beta=30^{\circ}$. Determine its focal length. | Answer: $\approx 13.7$ cm.
Solution. Parallel rays intersect at the focus, so $F$ is the focus of the given lens.
In triangle $O A F$: angle $F A O=30^{\circ}$, angle $O F A=15^{\circ}$, angle $A O F=135^{\circ}$.
Therefore, $\frac{O F}{\sin 30^{\circ}}=\frac{A O}{\sin 15^{\circ}}$.
We obtain that the focal length:
$F C=O F \sin 45^{\circ}=A O \frac{\sin 30^{\circ}}{\sin 15^{\circ}} \sin 45^{\circ} \approx 13.7 \mathrm{~cm}$.
## Multidisciplinary Engineering Olympiad "Star" in Natural Sciences
10th grade
Final Round
$2022-2023$
Version 2
Problems, answers, and evaluation criteria | 13.7 | Geometry | math-word-problem | olympiads | 78b0db16-409f-5e5d-9861-d426afbf5dc6 | 1 |
8. If $n$ is a natural number less than 50, find all values of $n$ such that the values of the algebraic expressions $4n+5$ and $7n+6$ have a common divisor greater than 1. | 8. Solution: Let $(4 n+5,7 n+6)=d>1$, then $d|(4 n+5), d|(7 n+6)$.
Thus $d \mid(7 n+6-(4 n+5))=3 n+1$,
$$
\begin{array}{l}
d \mid((4 n+5)-(3 n+1))=n+4, \\
d \mid((3 n+1)-2(n+4))=n-7, \\
d \mid((n+4)-(n-7))=11 .
\end{array}
$$
Since 11 is a prime number, then $d=11$.
Let $n-7=11 k$, then
$0<n=11 k+7<50$. Solving for $k$ gives $k=0,1,2,3$.
Thus $n=7,18,29,40$. | 7,18,29,40 | Number Theory | math-word-problem | olympiads | 9ecce5ef-1802-5883-8ec4-c9bab29b7647 | 2 |
5. 5.1. Find the largest natural number in which all digits are different and any two adjacent digits differ by 6 or 7. | Answer: 60718293.
## Solution.
We will map each digit from 0 to 9 to a vertex in a graph and connect the vertices with an edge if the corresponding digits differ by 6 or 7.
We see that the vertices corresponding to the digits 4 and 5 are not connected to anything, so the maximum number that can be obtained will be 8 digits long. The vertices corresponding to the digits 3 and 6 can only be the first or last in the number. Since we need to find the maximum number, it is uniquely determined by moving along the edges of the graph and will be equal to 60718293. | 60718293 | Number Theory | math-word-problem | olympiads | 7edacb04-aea8-5a67-b4b4-4a1930dcb4df | 2 |
# Problem 6. (3 points)
Three runners are moving along a circular track at constant equal speeds. When two runners meet, they instantly turn around and start running in opposite directions.
At some point, the first runner meets the second. After 15 minutes, the second runner meets the third for the first time. Another 25 minutes later, the third runner meets the first for the first time.
How many minutes does it take for one runner to complete the entire track? | # Answer: 80
## Solution:
Let the first runner meet the second, then after $a$ minutes the second runner meets the third for the first time, and after another $b$ minutes the third runner meets the first for the first time.
Let the first and second runners meet at point $A$, the second and third at point $B$, and the first and third at point $C$. Additionally, let the first runner be at point $D$ at the moment the second and third runners meet.
Then the points are arranged on the circle in the following order: $D, A, B, C$, where the first runner ran from $D$ to $A$ in $a$ minutes, the second runner ran from $A$ to $B$ in $a$ minutes, the third runner ran from $B$ to $C$ in $b$ minutes, and the first runner ran from $D$ to $C$ in $b$ minutes. These four segments together form the entire track, so it is covered in $2a + 2b$ minutes. | 80 | Logic and Puzzles | math-word-problem | olympiads | 62791263-04f8-508f-9886-2dd0d0a262be | 2 |
## Task 2 - 080522
In a warehouse, there are three times as many kilograms of wheat as in a second one. After removing $85000 \mathrm{~kg}$ from the first and $5000 \mathrm{~kg}$ from the second, the stocks were equal.
How many tons of wheat were in the first and how many in the second warehouse before the removal? | It is given: $85000 \mathrm{~kg}=85 \mathrm{t}$ and $5000 \mathrm{~kg}=5 \mathrm{t}$.
If after removing $85 \mathrm{t}$ of wheat from the first and $5 \mathrm{t}$ from the second, the stocks in both storage rooms are equal, there were $80 \mathrm{t}$ more in the first storage room; because $85 \mathrm{t} - 5 \mathrm{t} = 80 \mathrm{t}$.
Since the stock in the first storage room was initially three times as large as in the second, and after the removal, the stocks in both are equal, the initial surplus in the first room must have been twice as large as the stock in the second room.
Therefore, there were 40 t in the second room, which is half of 80 t, and 120 t in the first; because $40 \mathrm{t} \times 3 = 120 \mathrm{t}$. | 120 | Algebra | math-word-problem | olympiads | ba6907c3-ccde-571d-bb32-9a39b1991602 | 2 |
$17 \cdot 27$ As shown in the figure, in $\triangle A B C$, $\angle A=42^{\circ}$. If the trisectors of $\angle B$ and $\angle C$ intersect at $D, E$. Then the degree measure of $\angle B D C$ is
(A) $67^{\circ}$.
(B) $84^{\circ}$.
(C) $88^{\circ}$.
(D) $110^{\circ}$.
(China Beijing Junior High School Mathematics Competition, 1990); | [Solution] Extend $A D$. By the exterior angle property of a triangle,
$$
\angle 1=\angle 3+\angle 5, \angle 2=\angle 4+\angle 6,
$$
Adding these two equations, we get
$$
\begin{aligned}
\angle B D C & =\angle A+\frac{1}{3} \angle B+\frac{1}{3} \angle C \\
& =\angle A+\frac{1}{3}(\angle B+\angle C) \\
& =\angle A+\frac{1}{3}\left(180^{\circ}-\angle A\right)=60^{\circ}+\frac{2}{3} \angle A=88^{\circ} .
\end{aligned}
$$
Therefore, the answer is $(C)$. | 88 | Geometry | MCQ | olympiads | dd729366-2c6f-5255-8db0-20e8ec10f9a3 | 2 |
Example 5. Expand the function
$$
f(z)=\frac{z}{z^{2}-2 z-3}
$$
into a Taylor series in the neighborhood of the point $z_{0}=0$ using expansion (12), and find the radius of convergence of the series. | Solution. Let's decompose the given function into partial fractions:
$$
\frac{z}{z^{2}-2 z-3}=\frac{1}{4} \frac{1}{z+1}-\frac{3}{4} \frac{1}{z-3}
$$
Transform the right-hand side as follows:
$$
f(z)=\frac{1}{4} \frac{1}{1+z}-\frac{1}{4} \frac{1}{1-\frac{2}{3}}
$$
Using the expansion (12) of the function $\frac{1}{1+z}$, we get
$$
\begin{aligned}
f(z) & =\frac{1}{4}\left(1-z+z^{2}-z^{3}+\ldots\right)-\frac{1}{4}\left(1+\frac{z}{3}+\frac{z^{2}}{9}+\ldots\right)= \\
& =\frac{1}{4}\left(-\frac{4}{3} z+\frac{8}{9} z^{2}-\frac{28}{27} z^{3}+\ldots\right)=-\frac{z}{3}+\frac{2}{3^{2}} z^{2}-\frac{7}{3^{3}} z^{3}+\ldots
\end{aligned}
$$
The nearest singular point to the point $z_{0}=0$ of the given function is the point $z=-1$. Therefore, the radius of convergence of the obtained series is $R=1$. | 1 | Algebra | math-word-problem | olympiads | 632d4671-91dd-5893-a0e2-fcd043c434af | 2 |
Let $a_1,a_2,a_3,a_4,a_5$ be distinct real numbers. Consider all sums of the form $a_i + a_j$ where $i,j \in \{1,2,3,4,5\}$ and $i \neq j$. Let $m$ be the number of distinct numbers among these sums. What is the smallest possible value of $m$? | 1. **Define the problem and the sums:**
We are given five distinct real numbers \(a_1, a_2, a_3, a_4, a_5\). We need to consider all possible sums of the form \(a_i + a_j\) where \(i \neq j\). There are \(\binom{5}{2} = 10\) such sums since each pair \((i, j)\) is considered once.
2. **Calculate the number of distinct sums:**
We need to find the smallest possible number of distinct sums, denoted as \(m\).
3. **Example with specific values:**
Let's try the values \(a_1 = 0, a_2 = 1, a_3 = 2, a_4 = 3, a_5 = 4\). The sums are:
\[
\begin{aligned}
&0+1 = 1, \\
&0+2 = 2, \\
&0+3 = 3, \\
&0+4 = 4, \\
&1+2 = 3, \\
&1+3 = 4, \\
&1+4 = 5, \\
&2+3 = 5, \\
&2+4 = 6, \\
&3+4 = 7.
