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Source: IMO Shortlist 2000, Problem N2

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orl

#1 h Sep 6, 2003, 7:56 PM • 4 Y

Y by Adventure10, centslordm, Mango247, MarioLuigi8972

For a positive integer $n$ , let $d(n)$ be the number of all positive divisors of $n$ . Find all positive integers $n$ such that $d(n)^3=4n$ .

This post has been edited 2 times. Last edited by djmathman, Jul 22, 2016, 11:16 PM

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#2 h Oct 30, 2005, 8:03 AM • 3 Y

Y by centslordm, Adventure10, Adventure10

Of course $n > 1$ . So let $n = \prod_{i=1}^r p_i^{a_i}$ , where $r$ is a positive integer, $p_1 < p_2 < \ldots < p_r$ are pairwise distinct natural primes and $a_i \in \mathbb{Z}^+$ , for any $i = 1, 2, \ldots, r$ . We need $\prod_{i=1}^r (1 + a_i)^3 = 4\prod_{i=1}^r p_i^{a_i}$ , and $v_2(\mbox{LHS}) = v_2(\mbox{RHS}) > 0$ , where $v_2(\cdot)$ is the $2$ -adic valuation of its argument.

Well, actually $3 \mid v_2(\mbox{LHS})$ , so a fortiori $p_1 = 2$ and $a_1 = 3b_1 + 1$ , for a certain $b_1 \in \mathbb{N}$ . Moreover similarly $a_i = 3b_i$ , where $b_i \in \mathbb{Z}^+$ , if $i = 2, 3, \ldots, r$ . If $r = 1$ , this leads uniquely to $2+3b_1 = 2^{b_1+1}$ and $b_1 = 1$ or $b_1 = 2$ , viz $n = 2$ or $n = 2^7$ . Hence let $r > 1$ . You get $(2+3b_1)\prod_{i=2}^r (1+3b_i) = 2^{b_1 + 1} \prod_{i=2}^r p_i^{b_i}$ . So necessarily $p_2 \geq 5$ , as $\mbox{LHS} \equiv 2 \bmod 3$ .

If $b_1 = 1$ , this means $5\cdot \prod_{i=2}^r (1+3b_i) = 2^2 \prod_{i=2}^r p_i^{b_i}$ , leading to $p_2 = 5$ , $b_2 = 1$ and $\prod_{i=3}^r (1 + 3b_i) = \prod_{i=3}^r p_i^{b_i}$ , which is possible iff both the products in the left and right sides are empty, since by Bernoulli's inequality: $p^a > (1+3)^a \geq (1+3a)$ , for any $a \in \mathbb{N}$ and any real $p > 4$ . The same argument proves there's no further solution for $b_1 \geq 2$ , because that implies $2+3b_1 \leq 2^{b_1 + 1}$ , by induction. Then $d^3(n) = 4n$ iff $n = 2$ , $n = 2^7$ or $n = 2^4 \cdot 5^3$ .

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#3 h Jun 13, 2006, 4:30 PM • 2 Y

Y by centslordm, Adventure10

hitlEULER wrote:

Of course $n > 1$ . So let $n = \prod_{i=1}^r p_i^{a_i}$ , where $r$ is a positive integer, $p_1 < p_2 < \ldots < p_r$ are pairwise distinct natural primes and $a_i \in \mathbb{Z}^+$ , for any $i = 1, 2, \ldots, r$ . We need $\prod_{i=1}^r (1 + a_i)^3 = 4\prod_{i=1}^r p_i^{a_i}$ , and $v_2(\mbox{LHS}) = v_2(\mbox{RHS}) > 0$ , where $v_2(\cdot)$ is the $2$ -adic valuation of its argument.

Well, actually $3 \mid v_2(\mbox{LHS})$ , so a fortiori $p_1 = 2$ and $a_1 = 3b_1 + 1$ , for a certain $b_1 \in \mathbb{N}$ . Moreover similarly $a_i = 3b_i$ , where $b_i \in \mathbb{Z}^+$ , if $i = 2, 3, \ldots, r$ . If $r = 1$ , this leads uniquely to $2+3b_1 = 2^{b_1+1}$ and $b_1 = 1$ or $b_1 = 2$ , viz $n = 2$ or $n = 2^7$ . Hence let $r > 1$ . You get $(2+3b_1)\prod_{i=2}^r (1+3b_i) = 2^{b_1 + 1} \prod_{i=2}^r p_i^{b_i}$ . So necessarily $p_2 \geq 5$ , as $\mbox{LHS} \equiv 2 \bmod 3$ .

