IMO Shortlist
Divisors
equation
number theory

It was also in german IMO selection test (I have a clue where it's from...  ;)  )
I didn't find the solution in test itself :(

I'm having a little problem with this stuff, the problem is that I can prove:
(1) for all integer n there exist infinitely many a such that t(a*n) != n
But I cannot prove (since i have counterexamples :D)
(2) there exist infinitely many a such that for all integers n : t(a*n) !=n

I'm not sure about the bracket :  is it p^p-1 or p^(p-1) but i have counterexample for both case  (assuming k=p^p-1 or k=p^(p-1))
In fact by putting p=2, i get a=p^(p-1) = 2, and then by putting n=4
I get t(2*4) = 4
And the other way by putting p=2 , i get a=p^p-1=2^2-1=3, and now by putting n=3
I get t(3*3) = 3...

I'm really confused... Are you sure the statement of the problem is exactly the thing you typed ?
Or is the hint k=p^p-1 wrong ? Or simply I didn't understand the problem  ;) ?

The hint is: $a=p^{p-1}$, and $p$ has to be big enough

ok, just choose $a=p^{p-1}$ it is easy to show $p$ must divides $n$, so we get:

$d(p^m)d(a')=a'p^{m-p+1}$ from here, use the fact that $d(a')<a/2+1$ with equality only for $a=2$, and do some inequalities based on induction on $m$, to show rhs is always bigger!

As for the other problem posted here: iff $n=p_1^{a_1}p_2^{a_2}...p_k^{a_k}$ put $a=p_1^{p_1^{a_1}-a_1-1}p_2^{p_2^{a_2}-a_2-1}...p_k^{p_1^{a_k}-a_k-1}$ A simple induction in $a_k$ shows the exponents are indeed nonegative integers, so thats it.

Finally for the problem 2, the statement must be wrong, taking $n=1$! we get $a=1$ but it doesnt works! Iff 1 is not allowed, taking primes wich doesnt divide $a$ we get $d(a)=1$!

Let $a=p^{p-1}$, $p>2$. So suppose $p$ does not divide $n$ and
$\tau(an) = p\tau(n)=n$, which is contradiction.
So let $n=p^{a_1}p_2^{a_2}...p_k^{a_k}$
hence
$\tau(an)=(a_1+p)(a_2+1)....(a_k+1) = p^{a_1}p_2^{a_2}...p_k^{a_k}$

We can easily prove from induction that
$p^{m} > m+1$, $p>2$ and
$p^m \geq m+p$ for $m>1$ , $p>2$.

So LHS<RHS if $a_1>1$, so $a_1=1$.

If $k=2$ and $p_2=2$, then we have $1+p=p$, which is false.

Otherwise
let $p_i \geq 3$ be a prime factor of $n$.
Consider $\frac{p^m}{m+1}$ for postive integers $m \geq 1$, $p \geq 2$
If $p>q$ then $\frac{p^m}{m+1} > \frac{q^m}{m+1}$
If $j>m$ then $\frac{p^j}{j+1} > \frac{p^m}{m+1}$(this can be proven from induction, too lazy to write it out but it's straight forward)
Thus
$\frac{p(p_i^{a_i})}{(p+1)(a_i+1)} > \frac{p(3^1}{(p+1)(2)} >1$, since $p>2$ so $3p > 2p+2$. Hence $p*p_i^{a_i} < (p+1)(a_i+1)$,
so $\tau(an)=(1+p)(a_2+1)....(a_k+1) < pp_2^{a_2}...p_k^{a_k}$,

hence there's no positive integer solution to the given equation when
$a=p^{p-1}$ for all $p>2$, which is qed.

Its well known that $T(n) <= 2\sqrt{n}$ so taking $a = p^{p-1}$ we assume there is some x solving the equation and say $x = p^yz \rightarrow (p+y)T(z) = p^yz \rightarrow 2\sqrt{z}(p+y) \ge p^yz \rightarrow 2*(p+y) \ge p^y\sqrt{z}$ and this last inequality is clearly false for large enough $p$ is $y \ge 2$ and if $y = 1$ then $z \le 4$ and brute forcing these values of $z$ shows failure as desired.

Let $n=\prod_p p^{\alpha_p}$ and $a=\prod_p p^{\beta_p}$, where $\alpha_p=v_p(n)$, $\beta_p=v_p(a)$ for each prime $p$. Let $Q$ be some non-empty set of primes of the form $4k-1$.

We have
$$\frac{\tau(an)}{n}=\prod_p \frac{\alpha_p+\beta_p+1}{p^{\alpha_p}}.\qquad (1)$$
Also, the function $\alpha\mapsto \frac{\alpha+\beta+1}{p^{\alpha}}$ is non-increasing over the non-negative integers $(*)$, as its value at $\alpha-1$ divided by its value at $\alpha$ is at least $1$.

Let $\beta_q=q-1$ for $q\in Q$ and $\beta_q=0$ otherwise. If $\tau(an)=n$, then we cannot have $\alpha_q=0$ for and $q\in Q$, because then the numerator of one of the fractions in $(1)$ is $q$, but none of the denominators are divisible by $q$, so the product cannot be $1$. So we have $\alpha_q\ge 1$ for each $q\in Q$. Let there be $x$ of them with $\alpha_q=1$ and $y$ of them with $\alpha_q\ge 2$. Note that by $(*)$, if $\alpha_q\ge 2$, then
$$\frac{\alpha_q+\beta_q+1}{p^{\alpha_q}}\le \frac{q+2}{q^2}<\frac{2q}{q^2}<1.$$
Also, if $\alpha_q=1$, then
$$\frac{\alpha_q+\beta_q+1}{p^{\alpha_q}}=\frac{q+1}{q}< 2.$$
Now note that here, $4\mid q+1$ for $q\in Q$. So in product $(1)$, we have $x+y=|Q|$ factors of $4$ in the numerator, which need be compensated by $\alpha_2\ge 2|Q|$. We get from $(*)$ that
$$\frac{\alpha_2+\beta_2+1}{2^{\alpha_2}}\le \frac{2|Q|+1}{4^|Q|}< (1/2)^{|Q|},$$
where the last inequality is due to $2|Q|+1<2^{|Q|}$, [i]assuming[/i] $|Q|\ge 3$.
Also, for $p\notin Q$, the corresponding fraction is at most $1$.

Hence our product $(1)$ is less than $2^x\cdot (1/2)^{|Q|}<1$, and thus cannot be equal to $1$. This proves that for any such set $Q$ with $|Q|\ge 3$, we can find a new $a$ which works. There are infinitely many such sets as there cannot be finitely many primes of the form $4k-1$ (if there were only the ones $q_1,q_2,\dots,q_t$, then the number $4q_1q_2\dots q_t-1$ would have a prime factor $4k-1$, yet it is not divisible by any of the $q_i$, contradiction).

Oops. Here is a slight overkill. Oops. The "slight overkill" is also wrong. :(

We begin with a claim: There is a constant $C>0$ so that\[
\tau(n) < C\sqrt[3]{n}
\]for all $n$. Indeed, notice that for $p = 2,3,5,7$, there is a constant $c_p>0$ so that $(\alpha+1)^3 \leq c_pp^{\alpha}$ for all integer $\alpha \geq 0$; further note that if we set $c_p = 1$ for primes $p>8$ the inequality also holds. So, let $C = \sqrt[3]{c_2c_3c_5c_7}+1$. Then, write $n = \prod p_i^{\alpha_i}$, and then we want \[
\prod (\alpha_i+1)^3 < C^3 \prod p_i^{\alpha_i}.
\]But by assumption, \[
\prod (\alpha_i+1)^3 \leq \prod c_{p_i}{p_i}^{\alpha_i} < C^3 \prod p_i^{\alpha_i},
\]since $c_p = 1$ for $p > 8$. Hence the claim is true.

Notice that if $n$ is a solution to $\tau(an) = n$, then \[
n = \tau(an) < C\sqrt[3]{an},
\]so $n^3 < C^3an$, or $n< C^{\frac 32} \sqrt a$.

Now, suppose by contradiction that only finitely many $a$ exist; then, for $a>A$, there is a solution $n_a$ to \[
\tau (an_a) = n_a.
\]Pick some $N\in \mathbb N$, and consider \[
n_{A+1}, \ldots, n_{A+N}.
\]We have $n_{A+i} < C^{\frac 32} \sqrt{A+i} \leq C^{\frac 32} \sqrt{A+N}$. However, notice that the $n_{A+i}$ are all distinct, so there must be at least $N$ positive integers $x$ (namely our $n_{A+i}$) satisfying $x < C^{\frac 32} \sqrt{A+N}$; this implies that $N < C^{\frac 32} \sqrt{A+N}$, which is clearly a contradiction for $N$ big. $\blacksquare$

Here's my solution.
[hide=Solution]
For the sake of contradiction, assume that there does not exist any $a$ such that the equation $\tau(an)=n$ does not have a positive integer solution $n$.$\newline$

We will be using the fact that $\tau$ is a multiplicative function.

Let $a=p^{p-1}$, where $p$ is a prime. Also assume that $p$ does not divide $n$. $\Rightarrow (a,n)=1$$\newline$
$\Rightarrow \tau(an)=n \Rightarrow \tau(a)\tau(n)=n\Rightarrow p\cdot\tau(n)=n \Rightarrow p|n$, which is a contradiction, since we have assumed that $(p,q)=1$.$\newline$
Therefore, $p|n$ for $\tau(an)=n$ to hold.$\newline$

Therefore, let $n=p^\alpha q$ such that $(p,q)=1$. Then $\tau(an)=\tau(p^{p+\alpha-1}q)=n= p^\alpha q $$\newline$
$ \Rightarrow \tau(p^{p+\alpha-1})\tau(q)=p^\alpha q$$\newline$
$\Rightarrow p^{p+\alpha}\tau(q)=p^\alpha q \Rightarrow p^p\tau(q)=q \Rightarrow p|q$, which contradicts the fact that $(p,q)=1$.
So we have that $n$ is a power of $p$.$\newline$

Let $n=p^x$ for some integer $x\ge 1$. $\Rightarrow an=p^{p+x-1} \Rightarrow \tau(an)=p+x=p^x $$\newline$
$\Rightarrow p(p^{x-1}-1)=x$.$\newline$
However, it is understandable that for large enough $p$, $p^{x-1}-1>x$, meaning the last relation cannot always hold. Therefore, there exist infinitely many primes $p$ such that we can never find an $n$ which can satisfy the relation $p+x=p^x$, whatever the value of $x$ may be.

Therefore exist infinitely many $a$'s of the form $p^{p-1}$(where $p$ is large enough) which satisfy the condition given in the question i.e. there exist infinitely many positive integers $a$ such that the equation $\tau(an)=n$ does not have a positive integer $n$.

I think that for post #3, the fact that for $a=2$, $n=4$ is a solution can be explained by using the equation $p+x=p^x$
where $p=2,x=2$ which was derived in the solution.

[quote=MathStudent2002]
...Now, suppose by contradiction that only finitely many $a$ exist; then, for $a>A$, there is a solution $n_a$ to \[
\tau (an_a) = n_a.
\]Pick some $N\in \mathbb N$, and consider \[
n_{A+1}, \ldots, n_{A+N}.
\]We have $n_{A+i} < C^{\frac 32} \sqrt{A+i} \leq C^{\frac 32} \sqrt{A+N}$. However, notice that the $n_{A+i}$ [color=#f00]are all distinct[/color],...[/quote]

Why are $n_{A+i}$ all distinct?
No! Not so simple, but I like your approach. A natural one. I also like very much to solve NT problems without using NT at all (doesn't mean I don't like NT). So,  $n_a$ may not be distinct, but it could be fixed, I think. When $a$ runs over some interval of length, say $n$, $n_a$ runs over an interval of length $\sqrt{n}$, thus for some $\sqrt{n}$ amongst $a$'s , $n_a$ would take the same value. Could we get a contradiction out of it?

[quote=FISHMJ25]Why dont we use $p-3$ instead of $p-1$ ? I think it solves problem immediately. ($p>5$)[/quote]

I don't think that immediately solves the problem, because if $p-2$ is prime, we can let $n=(p-2)\cdot 2^3$.

[b]Solution:[/b]
Let $n=p_1^{k_1}p_2^{k_2}\dots p_m^{k_m}$ for distinct primes $p_i$ and positive $k_i$. We see that $k_i+1\le p_i^{k_i}$ $(*)$ for all primes $p_i$ and positive integers $k_i$ with equality only when $(p_i,k_i)=(2,1)$. Thus, $\tau(n) = (k_1+1)\dots(k_m+1)\le p_1^{k_1}\dots p_m^{k_m}=n$ with equality if and only if $n=2$.

Consider $a=p^{p-1}$ for prime $p\ge 5$. I claim that there are no solutions to $\tau(an)=n$.

Suppose, for the sake of contradiction, that there exists a solution $n$ to $\tau(an)=n$. If $p\nmid n$, then $p\mid \tau(an)=n$, a contradiction. So, we can let $n=p^{k_1}\cdot p_2^{k_2}\dots p_m^{k_m}$ and $\tau(p^{p-1}\cdot n)=(p+k_1)(k_2+1)\dots(k_m+1)$. If $k_1\ge 2$, then $p+k_1<p^{k_1}$ and $(k_2+1)\dots(k_m+1)\le p_2^{k_2}\dots p_m^{k_m}$ by $(*)$, so the RHS would be greater than the LHS. So, we must have $k_1=1$, and our equation becomes $$(p+1)(k_2+1)\dots(k_m+1)=pp_2^{k_2}\dots p_m^{k_m}.$$

This means that there exists some $k_j$ such that $k_j+1=qp$ for a positive integer $q$.
$$(p+1)(k_2+1)\dots qp\dots (k_m+1)=pp_2^{k_2}\dots p_m^{k_m}$$
$$(p+1)(k_2+1)\dots q\dots (k_m+1)=p_2^{k_2}\dots p_m^{k_m}$$

Consider $q(p+1)=(k_j+1)\cdot \frac{p+1}{p}$. For $p>2$, we have that $(k_j+1)\cdot \frac{p+1}{p}<p_j^{k_j}$ for all primes $p_j\ge 3$ and positive integers $k_j$. Combined with $(*)$, if $p_j\ge 3$, there is a contradiction, as the RHS would be greater than the LHS. So, we need $p_j=2$ and $(k_j+1)\cdot \frac{p+1}{p}\ge 2^{k_j}$. Remembering that $p\ge 5$, this leads to $k_j=1$ and $\frac{p+1}{p}=1$, a contradiction.

Thus, for all $a=p^{p-1}$ for prime $p\ge 5$, there are no solutions to $\tau(an)=n$.

what was the motivation behind taking a^p-1.