\end{aligned}
\]
The distinct sums are \{1, 2, 3, 4, 5, 6, 7\}, so \(m = 7\).
4. **Check if \(m = 6\) is possible:**
Suppose \(a_1 = 0\) (without loss of generality, since adding a constant to each \(a_i\) does not change the number of distinct sums). Let the remaining numbers be \(a, b, c, d\) in increasing order. The sums are:
\[
\begin{aligned}
&0 + a = a, \\
&0 + b = b, \\
&0 + c = c, \\
&0 + d = d, \\
&a + b, \\
&a + c, \\
&a + d, \\
&b + c, \\
&b + d, \\
&c + d.
\end{aligned}
\]
For \(m = 6\), we need some sums to be equal. Assume \(a + b = d\). Then the sums are:
\[
\{a, b, c, d, a + c, a + d, b + c, b + d, c + d\}.
\]
Since \(a + b = d\), we have:
\[
\{a, b, c, d, a + c, d + a, b + c, b + d, c + d\}.
\]
We need to check if we can have only 6 distinct sums. If \(a + c = b + d\), then \(a + c = b + (a + b) = a + 2b\), so \(c = 2b\). This leads to contradictions as we need all numbers to be distinct.
5. **Conclusion:**
After trying different configurations and ensuring all numbers are distinct, we find that the smallest possible value of \(m\) is indeed 7.
The final answer is \(\boxed{7}\). | 7 | Combinatorics | math-word-problem | aops_forum | 93026f85-b5b7-5474-8263-958bfe22fd1c | 2 |
3. There are ten small balls of the same size, five of which are red and five are white. Now, these ten balls are arranged in a row arbitrarily, and numbered from left to right as $1,2, \cdots, 10$. Then the number of arrangements where the sum of the numbers of the red balls is greater than the sum of the numbers of the white balls is. $\qquad$ | 3. 126.
First, the sum of the numbers is $1+2+\cdots+10=$ 55. Therefore, the sum of the numbers on the red balls cannot be equal to the sum of the numbers on the white balls.
Second, if a certain arrangement makes the sum of the numbers on the red balls greater than the sum of the numbers on the white balls, then by swapping the positions of the red and white balls, we get an arrangement where the sum of the numbers on the red balls is less than the sum of the numbers on the white balls, and vice versa.
Therefore, the number of arrangements where the sum of the numbers on the red balls is greater than the sum of the numbers on the white balls is equal to the number of arrangements where the sum of the numbers on the red balls is less than the sum of the numbers on the white balls. Thus, the number of arrangements we are looking for is $\frac{1}{2} \mathrm{C}_{10}^{5}=126$. | 126 | Combinatorics | math-word-problem | cn_contest | f9e7a3d8-0093-524b-a839-37ca3533f83e | 4 |
|
There are 85 balloons in the room - red and blue. It is known that: 1) at least one of the balloons is red; 2) in every arbitrarily chosen pair of balloons, at least one is blue. How many red balloons are in the room?
# | Think about whether there can be two red balls in the room.
## Solution
Since among any two balls, one is blue, there cannot be two red balls in the room. Therefore, there are 84 blue balloons and 1 red balloon in the room.
## Answer
1 ball. | 1 | Logic and Puzzles | math-word-problem | olympiads | 4671698a-54f7-52fb-988c-460378917299 | 2 |
3. Given positive numbers $a, b, c$ satisfying $a+b+c=1$,
then $\sqrt{\frac{(a+b c)(b+c a)}{c+a b}}+\sqrt{\frac{(b+c a)(c+a b)}{a+b c}}+\sqrt{\frac{(c+a b)(a+b c)}{b+c a}}=$ $\qquad$ | $2$ | 2 | Algebra | math-word-problem | olympiads | 68430211-520d-5b7d-bfb9-956fcf8ed26d | 2 |
# Problem 3.
A polynomial $P(x)$ with integer coefficients has the properties
$$
P(1)=2019, \quad P(2019)=1, \quad P(k)=k,
$$
where the number $k$ is an integer. Find this number $k$.
# | # Solution.
Since the polynomial $P(x)$ has integer coefficients, $P(a)-P(b)$ is divisible by $a-b$ for any integers $a$ and $b$.
We get that
$$
\begin{gathered}
P(k)-P(1)=(k-2019) \text { is divisible by }(k-1), \\
P(k)-P(2019)=(k-1) \text { is divisible by }(k-2019) .
\end{gathered}
$$
This can only be true if $|k-1|=|k-2019|$.
The solution to the obtained equation is $k=1010$.
Answer: $\quad k=1010$. | 1010 | Algebra | math-word-problem | olympiads | e3651185-feff-5ded-b50a-063f8af01839 | 2 |
15.15 A paper punch can be placed at any point in the plane, and when it operates, it can punch out points at an irrational distance from it. What is the minimum number of paper punches needed to punch out all points in the plane?
(51st Putnam Mathematical Competition, 1990) | ［Solution］Obviously, two hole punchers are not
enough.
Below we prove that three hole punchers are sufficient.
For example, in a Cartesian coordinate system, place the three hole punchers at points $A(0,0), B(d, 0)$, $C(2d, 0)$, where $d=\sqrt[4]{2}$.
Let $P$ be any point on the plane, and the distances from $P$ to $A, B, C$ be $a, b, c$ respectively. Thus, we have
$$
\begin{array}{l}
a^{2}+c^{2}=2 b^{2}+2 d^{2}, \\
a^{2}-2 b^{2}+c^{2}=2 \sqrt{2} .
\end{array}
$$
$a, b, c$ cannot all be rational numbers, so point $P$ must be punched by at least one hole puncher. | 3 | Geometry | math-word-problem | olympiads | d1c8a408-369d-5473-88ff-e15f0f7ca720 | 2 |
$7 \cdot 74$ There are several locks, and now six people each hold a part of the keys. It is known that any two people trying to open the locks together will have exactly one lock that they cannot open, while any three people can open all the locks. How many locks are there at least?
---
Note: The translation keeps the original format and structure of the text, including the mathematical notation and the problem statement, to ensure the context and meaning are accurately conveyed. | [Solution 1] Since every pair of people has one lock they cannot open, let the total number of these locks be $k$, then
$$
k \leqslant C_{6}^{2}=15.
$$
But $k<C_{6}^{2}$ is impossible, because if $k<C_{6}^{2}$, there must be two different pairs of people who cannot open the same lock, and two different pairs of people must include at least three people, which contradicts the fact that any three people can open all the locks.
Therefore, $k=C_{6}^{2}=15$.
That is, there are at least 15 locks.
[Solution 2] Let any pair of people be denoted as $(i, j)$ (numbering the 6 people as $1,2,3,4,5,6$), where $i \neq j$.
Clearly, $(i, j)=(j, i)$.
There are $C_{6}^{2}=15$ such pairs.
Let the lock that $(i, j)$ cannot open be denoted as $a_{ij}$.
We will now prove that when $(i, j) \neq (k, l)$, $a_{ij}$ and $a_{kl}$ are different.
If not, then the pairs $(i, j)$ and $(k, l)$ cannot open the same lock, and $(i, j)$ and $(k, l)$ must include at least three people, which contradicts the fact that any three people can open all the locks.
Therefore, the number of locks that pairs cannot open is the same as the number of pairs, so $C_{6}^{2}=15$ is the minimum number of locks. | 15 | Combinatorics | math-word-problem | olympiads | 3711a722-b1d5-5b30-8d25-58a864ddca63 | 2 |
5. Let the complex numbers $z_{1}, z_{2}$ satisfy $\left|z_{1}\right|=\left|z_{1}+z_{2}\right|=3,\left|z_{1}-z_{2}\right|=3 \sqrt{3}$, then $\log _{3}\left|\left(z_{1} \overline{z_{2}}\right)^{2000}+\left(\overline{z_{1}} z_{2}\right)^{2000}\right|=$ $\qquad$ . | 4000
5. 【Analysis and Solution】Solution 1 From the given conditions, we have
$$
\begin{array}{l}
9=\left|z_{1}+z_{2}\right|^{2}=\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}+z_{1} \bar{z}_{2}+\bar{z}_{1} z_{2}, \\
27=\left|z_{1}-z_{2}\right|^{2}=\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}-\left(z_{1} \bar{z}_{2}+\bar{z}_{1} z_{2}\right) .
\end{array}
$$
Since $\left|z_{1}\right|=3$, it follows that $\left|z_{2}\right|=3, z_{1} \bar{z}_{2}+\bar{z}_{1} z_{2}=-9$, and $\left|z_{1} \bar{z}_{2}\right|+| \bar{z}_{1} z_{2} \mid=9$.