If $b_1 = 1$ , this means $5\cdot \prod_{i=2}^r (1+3b_i) = 2^2 \prod_{i=2}^r p_i^{b_i}$ , leading to $p_2 = 5$ , $b_2 = 1$ and $\prod_{i=3}^r (1 + 3b_i) = \prod_{i=3}^r p_i^{b_i}$ , which is possible iff both the products in the left and right sides are empty, since by Bernoulli's inequality: $p^a > (1+3)^a \geq (1+3a)$ , for any $a \in \mathbb{N}$ and any real $p > 4$ . The same argument proves there's no further solution for $b_1 \geq 2$ , because that implies $2+3b_1 \leq 2^{b_1 + 1}$ , by induction. Then $d^3(n) = 4n$ iff $n = 2$ , $n = 2^7$ or $n = 2^4 \cdot 5^3$ .

Here you have used fortori 2-adic what does these words mean
Abdurashid

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#4 h Jul 13, 2006, 7:45 AM • 3 Y

Y by centslordm, Adventure10, Seungjun_Lee

Well $p$ -adic numbers are completion of $\mathbb Q$ with the metric $\rho(x,y)=\parallel x-y\parallel _{p}$ which if $x=\frac{a}{b}$ which $(a,b)=1$ then if $p|a$ $\parallel x\parallel _{p}=\parallel a\parallel _{p}$ if not then $\parallel x\parallel _{p}=-\parallel b\parallel _{p}$ . You could see they are like infinite digit numbers $\overline{\dots a_{2}a_{1}a_{0}.b_{1}b_{2}\dots b_{k}}$

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va2010

#5 h May 11, 2015, 6:23 PM • 3 Y

Y by centslordm, Adventure10, Mango247

The intended solution is probably like the following. Let $f(x) = \frac{d(x)^3}{x}$ . Observe that $f(x)$ is multiplicative. Also, $n$ is two times a perfect cube. Hence, $v_2(n) \equiv 1 \pmod{3}$ and $v_p(n) \equiv 0 \pmod{3}$ for odd primes $p$ .

Lemma: $v_3(n) = 0$

Proof: Assume otherwise, then using multiplicativity of $f(x)$ , an exponent of $3$ gets introduced to the denominator, and in fact it is not possible for this to be eliminated. $\blacksquare$ .

Now we can determine through tedious casework that $f(x) < 1$ if $x$ is a prime power of a prime greater than $3$ . Hence, the only way for a prime beyond this threshold to work is if $f(2^k)$ makes up for it. We can determine that $f(2^7) = 4$ and $f(2) = 4$ , and $f(2^{3k+1}) < 4$ for $k > 2$ . For $k=1$ , the only other possibility is multiplication by $5^3$ . Hence the solutions are $n = 2^7$ , $n = 2$ , $n = 2^4 * 5^3$ , which all work.

Comments: For similar problems, observe HMIC 2015.1, Spring OMO 2014.19, and Canada 1999.3

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#6 h Jul 22, 2016, 4:04 AM • 3 Y

Y by centslordm, Adventure10, Mango247

I believe the problem should read as follows,

"For a positive integer $n$ , let $d(n)$ be the number of all positive divisors of $n$ . Find all positive integers $n$ such that $d(n)^3=4n$ ."

I would appreciate if a moderator of this forum or site admin could edit the OP to match it so it'll read correctly on the contests section as well

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efang

#7 h Jul 22, 2016, 3:37 PM • 4 Y

Y by samuel, centslordm, Adventure10, EquationTracker

p-adic evaluations are not necessary for this problem; you can solve it with some pretty silly bounds

Clearly, the prime factorization of $n$ must be of the form $n = 2^{3x-2}*\prod_{i=1}^r p_i^{3e_i}$ .

Now from $d(n) = \sqrt[3]{4n}$ we have $(3x-1)*\prod_{i=1}(3e_i+1) = 2^x*\prod_{i=1}^r p_i^{e_i}$

First, the LHS of this equation clearly has no factors of 3 and thus none of the $p_i$ of the RHS can be 3. (1)

Second, $\forall p_i > 4$ we must have $p_i^{e_i} > 3e_i+1$ (easy to prove with induction) (2)

Third, for $x > 3$ we have $2^x > 3x-1$

Thus, we must have $x \le 3$ .