[quote=Fermat_Theorem] Then $\tau(an)=\tau(p^{p+\alpha-1}q)=n= p^\alpha q $$\newline$
$ \Rightarrow \tau(p^{p+\alpha-1})\tau(q)=p^\alpha q$$\newline$
$\Rightarrow p^{p+\alpha}\tau(q)=p^\alpha q \Rightarrow p^p\tau(q)=q \Rightarrow p|q$, which contradicts the fact that $(p,q)=1$.
[/quote]
I think there's a mistake probably :maybe: which can be fixed tho

[hide=Solution][quote=ISL 2004 N1]Let $\tau(n)$ denote the number of positive divisors of the positive integer $n$. Prove that there exist infinitely many positive integers $a$ such that the equation $ \tau(an)=n $ does not have a positive integer solution $n$.[/quote]
[b][color=#000]Solution:[/color][/b]  Choose a prime $p$, sufficiently big (to avoid edge cases). We claim $a-p^{p-1}$ works! As $p$ varies, we get infinitely many $a$. Now fix $p$. If $p~  \cancel{\mid} ~ n$
\begin{align*} \tau (an ) =n \implies \prod p_i ^{e_i} =p \prod (e_i +1) \end{align*}
Since, $p ~ \cancel{\mid} ~ n $ gives a contradiction! We'll use the following Lemma
------------------------------
[b][color=#000]Lemma 1:[/color][/b] For every $n \in \mathbb{N}$, $$ \tau (n) \le 2 \sqrt{n} $$
[b][color=#000]Proof:[/color][/b] If $\nu _3 (n) \geq 2 $ and $\nu _2 (n) \geq 6$, then \begin{align*} p_i ^{\tfrac{e_i}{2} }\geq e_i +1 \implies 2 \prod p_i ^{\tfrac{e_i}{2}} \geq (e_i +1 ) \end{align*}
When, $\nu _3 (n) =1$ and $\nu _2 (n) =1,2,3,4,5$, still the result holds and we're done $\qquad \square$
-----------------------------
If $ p \mid n$, Let $\nu_p (n) =e$ and $n=\prod p_i ^{e_i}$
\begin{align*} \tau (an) =n \implies (e+p) \prod (e_i +1) =p^e \prod p_i^{e_i} \end{align*}
If $e>1$, then $p^e >e+p$ and $p_i^{e_i} \geq e_i +1$ $\implies$ RHS is greater than LHS. If $e=1$,
\begin{align*} \tau (an)=n \implies (p+1) \prod (e_i +1) =p \prod p_i ^{e_i} \end{align*}
Using $\tau (n) \le 2 \sqrt{n}$,
\begin{align*} \frac{p}{p+1} \prod p_i^{e_i} = \prod (e_i +1) \le 2 \prod p_i ^{\tfrac{e_i}{2}} \implies \prod p_I^{e_i} \le 4 + \frac{8}{p} + \frac{4}{p^2} \end{align*}
Since, $p$ is sufficiently large, Thus,
\begin{align*} \prod p_i ^{e_i} \le 4 \end{align*}
Further checking each case yields no solution, thus $a=p^{p-1}$ works! $\qquad \blacksquare$ [/hide]

Solution from [i]Twitch Solves ISL[/i]:

The construction is quite specific: we claim that \[ a = p^{p-1} \] works for sufficiently large primes $p > 100$.
Indeed, assume for contradiction $\tau(an) = n$ for some integer $n$. Then: [list] 	[*]If $p \nmid n$, then $\nu_p(an) = p-1$ so $p \mid \tau(an)$, 	a contradiction. 	Thus we must have $p \mid n$.
[*]Now let $n = p^e t$ for some $e > 0$. 	Then \[ \underbrace{(e+p) \cdot \tau(t)}_{=\tau(an)} 		= \underbrace{p^e \cdot t}_{=n}. \] 	Now we split into two further cases. 	[list] 		[*]If $e > 1$ then $p^e > p+e$ and $t \ge \tau(t)$, 		give a size contradiction together. 		[*]If $e = 1$ we get 		$\tau(t) = \frac{p}{p+1} t \ge \frac{101}{102} t$.
However, it's known (and easy to show) 		that $\tau(t) < 2\sqrt t$ for all integers $t$. 		Hence we get $\frac{101}{102} t < 2 \sqrt t$, so $t < 5$.
But the equation $\tau(t) = \frac{p}{p+1} t$ 		gives no solution for $t = 1, 2$, 		$p = 2$ for $t = 3$, and $p = 3$ for $t = 4$. 		So when $p > 100$ the situation $t < 5$ cannot occur either. 	[/list] [/list] Having checked everything, the proof is complete.

This was more algebra than number theory.

Note (Since I was a bit lazy in writing everything down): All the inequalities I have left unproved are easily solved by a straightforward induction with suitable base cases.

We show that for each prime $p\geq 5$ and $a=p^{p-1}$, the equation $\tau(an)=n$ has no integer solution $n$. This would solve the problem.
Call a positive integer $n$ [i]rocking[/i], if it is a solution to $\tau(ax)=x$ for $a$ of the above mentioned form.
First, if $\gcd(p,n)=1$, then $n$ is not rocking since otherwise $\tau(an)=\tau(a)\tau(n)=p\cdot\tau(n)=n$ and so $p|n$ which is impossible.
Thus we may write $n=p^ep_1^{e_1}\prod_{i=2}^m p_i^{e_i}$ for distinct primes $p,p_1=2,p_2,\dots,p_m$ and positive integers $e,e_2,\dots,e_m$ and non-negative $e_1$. If $n$ is rocking, we have $\tau(an)=(e+p)\prod_{i=1}^m(e_i+1)=p^e\prod_{i=1}^m p_i^{e_i}$. Now we can easily show that for $p_i\geq 2$ and $e_i\geq 0$, we have $p_i^{e_i}\geq e_i+1$. So $\prod_{i=1}^m p_i^{e_i}\geq \prod_{i=1}^m(e_i+1)$. So we must have $p^e\leq p+e$. Now since $p\geq 5$, so for $e>1$ we can again show that $p^e>p+e$. Thus we must have $e=1$. This means $(p+1)\prod_{i=1}^m(e_i+1)=p\prod_{i=1}^m p_i^{e_i}$. Now for $p_i>2$, we have $p_i^{e_i}\geq 3^{e_i}>\frac{4}{3}(e_i+1)$ for all positive integers $e_i$. We also have $2^{e_1}\geq e_1+1$ for all non-negative integers $e_1$. Hence we have $\left(1+\frac{1}{p}\right)\cdot1\geq \left(1+\frac{1}{p}\right)\left(\frac{e_1+1}{2^{e_1}}\right)=\frac{\prod_{i=2}^m p_i^{e_i}}{\prod_{i=2}^m(e_i+1)}\geq \left(\frac{4}{3}\right)^{m-1}$. But for $p\geq 5$ this is only possible if $m-1=0$ or $m=1$. Thus $n=p\cdot 2^{e_1}$. This gives $(p+1)(e_1+1)=p\cdot2^{e_1}$. This means that $e_1>0$. Also, $e_1=1$ doesn't work. Now if $e_1>1$ then $2^{e_1}>\frac{5}{4}(e_1+1)$, so we get $1+\frac{1}{p}=\frac{2^{e_1}}{e_1+1}>\frac{5}{4}=1+\frac{1}{4}$ which contradicts the fact that $p\geq 5$. Hence $n$ cannot be rocking, and we are done. $\square$

We claim that $a=p^{p-1}$ works, where $p\ge 5$ is a prime number.

Assume that there exist $n=\tau(an)$. If $p$ doesn't divided $n$, then $n=\tau(p^{p-1}n)=p\tau(n)$, contradiction. So let $n=p^km$, we get $$0=n-\tau(p^{p-1}n)=p^{k}m-(p+k)\tau(m)=\sum_{d|m} p^k\varphi(d) - \sum_{d|m} (p+k)$$ If $k> 1$, then $p^k\varphi (d) -(p+k) > 0$ contradiction, so consider that $k=1$, that means $0=\sum_{d|m} p(\varphi(d) - 1)-1$ . $n \neq 1, 2$ of course, and for $d> 2$, we have $p(\varphi(d)-1)-1\ge p-1> 3$, only when $d=1, 2$, $p(\varphi(d)-1)-1= -1$, hence the summation is positive, contradiction. So we're done.

[quote=HKIS200543]I think that this solution is new.
[b]Claim 1. [/b] There exist no solutions to the equation
\[ 6 \tau(n) = n . \]
[i]Proof [/i] Obviously $6$ must divide $n$, so write the prime factoirsation $n = 2^{1 +e_1} \cdot 3^{1+e_2} + p_3^{e_3} \cdots p_k^{e_k}. $ Then
\[  \frac{e_1+2}{2^{e_1}} \cdot \frac{e_2 + 2}{3^{e_2}} = \prod_{i=3}^k \frac{p_i^{e_i}}{e_i+1}. \]
Observe that $x \to \frac{x+2}{2^x}$ and $\frac{x+2}{3^x}$ are both strictly decreasing functions, so the lefthand side is at most 4. Now we place some lower bounds on the righthand side. Note that the function $\frac{p^x}{x+1}$ is increasing for any prime $p > 3$.
[list]
[*] If $n$ is divisible by a prime $p \geq 11$, then $\frac{p^{e_i}}{e_i + 1} \geq p/2 > 4$. Thus there are no solutions if $n$ is divisible by a prime $p \geq 11$.
[*] If $n$ is divisible by $7^2$, then the righthand side is at least $\frac{49}{3} > 4$, so there are no solutions with $n$ divisible by 49.
[*] If $n$ is divisible by $5^2$, then the righthand side is at least $\frac{25}{3} > 4$, so there are no solutions with $n$ divisible by 25.
[*] If $n$ is divisible by $5 \cdot 7$, then the rightside is at least $\frac{5 \cdot 7}{4} > 4$, so there are no solutions here.
[*] If $n = 2^{a+1} \cdot 3^{b+1} \cdot 5$, then
\[ \frac{a+2}{2^a} \cdot \frac{b+2}{3^b} = \frac{5}{2}. \]
This has no solutions for size reasons when $a \geq 2$ or $b \geq 1$, so we just have to manually check $n=60$.
[*] if $n = 2^{a+1} \cdot 3^{b+1} \cdot 7$, then
\[ \frac{a+2}{2^a} \cdot \frac{b+2}{3^b} = \frac{7}{2} \]
which likewise has no solutions when $a \geq 2$ or $b \geq 1$ for size reasons. Thus we check $n = 84$, and we are done.
This establishes the claim.
[/list]
------
[b]Claim 2. [/b] Let $C$ be a constant. Then
\[ C\tau(n) <  n \]
for sufficiently large $n$.
[i]Proof.[/i] This is more dumb bounding. Note that
\[ \frac{n}{\tau(n)} = \prod_{p \mid n} \frac{ p^{\nu_p(n)}}{\nu_p(n) + 1} \geq \frac{ q^{\nu_p(n)}}{\nu_q(n) + 1} \]
for any fixed prime $q$. Obviously, $x \to \frac{q^x}{x+1}$ is unbounded and increasing. Thus $\nu_q(n)$ is bounded for all primes $q$. Hence $n$ is bounded.
------
Now let $N$ be the integer such that $\frac{7}{2} \tau(n) < n$ for all $n \geq N$ and choose a prime $p \geq N$. Then pick $a = p^5$. If $p$ does not divide $n$, then $\tau(p^5n) = 6\tau(n)$, which has no solutions by [b]Claim 1.[/b] Thus assume $n$ is divisible by $p$ and write $n = p^k m$ where $m$ is not divisibly by $p$. Then
\[ \tau(p^5n) = \tau(n) \left(1 + \frac{5}{k+1}\right) \leq \frac{7}{2} \tau(n) \]
Since $n \geq p > N$, we have that this is less than $n$, so we are done.[/quote]

I had a similar idea$,$ but you have to go a bit further than $6,$ since $72=6\cdot \tau(72).$

[b]Claim 1:[/b] The equation $n=18\tau(n)$ has no solution in positive integers$.$

[u]Proof:[/u] Consider the following$:$
$$A_2=\bigg\{\frac{2^k}{k+1}\mid k\in \mathbb{Z}^{+}_{0}\bigg\}=\bigg\{1, 1, \frac{4}{3}, 2, \frac{16}{5}, \cdots \bigg\}$$
$$A_3=\bigg\{\frac{3^k}{k+1}\mid k\in \mathbb{Z}^{+}_{0}\bigg\}=\bigg\{1, \frac{3}{2}, 3, \frac{27}{4}, \frac{81}{5} \cdots \bigg\}$$
$$A_5=\bigg\{\frac{5^k}{k+1}\mid k\in \mathbb{Z}^{+}_{0}\bigg\}=\bigg\{1, \frac{5}{2}, \frac{25}{3}, \cdots \bigg\}$$ and so on$.$ If it were true that $$18=\frac{n}{\tau(n)}=\prod_{p\mid n}\bigg(\frac{p^{v_p(n)}}{v_p(n)+1}\bigg) \hspace{1cm}   (*)$$ it must be true that the corresponding element from $A_3$ is either $\frac{27}{4}$ or $\frac{81}{5},$ since the numerator of the $\text{RHS  }$ in $(*)$ must be divisible by $9.$
If that element was $\frac{81}{5},$ it should be the case that $5\mid n,$ since one of the numerators in the product in the right must be divisible by $5.$ But now the smallest element of $A_5$ (given that $v_5(n)\geq 1$) is $\frac{5}{2}$ and $\frac{81}{5}\cdot \frac{5}{2}>18,$ contradiction$.$

Therefore we must have $v_3(n)=3$ and we must have taken $\frac{27}{4}$ from $A_3.$ Now since the denominator of $\frac{27}{4}$ is divisible by $4,$ in the product on the $\text{RHS  }$ of $(*)$ we must have an element from $A_2,$ which is divisible by at least $8.$ But the smallest element of $A_2$ with that property is $\frac{16}{5}$ and $\frac{16}{5}\cdot \frac{27}{4}>3\cdot 6=18,$ contradiction$.$ Therefore the equation has no solution in positive integers$.$

[b]The original problem:[/b]

Now assume that $p_1, p_2$ are very large primes and let $a=p_1^2\cdot p_2^5.$ Suppose that the equation $$\tau(an)=n$$ has a solution $n\in \mathbb{N}$ and let $n=p_1^{\alpha}p_2^{\beta}m,$ where $\alpha, \beta$ are nonnegative and $\gcd(p_1p_2, m)=1.$ The equation implies $$p_1^{\alpha}p_2^{\beta}m=(\alpha+3)(\beta+6)\tau(m).$$ It is is easy to see that $\tau(m)\leq m$ for all positive integers $m,$ so this yields $$p_1^{\alpha}p_2^{\beta}\leq (\alpha+3)(\beta+6).$$ Now if $\alpha \beta> 0,$ since we stated that $p_1, p_2$ are very large primes$,$ it is easy to see that $p_1^{\alpha}>\alpha +3$ and $p_2^{\beta}>\beta +6,$ as both statements can be proven by induction$,$ the base cases being $p_1>4$ and $p_2>7.$ But this is clearly a contradiction$,$ so suppose that $\beta=0$ (the case $\alpha=0$ can be dealt with in the same way)$.$ Now we have $p_1^{\alpha}\leq 6(\alpha +3),$ but it can again be easily shown that the reversed inequality holds for large $p_1,$ provided that $\alpha\geq 1.$ Thus it must be the case that $\alpha=\beta=0.$ But now $$m=18\tau(m)$$ and this clearly contradicts [b]Claim 1[/b]$.$

We have shown that for all sufficiently large primes $p_1, p_2,$ the number $a=p_1^2p_2^5$ works$.$ Done$.\hspace{1.5cm} \blacksquare$

(I'm sorry if the solution seems strange$,$ I'm just very tired to write it adequately)

We claim that all $a$ of the form $p^{p-1}$ for a prime $p\ge 3$ work. Suppose FTSOC that there exists $n$ satisfying $\tau(an)=n.$\\\\ If $p\nmid n,$ then $\nu_p(an)=p-1,$ so $p\mid\tau(an)$ and thus $p\mid n,$ contradiction. Thus we may write $n=p^k(q_{1}^{e_1}q_{2}^{e_2}\dots q_{m}^{e_m}),$ where $k,m$ are positive integers and $q_1,q_2,\dots,q_m$ are primes. Now our equation yields $$p^k(q_{1}^{e_1}q_{2}^{e_2}\dots q_{m}^{e_m})=(k+1)(e_1+1)(e_2+1)\dots (e_m+1).$$ Note that we always have $\nu_p(k+1)<\nu_p(p^k),$ so $p\mid (e_1+1)(e_2+1)\dots (e_m+1).$ Assume WLOG that $p\mid e_1+1;$ then, $e_1\ge p-1\ge 2,$ so it is easy to check that $e_1+1<q_1^{e_1}.$ Since we have $e_i+1\le q_{i}^{e_i}$ for all other $i,$ $$(k+1)(e_1+1)(e_2+1)\dots (e_m+1)<(k+1)(q_{1}^{e_1}q_{2}^{e_2}\dots q_{m}^{e_m})\implies p^k<k+1,$$ which is clearly impossible.

Take $a=p^{p-1}$ for primes $p\geq 13$. Then, if $p\nmid n$, but we still have $p\mid \tau(an)$, a contradiction. Thus $p\mid n$.

Then, we have that $p\mid n \Longrightarrow p\mid \tau(an)$. This happens if $v_q(an) \equiv -1 \pmod{p}$ for some $q$. If $q=p$, then $p^{2p-1} \mid an$ and $p^p \mid n$. If $q\neq p$, then $q^{p-1} \mid n$. Thus, we always have that $n\geq 2^{p-1}$.