Let $z_{1} \bar{z}_{2}=9(\cos \theta+i \sin \theta)$,
$$
\bar{z}_{1} z_{2}=9(\cos \theta-i \sin \theta) \text {. }
$$
$-9=z_{1} \bar{z}_{2}+z_{1} \bar{z}_{2}=18 \cos \theta$, which gives
$$
\cos \theta=-\frac{1}{2} \text {. }
$$
Thus, $z_{1} \bar{z}_{2}=9 \omega$ or $z_{1} \bar{z}_{2}=9 \omega^{2}$,
where $\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} i$.
When $z_{1} \bar{z}_{2}=9 \omega$, we get
$$
\left(z_{1} \bar{z}_{2}\right)^{2000}+\left(\bar{z}_{1} z_{2}\right)^{2000}=-9^{2000} \text {, }
$$
Therefore, $\log _{3}\left|\left(z_{1} \bar{z}_{2}\right)^{2000}+\left(\bar{z}_{1} z_{2}\right)^{2000}\right|=4000$.
When $z_{1} \bar{z}_{2}=9 \omega^{2}$, we get the same result, so the answer is 4000.
Solution 2 Consider the geometric meaning of complex numbers. Let the points corresponding to the complex numbers $z_{1}, z_{2}$ in the complex plane be $A, B$, and let $C$ be the midpoint of $A B$. From the conditions,
$$
\begin{array}{l}
|O A|=\left|z_{1}\right|=3, \\
|O C|=\frac{1}{2}\left|z_{1}+z_{2}\right|=\frac{3}{2},
\end{array}
$$
$|A C|=\frac{1}{2}\left|z_{1}-z_{2}\right|=\frac{3}{2} \sqrt{3}$, thus it is easy to see that $\triangle O A B$ is an equilateral triangle. Without loss of generality, assume $z_{2}=z_{1} e^{i \frac{\pi}{3}}$, substituting this in gives $\log _{3}\left|\left(z_{1} \bar{z}_{2}\right)^{2000}+\left(\bar{z}_{1} z_{2}\right)^{2000}\right|=4000$. | 4000 | Algebra | math-word-problem | olympiads | 33d55892-4916-5da3-a46d-d4c242e74b6b | 2 |
7. Given positive integers $a, b$ satisfy
$$
\sqrt{\frac{a b}{2 b^{2}-a}}=\frac{a+2 b}{4 b} \text {. }
$$
Then $|10(a-5)(b-15)|+2=$ | 7.2012.
Notice,
$$
\begin{array}{l}
\sqrt{\frac{a b}{2 b^{2}-a}}=\frac{a+2 b}{4 b} \\
\Leftrightarrow 16 a b^{3}=\left(2 b^{2}-a\right)\left(a^{2}+4 a b+4 b^{2}\right) \\
\Leftrightarrow a\left(a^{2}+4 a b+4 b^{2}\right)=2 b^{2}\left(a^{2}-4 a b+4 b^{2}\right) \\
\Leftrightarrow a(a+2 b)^{2}=2 b^{2}(a-2 b)^{2} .
\end{array}
$$
Let $d=(a, b), a=a_{0} d, b=b_{0} d,\left(a_{0}, b_{0}\right)=1$.
Then equation (1) $\Leftrightarrow a_{0}\left(a_{0}+2 b_{0}\right)^{2}=2 d b_{0}^{2}\left(a_{0}-2 b_{0}\right)^{2}$.
Since $b_{0} \mid a_{0}\left(a_{0}+2 b_{0}\right)^{2}$, we get
$$
b_{0} \mid a_{0}^{3} \Rightarrow b_{0}=1 \text {. }
$$
Then $a_{0}\left(a_{0}+2\right)^{2}=2 d\left(a_{0}-2\right)^{2}$
$$
\Rightarrow 2 d=a_{0}+\frac{8 a_{0}^{2}}{\left(a_{0}-2\right)^{2}}=a_{0}+2\left(\frac{2 a_{0}}{a_{0}-2}\right)^{2} \text {. }
$$
Thus $\left(a_{0}-2\right)\left|2 a_{0} \Rightarrow\left(a_{0}-2\right)\right| 4$
$$
\Rightarrow a_{0}-2= \pm 1,2,4 \text {. }
$$
From equation (2), we know $2 \mid a_{0}$. Therefore, $a_{0}=4,6$.
Substituting into equation (2) gives $d=18,12$.
Accordingly, $(a, b)=(72,18),(72,12)$.
Thus, $|10(a-5)(b-15)|+2=2012$. | 2012 | Algebra | math-word-problem | cn_contest | 2e8a9edd-dfb2-5c52-a7c3-ce50ae674f3a | 2 |
1. A three-digit number is 29 times the sum of its digits. Then this three-digit number is $\qquad$ | Let the three-digit number be $\overline{a b c}$. Then
$$
\begin{array}{l}
100 a+10 b+c=29(a+b+c) \\
\Rightarrow 71 a=19 b+28 c \\
\Rightarrow b=4 a-2 c+\frac{5}{19}(2 c-a) .
\end{array}
$$
Since $a$, $b$, and $c$ are positive integers, 19 divides $(2 c - a)$.
And $-9 \leqslant 2 c-a \leqslant 17$, then
$$
2 c-a=0 \Rightarrow a=2 c, b=6 c \text {. }
$$
From $0<b<10$, we know $c=1$.
Thus, $a=2, b=6$. | 261 | Number Theory | math-word-problem | cn_contest | 3a30a6d4-623e-51b6-ace9-e3bda28c7e5a | 2 |
Variant 9.5.2. Sides $A D$ and $D C$ of the inscribed quadrilateral $A B C D$ are equal. A point $X$ is marked on side $B C$ such that $A B = B X$. It is known that $\angle B = 32^{\circ}$ and $\angle X D C = 52^{\circ}$.
(a) (1 point) How many degrees does the angle $A X C$ measure?
(b) (3 points) How many degrees does the angle $A C B$ measure?
 | Answer: (a) $106^{\circ}$. (b) $48^{\circ}$. | 106 | Geometry | math-word-problem | olympiads | cfc8164c-37e4-54f2-970f-e1f34174ff46 | 1 |
22. A year is called a leap year if it is either divisible by 4 but not divisible by 100 , or divisible by 400 . Hence, the years 2000, 2004 and 2400 are leap years while the years 2006, 2100 and 2200 are not. Find the number of leap years between 2000 and 4000 inclusive. | 22. Ans: 486
Let $S=\{x \in \mathbb{Z} \mid 2000 \leq x \leq 4000\}, A=\{x \in S \mid x$ is divisible by 4$\}, B=\{x \in S \mid$ $x$ is divisible by 100$\}, C=\{x \in S \mid x$ is divisible by 400$\}$. The required answer is
$$
|A|-|B|+|C|=\left(\frac{4000}{4}-\frac{2000}{4}+1\right)-\left(\frac{4000}{100}-\frac{2000}{100}+1\right)+\left(\frac{4000}{400}-\frac{2000}{400}+1\right)=486
$$ | 486 | Number Theory | math-word-problem | olympiads | 0c23769a-24e9-5abf-a9e0-0b24562a8ef5 | 2 |
Example 7 How many positive integer factors does 20! have? | Analyze writing 20! in its standard factorization form $n=\beta_{1}^{a_{1}} \beta_{2}^{a_{2}} \cdots \beta_{k}^{q_{k}}$, and then using $r(n)=\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \cdots$ $\left(\alpha_{k}+1\right)$ to calculate the result.
Solution Since the prime numbers less than 20 are $2,3,5,7,11,13,17,19$, in the standard factorization of 20!, the highest power of 2 $=\left[\frac{20}{2}\right]+\left[\frac{20}{4}\right]+\left[\frac{20}{8}\right]+\left[\frac{20}{16}\right]=18$,
the highest power of 3 $=\left[\frac{20}{3}\right]+\left[\frac{20}{9}\right]=8$,
the highest power of 5 $=\left[\frac{20}{5}\right]=4$,
the highest powers of $7,11,13,17,19$ are $2,1,1,1,1$ respectively.
Thus, the standard factorization of 20! is
$$20!=2^{18} \cdot 3^{8} \cdot 5^{4} \cdot 7^{2} \cdot 11 \cdot 13 \cdot 17 \cdot 19$$
Therefore, the number of positive divisors $r(20!)$ of 20! is:
$$\begin{aligned}
r(20!) & =(18+1)(8+1)(4+1)(2+1)(1+1)(1+1)(1+1)(1+1) \\
& =19 \cdot 9 \cdot 5 \cdot 3 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \\
& =41040
\end{aligned}$$
Thus, the number of positive divisors of $20!$ is 41040. | 41040 | Number Theory | math-word-problem | number_theory | 38902e28-3cc8-5f40-a032-89bdea866710 | 2 |
The area of the region in the $xy$-plane satisfying the inequality \[\min_{1 \le n \le 10} \max\left(\frac{x^2+y^2}{4n^2}, \, 2 - \frac{x^2+y^2}{4n^2-4n+1}\right) \le 1\] is $k\pi$, for some integer $k$. Find $k$.