Case 1: $x=1$

Then, we have $2^x = 3x-1$ . Thus, from (2) we conclude that there can be no other prime factors in this case. Thus we have $n = 2^{3x-2} = 2$

Case 2: $x=2$

Then, we have $2^x < 3x-1$ . This makes it possible to have other prime factors. Notice that $3x-1 = 5$ thus 5 must be one of the prime factors. So we let $p_1 = 5$ . Now, if $e_1 > 1$ then we have that $2^2*5^{e_1} > 5(3e_1+1)$ . Thus, concluding from (2) again we see that $e_1 = 1$

When $e_1 = 1$ we have equality. Thus, by observation (2) again n cannot have any additional prime factors and this must be the only solution in this case.

The solution is $n = 2^{3x-2}*5^{3e_1} = 2^4*5^3$

Case 3: $x=3$

In this case we have $2^x = 3x-1$ and again like in case 1 we may conclude that n has no other prime factors. Thus the only solution in this case is $n = 2^{3x-2} = 2^7$

Thus, our solutions are $\boxed{2, 2^4*5^3, 2^7}$

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#8 h Feb 20, 2018, 6:40 PM • 7 Y

Y by UzbekMathematician, Not_real_name, Mathasocean, v4913, centslordm, Adventure10, Mango247

The answer is $n=2$ , $n=128$ , and $n = 2000$ . These are checked to work.

In general, we can let $n = 2^{3r+1} \cdot p_1^{3e_1} \dots p_k^{3e_k}$ where $p_i$ are distinct odd primes, and $e_i > 0$ . We assume $n > 1$ throughout (as $n=1$ fails). Then $d(n)^3 = 4n$ rearranges as $2^{r+1} p_1^{e_1} \dots p_k^{e_k} = (3r+2)(3e_1+1) \dots (3e_k+1).$ We now split into several cases.

Suppose $r = 0$ or $r = 2$ , so the equation becomes $p_1^{e_1} \dots p_k^{e_k} = (3e_1+1) \dots (3e_k+1).$ Then $e_i \neq 1$ for all $i$ , since the right-hand side is odd.

But then $p_i^{e_i} \ge 3^{e_i} > 3e_i + 1$ for $e_i \ge 2$ . Multiplying all these gives a contradiction unless $k = 0$ . These two cases give $n = 2 \cdot 1^3 = 2$ and $n = 2 \cdot 4^3 = 128$ respectively.
Suppose $r = 1$ , so we have Then is among the $p_i$ , so let us assume $p_1 = 5$ . We have two cases:
- Suppose $e_1 = 1$ . Then the equation becomes $p_2^{e_2} \dots p_k^{e_k} = (3e_2+1) \dots (3e_k+1)$ . Then and as in the first case, we get $e_i \neq 1$ for any $i \ge 2$ , and thus there can be no other primes. This gives the solution $n = 2 \cdot 10^3 = 2000$ .
- Now suppose $e_1 \ge 2$ . Then we multiply the estimates $\begin{align*} 5^{e_1} &\ge \tfrac{25}{7} (3e_1 + 1) \quad \text{since } e_i \ge 2\\ 3^{e_i} &\ge \tfrac34(3e_i+1) \quad \text{if } p_i = 3 \\ p_i^{e_i} &\ge 3e_i +1 \qquad \text{if } p_i \ge 7. \end{align*}$ to get a contradiction since $\frac{25}{7} \cdot \frac{3}{4} > \frac54$ .
Now suppose $r \ge 3$ . We then multiply the estimates $\begin{align*} 2^{r+1} &\ge \tfrac{16}{11} (3r+2) \qquad r \ge 3 \\ 3^{e_i} &\ge \tfrac34(3e_i+1) \quad \text{if } p_i = 3 \\ p_i^{e_i} &\ge 3e_i+1 \qquad \text{if } p_i \ge 5. \end{align*}$ This gives another contradiction since $\frac{16}{11} \cdot \frac34 > 1$ .

This post has been edited 1 time. Last edited by v_Enhance, Feb 20, 2018, 6:59 PM

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#9 h Aug 14, 2019, 3:02 PM • 3 Y

Y by centslordm, Adventure10, Mango247

A little tedious casework, but otherwise decent problem... here's the sketch of my proof.
First observe that, given $n=p_1^{\alpha_1}p_2^{\alpha_2}...p_k^{\alpha_k}$ , $p_1<p_2<...<p_k$ , $p_1$ must equal $2$ and $\alpha_1=3 \beta_1 +1$ for some integer $\beta_1$ - using similar logic we get $\alpha_i =3 \beta_i$ for $2 \leq i \leq k$ . Thus our equation simplifies to:
$2^{\beta_1+1}p_2^{\beta_2}p_3^{\beta_3}...p_k^{\beta_k}=(3 \beta_1+2)(3\beta_2+1)(3 \beta_3 +1)...(3 \beta_k+1)$
Here, a really important thing to reduce unnecessary casework is noticing that $p_2 \neq 3$ (trivial because $3$ can't divide LHS)- this will now be extremely useful because after playing around with the problem a bit, we can immediately see there'd be a heavy usage of bounding in the problem.