We now bound
\[\tau(an) \leq \tau(a) \cdot \tau(n) \leq p\cdot \left(\frac{p}{2^{p-1}}\cdot n\right) <n\]

Probably the same as everybody else.

We claim that $a=p^{p-1}$ for any sufficiently large $p$ will fail to have a corresponding $n$.
[b][color=red]Claim:[/b][/color] $p$ must divide $n$.
[i]Proof.[/i] Suppose that $p$ does not divide $n$. Then $$n=\tau(np^{p-1})=p\tau(n),$$ contradiction. $\blacksquare$.

Suppose that $v_p(n)=k$, and $\frac{n}{p^k}=m$. This means that $$mp^k=n=\tau(np^{p-1})=\tau(mp^{p-1+k})=(p+k)\tau(m).$$ This implies that $\tau(m)=\frac{mp^k}{p+k}$. It is straightforward to check that $m<4$ does not yield a solution for large enough $p$. It is well-known that $\tau(m)\le 2\sqrt{m}\le \frac{m}{2}$ for $m\ge 4$, so $$\frac{p^k}{p+k}\le \frac{1}{2},$$ which is impossible for large $p$ and positive $k$.

[b][color=red]Comments:[/b][/color] I worked on this problem for two and a half hours, getting mostly nowhere because I couldn't find a SINGLE value of $a$ that worked. I found that $p^{p-1}$ caused $n$ to be a multiple of $p$, but unfortunately didn't pursue this route because $2^1,3^2$ both had corresponding $n$. However, after I read the hint on post 4, I solved the problem in five minutes.

Is there any further motivation to do $p^{p-1}$?

Consider $a=p^{f(p)}$ for some prime $p.$ Then, let $n=p^x\cdot y.$ If $n$ satisfies $\tau(an)=n$ then \[\tau\big(p^{x+f(p)}\big)\cdot\tau(y)=p^x\cdot y.\iff \frac{x+f(p)+1}{p^x}=\frac{y}{\tau(y)}\geq\frac{\sqrt{y}}{2}.\]We used the well-known $\tau(k)\leq 2\sqrt{k}$ lemma. By Bernoulli, $p^x\geq x(p-1)+1$ so we can let $f(p)=p-1$ to get \[1>\frac{x+p}{x(p-1)+1}\geq\frac{\sqrt b}{2}\] so $b\leq 3.$ Choosing $p\geq 5$ and bashing the cases $b\in\{1,2,3\}$ will, indeed, yield that there are no solutions.

In conclusion, $p^{p-1}$ works.

We take $a=p^{p-1}$ for big enough $p$, and assume that there is such $n$. $p|n$. Let $n=p^tq_1^{a_1}\cdot ... \cdot q_k^{a_k}$. $n=p^tq_1^{a_1}\cdot ... \cdot q_k^{a_k}=n=\tau(p^{p-1}p^tn)=(p+t)(a_1+1)\cdot ... \cdot (a_k+1)$. We have $q^x \ge x+1$ and $p^t \ge p+t$ for $t\ge 2$(equality holds for $p=2=t=2$). Thus there is no such $n$. For $t=1$ there exists $a_i$ such that $a_i=pk$. It is true that $pq_1^{pk}>(p+1)pk$ for big enough $p$. Thus the only possibility is $n=p$(because there must not be $q_i$) which is not a solution.

[quote=jasperE3]Does this work?

We choose $a=p^{p-1}$ for any prime $p$ such that $p\nmid n$. Note that infinitely many such values exist for any value of $n$ by the infinitude of primes (since $n$ cannot be the product of infinitely many primes). Assume FTSOC that some value of $n$ satisfies $\tau\left(p^{p-1}n\right)=n$. Then we get $n=\tau\left(p^{p-1}n\right)=\tau\left(p^{p-1}\right)\tau(n)=p\tau(n)$, so we must have $p\mid n$, contradiction.

[b]Motivation:[/b] Try $a=p$, then $a=p^k$, we get $\tau\left(p^kn\right)=n$. To use multiplicity of $\tau$, we enforce $p\nmid n$, and see that $(k+1)\tau(n)=n$, so letting $k=p-1$ is a contradiction.[/quote]

$p|n$ is not a contradiction I guess. Because $n$ can take any value, it is variable.

[quote=SerdarBozdag]$p|n$ is not a contradiction I guess. Because $n$ can take any value, it is variable.[/quote]

But there are infinitely many $p$ for which $p\nmid n$.

There can be a solution n such that p is not a divisor of n which are infinite. You can see other solutions, they first prove that p cannot divide n then continue the solution. See posts 27,29,30...

We claim that $a=p^{p-1}$ works for all primes $p$ sufficiently large.
Lemma: $p \mid n$.
Assume not. Then we have that $$\tau (p^{p-1}n) = p \cdot \tau (n) = n,$$ contradiction.

Let $n=p^km$ for some $m$ such that $p \nmid m$. Then $$\tau (p^{p+k-1}m) = (p+k) \tau(m) = p^k m$$. Note that when $k=1$, $v_p(p)=1$ and $v_p(p+1)=0$, and for $k>1$, $p^k>p+k$ and $p^k$ has only $p$'s in its prime factorization, we get that $v_p(p^k)>v_p(p+k) \implies p \mid m$. This shows that $m$ is sufficiently large. Using $\tau (m) \le 2\sqrt m \le \frac{m}{2}$, because $m$ is sufficiently large, we get that $$p^km=(p+k) \tau(m) < \frac{m(p+k)}{2} \implies 2p^k<p+k,$$ an obvious contradiction.

Note our problem is equivalent to showing that the sequence
$$ \frac{1}{\tau(1)}, \frac{2}{\tau(2)}, \frac{3}{\tau(3)}, \ldots  $$
does not contain infinitely many positive integers, since if
$$ \tau(an) = n \implies \frac{an}{\tau(an)} = a $$
Conversely, if
$$ \frac{m}{\tau(m)} = a \implies m = an ~~ \text{for some } n \in \mathbb Z_{>0} $$
Fix prime $p \ge 5$. We claim numbers of the form $p^{p-1}$ are not in the sequence. FTSOC
$$ \frac{k}{\tau(k)} = p^{p-1} $$
Then $\nu_p(k) \ge p-1$. Note $\nu_p(k)$ cannot equal $p-1$ (otherwise $\nu(p)$ of LHS would equal $p-2$, as $p \mid (p-1) + 1$, so a factor of $p$ would be in denominator also). Thus $\nu_p(k) \ge p$. Now if $e = \nu_p(k) \ge p+1$, then
$$ \frac{k}{\tau(k)} \ge \frac{p^e}{e+1} > p^{p-1} $$
which gives an contradiction. Thus $\nu_p(k) = p$. But then for some prime $q$, $\nu_q(k)+1 \mid p$, more particularly $f = \nu_p(k) \ge p-1$. Thus,
$$ \frac{k}{\tau(k)} \ge \frac{p^p}{p+1} \cdot \frac{q^f }{f+1} \ge p^{p-1} \cdot \frac{q^{p-1}}{p+1} \ge p^{p-1} \cdot \frac{2^{p-1}}{p+1} > p^{p-1} $$
which gives a contradiction. $\blacksquare$

I have a potential patch:
Let $p$ be sufficiently large. From #31, we deduce that $p\mid n$. Let $m=\frac n{p^{v_p(n)}}$.
First, note that:
$$p^{v_p(n)}m=n=\tau(p^{p-1}n)=\tau(p^{v_p(n)+p-1}m)=(v_p(n)+p-1)\tau(m).$$
Assume that $m\ge4$, then:
$$p^{v_p(n)}m=(v_p(n)+p-1)\tau(m)\le2(v_p(n)+p-1)\sqrt m\le\frac m2(v_p(n)+p-1)$$
so $2p^{v_p(n)}\le v_p(n)+p-1$. By Bernoulli's inequality, $2+(2p-2)v_p(n)\le2v_p(n)+2p-2$ which rewrites as $(2p-4)v_p(n)\le2p-4$. If $v_p(n)>1$ then we have $2p-4>2p-4$, a contradiction. So $v_p(n)$ is exactly $1$. But then $2p\le p$, so no $n$ exists.
We can similarly eliminate $m=1,2,3$.

We claim that for $a=5^4$ there is no solution.

Assume towards a contradiction there was. Let $n=5^kX$ where $5$ does not divide $X$. Then we have
$$\tau(5^{k+4}X)=5^kX$$
$$(k+5)\tau(X)=5^kX$$
We know $k\neq 0$ since otherwise the LHS would have a factor of 5 and the RHS wouldn't. Also, since $X\geq \tau(X)$,
$$k+5\geq 5^k$$
So, $0<k\leq 1$, so $k=1$. So, $6\tau(X)=5X$. So, $\frac56=\frac{\tau(X)}{X}$.

Since only $X$ and the numbers less than or equal to $\frac{X}2$ can be divisors of $X$, we know $\tau(X)\leq 1+\frac{X}2$. So,
$$\frac{\tau(X)}{X}\leq \frac1{X}+\frac12$$
$$\frac56\leq\frac1{X}+\frac12$$
$$\frac13\leq\frac1{X}$$
$$3\geq X$$
However, $X$ must be a multiple of $6$, a contradiction.

Now we will show there are infinitely many other numbers for which there is no solution. Let $a=p^{5^4-1}$ for a large prime $p$. Assume towards a contradiction $n$ is a positive integer such that $\tau(p^{5^4-1}n)=n$. Let $n=p^aX$ where $p$ doesn't divide $X$. Then,
$$(5^4+a)\tau(X)=p^aX$$
So,
$$5^4+a\geq p^a$$
For sufficiently large $p$, this implies $a=0$. So, plugging in $a=0$, we get
$$5^4\tau(X)=X$$
Let $k=\tau(X)$. Then,
$$5^4\tau(5^4k)=5^4k$$
$$\tau(5^4k)=k$$
a contradiction.

We claim $\boxed{a=p^{p-1}}$ for any prime $p>547$ satisfies that $\tau(an)=n$ has no positive integer solution $n$.

Suppose there existed $n$ with $\tau(p^{p-1}n) = n$.

Claim: $p$ divides $n$.
Proof: Suppose not.

Then $\nu_p(p^{p-1} n )=p-1$, so $p\mid \tau(p^{p-1}n)=n$, contradiction. $\square$

Let $n=c\cdot p^k$, where $p\nmid c$ and $k=\nu_p(n)>0$.

So we have $\tau(c\cdot p^{k+p-1}  ) = c\cdot p^k$

Thus, $\tau(c) \cdot (k+p) = c\cdot p^k$.

If $k>1$, we have $c\ge  \tau(c) $ and $p^k > p+k$, so $c\cdot p^k > \tau(c)\cdot (p+k)$, contradiction.

So $k=1$.

Thus, $\tau(c) \cdot (p+1) = c\cdot p$, so $\tau(c) = \frac{p}{p+1}\cdot c$.

Claim: $\tau(c) \le 2\sqrt{c}$.
Proof: If not, then the number of unordered factor pairs multiplying up to $n$ is greater than $\sqrt{c}$. However, the least number in the factor pair has to be at most $\sqrt{c}$, a contradiction because each positive integer is in at most one ordered factor pair of $c$. $\square$

So we have $c\cdot \frac{p}{p+1} \le 2\sqrt{c}$. Squaring both sides gives \[c^2\cdot \left(\frac{p}{p+1}\right)^2 \le 4c\implies c\cdot \left(\frac{p}{p+1}\right)^2\le 4 \]

However, since $p>187$, we have $4> c\cdot \left(\frac{187}{548}\right)^2=  \frac{299209c}{300304}\implies c<\frac{1201216}{299209}<5$

Case 1: $c=4$.
Then $3=\frac{4p}{p+1}$. This implies $3\mid p$, not possible as $p>187$.

Case 2: $c=3$.
Then $2=\frac{3p}{p+1}$, which implies $p$ is even, not possible as $p>187$.

Case 3: $c=2$.
Then $2=\frac{2p}{p+1}$, so $\frac{p}{p+1} = 1$, not possible.

Case 4: $c=1$.
Then $\frac{p}{p+1} = 1 $, not possible.

We have exhausted all cases, so we are done.

Same as above solutions; posting for storage.

We claim $a=p^{p-1}$ works for $p>1234.$ Suppose FTSOC the equation has solutions. We claim $p\mid n$; assume the contrary. Then, $$n=\tau(p^{p-1}n)=\tau(p^{p-1})\tau(n)=p\tau(n),$$ a contradiction. Let $n=p^{\alpha}m$ where $\alpha>0$ and $p\nmid m.$ If $\alpha>1,$ clearly $p+\alpha<p^{\alpha}.$ Since $\tau(m)\le m,$ we see $$\tau(an)=\tau(p^{p-1+\alpha}m)=(p+\alpha)\tau(m)<p^{\alpha}m=n,$$ which is absurd. Thus, $\alpha=1.$ It is well known that $\tau(m)<2\sqrt{m},$ so if $m\ge 5,$ we have $$m>4(1+1/p)^2\implies n=pm>(p+1)2\sqrt{m}>\tau(m)\tau(p),$$ which leads to no solutions. If $m=4,$ then $(p+1)3=4p,$ or $p=3.$ If $m=3,$ then $(p+1)2=3p,$ or $p=2.$ If $m=2,$ then $(p+1)2=2p,$ which yields no solutions. If $m=1,$ then $p+1=p,$ which is absurd. Since $p>1234,$ we have no solutions in any of these cases. $\square$

Let $n=p^{e} \cdot \prod_{i=1}^{j}{p_i^{e_i}}$ where $p_1<p_2<\dots<p_j$ and $p$ is distinct from all the other primes and is odd.

Let $a=p^{p-1}$ then $\tau(an)=(e+p)(e_1+1)(e_2+1)\dots (e_n+1).$ If $e=0$ then $p\nmid n$ but then $n=\tau(an)=p\tau(n)$ so there aren't solutions. Now, $p\mid n.$ If $\nu_{p}{n}>2$ then for each $i$ we have $p_i^{e_i}\ge e_i+1$ with equality iff $p_i=2.$ Then, $e>2\implies p^e>p+e$ so we have no solutions.

Thus, if there is a solution then $e=1.$ We have $(p+1)(e_1+1)\dots(e_j+1)=p\cdot p_1^{e_1}p_2^{e_2}\dots p_j^{e_j}.$ Obviously we have $p\mid e_i+1$ for some $i$ then $e_i\ge p-1.$ Then, we have $p_i^{p-1}>p+1$ whenever $p>3$ and we use $p_i^{e_i}\ge e_i+1$ for all the other $i$ and we win.

[hide=Solution]
First, we begin with a few estimates.

[b]Claim 1:[/b] For any integer $b \geq 2$, $5^b > b+5$.

[b]Proof:[/b] Induct on $b$. We know that $25 = 5^2 > 2+5 = 7$, so the inequality holds for $b = 2$. Now, assume the inequality is true for some $b\geq 3$, i.e., $3^b > b+3$. Then,
\begin{align*}
5^{b+1} &= 5\cdot 5^b \\
        &> 5(b+5) \\
        &= 5b+25 \\
        &> b+6 \\
        &= (b+1)+5,
\end{align*}
so the inductive step is complete and so the claim is true. $\square$

[b]Claim 2:[/b] For any prime $p > 1000$, and any integer $a\geq 1$, we have that $\frac{a+5}{p^a} < \frac{1}{125}$.
[b]Proof:[/b] Clearly, it suffices to show this result for $p = 1000$ (even though it's not a prime), since for $p > 1000$, $\frac{a+5}{p^a} < \frac{a+5}{1000^a}$. Thus, we want to show that $$\frac{1}{125}\cdot 1000^a > a+5$$ for all integers $a\geq 1$. We induct on $a$. For $a = 1$, we have $8 = \frac{1000}{125} = \frac{1}{125}\cdot 1000 > 1 + 5 = 6$. Now, assume the inequality is true for some $a\geq 1$, i.e., $\frac{1}{125}\cdot 1000^a > a+5$. Then,
\begin{align*}
\frac{1}{125}\cdot 1000^{a+1} &= 1000\cdot \left(\frac{1}{125}\cdot 1000^a\right) \\
                              &> 1000(a+5) \\
                              &= 1000a + 5000 \\
                              &> a + 6 \\
                              &= a+5,
\end{align*}
so the inductive step is complete and so the claim is true. $\square$

Now, for our main claim.