[i]Proposed by Michael Tang[/i] | To find the area of the region in the $xy$-plane satisfying the inequality
\[
\min_{1 \le n \le 10} \max\left(\frac{x^2+y^2}{4n^2}, \, 2 - \frac{x^2+y^2}{4n^2-4n+1}\right) \le 1,
\]
we need to analyze the given inequality step by step.
1. **Understanding the Inequality:**
The inequality involves two expressions:
\[
\max\left(\frac{x^2 + y^2}{4n^2}, 2 - \frac{x^2 + y^2}{4n^2 - 4n + 1}\right) \le 1.
\]
This means that for some integer $n$ in the range $1 \le n \le 10$, the maximum of the two expressions must be less than or equal to 1.
2. **Breaking Down the Inequality:**
For the maximum of the two expressions to be less than or equal to 1, both expressions must individually be less than or equal to 1:
\[
\frac{x^2 + y^2}{4n^2} \le 1 \quad \text{and} \quad 2 - \frac{x^2 + y^2}{4n^2 - 4n + 1} \le 1.
\]
3. **Simplifying Each Expression:**
- For the first expression:
\[
\frac{x^2 + y^2}{4n^2} \le 1 \implies x^2 + y^2 \le 4n^2.
\]
- For the second expression:
\[
2 - \frac{x^2 + y^2}{4n^2 - 4n + 1} \le 1 \implies \frac{x^2 + y^2}{4n^2 - 4n + 1} \ge 1 \implies x^2 + y^2 \ge 4n^2 - 4n + 1.
\]
4. **Combining the Conditions:**
Therefore, for some $n$ in the range $1 \le n \le 10$, the point $(x, y)$ must satisfy:
\[
4n^2 - 4n + 1 \le x^2 + y^2 \le 4n^2.
\]
This describes a ring (annulus) in the $xy$-plane with inner radius $\sqrt{4n^2 - 4n + 1}$ and outer radius $2n$.
5. **Calculating the Area of Each Annulus:**
The area of an annulus with inner radius $r_1$ and outer radius $r_2$ is given by:
\[
\pi (r_2^2 - r_1^2).
\]
For our specific case:
\[
r_2 = 2n \quad \text{and} \quad r_1 = \sqrt{4n^2 - 4n + 1}.
\]
Therefore, the area of the annulus is:
\[
\pi \left((2n)^2 - (4n^2 - 4n + 1)\right) = \pi \left(4n^2 - (4n^2 - 4n + 1)\right) = \pi (4n - 1).
\]
6. **Summing the Areas for $n$ from 1 to 10:**
We need to sum the areas of the annuli for $n = 1, 2, \ldots, 10$:
\[
\sum_{n=1}^{10} \pi (4n - 1) = \pi \sum_{n=1}^{10} (4n - 1).
\]
The sum inside the parentheses is:
\[
\sum_{n=1}^{10} (4n - 1) = 4 \sum_{n=1}^{10} n - \sum_{n=1}^{10} 1 = 4 \left(\frac{10 \cdot 11}{2}\right) - 10 = 4 \cdot 55 - 10 = 220 - 10 = 210.
\]
7. **Final Area:**
Therefore, the total area is:
\[
210\pi.
\]
The final answer is $\boxed{210}$. | 210 | Inequalities | math-word-problem | aops_forum | 36667977-8bb5-552e-8fcc-d328f2a797c0 | 2 |
2. On a line, several points were marked, including points $A$ and $B$. All possible segments with endpoints at the marked points are considered. Vasya calculated that point $A$ is inside 50 of these segments, and point $B$ is inside 56 segments. How many points were marked? (The endpoints of a segment are not considered its internal points.) | Answer: 16.
Solution. Let there be $a_{1}$ points on one side of point $A$ and $a_{2}$ points on the other side; $b_{1}$ points on one side of point $B$ and $b_{2}$ points on the other side. We can assume that $a_{1} \leqslant a_{2}, b_{1} \leqslant b_{2}$. Then $a_{1} a_{2}=50, b_{1} b_{2}=56$. At the same time, $a_{1}+a_{2}=b_{1}+b_{2}$. Let's consider all possible factorizations of the numbers 50 and 56:
$$
50=1 \cdot 50=2 \cdot 25=5 \cdot 10 ; \quad 56=1 \cdot 56=2 \cdot 28=4 \cdot 14=7 \cdot 8
$$
Only in one case do the sums of the divisors of these two numbers match: $5+10=7+8$. Therefore, $a_{1}=5, a_{2}=10, b_{1}=7, b_{2}=8$, and there are a total of 16 points.
Evaluation. 12 points for a correct solution. 2 points for a correct answer (without justification). If the correctness of the answer is shown but the uniqueness is not justified, 6 points. | 16 | Combinatorics | math-word-problem | olympiads | 986c0e87-ff9d-53be-8035-c9ef57638b9e | 2 |
9. In 1978, someone introduced their family by saying: "I have a son and a daughter, they are not twins. The cube of my son's age plus the square of my daughter's age is exactly my birth year. I was born after 1900. My children are both under 21 years old. I am 8 years older than my wife." Calculate the age of each family member in 1978. (15 points) | 9. In 1978, the son's age was 12 years old, the daughter's age was 14 years old, the mother's age was 46 years old, and the father's age was 54 years old.
Solution: Since \(13^3 = 2197\), the son's age must be less than 13 years; also, since \(11^3 = 1331\), even if we add \(21^2 = 441\), it is only \(1331 + 441 = 1772\), so \(x^2 > 1900 - 12^3\), \(x^2 > 172\), which means the daughter's age is greater than 13 years. Given that this age is less than 21 years, the possible ages for the daughter are: 14, 15, 16, 17, 18, 19, 20.
If \(x = 15\), then the father's birth year would be: \(12^3 + 15^2 = 1953\), which is clearly unreasonable. Similarly, the daughter's age cannot be greater than 15, so it can only be 14 years old.
At this time, the father's birth year is: \(12^3 + 14^2 = 1924\), and his age in 1978 is: \(1978 - 1924 = 54\) (years old).
The mother's age in 1978 is: \(54 - 8 = 46\) (years old).
Scoring reference: (1) Correctly providing the answer, 5 points; (2) For each formula, explaining the reasoning, 10 points; (3) Even if the final answer is incorrect, for correctly reasoned formulas, appropriate points should be given. | 12,14,46,54 | Algebra | math-word-problem | olympiads | 70640b9f-539a-5b1c-9662-36bedb831ce5 | 2 |
3. (8 points) When Tongtong was calculating a division with a remainder, she mistakenly read the dividend 472 as 427, resulting in a quotient that was 5 less than the original, but the remainder was exactly the same. What is this remainder?
A. 4
B. 5
C. 6
D. 7 | 【Answer】Solution: $(472-427) \div 5$
$$
\begin{array}{l}
=45 \div 5 \\
=9
\end{array}
$$
$$
472 \div 9=52 \cdots 4
$$
Answer: The remainder is 4.
Therefore, the correct choice is: A. | 4 | Number Theory | MCQ | olympiads | b0d7cb9b-76ed-5635-91e1-e8eefebf47d4 | 2 |
4.9 On a $10 \times 10$ grid paper, there are 11 horizontal grid lines and 11 vertical grid lines. A line segment connecting two adjacent nodes on the same straight line is called a "link segment". How many link segments need to be erased at least, so that at most 3 link segments remain at each node? | [Solution] Removing the dotted links in the right figure satisfies the requirements of the problem, and there are 41 dotted links in the figure, so the minimum value required does not exceed 41.
On the other hand, there are 81 internal nodes on the grid paper, each with 4 links. Therefore, at least one link must be erased from each node to meet the requirement. Since each link connects two nodes, at least 41 links must be erased.
In summary, the minimum number of links that need to be erased is 41. | 41 | Combinatorics | math-word-problem | olympiads | 8253de0b-ede9-5630-a1d1-7d0ae1ae5cd2 | 2 |
Example 13 Given that $a$, $b$, and $c$ are all positive integers, and the parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at two distinct points $A$ and $B$. If the distances from $A$ and $B$ to the origin are both less than 1, find the minimum value of $a+b+c$.
(1996, National Junior High School Mathematics League) | Solution: Let $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$, and $x_{1} < 0 < x_{2}$,
then $x_{1} < 0, \\
\therefore b > 2 \sqrt{a c} . \\
\text{Also, } \because |O A| = |x_{1}| > 1$. Therefore, the parabola opens upwards, and when $x = -1$, $y > 0$, so $a(-1)^{2} + b(-1) + c > 0$, which means $b < a + c + 1$. Since $b > 2 \sqrt{a c}$, we have $2 \sqrt{a c} + 1 < a + c$,
i.e., $(\sqrt{a} - \sqrt{c})^{2} > 1$.