The pure "number theoretical" part of the problem is now over, and all that's left is a little casework... first we see that $\frac{p_i^{\beta_i}}{3 \beta_i +1} > 1$ for $p_i \geq 5$ (the proof is trivial), hence we can directly analyse when $\frac{2^{\beta_1+1}}{3 \beta_1 +2} \leq 1$ (as we need $\frac{2^{\beta_1+1}p_2^{\beta_2}p_3^{\beta_3}...p_k^{\beta_k}}{(3 \beta_1+2)(3\beta_2+1)(3 \beta_3 +1)...(3 \beta_k+1)}=1$ ), and from this we get $\beta_1 \leq 2$ . Now plugging in $\beta_1=2$ directly gives you the only solution of its type as $2^{3 \cdot 2 +1}=2^7$ , $\beta=0$ again gives an only solution as $n=2$ . We have no choice but to put in $\beta_1=1$ and that'll again give an only solution (with a little more calculation) as $2^{3+1}5^3=2000$ , and we're done.

This post has been edited 2 times. Last edited by ubermensch, Aug 14, 2019, 4:15 PM

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#10 h Oct 1, 2019, 7:31 PM • 3 Y

Y by centslordm, Adventure10, Mango247

Solution

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#11 h May 18, 2020, 9:34 PM • 4 Y

Y by centslordm, Mango247, Mango247, Mango247

Let $\tau(n)$ denote $d(n)$
Then, let's say that $n=2t^3$ , then we have that $\tau(2t^3)^3=8t^3$ , that implies $\tau(2t^3)=2t$ .Now let's say that $t=p_1^{\alpha_1}p_2^{\alpha_2}\dots p_k^{\alpha_k}$
We encounter two cases.The first case is when $2 \mid t$ , let's say that the primes $p_1,p_2,\dots,p_k$ are all ordered in the following way $p_1 < p_2 < p_3 < \cdots p_k$ .Then we have that $p_1=2$ , from here we have the relation:
$(3\alpha_1+2)\prod_{i=2}^{k} (3\alpha_i +1) = 2^{\alpha_1+1}\prod_{i=1}^{k}p_i^{\alpha_i}$ We first examine the inequality $2^{\alpha_1+1} > 3\alpha_1+2$ , we see that this holds for every $\alpha_1 > 2$ , so we will examine two cases.

The first case is when $\alpha_1 = 1$ , then we have that:
$5\prod_{i=2}^{k}(3\alpha_i+1) = 4\prod_{i=1}^{k}p_i^{\alpha_i}$ We will have that one prime is $5$ , and since for every prime number greater than $5$ we have the following inequality:
$p^{a} > (3a+1)$ This holds for any $a \in N$ we choose, now we have the following:
$\prod_{i=2}^{k} (3\alpha_i+1)=4.5^{\alpha_2-1}\prod_{i=2}^{k} p_i^{\alpha_i}$ $(3\alpha_2+1)=4.5^{\alpha_2-1}$ From here we see that $\alpha_2=1$ , and from all of this we get that $n=2^4.5^3=2000$

The second case is $\alpha_1=2$ , then we have that:
$8\prod_{i=2}^{k}(3\alpha_i+1)=2^3\prod_{i=2}^{k}p_i^{\alpha_i}$ $\prod_{i=2}^{k}(3\alpha_i+1)=\prod_{i=2}^{k}p_i^{\alpha_i}$ Because of the inequality $p^a>3a+1$ , for every prime greater than $5$ we have that $p_2=3$ , but this won't hold up since we get $2 \equiv 0 \; (mod \; 3)$ .So $3 \nmid t$ , we have that $n=2.2^6=128$

The second general case is $2 \nmid t$ , but this is easy since of the inequality $p^a > 3a+1$ , we get that any prime greater than $5$ doesn't divide $t$ , and especially we get that $3 \nmid n$ , so we have that $t=1$ , and so we get that $n=2$

To summarize the only solutions are $n \in \{ 2,128,2000 \}$ . . .