[b]Main Claim:[/b] The equation $n = \tau(625n)$ has no solutions over positive integers.

[b]Proof:[/b] If we had $5\nmid n$, then we would have $n = \tau(625n)=\tau(625)\tau(n) = 5\tau(n),$ meaning that $5\mid n$, absurd. Thus, we have $5\mid n$. Let $r = \nu_5(n) \geq 1$. Then, letting $n = 5^rk$ for some positive integer $k$, we have that $$5^rk = n = \tau(625n) = \tau(5^{r+4}k) = (r+5)\tau(k),$$ so $$\frac{k}{\tau(k)} = \frac{r+5}{5^r}.$$ If $r\geq 2$, then by Claim $1$, this would mean $\frac{k}{\tau(k)} < 1$, absurd, since clearly $k\geq \tau(k)$, since the set of positive integer divisors of $k$ is a subset of $\{1, 2, \cdots, k\}$. Thus, we have $r = 1$, so $$\frac{k}{\tau(k)} = \frac{6}{5},$$ so $$k = \frac{6}{5}\tau(k).$$ Since $\tau(k) \leq 2\sqrt{k}$ (well-known), this means $$k \leq \frac{12}{5}\sqrt{k},$$ we have $$\sqrt{k}\leq \frac{12}{5},$$ so $$k\leq \frac{144}{25} < 6,$$ however, we know that $6\tau(k) = 5k$, so $6\mid 5k$, so $6\mid k$, meaning $k\geq 6$, a contradiction. Thus, the equation has no solutions over positive integers, as desired. $\square$

Now, let $p,q,r,s$ be distinct primes greater than $1000$. Then, I claim that if $a=p^4q^4r^4s^4$, then $n = \tau(an)$ has no solutions in positive integers. This will immediately imply our desired result, since there are infinitely many primes (and therefore there are infinitely many priems greater than $1000$).

Assume for the sake of contradiction that there existed a positive integer solution $n$. If $n$ is not divisible by $p$, $q$, $r$, $s$, then we have that $n = \tau(an) = \tau(a)\tau(n) = 625\tau(n),$ which by our main claim has no solutions. Thus, $n$ is divisible by at least one of $p, q, r, s$. WLOG, suppose $n$ was divisible by $p$. Let $n = p^aq^br^cs^dA$ for some nonnegative integers $a, b, c, d$ and a positive integer $A$ with $A$ not divisible by $p,q,r$, or $s$. Note that $a\geq 1$. Then, we have that $$p^aq^br^cs^dA = \tau(p^{a+4}q^{b+4}r^{c+4}s^{d+4}A) = (a+5)(b+5)(c+5)(d+5)\tau(A),$$ so $$\frac{A}{\tau(A)} = \frac{a+5}{p^a}\cdot \frac{b+5}{q^b} \cdot \frac{c+5}{r^c}\cdot \frac{d+5}{s^d}.$$ By Claim $2$, since $a\geq 1$, we have $\frac{a+5}{p^a} < \frac{1}{125}$. If $b\geq 1$, then $\frac{b+5}{q^b} < \frac{1}{125}$, and if $b = 0$, then $\frac{b+5}{q^b} = 5$, so $\frac{b+5}{q^b} \leq 5$. Similarly, $\frac{c+5}{r^c} \leq 5$ and $\frac{d+5}{s^d} \leq 5$. Thus, $$\frac{A}{\tau(A)} < \frac{1}{125}\cdot 5\cdot 5\cdot 5 = 1,$$ which is absurd since $A \geq \tau(A)$ as discussed earlier. Thus, $n = \tau(an)$ has no positive integer solutions for $n$ if $a = p^4q^4r^4s^4$ for pairwise distinct primes $p, q, r, s > 1000$ and therefore $n = \tau(an)$ has has no positive integer solutions for $n$ for infinitely many positive integers $a$, as desired. $\blacksquare$

[quote=amuthup]We claim that all $a$ of the form $p^{p-1}$ for a prime $p\ge 3$ work.[/quote]
[quote=beta]Let $a=p^{p-1}$, $p>2$.[/quote]
For $p = 3$, $p^{p-1} = 9$, and $\tau(9\cdot 12) = 12$.
[quote=HKIS200543]Claim 1. There exist no solutions to the equation
\[ 6 \tau(n) = n . \][/quote]
[Nevermind this was already mentioned by someone else earlier in the thread] $n = 72$. (Your proof of the claim fails to check $n = 2^a3^b$.)

Solved with motivation for bashing from [b]my classmates.[/b]

I claim that $a=p^5q^2$ where $p$ and $q$ are distinct primes greater than $100$ always works, which obviously finishes.

[b]Claim.[/b] $\gcd(n,a)>1$.
[b]Proof.[/b] Assume FTSOC that $\gcd(n,a)=1$. Then $d(an)=18d(n)$. So $d(n)=n/18$.
If $n>18\cdot 72$, then
\[d(n)\le 2\sqrt n <\frac{n}{18},\]
so $n\le 18\cdot 72$. $1680$ is a highly composite number greater than $18\cdot 72$, and it has $40$ positive divisors, so $n\le 18\cdot 40$. I tested all positive multiples of $18$ less than or equal to $18\cdot 40$, and none of them worked. $\blacksquare$

Now let $n=p^x q^y m$ where $p\nmid m$ and $q\nmid m$. Then
\[p^x q^y m = n = d(pqn) =
d\left(p^{x+5}q^{y+2}m\right) = (x+6)(y+3)d(m), \]
meaning that
\[\frac{x+6}{100^x}\cdot\frac{y+3}{100^y} \ge \frac{(x+6)(y+3)}{p^x q^y} = \frac{m}{d(m)} \ge 1.\]
But since $x+y>0$ by the previous claim, the LHS is less than $1$, contradiction. $\blacksquare$

We claim all $a=p^{p-1}$ for prime $p\geq 5$ work.

Suppose $n$ is a solution to $\tau(p^{p-1}n)=n$. Suppose $p\nmid n$. Then $p\tau(n)=\tau(p^{p-1}n)=n$, which is a contradiction. So, $p\mid n$ and $n=p^kn'$ for some positive integer $k$ and positive integer $n'$ not divisible by $p$. We now have the equation \[(p+k)\tau(n')=\tau(p^{p-1}p^kn')=\tau(p^{p-1}n)=n=p^kn'\]
 We now use the bound \[\tau(m)\leq 2\sqrt{m}\]
Which can be shown using the fact that factors of $m$ come in pairs, one being $\geq \sqrt{m}$ and one being $\leq \sqrt{m}$.

Applying the bound, we get $\frac{p^k}{p+k}n'=\tau(n')\leq 2\sqrt{n'}\implies \frac{p^k}{p+k}\leq \frac{2}{\sqrt{n'}}$.

Case 1: $k=1$
Using $(p+1)\tau(n')=pn'$, we get that $p\mid \tau(n')$. So, $n'$ has some divisor of the form $q^{dp-1}$, which is at least $2^4=16$. So $n'\geq 16$ and we have $\frac{1}{2}<\frac{p}{p+1}\leq \frac{2}{\sqrt{n'}}\leq \frac{1}{2}$, which is a contradiction.

Case 2: $k>1$
$n'=1$ or $n'=2$ gives $p+k=p^k$, and $n'=3$ gives $p+k=\frac{3}{2}p^k$. Each of these are contradictions since $p^k>p+k$. So, $n'\geq 4$. We then have $1<\frac{p^k}{p+k}\leq \frac{2}{\sqrt{n'}}\leq 1$, a contradiction.

So, no such $n$ exist and the infinitely many $a$ of the form $p^{p-1}$ for prime $p\geq 5$ work.

>consider $p^{p-1}$ immediately
>do not find a proof
>come up with whatever this is

I claim that $\boxed{p^{14}}$ does the job for sufficiently large primes $p$.

Suppose we prime factorize $n=p^{f}p_1^{f_1}\ldots p_k^{f_k}$ for positive integers $f_i$. Then we need $(f+15)(f_1+1)\ldots(f_k+1)=p^{f}p_1^{f_1}\ldots p_k^{f_k}$. Let $\omega(m)$ denote the number of prime powers greater than $1$ dividing $m$; note that $\omega$ is completely multiplicative. Applying it to both sides of the equation, we get
$$\omega(f+15)+\omega(f_1+1)+\cdots+\omega(f_k+1)=f+f_1+\cdots+f_k \implies (f_1-\omega(f_1+1))+\cdots+(f_k-\omega(f_k+1))=\omega(f+15)-f$$

Since $\omega(m) \leq \tau(m)\leq 2\sqrt{m}$, where the last inequality is well-known and easy to prove, it follows that for fixed positive integers $r$, $\omega(m+r)-m$ is bounded above and eventually negative as $m$ varies. Furthermore, we can check (by examining small $m$ manually) that we must always have $m-\omega(m+1) \geq 0$.

In particular, this means that, by appealing to the fact that $\omega(f+15)-f$ is bounded above, $f_i-\omega(f_i+1)$ is bounded above as well, hence the $f_i$ are all bounded above. Furthermore, since $f-\omega(f+15)$ should be nonpositive, $f$ is bounded above as well. Hence if $p$ is sufficiently large, we are forced to have $f=0$, else $p$ divides $n$ but $p \nmid (f+15)(f_1+1)\ldots(f_k+1)$ for size reasons.

Hence the RHS of the equation just equals $2$, forcing $f_i-\omega(f_i+1) \leq 2$ for all $i$. We may manually verify that for positive integers $m$, $m-\omega(m+1) \leq 2 \implies m \in \{1,2,3\}$, hence $f_i \in \{1,2,3\}$ for all $i$. Now we do casework.
[list]
[*] Because $15 \mid n$, we should have $\nu_5(n) \in \{1,2,3\}$. On the other hand, if $\nu_5(n) \geq 2$ then we need some prime $q$ to have $\nu_q(n) \equiv 4 \pmod{5}$, but this contradicts the fact that it should be at most $3$ always. Thus $\nu_5(n)=1$.
[*] We should also have $\nu_3(n) \in \{1,2,3\}$. We consider these three cases.
[list]
[*] If $\nu_3(n)=1$, then $\nu_2(n) \in \{2,3\}$, since now $(1+1)(1+1) \mid \tau(an)$. But $\nu_2(n)=2$ would imply $9 \mid \tau(an)$ and $\nu_2(n)=3$ would imply $16 \mid \tau(an)$, so this case is impossible.
[*] If $\nu_3(n)=2$, then $\nu_2(n) \in \{1,2,3\}$. But $\nu_2(n)=1$ would imply $4 \mid \tau(an)$: contradiction. On the other hand, if $\nu_2(n) \in \{2,3\}$, we are faced with a problem of having "too many" factors of $2$, so there has to be some prime $q \mid n$ which we haven't considered yet that has odd exponent. But then $q \geq 7$, so $q \nmid 15$ and $q \nmid f_i+1$ due to size reasons: also a contradiction.
[*] If $\nu_3(n)=3$, then $\nu_2(n)=3$ is forced since $8 \mid \tau(an)$. But then $32 \mid \tau(an)$, which is again impossible.
[/list]
[/list]
This finishes the problem. $\blacksquare$

[b]Remark:[/b] $a=p^{14}$ is like the 11th thing I considered of this form, but there's actually a reason it works and everything I considered beforehand doesn't. In particular I think $a=p^{3q-1}$ should work as well for $q\geq 5$, if $p$ is taken to be sufficiently large relative to $q$.


geometry
calculus
inequalities
topology
real analysis
real analysis unsolved

In the book: "Introduction to Calculus and Analysis. Volume2" Author: Richard Courant and Fritz John.

in the forth chapter,they talk about the area of the graph on the plane: $ A_{n}^{\plus{}}(S)\minus{}A_{n}^{\minus{}}(S)\leq A_{n}^{\plus{}}(\partial S)$

I think the equality will always hold: $ A_{n}^{\plus{}}(S)\minus{}A_{n}^{\minus{}}(S)\equal{} A_{n}^{\plus{}}(\partial S)$.

[color=red]can anyone give an example that the inequality holds strictly: give $ S$ ,such that:$ A_{n}^{\plus{}}(S)\minus{}A_{n}^{\minus{}}(S)< A_{n}^{\plus{}}(\partial S)$[/color]

where: $ S$ denotes the graph on the plane.  and $ A_{n}^{\plus{}}(S)$ denotes the number of the squares that has common points with $ S$,

$ A_{n}^{\minus{}}(S)$ denotes the number of the squares that  in the internal of $ S$, $ A_{n}^{\plus{}}(\partial S)$ denotes the number of the squares

that has common points with the boundary of $ S$

Take an open set $ S$ and the grid that has a few squares just touching the boundary from outside. Those squares will be counted in $ A_n(\partial S)$ but not in $ A^\plus{}_n(S)$.


Αν $ 0<a<b<c$ να βρεθει το min της παραστασης

$ \frac{4(c^3\minus{}a^3)\plus{}27(a^2b\minus{}c^2b)\plus{}24(cb^2\minus{}ab^2)}{6b^3}$

Καλημέρα σε όλους.
Στέλνω τις σκέψεις μου για αυτή την πολύ ωραία άσκηση.
Ονομάζουμε Α το κλάσμα. Θέτουμε $ x\equal{}\frac{a}{b}$, $ y\equal{}\frac{c}{b}$. Τότε 6A=f(y)-f(x) όπου 0<x<1 και y>1 με $ f(t)\equal{}4t^3\minus{}27t^2\plus{}24t$. Το ελάχιστο επιτυγχάνεται όταν ελαχιστοποιήσουμε το f(y) και μεγιστοποιήσουμε το f(x). Με τη βοήθεια παραγώγων βρίσκουμε ότι αυτό συμβαίνει όταν y=4 και $ x\equal{}\frac{1}{2}$. ΄Αρα $ A_{min}\equal{}\frac{f(4)\minus{}f(0.5)}{6}$.

καλως ηρθες στην παρεα μας
Ωραια η λυση σου
Δες τηνε και με ολοκληρωμα η και αλλιως...

Χρήστο η άσκηση σου μου εδωσε την ιδέα για το παρακάτω

1.	Να κατασκευάσετε την γραφική παράσταση της συνάρτησης  $ f(x)=x(x-3)^2^$
2.	Να βρείτε για ποιες τιμές του m η εξίσωση $ f(x)=m$ έχει 3 λύσεις a,b,c με a<b<c
3.	Να δείξετε ότι 0<a<1<b<3<c<4
4.	Να δείξετε ότι η εξίσωση $ f(x)=f(x+y),0<x<1$ με άγνωστο τον y έχει δυο θετικές λύσεις
5.	Αν z η μεγαλύτερη από δυο προηγούμενες τότε δείξτε ότι  $ 3-x<z<4-x$
6.	Να εκφράσετε το z ως συνάρτηση του x
7.	Να βρείτε το μέγιστο μήκος οριζόντιας χορδής της Cf  η οποία εκτός από τα δυο άκρα της έχει τουλάχιστον άλλο ένα κοινό σημείο με την Cf
8.	Σε πόσο «ύψος» από τον άξονα x βρίσκεται αυτή η χορδή?
9.	Αν $ 0<p<q<r<(4q)/3$ να βρεθεί η μέγιστη τιμή της παράστασης $ \frac{(2p-3q)(p^2+9(q-r)^2-3pr)}{q^3}$

Πολυ πολυ καλη....
Ροδολφε να σου δινουμε ιδεες να μας κανεις ομορφες ασκησεις  ......
Το (6) το βγαζω αλγεβρικα, εσυ;
Αν την εχεις λυμενη βαλε την λυση της γιατι με $ Latex$ θελω μερες...