From equation (2), we get $\sqrt{a} - \sqrt{c} > 1$,
i.e., $\sqrt{a} > \sqrt{c} + 1$.
Thus, $a > (\sqrt{c} + 1)^{2} \geq (\sqrt{1} + 1)^{2} = 4$.
Solving, we get $a \geq 5$.
$$
\begin{array}{l}
\text{Also, } \because b > 2 \sqrt{a c} \geq 2 \sqrt{5 \times 1} > 4, \\
\therefore b \geq 5 .
\end{array}
$$
When $a = 5, b = 5, c = 1$, the parabola $y = 5 x^{2} + 5 x + 1$ satisfies the given conditions.
Therefore, the minimum value of $a + b + c$ is 11. | 11 | Algebra | math-word-problem | cn_contest | 145da507-b5fa-5f63-b6dd-5097b644c349 | 2 |
13. (15 points) As shown in the figure, ten points are evenly marked on the circumference of a circle. Place the ten natural numbers $1 \sim 10$ on these ten points. Rotate a line passing through the center of the circle. When the line does not pass through any marked points, it divides the numbers 1 to 10 into two groups. For each arrangement, there are five ways to divide the numbers as the line rotates. For each division, there is a product of the sums of the two groups, and the smallest of these five products is denoted as $K$. What is the maximum value of $K$ among all possible arrangements? | 【Analysis】No matter how these 10 numbers are placed, their sum remains unchanged, $1+2+3+4+5+6+7+8+9+10=55$, divided into two groups, the closer the sums of the two groups, the larger the required product will be.
【Solution】
$$
\begin{array}{l}
1+2+3+4+5+6+7+8+9+10=55 \\
55=27+28
\end{array}
$$
Therefore, as long as the 10 numbers are divided into two parts with sums of 27 and 28, the product at this time is $27 \times 28=756$. Answer: In all the arrangements, $K$ is maximized at 756. | 756 | Combinatorics | math-word-problem | olympiads | cbdaf71c-479e-5ef7-b26b-7b66cd85e7cf | 1 |
7. It is known that the number $a$ satisfies the equation param1, and the number $b$ satisfies the equation param2. Find the smallest possible value of the sum $a+b$.
| param1 | param2 | |
| :---: | :---: | :--- |
| $x^{3}-3 x^{2}+5 x-17=0$ | $x^{3}-6 x^{2}+14 x+2=0$ | |
| $x^{3}+3 x^{2}+6 x-9=0$ | $x^{3}+6 x^{2}+15 x+27=0$ | |
| $x^{3}-6 x^{2}+16 x-28=0$ | $x^{3}+3 x^{2}+7 x+17=0$ | |
| $x^{3}+6 x^{2}+17 x+7=0$ | $x^{3}-3 x^{2}+8 x+5=0$ | | | 7. It is known that the number $a$ satisfies the equation param1, and the number $b$ satisfies the equation param2. Find the smallest possible value of the sum $a+b$.
| param1 | param2 | Answer |
| :---: | :---: | :---: |
| $x^{3}-3 x^{2}+5 x-17=0$ | $x^{3}-6 x^{2}+14 x+2=0$ | 3 |
| $x^{3}+3 x^{2}+6 x-9=0$ | $x^{3}+6 x^{2}+15 x+27=0$ | -3 |
| $x^{3}-6 x^{2}+16 x-28=0$ | $x^{3}+3 x^{2}+7 x+17=0$ | 1 |
| $x^{3}+6 x^{2}+17 x+7=0$ | $x^{3}-3 x^{2}+8 x+5=0$ | -1 | | -3 | Algebra | math-word-problem | olympiads | 39d41171-a660-587a-a08a-986f4856de1a | 1 |
. In the decimal writing of $A$, the digits appear in (strictly) increasing order from left to right. What is the sum of the digits of $9 A$? | . Let $A=\overline{a_{1} a_{2} \cdots a_{k}}$ be the decimal representation of $A$. By performing the subtraction
$$
\begin{aligned}
& \begin{array}{cccccc}
a_{1} & a_{2} & a_{3} & \cdots & a_{k} & 0
\end{array} \\
& \begin{array}{llllll}
- & a_{1} & a_{2} & \cdots & a_{k-1} & a_{k} \\
\hline
\end{array}
\end{aligned}
$$
we find that the digits of $9 A=10 A-A$ are $a_{1}, a_{2}-a_{1}, a_{3}-a_{2}, \ldots, a_{k-1}-a_{k-2}, a_{k}-a_{k-1}-1,10-a_{k}$. Their sum is $10-1=9$. | 9 | Number Theory | math-word-problem | olympiads | 8a479e05-2022-544b-9c16-55e549de50e3 | 2 |
30. 20 identical balls are placed into 4 identical boxes, with no box left empty. How many ways are there to do this? If empty boxes are allowed, how many ways are there? | 30. 20 identical balls, 4 identical boxes, solutions for no empty boxes and allowing empty boxes, i.e., find
$$
f(x)=\frac{1}{(1-x)\left(1-x^{2}\right)\left(1-x^{3}\right)\left(1-x^{4}\right)},
$$
expanding the coefficients of $x^{16}$ and $x^{20}$. This expansion is not easy, so we let
$$
G(x)=(1-x)\left(1-x^{2}\right)\left(1-x^{3}\right)\left(1-x^{4}\right),
$$
then
$$
\begin{array}{c}
f(x) \cdot G(x)=1, \\
f(x)=f_{0}+f_{1} x+\cdots+f_{n} x^{n}+\cdots, \\
G(x)=1-x-x^{2}+2 x^{5}-x^{8}-x^{9}+x^{10},
\end{array}
$$
From $\left(f_{0}+f_{1} x+\cdots+f_{n} x^{n}+\cdots\right)\left(1-x-x^{2}+2 x^{5}-x^{8}-x^{9}+x^{10}\right)=1$, we get the recurrence relation for $f_{n}$: $f_{n}-f_{n-1}-f_{n-2}+2 f_{n-5}-f_{n-8}-f_{n-9}+f_{n-10}=0$,
$$
f_{n}=f_{n-1}+f_{n-2}-2 f_{n-5}+f_{n-8}+f_{n-9}-f_{n-10} .
$$
Using the recurrence relation, we can compute the values of $f_{16}=64$ and $f_{20}=108$.
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline$n$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & $\cdots$ & 16 & $\cdots$ & 20 \\
\hline$f_{n-1}$ & 1 & 1 & 2 & 3 & 5 & 6 & 9 & 11 & 15 & 18 & $\cdots$ & 54 & $\cdots$ & 94 \\
\hline$f_{n-2}$ & & 1 & 1 & 2 & 3 & 5 & 6 & 9 & 11 & 15 & & 47 & & 84 \\
\hline$f_{n-5}$ & & & & & 1 & 1 & 2 & 3 & 5 & 6 & & 27 & & 54 \\
\hline$f_{n-8}$ & & & & & & & & 1 & 1 & 2 & & 15 & & 34 \\
\hline$f_{n-9}$ & & & & & & & & & 1 & 1 & & 11 & & 27 \\
\hline$f_{n-10}$ & & & & & & & & & & 1 & & 9 & & 23 \\
\hline$f_{n}$ & 1 & 2 & 3 & 5 & 6 & 9 & 11 & 15 & 18 & 23 & $\cdots$ & 64 & $\cdots$ & 108 \\
\hline
\end{tabular} | 108 | Combinatorics | math-word-problem | olympiads | 4796e90e-e612-5f63-9707-337f78882237 | 1 |
1.57. The vertices of a rectangle inscribed in a circle divide it into four arcs. Find the distance from the midpoint of one of the larger arcs to the vertices of the rectangle, if its sides are 24 and 7 cm. | 1.57. Since $A C$ is the diameter of the circle (Fig. 1.44), then $R=0.5 \sqrt{24^{2}+7^{2}}=12.5 \quad$ (cm). In $\triangle B O F$ we have $O F=$ $=\sqrt{O B^{2}-B F^{2}}=\sqrt{12.5^{2}-12^{2}}=3.5$ (cm); hence, $M F=12.5-3.5=9$ (cm), $M K=12.5+3.5=16$ (cm). From $\triangle M B F$ and $\triangle M A K$ we find the required distances: $\quad M B=\sqrt{M F^{2}+B F^{2}}=\sqrt{9^{2}+12^{2}}=15(\mathrm{~cm})$, $M A=\sqrt{M K^{2}+K A^{2}}=\sqrt{16^{2}+12^{2}}=20(\mathrm{~cm})$.