[quote="r_boris"]Χρήστο η άσκηση σου μου εδωσε την ιδέα για το παρακάτω

1.	Να κατασκευάσετε την γραφική παράσταση της συνάρτησης  $ f(x) = x(x - 3)^2^$
2.	Να βρείτε για ποιες τιμές του m η εξίσωση $ f(x) = m$ έχει 3 λύσεις a,b,c με a<b<c
3.	Να δείξετε ότι 0<a<1<b<3<c<4
4.	Να δείξετε ότι η εξίσωση $ f(x) = f(x + y),0 < x < 1$ με άγνωστο τον y έχει δυο θετικές λύσεις
5.	Αν z η μεγαλύτερη από δυο προηγούμενες τότε δείξτε ότι  $ 3 - x < z < 4 - x$
6.	Να εκφράσετε το z ως συνάρτηση του x
7.	Να βρείτε το μέγιστο μήκος οριζόντιας χορδής της Cf  η οποία εκτός από τα δυο άκρα της έχει τουλάχιστον άλλο ένα κοινό σημείο με την Cf
8.	Σε πόσο «ύψος» από τον άξονα x βρίσκεται αυτή η χορδή?
9.	Αν $ 0 < p < q < r+p < (4q)/3$ να βρεθεί η μέγιστη τιμή του λόγου r/q όταν $ 3rp^2+r(r-3q)^2=3pr(4q-r)^2$[b]
ΣΥΓΝΩΜΗ ΓΙΑ το 9ο το ΔΙΟΡΘΩΣΑ[/b]
Στέλνω σε συνημμένο την λύση

[quote="r_boris"][quote="r_boris"]Χρήστο η άσκηση σου μου εδωσε την ιδέα για το παρακάτω

1.	Να κατασκευάσετε την γραφική παράσταση της συνάρτησης  $ f(x) = x(x - 3)^2^$
2.	Να βρείτε για ποιες τιμές του m η εξίσωση $ f(x) = m$ έχει 3 λύσεις a,b,c με a<b<c
3.	Να δείξετε ότι 0<a<1<b<3<c<4
4.	Να δείξετε ότι η εξίσωση $ f(x) = f(x + y),0 < x < 1$ με άγνωστο τον y έχει δυο θετικές λύσεις
5.	Αν z η μεγαλύτερη από δυο προηγούμενες τότε δείξτε ότι  $ 3 - x < z < 4 - x$
6.	Να εκφράσετε το z ως συνάρτηση του x
7.	Να βρείτε το μέγιστο μήκος οριζόντιας χορδής της Cf  η οποία εκτός από τα δυο άκρα της έχει τουλάχιστον άλλο ένα κοινό σημείο με την Cf
8.	Σε πόσο «ύψος» από τον άξονα x βρίσκεται αυτή η χορδή?
9.	Αν $ 0 < p < q < (r + p)/3< (4q)/$ να βρεθεί η μέγιστη τιμή του λόγου r/q όταν $ 3rp^2 + r(r - 3q)^2 = 3pr(4q - r)^2$[b]
ΣΥΓΝΩΜΗ ΓΙΑ το 9ο το ΔΙΟΡΘΩΣΑ[/b]
Στέλνω σε συνημμένο την λύση[/quote][/quote]


[hide="chetjan read this"]There are as many 4-digit palindromes as the number of two digit numbers. The sum of all the 2-digit numbers is 4905.

This times 101 is [b]495405[/b], thus our answer.[/hide]

shine09 is right.

[hide="Proof"]There exist $ 90$ palindromes, each of which can be obtained by taking a positive integer not divisible by $ 10$ and reversing. So, $ 5$ would obtain $ 5005$, and $ 47$ would obtain $ 7447$.

The method of generation is pointless. Consider an arbitrary palindrome $ abba$. The number $ 11000 \minus{} abba$ is a palindrome. This is also easy to prove: Let the number be $ cdef$. We have $ a\plus{}f \equal{} 10$, $ b\plus{}e \equal{} 9$, $ b \plus{} d \equal{} 9$, and $ a\plus{}c \equal{} 10$. So $ c\equal{}f$ and $ d\equal{}e$.

$ 11000 \times 45 \equal{} \boxed{495000}$.

[/hide]

wait..which one??which answer??kingsmasher's or uh..dark's?

[quote="AndrewTom"]A palindrome is a word or number that reads the same backwards as forwards, e.g. LEVEL or $ 9889$. Find the sum of all the four-digit positive integer palindromes.[/quote]
[hide]Note that adjacent palindromes like $ 1001$ and $ 9999$ add to $ 11000$. Then, each $ 2$ digits on the right/left can be different values, so to find the toal number, there is $ 9$ possibilities for the 1st one, and $ 10$ for the 2nd one (which includes 0). Now, $ 9(10) \equal{} 90$. Divide by $ 2$ to find the number of terms, yielding $ 45$. Multiply the values, getting $ 45(11000) \equal{} 495000$.[/hide]

[quote="cmtappu96"]shine09 is really right
see...
a palindrome will have the general form abba.
we can have 9 values for a (1-9) and 10 for b (0-9).
Let a=1
general form 1bb1.
ten possibilities here (1001-1991)
hence sum of all 1s (with place value) which are at a in all possibilities
is 10 times 1001=10010
similarly , if a = 2, then this sum is 20020
when we add up all this from 1 to 9, we get
the sum of all possible a's is 450450.
now look at the b's
if b= 1, then general form a11a.
9 possibilities (1111-9119)
so sum of all 1s is 9*0110=990
similarly, for a=2, then this sum is 1980.
when we add up, sum of all possible b's is 44550.
so sumof all palindromes = sum of a's + sum of b's
                                 =450450 + 44550
                      sum      =495000[/quote]

Yes shine09 is definitely rite..

The easiest way to do it using Free For All's solution. Although they aren't exactly an arithmetic series, palindromes come in pairs that sum to $ 11000$. There are $ 9(10) \equal{} 90$ 4-digit palindromes. To sum an arithmetic series we must divide the total number by 2. $ \frac{90}{2}\equal{}45$. $ 11000(45) \equal{} \boxed{495000}$

I really liked cmtappu96's solution. I think it's the simplest way of doing it...

[quote="darkdieuguerre"]kingsmasher's solution doesn't generate palindromes. In fact, it generates integers divisible by $ 101$, which almost always are not palindromic. As a simple example, $ 1010$ is not palindromic.

I checked my solution by writing a program to generate all palindromes. My program generates the same sum, so I'm going to assume it's correct.[/quote]
i agree, all are easy ways, nice try king :rotfl:
jk :D

495000 is correct

[hide="Solution 1"]

List the first few palindromes and the last few. Notice anything?

$ 1001,1111,1221,1331.......9779,9889,9999.$

$ 1001 \plus{} 9999 \equal{} 11000.$
$ 1111 \plus{} 9889 \equal{} 11000.$
$ 1221 \plus{} 9779 \equal{} 11000$.

There are $ 9$ choices for the first digit, $ 10$ for the second, and $ 1$ for the third and last. Thus there are $ 9*10 \equal{} 90$ palindromes and $ 45$ pairs.

$ 11000*45 \equal{} \boxed{495000}$

[/hide]
[hide="Solution 2"]

As shown previously, there are $ 90$ palindromes. We need to figure out the average of these and multiply by $ 90$.

Unfortunately, the palindromes do not go in an arithmetic progression. However, we can find the average through the digits.

Thousands: $ 10\ 1's, 10\ 2's...10\ 9's.$

Average: $ \dfrac{10(1 \plus{} 2... \plus{} 9)}{90} \equal{} \dfrac{10*45}{90} \equal{} 5.$

Hundreds: $ 9\ 1's, 9\ 2's...9\ 9's.$

Average: $ \dfrac{9(1 \plus{} 2 \plus{} ... \plus{} 9)}{90} \equal{} \dfrac{9*45}{90} \equal{} 4.5$

The tens is the same as the hundreds so its average is $ 4.5$ Similarly, the unit's average is $ 5$.

Thus, $ 5*1000 \plus{} 4.5*100 \plus{} 4.5*10 \plus{} 5*1 \equal{} 5000 \plus{} 450 \plus{} 45 \plus{} 5 \equal{} 5500$.

$ 5500*90 \equal{} \boxed{495000}$[/hide]

[hide="The Worst Solution"]

Add all the palindromes from scratch. See I told you it was "the worst solution" :D[/hide]


I was wondering whether it is possible to make a contest similar to the USAMTS but for college students?

You mean something different from the Putnam contest, right?

In a sense, doing research as an undergraduate [b]is[/b] like the USAMTS for undergraduates. And that's not just a contest, but the real thing.

I mean like five problems and about a month to solve them. Maybe it should include some calculus too but should be on a similar level to the USAMTS for undergrads.

There are somehow similar conquests around the web, even when may be the topics are not quite in pure math, but in "Ludic math" (Look for PQRST). I suppose it could be great. Of course, to do it interesting, it needs time and effort, and some money. Someone wants to try to talk with the NSF or NSA.

Best regards,


geometry
3D geometry
calculus
projective geometry
pyramid
sphere

[quote="davidlizeng"]The formula of the volume of a cone is 1/3 pi r^2. How
do you prove this w/o calculus?[/quote]

I think you forgot the height  :wink:

Can't.  Not really.  Archimedes proved it using a non-rigorous limit argument, I think.

The volumes and surfaces of differnt solids can be calculated with the aid of Pappus Theoreme, but it is stil proved with calculus
http://mathworld.wolfram.com/PappussCentroidTheorem.html

Correct me if I am wrong, but Archimedes used what is known today as Reimann sums to find the volume correct?

T0r: Doesn't he do this by assuming a cone is a pyramid with an arbitrarily large number of sides?

Which is doing it with Riemann Sums, which does happen to be Calculus...

Here is a link on the method finding the volume of the sphere.
Perhaps it could be "reversed"?
[url]http://www.as.ysu.edu/~faires/Courses/2000-2001/01M3750/ASV.pdf[/url]
and this might be good
[url]http://www.mathpages.com/home/kmath343.htm[/url]
[url]http://mathdl.maa.org/convergence/1/?pa=content&sa=viewDocument&nodeId=631&pf=1[/url]

Yes, but then you'd still need to find the volume of a sphere without calculus  :P

There are two problems:
1) What's the area of a circle?
2) How do we know that pointy things (cones, pyramids, etc.) get a factor of 1/3?

If you're not worried about (1), (2) can be done using [url=http://mathworld.wolfram.com/CavalierisPrinciple.html]Cavalieri's Principle[/url], which is sort-of calculus.


geometry
3D geometry
tetrahedron
inequalities
function
trigonometry
pyramid

[url=http://www.mathlinks.ro/Forum/download.php?id=1957]PDF Version[/url]

Please click on each grade to see the problems for that grade. To see the solutions and discussion over each problem please click on the problem number.

[hide="7th Grade"]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31913]Problem 1[/url]
Let $ABCD$ be a parallelogram. The interior angle bisector of $\angle ADC$ intersects the line $BC$ in $E$, and the perpendicular bisector of the side $AD$ intersects the line $DE$ in $M$. Let $F= AM \cap BC$. Prove that:

a) $DE=AF$;
b) $AD\cdot AB = DE\cdot DM$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31914]Problem 2[/url]
Let $a,b$ be two integers. Prove that

a) $13 \mid 2a+3b$ if and only if $13 \mid 2b-3a$;
b) If $13 \mid a^2+b^2$ then $13 \mid (2a+3b)(2b+3a)$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31915]Problem 3[/url]
Let $ABCD$ be a quadrilateral with $AB\parallel CD$ and $AC \perp BD$. Let $O$ be the intersection of $AC$ and $BD$. On the rays $(OA$ and $(OB$ we consider the points $M$ and $N$ respectively such that $\angle ANC = \angle BMD = 90^\circ$. We denote with $E$ the midpoint of the segment $MN$. Prove that

a) $\triangle OMN \sim \triangle OBA$;
b) $OE \perp AB$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31916]Problem 4[/url]
On a circle there are written 2005 non-negative integers with sum 7022. Prove that there exist two pairs formed with two consecutive numbers on the circle such that the sum of the elements in each pair is greater or equal with 8.
[/hide]
[hide="8th Grade"]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31917]Problem 1[/url]
We consider a cube with sides of length 1. Prove that a tetrahedron with vertices in the set of the vertices of the cube has the volume $\dfrac 16$ if and only if 3 of the vertices of the tetrahedron are vertices on the same face of the cube.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31918]Problem 2[/url]
For a positive integer $n$, written in decimal base, we denote by $p(n)$ the product of its digits.

a) Prove that $p(n) \leq n$;
b) Find all positive integers $n$ such that
\[ 10p(n) = n^2+ 4n - 2005. \]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31919]Problem 3[/url]
Let the $ABCA'B'C'$ be a regular prism. The points $M$ and $N$ are the midpoints of the sides $BB'$, respectively $BC$, and the angle between the lines $AB'$ and $BC'$ is of $60^\circ$. Let $O$ and $P$ be the intersection of the lines $A'C$ and $AC'$, with respectively $B'C$ and $C'N$.

a) Prove that $AC' \perp (OPM)$;
b) Find the measure of the angle between the line $AP$ and the plane $(OPM)$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31920]Problem 4[/url]
a) Prove that for all positive reals $u,v,x,y$ the following inequality takes place:
\[ \frac ux + \frac vy \geq \frac {4(uy+vx)}{(x+y)^2} . \]
b) Let $a,b,c,d>0$. Prove that
\[ \frac a{b+2c+d} + \frac b{c+2d+a} + \frac c{d+2a+b} + \frac d{a+2b+c} \geq 1.\]
[/hide]
[hide="9th Grade"]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31921]Problem 1[/url]
Let $ABCD$ be a convex quadrilateral and the points $E=AD \cap BC$ and $I = AC\cap BD$. Prove that the triangles $EDC$ and $IAB$ have the same centroid if and only if $AB \parallel CD$ and $IC^2 = IA \cdot AC$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31922]Problem 2[/url]
Find all functions $f:\mathbb{R}\to\mathbb{R}$ for which
\[ x(f(x+1)-f(x)) = f(x), \]
for all $x\in\mathbb{R}$ and
\[ | f(x) - f(y) | \leq |x-y| , \]
for all $x,y\in\mathbb{R}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31923]Problem 3[/url]
Prove that for all positive integers $n$ there exists a single positive integer divisible with $5^n$ which in decimal base is written using $n$ digits from the set {1,2,3,4,5}.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31924]Problem 4[/url]
Let $x_1,x_2,\ldots,x_n$ be positive reals. Prove that
\[ \frac 1{1+x_1} + \frac 1{1+x_1+x_2} + \cdots + \frac 1{1+x_1+\cdots + x_n} < \sqrt { \frac 1{x_1} + \frac 1{x_2} + \cdots + \frac 1{x_n}} . \]
[/hide]
[hide="10th Grade"]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31925]Problem 1[/url]
Let $n$ be a positive integer, $n\geq 2$. For each $t\in \mathbb{R}$, $t\neq k\pi$, $k\in\mathbb{Z}$, we consider the numbers
\[ x_n(t) = \sum_{k=1}^n k(n-k)\cos{(tk)} \textrm{ and } y_n(t) =  \sum_{k=1}^n k(n-k)\sin{(tk)}. \]
Prove that if $x_n(t) = y_n(t) =0$ if and only if  $\tan {\frac {nt}2} = n \tan {\frac t2}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31926]Problem 2[/url]
The base $A_1A_2\ldots A_n$ of the piramid $VA_1A_2\ldots A_n$ is a regular polygon. Prove that if
\[ \angle VA_1A_2 \equiv \angle VA_2A_3 \equiv \cdots \equiv \angle VA_{n-1}A_n \equiv VA_nA_1 , \] then the piramid is regular.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31927]Problem 3[/url]
a) Prove that there are no one-to-one (injective) functions $f:\mathbb{N} \to \mathbb{N}\cup \{0\}$ such that
\[ f(mn) = f(m)+f(n) , \ \forall \ m,n \in \mathbb{N}. \]

b) Prove that for all positive integers $k$ there exist one-to-one functions $f:\{1,2,\ldots,k\}\to\mathbb{N}\cup \{0\}$ such that $f(mn) = f(m)+f(n)$ for all $m,n\in \{1,2,\ldots,k\}$ with $mn\leq k$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31928]Problem 4[/url]
For $\alpha \in (0,1)$ we consider the equation $\{x\{x\}\}= \alpha$.

a) Prove that the equation has rational solutions if and only if there exist $m,p,q\in\mathbb{Z}$, $0<p<q$, $\gcd(p,q)=1$, such that $\alpha = \left( \frac pq\right)^2 + \frac mq$.
b) Find a solution for $\alpha = \frac {2004}{2005^2}$.
[/hide]
[hide="11th Grade"]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31929]Problem 1[/url]
Let $n\geq 2$ a fixed integer. We shall call a $n\times n$ matrix $A$ with rational elements a [i]radical[/i] matrix if there exist an infinity of positive integers $k$, such that the equation $X^k=A$ has solutions in the set of $n\times n$ matrices with rational elements.

a) Prove that if $A$ is a radical matrix then $\det A \in \{-1,0,1\}$ and there exists an infinity of radical matrices with determinant 1;
b) Prove that there exist an infinity of matrices that are not radical and have determinant 0, and also an infinity of matrices that are not radical and have determinant 1.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31939]Problem 2[/url]
Let $f:[0,1)\to (0,1)$ a continous onto (surjective) function.

a) Prove that, for all $a\in(0,1)$, the function $f_a:(a,1)\to (0,1)$, given by $f_a(x) = f(x)$, for all $x\in(a,1)$ is onto;
b) Give an example of such a function.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31930]Problem 3[/url]
Let $X_1,X_2,\ldots,X_m$ a numbering of the $m=2^n-1$ non-empty subsets of the set $\{1,2,\ldots,n\}$, $n\geq 2$. We consider the matrix $(a_{ij})_{1\leq i,j\leq m}$, where $a_{ij}=0$, if $X_i \cap X_j = \emptyset$, and $a_{ij}=1$ otherwise. Prove that the determinant $d$ of this matrix does not depend on the way the numbering was done and compute $d$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31937]Problem 4[/url]
Let $f:\mathbb{R}\to\mathbb{R}$ be a convex function.

a) Prove that $f$ is continous;
b) Prove that there exists an unique function $g:[0,\infty)\to\mathbb{R}$ such that for all $x\geq 0$ we have
\[ f(x+g(x)) = f(g(x))  - g(x) . \]
[/hide]
[hide="12th Grade"]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31931]Problem 1[/url]
Prove that the group morphisms $f: (\mathbb{C},+)\to(\mathbb{C},+)$ for which there exists a positive $\lambda$ such that $|f(z)| \leq \lambda |z|$ for all $z\in\mathbb{C}$, have the form
\[ f(z) = \alpha z + \beta \overline{z} \] for some complex $\alpha$, $\beta$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31932]Problem 2[/url]
Let $G$ be a group with $m$ elements and let $H$ be a proper subgroup of $G$ with $n$ elements. For each $x\in G$ we denote $H^x = \{ xhx^{-1} \mid h \in H \}$ and we suppose that $H^x \cap H = \{e\}$, for all $x\in G - H$ (where by $e$ we denoted the neutral element of the group $G$).

a) Prove that $H^x=H^y$ if and only if $x^{-1}y \in H$;
b) Find the number of elements of the set $\bigcup_{x\in G} H^x$ as a function of $m$ and $n$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31834]Problem 3[/url]
Let $f:[0,\infty)\to(0,\infty)$ a continous function such that $\lim_{n\to\infty} \int^x_0 f(t)dt$ exists and it is finite. Prove that
\[ \lim_{x\to\infty} \frac 1{\sqrt x} \int^x_0 \sqrt {f(t)} dt = 0. \]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31933]Problem 4[/url]
Let $A$ be a ring with $2^n+1$ elements, where $n$ is a positive integer and let
\[ M = \{ k \in\mathbb{Z} \mid k \geq 2, \ x^k =x , \ \forall \ x\in A \} . \]
Prove that the following statements are equivalent:

a) $A$ is a field;
b) $M$ is not empty and the smallest element in $M$ is $2^n+1$.
[/hide]

All the questions for the NMO are posted. The TSTs will follow tomorrow ;) Have fun! :P

PDF Version posted, topic de-stickyfied :P

This is definitely an old topic and my question may be off topic.. But I wanted to ask, which grade corresponds to the Olympiad level. 12 grade is really high(i don't even understand the terminlogy!!!)

What are these grades?? I'll appreciate your clarification!!!

[quote="Rushil"]What are these grades?? I'll appreciate your clarification!!![/quote]High-school grades (12th grade are students between 17-18 years old for example).

[quote="MysticTerminator"]although it is likely that by 12th grade, many of the students will have seen the mathematics used on the olympiad (group theory, for example), what is the point of putting it on there? IMO is strictly precalculus.[/quote]The Math Olympiad in Romania is not designed to be the IMO trambuline as in other countries, so it resembles the high-school curriculum until (inclusively) the national phase. The best in each grade are [b]then[/b] selected to participate in a series of 5-6 IMO Team Selection tests which determine the team.

By the way Valentin, do you know where NMO 2006 is going to take place?

[quote="DiaDi"]as far as I know it's going to be held at Iasi (am I wrong?).[/quote]I am not sure, but it is the most likely place to be hold.

Wel,if it isn't sure where it is going to be held,what about other places? :?


geometry
number theory
greatest common divisor
combinatorics unsolved
combinatorics

On a circle are given $ n$ distinct points. Let $ m$ be a positive integer with $ 3\leq m<\frac{n}{2}$ and $ (m,n)\equal{}1$. Each point is connected by a segment to the $ m$th point from it, counting counterclockwise. The obtained segments intersect in $ I$ different points inside the circle.

(a) Find the maximum value of $ I$ when the $ n$ points take different positions on the circle

(b) Prove that $ I\geq n$ and show that equality is possible for $ m\equal{}3$ and an arbitrary even number $ n\geq 8$.

$ (m, n)$ is common notation for the GCD of $ m$ and $ n$.


For a planar point set S, is the number of pointed pseudotriangulations always at least the number of triangulations?

 Will somebody help me?

i don't remember exactly what a pseudotriangulation is, but i think that each pseudotriangulation can be completed to a triangulation by tiangulating each pseudotriangle.
please post the exact definition of a pseudotriangulation (or an online reference)


geometry
3D geometry
sphere
calculus
trigonometry
ratio
area of a triangle

how do you find the volume of a cap given the radius of a sphere and the distance between the center of a sphere and the "cutting" plane that defines the cap region?

I don't think that is possible without calculus

It's easily done with calculus, but I don't think it's possible without.

If anyone wants to teach me a bit about 3D trig, I could easily find a solution utilizing that. (without calculus, assuming we don't consider 3D trig a calculus step).

Well I thought about moving from 2 dimensions to 3. We know that we can find the area of a region of circle cut off from a chord by subtracting the area of a triangle from a corresponding sector. This naturally makes you think of subtracting a cone (analogue of triangle) from a "lobe" (analogue of sector) of the sphere. However, finding the area of the lobe would be difficult.

What about this though. Suppose we have a circle of radius $r$ and a sector subtended by angle $\theta$. The area of the sector divided by the area of the circle is $\frac{\frac{1}{2}\theta r^2}{\pi r^2}=\frac{\theta}{2\pi}$. Now imagine a corresponding cap and sphere defined by rotating the figure. Maybe the ratio of the volume of the cap to the volume of the sphere is $(\frac{\theta}{2\pi})^2$ or $(\frac{\theta}{2\pi})^{\frac{3}{2}}$. Would that work?

The reason I brought up 3D trig was because it'd be easy with it. You see, a sphere is made up of $4\pi$ steradians. That is the surface area is $4\pi r^2$, kinda how a circle is made up of $2\pi$ radians because its circumference is $2\pi r$. So you see, if you could find the 3D angle of the top of the cone, it would become incredibly easy.

for volume of a spherical cap
[url=http://mathforum.org/dr.math/faq/formulas/faq.sphere.html]look here and scroll down a bit[/url]

[quote="drunner2007"]wow, that sounds really cool. so a steradian is the "angle" (i'm not sure what to call what I'm calling an "angle") that subtends a circle of area $r^2$ on the surface of the sphere?[/quote]

Well it's called a space angle. Think of a pyramid/cone/whatever (it doesn't have to be a circle. It could substend a spherical triangle/square/pentagon/anything). The top point is the space angle. If it substends an area of $r^2$, then, yes, it is 1 steradian.

About the acutal formula, that's really cool. And it's possible to get with the information we're given. Now I want to know how it's derived.

i believe that you can use the cavaleri principle to derive the volume of a spherical cap


It has been referred to as a genocide with the government helping in providing arms and staging joint attacks yet the UN does not move in because the Government does not want them to.

While Israel has help from the US, Lebanon has help from Hezbollah, no one dares to move in to Sudan when a genocide is occuring there?

Well Chad has gone to war with Sudan about a year ago, but I'm no expert on the subject.  They accused each other of supporting rebel groups.  Now they seem to have agreed on not doing that anymore.

In 1998, USA did do something against Sudan and bombed a factory in Khartoum (the main reason was their support for Osama Bin Laden who bombed two embassies).  It was said to be a factory producing medicines...

After the "black hawk down" incident in Somalia, I doubt USA will be very eager to do something about it.  In fact, I don't think many countries are.


I know that MikTex includes a lot of package files in the texmf\tpm\packages folder. When I try to build some code, and TeXnicCenter asks to install a certain package, and I direct it to the packages folder, it says "this does not appear to be a valid package repository." How do I fix this problem?

This is not a TeXnic Center problem, but a problem with TeX itself.  (TeXnic Center is merely an editor which happens to have some built-in TeX features.)

Assuming that you're running MikTeX under Windows, you need to run the MikTeX Package Manager (which is probably accessible using the Windows Start menu under All Programs -> MikTeX).  Then, in the Repository menu, select "Change Package Repository...", select "Internet" from the first screen, and choose a different site.  I use the tug.ctan.org FTP site and it seems to work fine.  There will then be a LONG pause as the system tries to reconfigure itself, but hopefully it will then work fine.


inequalities
trigonometry
inequalities unsolved

For positive real numbers $ a, b, c$, prove that

$ (a^{2}\plus{}b^{2})^{2} \ge (a\plus{}b\plus{}c)(a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)$.

If one of three numbers
$ (a \plus{} b \minus{} c)$, $ (b \plus{} c \minus{} a)$ and $ (c \plus{} a \minus{} b)$
is negative, we are done

If all three numbers
$ (a \plus{} b \minus{} c)$, $ (b \plus{} c \minus{} a)$ and $ (c \plus{} a \minus{} b)$ are positive, we can assume that $ a$ ,$ b$ and $ c$ are the sides of the triangle.
So
$ (a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)$
= 16$ S^2$=$ (2ab\sin c)^2$

It suffices to prove that
$ a^2  \plus{} b^2  \ge 2ab\sin c$

But which is true because
\[ \minus{} 1 \le \sin c \le 1\]
Th inequality is proved


Here is a problem that I composed.

Let $n \geq 2$ be an integer. Prove that there exists a positive integer $k \in \mathbb{N}$ such that $2k+1 \, | \, n \pm k!$ for a suitable choice of the $\pm$ sign.

Good luck with solving this problem!

Very nice problem. Let p a prime factor of $ n^2+1$. Then p is 1 modulo 4 and it-s easy to see from Wilson theorem that $ p| ((\frac{p-1}{2})!)^2+1 $. Now, take $k=\frac{p-1}{2}$. We have $ 2k+1| (k!)^2-n^2 $ since $ p| n^2+1, p| (k!)^2+1 $ and thus we have $ 2k+1| k!+n $ or $ 2k+1| k!-n $. I hope it's not another bad day and the solution is right.


Are you sure? It clearly diverges since $\lim_{n \rightarrow \infty}\cos{\frac{\pi}{2^{n}}}= 1$.

[quote="t0rajir0u"]I think you mean the product.

[hide="Huge hint"] $\cos \frac{x}{2}\cos \frac{x}{4}\cos \frac{x}{8}... = \frac{\sin x}{x}$ [/hide][/quote]

[hide="hope this is right..."]
Using that hint:
$\cos\frac{\pi}{4}\cos\frac{\pi}{8}\cdots=\frac{\sin\pi}{\pi\cos\frac{\pi}{2}}$
So I need to evaluate the limit of that
$\frac{\sin\pi}{\pi\cos\frac{\pi}{2}}=\frac{2\sin\frac{\pi}{2}\cos\frac{\pi}{2}}{\pi\cos\frac{\pi}{2}}=\frac{2\sin\frac{\pi}{2}}{\pi}=\boxed{\frac{2}{\pi}}$

Just wondering, how are you able to memorize identities like that?
[/hide]

Knowing the "proof" helps.

[hide="Demonstration"] Repeatedly applying the sine double-angle formula, we obtain the sequence of identities

\begin{eqnarray*}\sin x &=& 2 \sin \frac{x}{2}\cos \frac{x}{2}\\ &=& 4 \sin \frac{x}{4}\cos \frac{x}{4}\cos \frac{x}{2}\\ &=& 8 \sin \frac{x}{8}\cos \frac{x}{8}\cos \frac{x}{4}\cos \frac{x}{2}\\ &=& ... \\ &=& 2^{n}\sin \frac{x}{2^{n}}\cos \frac{x}{2^{n}}... \cos \frac{x}{2}\\ \end{eqnarray*}

Furthermore, $\lim_{n \to \infty}2^{n}\sin \frac{x}{2^{n}}= x$. [/hide]
The way you used the identity was not entirely correct.  Instead of using $x = \pi$, which makes both sides $0$, use $x = \frac{\pi}{2}$.

[quote="t0rajir0u"]
The way you used the identity was not entirely correct.  Instead of using $x = \pi$, which makes both sides $0$, use $x = \frac{\pi}{2}$.[/quote]

Oh duh.

[hide]
$\cos\frac{\pi}{4}\cos\frac{\pi}{8}\cos\frac{\pi}{16}\cdots=\frac{\sin\frac{\pi}{2}}{\pi/2}=\boxed{\frac{2}{\pi}}$
[/hide]


[quote="314.math"]Wolfram (http://mathworld.wolfram.com/Trapezoid.html) says that the height of a trapezoid is [img]http://mathworld.wolfram.com/images/equations/Trapezoid/NumberedEquation2.gif[/img].  What is the symbol in the numerator on the right hand side, and what does it stand for?

Thanks in advance![/quote]

I don't know what the name of that symbol is.  It is defined in the next
line of that webpage in terms of trapezoid side lengths a, b, c, and d.

Does anybody the answer for eta in this problem?

And one more question.
Do you know what will happen, if two diagonals of the trapezoid become perpendicular to each other?
I mean, will the above formula become simpler?
A book says in this case the height is:
h^2=AxB
h is the height, A and B are two parallel sides of the trapezoid.


I think a normal approach is to find k such that:
$A_{k-1} \le A_k \ge A_{k+1}$

Please post full solutions if you want .Do you agree?

[quote="silouan"]Let $A_k=\frac{2^k\cdot 1000!}{10^k\cdot k!(1000-k)!}$  for $k=1,2,...,1000$

For which $k$  the number $A_k$  is the biggest ?[/quote]

$\Rightarrow A_k = \left(\frac{1}{5}\right)^k \displaystyle\binom{1000}{k}$

Thus, $A_{k+1} = (A_k) \cdot \left(\frac{1}{5}\right) \cdot \left(\frac{1000-k}{k+1}\right)$.

So, we need to find an $n$ such that $\left(\frac{1}{5}\right) \cdot \left(\frac{1000-k}{k+1}\right)$ is more than 1 for $k = n$ and less than 1 for $k = n + 1$. This is essentially the switching off point for getting bigger and getting smaller.

$\frac{1000-k}{5k+5} \geq 1 \Rightarrow 995 \geq 6k \Rightarrow k \leq 165.83\ldots$
$\frac{1000-k}{5k+5} \leq 1 \Rightarrow 995 \leq 6k \Rightarrow k \geq 165.83\ldots$
(We can multiply by $5k+5$ because it will always be positive since $k$ must be between 0 and 1000)

So, when $k = 165$, the multiplier is greater than 1 and when $k = 166$ it is less than 1.