Answer: 15 and 20 cm. | 15 | Geometry | math-word-problem | olympiads | ff8623f2-0efa-534f-bd28-e6c8b7c6dae3 | 2 |
5. Let $a$ and $n$ be natural numbers, and it is known that $a^{n}-2014$ is a number with $2014$ digits. Find the smallest natural number $k$ such that $a$ cannot be a $k$-digit number. | Solution. Let $a$ be a $k$-digit number, then $10^{k-1} \leq a2013$, i.e., $k>45$.
We will check for $k$ starting from 46, whether the condition of the absence of an integer in the specified interval is met. We will find that
(for $k=46$) $2013 / 462013$ );
(for $k=47$) $2013 / 472013$ );
(for $k=48$) $2013 / 482013$ );
but (for $k=49$) $2013 / 49 > 41$ (since $\quad 41 \cdot 49 = 45^2 - 1^2 = 2025 - 16 < 2013$).
Thus, the interval $[2013 / 49, 2013 / 48)$ does not contain any integers, and for $k=49$, there are no suitable $n$.
## Answer: $\mathbf{k}=49$.
## Criteria.
The problem is complex, so don't be stingy with points for reasonable thoughts. Any answer in the vicinity of $\sqrt{2014}$, which is preceded by something reasonable, is worth points. A point is deducted for the absence of what is in small print. | 49 | Number Theory | math-word-problem | olympiads | 77cba4e7-5a90-5533-9f4b-220b85918184 | 1 |
What is the $21^{\varrho}$ term of the sequence
$$
1 ; 2+3 ; 4+5+6 ; 7+8+9+10 ; 11+12+13+14+15 ; \ldots ?
$$ | Observe that the $21^{\text{st}}$ term is the sum of 21 consecutive numbers. Taking the first term of each term, that is, 1, 2, 4, 7, 11, $16, \ldots$, we have that
$$
\begin{aligned}
2 & =1+1 \\
4 & =2+1+1 \\
7 & =3+2+1+1 \\
11 & =4+3+2+1+1 \\
16 & =5+4+3+2+1+1
\end{aligned}
$$
Thus, the first term of the $21^{\text{st}}$ term is
$$
20+19+\cdots+3+2+1+1=\frac{20 \times 21}{2}+1=211
$$
and the $21^{\text{st}}$ term is
$$
211+212+\cdots+230+231=\frac{(211+231) \times 21}{2}=221 \times 21=4641
$$ | 4641 | Number Theory | math-word-problem | olympiads | ac33a322-ff2a-5a42-8d4e-dc84713e0a1f | 2 |
2. Let the function $f(x)=x^{2}+a x+b$, for any $a, b \in \mathbf{R}$, there always exists $t \in[0,4]$, such that $|f(t)| \geqslant m$ holds, then the maximum value of the real number $m$ is $\qquad$. | Let $M=\max \left\{|f(0)|,|f(4)|,\left|f\left(-\frac{a}{2}\right)\right|\right\}=\max \left\{|b|,|16+4 a+b|,\left|b-\frac{a^{2}}{4}\right|\right\}$,
then $4 M \geqslant|b|+|16+4 a+b|+2\left|b-\frac{a^{2}}{4}\right| \geqslant\left|16+4 a+\frac{a^{2}}{2}\right|$ $=\frac{1}{2}\left|a^{2}+8 a+32\right|=\frac{1}{2}\left|(a+4)^{2}+16\right| \geqslant 8 \Rightarrow M \geqslant 2$, equality holds when $a=-4, b=2$. Therefore, the maximum value of $m$ is 2. | 2 | Algebra | math-word-problem | olympiads | 8ad404e9-ddbf-5f04-8a73-f2b24cc1607f | 2 |
Problem 2. In two fruit stores, there were a total of $365 \mathrm{~kg}$ of apples, which were sold at the same price. When the first store sold a certain amount of apples and received 4340 denars for it, and the second store received 8750 denars for its sold amount of apples, then in the first store, $102 \mathrm{~kg}$ remained, and in the second store, 76 $\mathrm{kg}$. How many kilograms of apples were there in each store at the beginning? | Solution. The two stores together sold: $365-(102+76)=187 \mathrm{~kg}$ of apples. For this, they received $4340+8750=13090$ denars. Therefore, the price of $1 \mathrm{~kg}$ of apples was: $13090: 187=70$ denars. At the first store, initially there were $4340: 70+102=164 \mathrm{~kg}$ of apples, and at the second store there were $365-164=201 \mathrm{~kg}$ of apples. | 164201 | Algebra | math-word-problem | olympiads | 63625e15-cc5d-51fe-917c-2d71d4c46e87 | 2 |
1. What are the last two digits of $1 \times 2 \times 3 \times 4 \times \cdots \times 2004 \times 2005 \times 2006$ ?
(A) 00;
(C) 30;
(D) 50 ;
(E) 60 . | 1. Ans: (A)
Because 100 is one of the factors. | 0 | Number Theory | MCQ | olympiads | fae900dc-d045-5733-a439-d2580c4250e9 | 2 |
Lines $l_1^{}$ and $l_2^{}$ both pass through the origin and make first-quadrant angles of $\frac{\pi}{70}$ and $\frac{\pi}{54}$ radians, respectively, with the positive x-axis. For any line $l^{}_{}$, the transformation $R(l)^{}_{}$ produces another line as follows: $l^{}_{}$ is reflected in $l_1^{}$, and the resulting line is reflected in $l_2^{}$. Let $R^{(1)}(l)=R(l)^{}_{}$ and $R^{(n)}(l)^{}_{}=R\left(R^{(n-1)}(l)\right)$. Given that $l^{}_{}$ is the line $y=\frac{19}{92}x^{}_{}$, find the smallest positive integer $m^{}_{}$ for which $R^{(m)}(l)=l^{}_{}$. | Let $l$ be a line that makes an angle of $\theta$ with the positive $x$-axis. Let $l'$ be the reflection of $l$ in $l_1$, and let $l''$ be the reflection of $l'$ in $l_2$.
The angle between $l$ and $l_1$ is $\theta - \frac{\pi}{70}$, so the angle between $l_1$ and $l'$ must also be $\theta - \frac{\pi}{70}$. Thus, $l'$ makes an angle of $\frac{\pi}{70}-\left(\theta-\frac{\pi}{70}\right) = \frac{\pi}{35}-\theta$ with the positive $x$-axis.
Similarly, since the angle between $l'$ and $l_2$ is $\left(\frac{\pi}{35}-\theta\right)-\frac{\pi}{54}$, the angle between $l''$ and the positive $x$-axis is $\frac{\pi}{54}-\left(\left(\frac{\pi}{35}-\theta\right)-\frac{\pi}{54}\right) = \frac{\pi}{27}-\frac{\pi}{35}+\theta = \frac{8\pi}{945} + \theta$.
Thus, $R(l)$ makes an $\frac{8\pi}{945} + \theta$ angle with the positive $x$-axis. So $R^{(n)}(l)$ makes an $\frac{8n\pi}{945} + \theta$ angle with the positive $x$-axis.
Therefore, $R^{(m)}(l)=l$ iff $\frac{8m\pi}{945}$ is an integral multiple of $\pi$. Thus, $8m \equiv 0\pmod{945}$. Since $\gcd(8,945)=1$, $m \equiv 0 \pmod{945}$, so the smallest positive integer $m$ is $\boxed{945}$. | 945 | Geometry | math-word-problem | amc_aime | 8791f647-68c1-5dec-b9f3-b235014b5116 | 4 |
2. Given the function $y=\frac{a-x}{x-a-1}$, the graph of its inverse function is symmetric about the point $(-1,4)$. Then the value of the real number $a$ is $\qquad$ . | 2.3.-
From the problem, we know that the graph of the function $y=\frac{a-x}{x-a-1}$ is centrally symmetric about the point $(4,-1)$.
$\because y=\frac{a-x}{x-a-1}=-1-\frac{1}{x-(a+1)}$, we have $(y+1)[x-(a+1)]=-1$,
$\therefore$ the graph of the function is a hyperbola with its center at $(a+1,-1)$.
Also, $\because$ the graph of the hyperbola is centrally symmetric about its center,
$\therefore a+1=4$, which means $a=3$. | 3 | Algebra | math-word-problem | cn_contest | ffccaa4e-8118-5206-b8c0-cbb5298378ed | 4 |
Example 4 Find the integer solution of the equation $\left[\frac{x}{1!}\right]+\left[\frac{x}{2!}\right]+\left[\frac{x}{3!}\right]+\cdots+$ $\left[\frac{x}{10!}\right]=1001$. | Solution: $\because$ the solution $x$ of the equation is a natural number, and $xx+\frac{x}{2}+\frac{x}{6}+\frac{x}{24}+\frac{x}{120}-4 \\
=\frac{103}{60} x-4, \therefore 583<x<585, x=584,
\end{array}
$$
Verification shows that $x=584$ is the unique solution of the equation. | 584 | Number Theory | math-word-problem | olympiads | b243ea4d-43cd-560b-919d-ac59cbae4cb9 | 2 |
321. $\sqrt{2 x+5}+\sqrt{x-1}=8$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
321. $\sqrt{2 x+5}+\sqrt{x-1}=8$. | Solution. Isolate one of the radicals and square both sides of the equation:
$$
(\sqrt{2 x+5})^{2}=(8-\sqrt{x-1})^{2} ; 2 x+5=64-16 \sqrt{x-1}+(x-1)
$$
Move $16 \sqrt{x-1}$ to the left side, and all other terms to the right side:
$$
16 \sqrt{x-1}=64+x-1-2 x-5 ; 16 \sqrt{x-1}=58-x
$$
Square both sides of the equation again:
$(16 \sqrt{x-1})^{2}=(58-x)^{2} ; 256 x-256=3364-116 x+x^{2} ; x^{2}-372 x+3620=0$.