So, when we go from 1 to 2, 3 to 4, etc. all the way until we make the step from 165 to 166 each term gets bigger, but when we go from 166 to 167 and so on, each term gets smaller, so $A_{166}$ is the largest.


quadratics Vieta algebra polynomial function

[quote="atomicwedgie"][quote="ckck"]The roots of the equation $2x^{2}-x-1$ are r,s,t. The roots of $x^{3}+ax^{2}+bx+c=0$ are $-\frac{2}{r},-\frac{2}{s},-\frac{2}{t}$. Find a+b+c.[/quote]

How does a quadratic possess three roots?[/quote]
Oh, my bad. Its supposed to be a cubic.  :blush:

[quote="Art of Owna"]a= -2
b= 0
c= 8

Vieta's Theorm.[/quote]

What exactly is Vieta's Theorem?

Google, Mathworld, and Wikipedia remain your stalwart friends.

http://mathworld.wolfram.com/VietasFormulas.html

Edit:

[hide="Without using Vieta"] The polynomial

$P(x) = 2x^{3}-x-1$

Has roots $r, s, t$.  Therefore the function

$P \left(-\frac{2}{x}\right) =-\frac{16}{x^{3}}+\frac{2}{x}-1$

Has roots $-\frac{2}{r},-\frac{2}{s},-\frac{2}{t}$.  Multiplying by $-x^{3}$ we get the polynomial

$Q(x) = x^{3}-2x^{2}+16$

Hence $\boxed{ a =-2, b = 0, c = 16}$. [/hide]

[quote="t0rajir0u"]][quoteGoogle, Mathworld, and Wikipedia remain your stalwart friends.

http://mathworld.wolfram.com/VietasFormulas.html

Edit:

The polynomial

$P(x) = 2x^{3}-x-1$

Has roots $r, s, t$.  Therefore the function

$P \left(-\frac{2}{x}\right) =-\frac{16}{x^{3}}+\frac{2}{x}-1$

Has roots $-\frac{2}{r},-\frac{2}{s},-\frac{2}{t}$.  Multiplying by $-x^{3}$ we get the polynomial

$Q(x) = x^{3}-2x^{2}+16$

Hence $\boxed{ a =-2, b = 0, c = 16}$. [\quote][/quote]

hmm I got a=-2, b=0 and c=8, but I just realized that rst=1/2...so -8/(1/2)=-c so c=16..so your right.

[quote="ckck"]The roots of the equation $2x^{3}-x-1$ are r,s,t. The roots of $x^{3}+ax^{2}+bx+c=0$ are $-\frac{2}{r},-\frac{2}{s},-\frac{2}{t}$. Find a+b+c.[/quote]

[hide="solution"]$\text{We know that}\ a\ \text{is the negative of the sum of the roots which is}\ \frac{2}{r}+\frac{2}{s}+\frac{2}{t}\ \text{or}\ \frac{2(st+rt+rs)}{rts}.$

$\text{Also,}\ b\ \text{is the sums of pairs which is}\ \frac{4}{rs}+\frac{4}{st}+\frac{4}{rt}\ \text{or}\ \frac{4(r+s+t)}{rst}$.

$\text{Lastly,}\ c\ \text{is the negative of the product of the roots which is}\ \frac{8}{rst}$.

$\text{By Vieta's theorem,}\ r+s+t=0,\ rs+st+rt=-\frac{1}{2},\ \text{and}\ rst=\frac{1}{2}$

$\text{Plugging these values in gives us}\ \boxed{a=-2, b=0, \text{and}c=16}$[/hide]


Hi,

I guess I don't need to explain the problem  :D . Could an admin please reinitialize my section "My Favorites" please - which is empty this time, lol - ? Thank you.

Bah basically I decided to clean my favorites and right after I pushed the delete button the same problem as before occurred. If you grant me another reinitialization of the favorites  :blush:  :D  :ninja: , please leave it blank (the problem may have its source in the handling of a previous post).

Anyway ...  :D


group theory
abstract algebra
number theory
relatively prime
superior algebra
superior algebra solved

(1)If A & B are subgroups of finite index in a groups G & their indices are relatively prime then prove that G=AB
   (2) If H is a subgroup of finite index in a group G then H contains a subgroup N which is of finite index & normal in G

For the first one, I think it's pretty clear that the number of sets of the form $AxB$ is a divisor of both the indices, and is thus one, meaning that $G=AB$.

It's maybe the fourth time I see the second one posted on this forum :).

oops sorry has the 2nd probem been discussed before.If so where will I find it.


Find all functions:
$ f: N$→$ N$
$ f(f(m)\plus{}f(n))\equal{}m\plus{}n$

I remember seeing this problem somewhere in Mathlinks.
I think the only solution is $ f(x)\equal{}x$

This looks suspiciously like a USAMTS problem from this year, which I solved by setting f(0) = C, using that to prove the equation was somehow linear, then proving C=0.

edit: 1000th post

[hide="half solution"]
If f(0)=C, then f(f(0)+f(m))=m, or f(C+f(m))=m (1).  Then f(C+f(m))+C=m+C and f(f(C+f(m))+C)=f(m+C).  Using (1) on the LHS, we get C+f(m)=f(C+m) (2).  This shows that unless C=0, the function is periodic, with f(-C)=0, f(C)=2C.  But then plugging back into (1) gives f(C+0)=-C.  Therefore C=0.  Now (1) reduces to f(f(m))=m.

i forgot where to go from here
it might involve setting n=f(m) and using the original equation to get x=f(m)+m such that f(x)=x, then doing something tricky from there
[/hide]

double edit: the USAMTS problem was for all nonnegative reals, not just natural numbers if i remember correctly.  The answer to that one is f(x)=x only.

This is exactly the USAMTS problem; it *was* on naturals, not nonnegative reals. If it was nonnegative reals, there are solutions other than $ f(x) \equal{} x$.

[hide="My solution from that round"]

Let $ m \equal{} n \equal{} 0$. We see that $ 2f(0)$ is the sole root of $ f(x)$.

Now, add $ f(2f(0))$ to both sides, take $ f$ of both sides, then apply the original functional identity. We get $ f(m) \plus{} f(n) \plus{} 2f(0) \equal{} f(m \plus{} n) (*)$ - we can use the same technique to get $ f(m) \minus{} f(n) \plus{} 2f(0) \equal{} f(m \minus{} n) (')$. Let's rearrange $ (*)$ and take $ f$ of both sides of it. We get $ f(f(m) \plus{} f(n) \minus{} 2f(0)) \equal{} f^2(m \plus{} n)$. But applying $ (*)$ to the LHS of this equation, we get $ f^2(m \plus{} n) \plus{} 2f(0) \equal{} m \plus{} n$. Now suppose $ 2f(0) > 0$. (It can't be less than $ 0$ for obvious reasons.) Then, adding $ (*)$ and $ (')$, we have $ 2f(m) > f(m \plus{} n) \plus{} f(m \minus{} n) \plus{} 8f(0) > f(n) \plus{} f(m)$. Hence $ f(m) > f(n)$ for any $ m,n$; clearly impossible. Thus, $ 2f(0) \equal{} f(0) \equal{} 0$. This means that $ f$ being additive (by $ (*),(')$) and cyclic of order 2 is a necessary condition for $ f$ to satisfy the problem. It of course is also a sufficient condition, since if $ f$ is additive and an involution, then $ f(f(m) \plus{} f(n)) \equal{} f(f(m \plus{} n)) \equal{} m \plus{} n$. We now wrap up our proof with the following lemma:

Lemma: The only functions on the nonnegative integers which are additive are functions of the form $ f(x) \equal{} cx$.

Proof: Let $ f(0) \equal{} a$. Then $ f(1) \equal{} f(0 \plus{} 1) \equal{} a \plus{} f(1)$, so $ a \equal{} 0$. Also, $ f(x) \equal{} f(1 \plus{} 1 \plus{} \ldots \plus{} 1) \equal{} xf(1)$, and hence if we let $ f(1) \equal{} a$, then we have $ f(x) \equal{} cx$, as desired.

Now we have $ f(f(x)) \equal{} c^2x \equal{} x$ for all x, thus $ c \equal{} \pm 1$, hence the only functions of that form which satisfy our second restriction are $ f(x) \equal{} x, \minus{} x$; but the latter isn't mapped to naturals; therefore our only answer is $ \boxed{f(x) \equal{} x}$.

$ \mathbb{QED.}$[/hide]

[hide="Solution"]$ f(f(m) \plus{} f(n)) \equal{} m \plus{} n$

Let $ m, n \equal{} 0$

$ f(f(0) \plus{} f(0)) \equal{} 0 \plus{} 0$

$ f(2f(0)) \equal{} 0$

Let $ m, n \equal{} 2f(0)$

$ f(f(2f(0)) \plus{} f(2f(0))) \equal{} 2f(0) \plus{} 2f(0)$

$ f(0) \equal{} 4f(0)$

$ f(0) \equal{} 0$

Let $ m \equal{} 0$

$ f(f(0) \plus{} f(n)) \equal{} 0 \plus{} n$

$ f(f(n)) \equal{} n$

$ f(n) \equal{} n$ or $ f(n) \equal{} \minus{}n$

But as this question is posed on positive numbers only, the only solution is $ f(x) \equal{} x$.[/hide]

This is not the USAMTS problem. On the USAMTS, the function was from non-negative integers to non-negative integers. The problem here is from natural numbers (positive integers) to natural numbers, so any solutions that involve letting $ n\equal{}m\equal{}0$ do not apply.

Oh. I thought N = natural numbers = nonnegative integers there for a second.  :oops: I think it's a national differences thing.

[hide="Real solution"]If $ f(a) \equal{} f(b)$, then $ f(f(a) \plus{} f(a)) \equal{} f(f(b) \plus{} f(b)) \equal{} 2a \equal{} 2b$ and $ a \equal{} b$. So $ f$ is injective. We have $ f(f(n) \plus{} f(1)) \equal{} f(f(n \minus{} 1) \plus{} f(2))$, which by injectivity is $ f(n) \minus{} f(n \minus{} 1) \equal{} f(2) \minus{} f( 1)$. The finite difference of $ f$ is thus constant, and therefore by induction $ f(x) \equal{} ax \plus{} b$. Letting $ m \equal{} n \equal{} x$ in the original equation, $ a(2(ax \plus{} b)) \equal{} 2x$, or $ 2a^2 \equal{} 2$ and $ 2ab \equal{} 0$. Since $ a\neq0$, $ b \equal{} 0$. If $ a \equal{} \minus{} 1$, $ f(x) \le 0$ for all $ x$, so $ a \equal{} 1$ and $ \boxed{f(x) \equal{} x}$, which is what we wanted.[/hide]

The symbol $ \mathbb{N}$ is used both for positive integers and for nonnegative integers, i.e., it's an ambiguous notation.


Write the net ionic equation for the reaction of Cu (N3)2 with acid.

Experimentally, such a reaction should be performed with great care, since azides are shock sensitive. Well, treating Cu(N3)2 with aqueous acid should yield the complex Cu(H2O)6 and hydrazoic acid HN3.


$ LHS \equal{} 3\sum_{cyc}\frac {a^2}{bc} \plus{} 7\sum_{cyc}\frac {a}{b} \plus{} 17\sum_{cyc}\frac{b}{a}\ge \frac {9(a \plus{} b \plus{} c)^2}{3(ab \plus{} bc \plus{} ca)} \plus{} 21 \plus{} 51\ge 81$

[quote="filomen"][quote="alex2008"]$ LHS \equal{} 3\sum_{cyc}\frac {a^2}{bc} \plus{} 7\sum_{cyc}\frac {a}{b} \plus{} 17\sum_{cyc}\frac {b}{a}\ge \frac {9(a \plus{} b \plus{} c)^2}{3(ab \plus{} bc \plus{} ca)} \plus{} 21 \plus{} 51\ge 81$[/quote]

dear Alex2008, can you post full proof? Thanks[/quote]
It's the full proof, filomen.
Dear my friend, I hope that you'll not ask the questions like this.
Some friends of me are thinking that you are spamer  :wink:


I think I know what it is but I don't understand why it true and how I can use it.

I think it states that "many radical propogation occurs before they recombinate".

Is this referring to the kinetic study of chain reactions?

I guess so, the original question involves the mechanism behind the pyrolysis of 1,3-diphenylpropane.


For every $ n$-digit number, we take the product of its digits. Then we add all the products together. What is the result?

All what products now? I only see one  :)

[hide="hintie"]start from small numbers and you will see a pattern [/hide]

[quote="seamusoboyle"]So if we took 4 digits say we'd end up adding up (1)(1)(1)(1) + (1)(1)(1)(2) + (1)(1)(1)(3) + .... +(9)(9)(9)(9) ???[/quote]

Strictly speaking you should start with $ 1000$ but it doesn't change much.


The polynomial $W(x)=x^{2}+ax+b$, with $a, b$ being integers, is given. For every prime number $p$, there exist integer $k$ such that $W(k)$ and $W(k+1)$ are divisible by $p$. Prove, that there exist integer $m$, such that:
\[W(m)=W(m+1)=0\]

This is easy problem. From hypothesis $a$ is odd, write $a=-2m-1$.

For all primes $p$ , we have exist  integer $n$ such that $a\equiv-2n-1(\mod{p})$ and $b\equiv n(n+1)(\mod{p})$, therefore $p|4b+1-a^{2}$ and so $b=\frac{a^{2}-1}{4}=m(m+1)$, Q.E.D


In a collection of $ 2n\plus{}1$ persons, for any group of $ n$ persons, there exists a person, not in the group, who knows each person of the group. Prove that there exists a person who knows everybody.

http://www.mathlinks.ro/viewtopic.php?t=51
http://www.mathlinks.ro/viewtopic.php?t=4949
http://www.mathlinks.ro/viewtopic.php?t=75715

  darij


How many integer side lengths are possible to complete a triangle
in which the other sides measure 6 units and 3 units?

Let the other side be $ x$.
$ 3\plus{}x>6 \iff x>3$
$ 6\plus{}3>x \iff x>9$
$ x\equal{}4,5,6,7,8 \Rightarrow$ 5 possibilities.

There's a rule for the lengths of triangle sides. The length of one side cannot be more or equal to the sum of the other two sides. So the last side cannot be $ 6 \plus{} 3 \equal{} 9$ or higher. The upper bound is $ 8$.

$ 6$ can't be more than $ 3$ added to the other side, which we will call $ x$. So $ 3 \plus{} x  > 6$, and we find that $ x > 3$. The lower bound is $ 4$. The possible lengths are: $ 8, 7, 6, 5, \text{and } 4$. We count these, and there are $ \boxed{5 \text{ possibilities}}$.


Το παρακάτω πρόβλημα μου γεννήθηκε σε σχετικά άσχετη στιγμή (pun intended :P) αλλά ίσως το βρείτε ενδιαφέρον. H απάντηση με εξέπληξε :)

Έστω $ C(\mathbb{R})$ το συνόλο των συνεχών συναρτήσεων $ f: \mathbb{R}\to\mathbb{R}$. Ποιός ο πληθυκός αριθμός του συνόλου αυτού?

Προς αποφυγή παρεξηγήσεων ορολογίας κλπ, αρκεί φυσικά να βρεθεί μια 1-1 αντιστοιχία του παραπάνω συνόλου με ένα "γνωστό" σύνολο, όπως π.χ. οι φυσικοί $ \mathbb{N}$, οι πραγματικοί $ \mathbb{R}$, το δυναμοσύνολο $ 2^\mathbb{R}$ των πραγματικών, κ.ο.κ.

Ελπίζω να βρείτε το πρόβλημα ενδιαφέρον. Cheers,

Durandal 1707

PS: Ως πρώτο πρόβλημα που γράφω ποτέ στο forum, αφιερωμένο στον δάσκαλό μου r_boris :)

[quote="Durandal"]Το παρακάτω πρόβλημα μου γεννήθηκε σε σχετικά άσχετη στιγμή (pun intended :P) αλλά ίσως το βρείτε ενδιαφέρον. H απάντηση με εξέπληξε :)

Έστω $ C(\mathbb{R})$ το συνόλο των συνεχών συναρτήσεων $ f: \mathbb{R}\to\mathbb{R}$. Ποιός ο πληθυκός αριθμός του συνόλου αυτού?

Προς αποφυγή παρεξηγήσεων ορολογίας κλπ, αρκεί φυσικά να βρεθεί μια 1-1 αντιστοιχία του παραπάνω συνόλου με ένα "γνωστό" σύνολο, όπως π.χ. οι φυσικοί $ \mathbb{N}$, οι πραγματικοί $ \mathbb{R}$, το δυναμοσύνολο $ 2^\mathbb{R}$ των πραγματικών, κ.ο.κ.

Ελπίζω να βρείτε το πρόβλημα ενδιαφέρον. Cheers,

Durandal 1707

PS: Ως πρώτο πρόβλημα που γράφω ποτέ στο forum, αφιερωμένο στον δάσκαλό μου r_boris:)[/quote]
Μια πρώτη ιδέα είναι : Τουλάχιστον όσες το R Γιατί αν σε κάθε σταθερή (συνεχής) αντιστοιχίσουμε έναν πραγματικό ,την τιμή της , υπάρχει αμφιμονοσήμαντη και επί αντιστοιχία του ζητούμενου συνόλου στο R Πιστεύω ότι δεν είναι περισσότερες από το R  αλλά ???

Καλωσόρισες στο υπέροχο αυτό forum ευχαριστώ για την αφιέρωση!!!

Σίγουρα ναι: $ \vert C(\mathbb{R})\vert \geq \vert\mathbb{R}\vert \equal{} \mathfrak{c}$.

Βάζω μια υπόδειξη μια και το ανέφερες...

[hide]Η παραπάνω προφανής ανισότητα ισχύει πράγματι ως [i]ισότητα[/i]: $ \vert C(\mathbb{R})\vert \equal{} \mathfrak{c}$ (εγώ έχασα μία μέρα επειδή ήμουν πεπεισμένος ότι η ανισότητα ήταν γνήσια, grrrrrr.... :mad:). Οπότε, το πρόβλημα είναι ισοδύναμο με το να κατασκευαστεί μία απεικόνιση $ \mathbb{R}\to C(\mathbb{R})$ που να είναι επί.

Η δομή του $ \mathbb{R}$ δίνει τη λύση :)

Επίσης, πρέπει να πω ότι βρίσκω πολύ εντυπωσιακή την τελική απάντηση ότι $ \vert C(\mathbb{R}) \vert \equal{} \mathfrak{c}$: δεν είχα ιδέα ότι οι συνεχείς συναρτήσεις είναι τόσο, μα τόσο σπάνιες (ή, ισοδύναμα, ότι οι πραγματικοί αριθμοί είναι τόσο, μα τόσο πολλοί)... :!:
[/hide]

Καλώς σας βρήκα στο forum! Cheerio,

Durandal 1707

Vasika den mporw na katalavw ti ennoeis me ton oro plithikos...Pantws genika oi sinexis sinartiseis einai para poli liges se sxesi me oles tis sinartiseis pou iparxoune....kai afto fenete proxira apo to gegonos oti o xoros C(R) den einai Banach me tin sinithi normiki .....

SorrowMan, με τον όρο πληθυκός αριθμός αναφέρομαι στο [url=http://en.wikipedia.org/wiki/Cardinality]cardinality[/url] ενός συνόλου, δηλαδή στον "αριθμό" των στοιχείων του. Επίτηδες δε θέλησα να δώσω έναν ακριβή ορισμό του πληθάριθμου επειδή δε χρειαζόταν για το πρόβλημα και επειδή δεν ήθελα να το μπλέξω. Sorry αν η διατύπωσή μου δεν ήταν αρκετά σαφής και προκάλεσε προβλήματα, ήλπιζα ότι η περαιτέρω περιγραφή του προβλήματος και η συζήτηση με τον r_boris θα αρκούσε για να γίνει ξεκάθαρο το τι χρειάζεται να αποδειχθεί.

Το "ζουμί" και το ωραίο κομμάτι του προβλήματος (και, βασικά, ο λόγος που πιστεύω ότι το κάνει ενδιαφέρον) είναι να δείξεις ότι υπάρχει μία αμφιμονοσήμαντη απεικόνιση $ C(\mathbb{R})\to\mathbb{R}$.

Μετά το post του r_boris, ουσιαστικά αρκεί να κατασκευάσεις μία απεικόνιση που να είναι "επί" (αυτό είναι το δύσκολο και όμορφο κομμάτι!).

Το ότι οι συνεχείς συναρτήσεις είναι ελάχιστες σε σχέση με όλες τις δυνατές συναρτήσεις $ f\in\mathbb{R}^\mathbb{R}$ είναι σίγουρα εμφανές, έχεις απόλυτο δίκιο. Ωστόσο, το ότι είναι [i]τόσο[/i] λίγες ώστε να τίθενται σε "1-1" αντιστοιχία με τους πραγματικούς αριθμούς... αυτό είναι που με εντυπωσιάζει ;)

Cheerio,

Durandal 1707

Hmmmm, ίσως χρειάζεται μία υπόδειξη το πρόβλημα...

[hide="Hint"]Αν $ f_1,f_2$ συνεχείς συναρτήσεις και $ f_1(r) \equal{} f_2(r)\,\forall r\in\mathbb{Q}$, τότε $ f_1\equiv f_2$. Άρα το πεδίο ορισμού των συνεχών συναρτήσεων είναι στην πραγματικότητα πολύ μικρότερο από το $ \mathbb{R}$.[/hide]
Cheerio,

Durandal 1707

Ok, αυτή είναι η λύση που είχα στο μυαλό μου:

[hide]
Προφανώς, μία συνεχής συνάρτηση $ f: \mathbb{R}\to\mathbb{R}$ καθορίζεται πλήρως από τις τιμές τις πάνω στους ρητούς αριθμούς $ \mathbb{Q}$. Δηλαδή, το σύνολο των συνεχών συναρτήσεων $ \mathbb{R}\to\mathbb{R}$ είναι υποσύνολο του συνόλου [i]όλων[/i] των απεικονίσεων $ \mathbb{Q}\to\mathbb{R}$. Έτσι, αν δείξουμε ότι το σύνολο αυτών των απεικονίσεων είναι το πολύ της τάξης του συνεχούς, τελειώσαμε...

Με τη σειρά τους, οι ρητοί αριθμοί τίθενται σε αμφιμονοσήμαντη αντιστοιχία με τους φυσικούς αριθμούς (π.χ. αν απλώσουμε το $ \mathbb{Q}\equal{}\mathbb{N}^2$ στο επίπεδο, μπορούμε να ξεκινήσουμε από την αρχή $ (0,0)$ και να διαγράψουμε όλο το δίκτυο με μία σπείρα που να μεγαλώνει).

Συνεπώς, αρκεί να δείξουμε ότι το σύνολο των απεικονίσεων $ \mathbb{N}\to\mathbb{R}$ (δηλ. των ακολουθιών) τίθεται σε αμφιμονοσήμαντη αντιστοιχία με τους πραγματικούς αριθμούς. Κοντεύουμε...

Προφανώς, είτε μιλάμε για ακολουθίες με τιμές στο $ \mathbb{R}$ είτε μιλάμε για ακολουθίες με τιμές στο $ (0,1)$, είναι το ίδιο πράγμα. Γι'αυτό, έστω μία ακολουθία $ {a_n}$ με τιμές στο $ (0,1)$ και ας την γράψουμε στη δεκαδική της αναπαράσταση:

\[ a_{1}	\equal{}	0.d_{1}^{1}d_{1}^{2}d_{1}^{3}\ldots d_{1}^{n}\ldots\notag\\
a_{2}	\equal{}	0.d_{2}^{1}d_{2}^{2}d_{2}^{3}\ldots d_{2}^{n}\ldots\notag\\
		\ldots\notag\\
a_{n}	\equal{}	0.d_{n}^{1}d_{n}^{2}d_{n}^{3}\ldots d_{n}^{n}\ldots\notag\\
		\ldots\]
Τώρα, ας αρχίσουμε να διαγράφουμε τα ψηφία της ακολουθίας με έναν zig-zag τρόπο ώστε να φτάσουμε στον πραγματικό αριθμό:
\[ a \equal{} 0.\,d_{1}^{1}\,\,d_{2}^{1}d_{1}^{2}\,\,d_{1}^{3}d_{2}^{2}d_{3}^{1}\,\,d_{4}^{1}d_{3}^{2}d_{2}^{3}d_{1}^{4}\ldots\in (0,1)\]
Αυτός ο αριθμός περιέχει όλην την ακολουθία (!!!) (ok, πρέπει να προσέξουμε λίγο για παγίδες του στυλ 0.99999... = 1, αλλά who cares ;) ). Έτσι, μόλις κατασκευάσαμε μία επί απεικόνιση από τους πραγματικούς αριθμούς στό σύνολο των ακολουθιών, ο.ε.π.

Συνοψίζοντας (και με το συμβολισμό ότι $ A^B$ είναι το σύνολο των απεικονίσεων $ B\to A$), αυτό που κάναμε είναι το εξής:
\[ \vert\mathbb{R}\vert \leq \vert C(\mathbb{R}) \vert \leq \left\vert\mathbb{R}^\mathbb{Q}\right\vert \equal{} \left\vert\mathbb{R}^\mathbb{N}\right\vert \equal{} \left\vert (0,1)^\mathbb{N}\right\vert \leq \vert\mathbb{R}\vert\]
[/hide]

Άρα, οι συνεχείς συναρτήσεις έχουν ακριβώς το πλήθος του συνεχούς, δηλ. των πραγματικών αριθμών. Ελάχιστες, eh? :)

Cheerio,

Durandal 1707

[quote="Durandal"]Ok, αυτή είναι η λύση που είχα στο μυαλό μου:

[hide]
Προφανώς, μία συνεχής συνάρτηση $ f: \mathbb{R}\to\mathbb{R}$ καθορίζεται πλήρως από τις τιμές τις πάνω στους ρητούς αριθμούς $ \mathbb{Q}$. Δηλαδή, το σύνολο των συνεχών συναρτήσεων $ \mathbb{R}\to\mathbb{R}$ είναι υποσύνολο του συνόλου [i]όλων[/i] των απεικονίσεων $ \mathbb{Q}\to\mathbb{R}$. Έτσι, αν δείξουμε ότι το σύνολο αυτών των απεικονίσεων είναι το πολύ της τάξης του συνεχούς, τελειώσαμε...

Με τη σειρά τους, οι ρητοί αριθμοί τίθενται σε αμφιμονοσήμαντη αντιστοιχία με τους φυσικούς αριθμούς (π.χ. αν απλώσουμε το $ \mathbb{Q} \equal{} \mathbb{N}^2$ στο επίπεδο, μπορούμε να ξεκινήσουμε από την αρχή $ (0,0)$ και να διαγράψουμε όλο το δίκτυο με μία σπείρα που να μεγαλώνει).

Συνεπώς, αρκεί να δείξουμε ότι το σύνολο των απεικονίσεων $ \mathbb{N}\to\mathbb{R}$ (δηλ. των ακολουθιών) τίθεται σε αμφιμονοσήμαντη αντιστοιχία με τους πραγματικούς αριθμούς. Κοντεύουμε...

Προφανώς, είτε μιλάμε για ακολουθίες με τιμές στο $ \mathbb{R}$ είτε μιλάμε για ακολουθίες με τιμές στο $ (0,1)$, είναι το ίδιο πράγμα. Γι'αυτό, έστω μία ακολουθία $ {a_n}$ με τιμές στο $ (0,1)$ και ας την γράψουμε στη δεκαδική της αναπαράσταση:
\[ a_{1} \equal{} 0.d_{1}^{1}d_{1}^{2}d_{1}^{3}\ldots d_{1}^{n}\ldots\notag \\
a_{2} \equal{} 0.d_{2}^{1}d_{2}^{2}d_{2}^{3}\ldots d_{2}^{n}\ldots\notag \\
\ldots\notag \\
a_{n} \equal{} 0.d_{n}^{1}d_{n}^{2}d_{n}^{3}\ldots d_{n}^{n}\ldots\notag \\
\ldots
\]
Τώρα, ας αρχίσουμε να διαγράφουμε τα ψηφία της ακολουθίας με έναν zig-zag τρόπο ώστε να φτάσουμε στον πραγματικό αριθμό:
\[ a \equal{} 0.\,d_{1}^{1}\,\,d_{2}^{1}d_{1}^{2}\,\,d_{1}^{3}d_{2}^{2}d_{3}^{1}\,\,d_{4}^{1}d_{3}^{2}d_{2}^{3}d_{1}^{4}\ldots\in (0,1)
\]
Αυτός ο αριθμός περιέχει όλην την ακολουθία (!!!) (ok, πρέπει να προσέξουμε λίγο για παγίδες του στυλ 0.99999... = 1, αλλά who cares ;) ). Έτσι, μόλις κατασκευάσαμε μία επί απεικόνιση από τους πραγματικούς αριθμούς στό σύνολο των ακολουθιών, ο.ε.π.

Συνοψίζοντας (και με το συμβολισμό ότι $ A^B$ είναι το σύνολο των απεικονίσεων $ B\to A$), αυτό που κάναμε είναι το εξής:
\[ \vert\mathbb{R}\vert \leq \vert C(\mathbb{R}) \vert \leq \left\vert\mathbb{R}^\mathbb{Q}\right\vert \equal{} \left\vert\mathbb{R}^\mathbb{N}\right\vert \equal{} \left\vert (0,1)^\mathbb{N}\right\vert \leq \vert\mathbb{R}\vert
\]
[/hide]

Άρα, οι συνεχείς συναρτήσεις έχουν ακριβώς το πλήθος του συνεχούς, δηλ. των πραγματικών αριθμών. Ελάχιστες, eh? :)

Cheerio,

Durandal 1707[/quote]
[b]ΘΑΥΜΑΣΙΑ απόδειξη Μπράβο Παναγιώτη[/b]


A regular tetrahedron is cut off each vertex of a cube in such a way that each original face of the cube becomes a regular octagon. The length of an edge of the cube is 1. Find the length of an edge of the new 14-sided solid.

A. $ \frac{2\minus{}\sqrt2}{2}$
B. $ \frac{1}{3}$
C. $ \frac{2\sqrt2\minus{}2}{3}$
D. $ \sqrt2\minus{}1$
E. Not Here

[hide="solution"]Let $ x$ denote the length of the side of a tetrahedron. We have $ 2x \plus{} x\sqrt {2} \equal{} 1\Rightarrow x \equal{} \frac {2 \minus{} \sqrt {2}}{2}$. So the answer is $ 1 \minus{} \frac {2 \minus{} \sqrt {2}}{2} \equal{} \sqrt {2} \minus{} 1$, which is D.[/hide]

edit: fixed.

[hide]I keep getting $ 1 \minus{} \frac {20 \minus{} 14\sqrt {2}}{12} \equal{} \frac {7\sqrt {2} \minus{} 4}{6}\Rightarrow E$, but I never check my work, so I probably did the arithmetic incorrectly.[/hide]

Edit: Whoops, side length? That makes everything a few billion times easier....

You can't cut off a regular tetrahedron (angle on a face being 60 degrees) off the vertices of a cube (face angle being 90 degrees). This question was on the [url=http://www.artofproblemsolving.com/Wiki/index.php/2007_AMC_12A_Problems/Problem_20]2007 AMC 12A[/url], the solution is there.

[hide="The answer is .."]
$ \frac {10 \minus{} 7\sqrt {2}}{3}\ \mathrm{E}$
[/hide]

Edit: Wait, the question asks for a side length, not the volume of the solid! The answer is then

[hide="Really .."]
Let an edge length of the solid be $ x$. Then on any face of the cube, the cut off tetrahedrons create 45-45-90 triangles, so one side of the face we get $ 2\frac{x}{\sqrt{2}} \plus{} x \equal{} 1 \Longrightarrow x \equal{} \frac{1}{1 \plus{} \sqrt{2}} \equal{} \sqrt{2} \minus{} 1\ \mathrm{D}$
[/hide]

(I hate reading errors ..)

Oops.. I thought it was volume too...  :D