Solve the resulting quadratic equation. Since $a=1, b=-372$, $c=3620, \quad D=b^{2}-4 a c=(-372)^{2}-4 \cdot 1 \cdot 3620=138384-14480=123904$; $\sqrt{D}=352$, then
$$
x_{1}=\frac{-b-\sqrt{D}}{2 a}=\frac{372-352}{2}=10 ; \quad x_{2}=\frac{-b+\sqrt{D}}{2 a}=\frac{372+352}{2}=362
$$
Check both values $x_{1}=10 ; x_{2}=362$.
If $x=10$, then $\sqrt{2 \cdot 10+5}+\sqrt{10-1}=8 ; \sqrt{25}+\sqrt{9}=8 ; 5+3=8$; $8=8$
If $x=362$, then $\sqrt{2 \cdot 362+5}+\sqrt{362-1}=8 ; \sqrt{729}+\sqrt{361}=8 ; 27+$ $+19 \neq 8$; therefore, $x=362$ is an extraneous root.
The root of the given equation is $x=10$. | 10 | Algebra | math-word-problem | olympiads | 512b5c06-d4f8-5506-afbf-d88196ff995d | 2 |
2、D Teacher has five vases, these five vases are arranged in a row from shortest to tallest, the height difference between adjacent vases is 2 centimeters, and the tallest vase is exactly equal to the sum of the heights of the two shortest vases, then the total height of the five vases is _ centimeters | 【Answer】50
Analysis: Let the five chimneys be $x-4, x-2, x, x+2, x+4$, then $x+4=x-2+x-4, x=10$, the sum is $5 x=50$ | 50 | Algebra | math-word-problem | olympiads | 2a1d164a-5ebe-5808-aa84-40134dfaca9b | 2 |
11. Choose three different digits from $0,1, \cdots, 9$ to form a four-digit number (one of the digits can appear twice), such as 5 224. Then the total number of such four-digit numbers is $\qquad$. | 11. 3888 .
Discuss in three scenarios.
Four-digit numbers that do not contain 0:
$$
\mathrm{C}_{9}^{3} \mathrm{C}_{3}^{1} \mathrm{C}_{4}^{2} \mathrm{~A}_{2}^{2}=3024 \text { (numbers); }
$$
Four-digit numbers that contain 0 and 0 appears only once:
$$
\mathrm{C}_{9}^{2} \mathrm{C}_{2}^{1} \times 3 \times 3=648 \text { (numbers); }
$$
Four-digit numbers that contain 0 and 0 appears twice:
$$
\mathrm{C}_{9}^{2} \times 3 \mathrm{~A}_{2}^{2}=216 \text { (numbers). }
$$
Thus, there are a total of $3024+648+216=3888$ numbers. | 3888 | Combinatorics | math-word-problem | olympiads | ebbabef4-5743-53b4-ab1d-398aaac715fb | 2 |
2. Function $f$ is defined on the set of integers, satisfying
$$
f(n)=\left\{\begin{array}{ll}
n-3 & n \geqslant 1000 \\
f[f(n+5)] & n<1000
\end{array},\right.
$$
Find $f(84)$. | 2. Let $f_{1}(n)=f(n), f_{m+1}(n)=f_{m}(f(n)) \quad m=1,2, \cdots$, then $f(84)=f_{2}(84+5)=\cdots=f_{184}(84 \times 183 \times 5)=f_{184}(999)$.
For natural number $n$, we have
$$
\begin{array}{c}
f_{n+2}(999)=f_{n+3}(1004)=\cdots=f_{n}(997)=f_{n+1}(1002)=f_{n}(999), \\
\text { so } f(84)=f_{184}(999)=f_{182}(999)=\cdots=f_{2}(999)=f_{3}(1004) \\
=f_{2}(1001)=f(998)=f_{2}(1003)=f(1000)=997 .
\end{array}
$$ | 997 | Number Theory | math-word-problem | olympiads | d87a3631-6f4b-5728-a85b-35120a555828 | 2 |
Problem 10.7. The graph of the function $f(x)=\frac{1}{12} x^{2}+a x+b$ intersects the $O x$ axis at points $A$ and $C$, and the $O y$ axis at point $B$, as shown in the figure. It turned out that for the point $T$ with coordinates $(3 ; 3)$, the condition $T A=T B=T C$ is satisfied. Find $b$.
 | Answer: -6.
Fig. 11: to the solution of problem 10.7
Solution. Let point $A$ have coordinates $\left(x_{1} ; 0\right)$, and point $C$ have coordinates $\left(x_{2} ; 0\right)$. From the condition, it is clear that $x_{1} < 0$ and $x_{2} > 0$. Since $x_{1}$ and $x_{2}$ are the roots of the quadratic trinomial $f(x)$, by Vieta's theorem, we have $x_{1} \cdot x_{2} = 12b$, from which we get $b = \frac{x_{1} \cdot x_{2}}{12} < 0$.
Let $H$ be the point with coordinates $(3 ; 0)$ (Fig. 11). Clearly, in the isosceles triangle $A T C$, the segment $T H$ is the height, and therefore it is also the median. Thus, $3 - x_{1} = A H = H C = x_{2} - 3$, from which we get $x_{1} = 6 - x_{2}$.
Let $M$ be the point with coordinates $(0, 3)$. Since $T H = T M = 3$ and $T A = T B$, the right triangles $A T H$ and $B T M$ are equal by the leg and hypotenuse. Therefore, $3 - x_{1} = H A = M B = 3 - b$, that is, $x_{1} = b = \frac{x_{1} \cdot x_{2}}{12}$ (by Vieta's theorem), from which we find $x_{2} = 12$. Finally, $x_{1} = 6 - x_{2} = 6 - 12 = -6$ and $b = x_{1} = -6$.
Another solution. As in the previous solution, let the abscissas of points $A$ and $C$ be $x_{1}$ and $x_{2}$, respectively; we will also use the fact that point $B$ has coordinates $(0 ; b)$. Immediately, we understand that $O A = |x_{1}| = -x_{1}$, $O C = |x_{2}| = x_{2}$, and $O B = |b| = -b$.
Let's find the second intersection of the circle with the y-axis, let this be point $D$ with coordinates $(0 ; d)$ (Fig. 12). The chords $A C$ and $B D$ of the circle intersect at the origin $O$; from the properties of the circle, we know that $O A \cdot O C = O B \cdot O D$. We get $-x_{1} \cdot x_{2} = -b \cdot d$, from which, replacing $x_{1} \cdot x_{2}$ with $12b$ by Vieta's theorem, we get $d = 12$.
Fig. 12: to the solution of problem 10.7
It remains to note that triangle $BTD$ is isosceles, and the midpoint of its base, point $M$, has coordinates $(0 ; 3)$. Reflecting point $D(0 ; 12)$ relative to it, we get $B(0 ; -6)$. | -6 | Algebra | math-word-problem | olympiads | 4f14788f-fd45-586e-a5e9-c4653ba17885 | 2 |
10.024. The lengths of the parallel sides of the trapezoid are 25 and 4 cm, and the lengths of the non-parallel sides are 20 and 13 cm. Find the height of the trapezoid. | ## Solution.
Given $B C=4 \text{ cm}, A D=25 \text{ cm}, A B=20 \text{ cm}, C D=13 \text{ cm}$ (Fig. 10.24). Draw $B E \perp A D$ and $C F \perp A D$. Let $B E=C F=h, A E=x$, $F D=y$. Then from $\triangle A B E$ and $\triangle C F D$ we find $h^{2}=20^{2}-x^{2}=13^{2}-y^{2}$. Considering that $y=25-4-x=21-x$, we have $20^{2}-x^{2}=13^{2}-(21-x)^{2}$ or $42 x=672$, from which $x=16$ (cm). Therefore, $h=\sqrt{20^{2}-16^{2}}=12$ (cm).
Answer: 12 cm. | 12 | Geometry | math-word-problem | olympiads | a065ae49-8a7c-5a6d-8c48-e652a2042578 | 2 |
6. (10 points) As shown in the figure, the area of square $A B C D$ is 840 square centimeters, $A E=E B, B F=2 F C, D F$ intersects $E C$ at $G$. Then the area of quadrilateral $A E G D$ is square centimeters. | 510
【Analysis】This figure easily leads one to think about finding the areas of triangles $BCE$, $CDF$, and $CGF$. The most challenging part is finding the area of $\triangle CGF$. According to the given conditions, we should draw $GM \perp BC$ with the intersection point at $M$. This forms two sets of similar triangles corresponding to $\triangle BCE$ and $\triangle CDF$. By using the property that the ratio of the areas of similar triangles is equal to the square of the similarity ratio, we can find the area of $\triangle CGF$. Finally, based on the figure, we can calculate the area of the required shape.
【Solution】Draw $GM \perp BC$ with the intersection point at $M$,
$$
\begin{array}{l}
\therefore \triangle FMG \sim \triangle FCD \Rightarrow FM: FC = MG: CD, \\
\because BF = 2FC \Rightarrow BC = 3FC, \\
\therefore MG = 3FM, \\
\because \triangle CGM: \triangle CEB \Rightarrow CM: CB = GM: BE, \\
BC = 2BE, \\
\therefore GM = \frac{1}{2} CM = 3FM \Rightarrow CM = 6FM, \\
\therefore FM: FC = 1: 7, \\
CM: CB = 2: 7, \\
S \triangle BCE = \frac{1}{4} \times \square ABCD = 210, \\
S \triangle CGM = 4 \div 49 \times 210 = 840 \div 49 = 120 \div 7, \\
S \triangle CDF = S \square ABCD \div 6 = 140, \\
S \triangle MGF = 140 \times 1 \div 49 = 140 \div 49 = 20 \div 7, \\
S \triangle CGM + \triangle MGF = 120 \div 7 + 20 \div 7 = 20, \\
840 - 210 - 140 + 20 = 510 \text{ (square centimeters). }
\end{array}
$$
Therefore, the area of quadrilateral $AEGD$ is 510 square centimeters. | 510 | Geometry | math-word-problem | olympiads | 2a224068-3c17-5d14-ad3a-426aec3cae2b | 2 |
Given the set $\{1,2,3,5,8,13,21,34,55\}$, how many integers between 3 and 89 cannot be written as the sum of exactly two elements of the set?
(A) 51
(B) 57
(C) 55
(D) 34
(E) 43 | Solution
First count the number of integers between 3 and 89 that can be written as the sum of exactly two elements. Since each element in the set is the sum of the two previous
elements, 55 can be added to each of the seven smallest elements to form seven unique integers smaller than 89 .
In the same way, 34 can be added to each of the seven smaller elements, 21 can be added to each of the six smaller elements, and so on.
The number of integers between 3 and 89 that can be written as the sum of two elements of the set is $7+7+6+5+4+3+2=34$. Since there are 85 integers between 3 and 89 , then $85-34=51$ integers cannot be written as the sum of exactly two elements in the set.
ANSWER: (A)
# | 51 | Number Theory | MCQ | olympiads | 5458dc44-05dc-54f0-b904-9a9981455324 | 2 |
9. Given that $M N$ is a moving chord of the circumcircle of equilateral $\triangle A B C$ with side length $2 \sqrt{6}$, $M N=4$, and $P$ is a moving point on the sides of $\triangle A B C$. Then the maximum value of $\overrightarrow{M P} \cdot \overrightarrow{P N}$ is | 9.4.
Let the circumcenter of $\triangle ABC$ be $O$.
It is easy to find that the radius of $\odot O$ is $r=2 \sqrt{2}$.
Also, $MN=4$, so $\triangle OMN$ is an isosceles right triangle, and
$$
\begin{aligned}
& \overrightarrow{O M} \cdot \overrightarrow{O N}=0,|\overrightarrow{O M}+\overrightarrow{O N}|=4 . \\
& \text { Let }|\overrightarrow{O P}|=x . \\
& \text { Then } \sqrt{2} \leqslant x \leqslant 2 \sqrt{2} \text {, and } \\
& \overrightarrow{M P} \cdot \overrightarrow{P N}=(\overrightarrow{O P}-\overrightarrow{O M}) \cdot(\overrightarrow{O N}-\overrightarrow{O P}) \\
& =-\overrightarrow{O P}{ }^{2}+(\overrightarrow{O M}+\overrightarrow{O N}) \cdot \overrightarrow{O P}-\overrightarrow{O M} \cdot \overrightarrow{O N} \\
= & -x^{2}+(\overrightarrow{O M}+\overrightarrow{O N}) \cdot \overrightarrow{O P} \\
\leqslant & -x^{2}+|\overrightarrow{O M}+\overrightarrow{O N}||\overrightarrow{O P}| \\
= & -x^{2}+4 x=4-(x-2)^{2} \leqslant 4,
\end{aligned}
$$
The equality holds if and only if $x=2$ and $\overrightarrow{O P}$ is in the same direction as $\overrightarrow{O M}+\overrightarrow{O N}$.
Therefore, the maximum value of $\overrightarrow{M P} \cdot \overrightarrow{P N}$ is 4. | 4 | Geometry | math-word-problem | cn_contest | fea23b7f-e22e-521d-be6d-4c469b94d965 | 1 |
For example, in the sequence $x_{1}, x_{2}, \cdots, x_{n}, \cdots$, the sum of any three consecutive terms is 20, and $x_{1}=9, x_{12}=7$. Find the value of $x_{2000}$. | Given that $x_{n+1}+x_{n+2}+x_{n+3}=20$
$$
x_{n}+x_{n+1}+x_{n+2}=20
$$
Subtracting the two equations, we get $x_{n+3}=x_{n}(n \geqslant 1)$. Therefore, $\left\{x_{n}\right\}$ is a periodic sequence with a period of 3, so $x_{4}=x_{1}=9, x_{12}=$ $x_{3}=7$. Since $x_{1}+x_{2}+x_{3}=20$, we have $x_{2}=4$. Hence, $x_{2000}=x_{2}=4$ | 4 | Algebra | math-word-problem | olympiads | f44cc721-ff62-5eee-b838-61e3c0ce38dc | 2 |
13.223. Point $C$ is located 12 km downstream from point $B$. A fisherman set out from point $A$, located upstream from point $B$, to point $C$. After 4 hours, he arrived at $C$, and the return trip took 6 hours. On another occasion, the fisherman used a motorboat, thereby tripling his own speed relative to the water, and reached from $A$ to $B$ in 45 minutes. It is required to determine the speed of the current, assuming it is constant. | Solution.
Let $v_{T}$ km/h be the speed of the current; $v_{,}$ km/h be the speed of the boat; $a$ km/h be the distance between $A$ and $B$. Then, according to the problem,
$$
\left\{\begin{array}{l}
\left(v_{T}+v_{T}\right) \cdot 4=12+a, \\
\left(v_{L}-v_{T}\right) \cdot 6=12+a, \\
a=\left(3 v_{T}+v_{T}\right) \cdot 0.75
\end{array}\right.
$$
Answer: 1 km/h. | 1 | Algebra | math-word-problem | olympiads | 3dd909aa-f8a5-53ac-915c-448904215fc2 | 2 |
# Problem 5. (3 points)
In trapezoid $A B C D$, the lateral side $B C$ is equal to the diagonal $B D$. On the smaller arc $A B$ of the circumscribed circle of triangle $A B C$, a point $E$ is chosen such that $B C=B E$. Find the angle $\angle A E D$. | Answer: $90^{\circ}$.
## Solution:
From the isosceles triangle $B C D$ and the parallelism of $A B$ and $C D$, we get $\angle B C D = \angle B D C = \angle D B A = \alpha$.
Let the line $C D$ intersect the circumcircle of triangle $A B C$ at point $F$. Then $B C F A$ is an inscribed, i.e., isosceles, trapezoid, from which the arcs $B C, B E$, and $A F$ are equal. Hence, $\angle B C E = \angle A C F = \beta$.
$\angle E B A = \angle E C A = \gamma$, since these angles subtend the same arc.
$\angle B C A = \angle B C E + \angle E C A = \beta + \gamma$.
$\angle E B D = \angle E B A + \angle D B A = \gamma + \alpha$. Therefore, in the isosceles triangle $E B D$, the equality $\angle B D E = \angle B E D = \frac{180^{\circ} - \alpha - \gamma}{2}$ holds.
Moreover, $\alpha = \angle B C D = \angle B C F = \angle B C A + \angle A C F = 2 \beta + \gamma$.
$\angle A E D = \angle B E A - \angle B E D = (180^{\circ} - \angle B C A) - \frac{180^{\circ} - \alpha - \gamma}{2} = 180^{\circ} - \beta - \gamma - 90^{\circ} + \frac{2 \beta + 2 \gamma}{2} = 90^{\circ}$ | 90 | Geometry | math-word-problem | olympiads | 02f1d413-9526-5c96-a94d-ea128382c247 | 2 |
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