This post has been edited 1 time. Last edited by EulersTurban, May 18, 2020, 9:49 PM

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#12 h Aug 21, 2020, 3:01 PM • 1 Y

Y by centslordm

The only solutions are $n = 2, 128, 2000$ , which are all seen to work. Since $2n$ is a perfect cube, we may write
$n = 2^{3a -2} p_1^{3e_1} p_2^{3e_2} \dots p_k^{3e_k} .$ where $p_1 < p_2 < \cdots < p_k$ and $a, e_1, e_2, \dots, e_k$ are positive integers.Then the $d(n)^3 = 4n$ is equivalent to
$(3a - 1)^3 (3e_1+1)^3 (3e_2 + 1)^3 \dots (3e_k+1)^3 = 2^{3a} p_1^{3p_1} p_2^{3p_2} \dots p_k^{3p_k} .$ We split into cases based on the value of $a$ .

If $a=1$ , then
$(3e_1 + 1)^3 (3e_2+1)^3 \dots (3e_k+1)^3 = p_1^{3p_1} p_2^{3p_2} \dots p_k^{3p_k}.$ However, it is easily checked $p_i^{3e_i} > (e_i + 1)^3$ for all $e_i \geq 1$ whenever $p_1 \geq 3$ .Hence the only solution in this case is $n = 2$ .
If $a = 2$ , then
$5^3 (3e_1 + 1)^3 (3e_2 + 1)^3 \dots (3e_k + 1)^3 = 2^6 p_1^{3e_1} p_2^{3e_2} \dots p_k^{3e_k}.$ Since 3 does not divide the left hand side, but 5 does, we must have $p_1 = 5$ . Clearly $n = 2^4 \cdot 5^3$ works. Now if there exist $e_i \geq 1$ with $i \geq 2$ , then we may apply our work from the previous case to deduce that $d(n)^3 < 4n$ . Thus $n = 16 \cdot 5^e$ . Then the equation simplifies to
$125 (3e_1+1)^3 = 64 5^{3e} .$ The right hand side increases much faster than the left hand side, so there is only one solution $e=1$ . (Increasing $e$ by 1 multiplies the left right hand side by 125, but the left hand side increases by a factor of at most 8.) Thus the only solution is $n = 2^4 \cdot5 ^3 = 2000$
If $a = 3$ , then
$(3e_1 + 1)^3 (3e_2+1)^3 \dots (3e_k+1)^3 = p_1^{3p_1} p_2^{3p_2} \dots p_k^{3p_k}.$ which collapses into the case $a = 2$ . Hence the only solution here is $n = 2^{7} = 128$ .
If $a \geq 4$ , then
$(3a-1)^3 < 2^{3a} .$ This implies
$(3e_1 + 1)^3 (3e_2 + 1)^3 \dots (3e_k + 1)^3 < p_1^{3e_1} p_2^{3e_2} \dots p_k^{3e_k},$ which we have already shown to be impossible.

Having exhausted all possible cases, we may conclude that these are the only solutions.

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#13 h Nov 18, 2020, 2:53 PM • 1 Y

Y by centslordm

Let $n=p_1^{\alpha_1}p_2^{\alpha_2}...p_k^{\alpha_k}$ for $p_1<p_2<...<p_k$ where $p_1=2$ . As $4n$ is a perfect cube, we write $\alpha_1=3 \beta_1 -2$ and $\alpha_i =3 \beta_i$ for $2 \leq i \leq k$ .
So our final equation is
$2^{\beta_1}p_2^{\beta_2}p_3^{\beta_3}...p_k^{\beta_k}=(3 \beta_1-1)(3\beta_2+1)(3 \beta_3 +1)...(3 \beta_k+1)$ We notice that $p_2 \neq 3$ since $3$ does not divide the RHS. Now we rearrange our equation in this format:
$\frac{2^{\beta_1}}{(3{\beta_1}-1)}=\frac{(3{\beta_2}+1)}{p_2^{\beta_2}}\cdot\frac{(3{\beta_3}+1)}{p_3^{\beta_3}}\cdots\frac{(3{\beta_k}+1)}{p_k^{\beta_k}}$ If $p_2=5$ , we can see that $5^{\beta_2}=25^{(\beta_2)/2}=(1+24)^{(\beta_2)/2}\geq(1+12\beta_2)$ by Bernoulli's inequality. Thus the first fraction on RHS is $\leq1$ . Similarly we can prove this for other primes. Hence the RHS is $\leq1$ . Thus we obtain that $\beta_1={1,2,3}$ . For each of the cases, we get $n=128,2,2000$ respectively [after some calculation]

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#14 h Nov 19, 2020, 9:04 PM • 1 Y

Y by centslordm

orl wrote:

For a positive integer $n$ , let $d(n)$ be the number of all positive divisors of $n$ . Find all positive integers $n$ such that $d(n)^3=4n$ .

Let $n=\prod_{i=1}^{r} p_i^{a_i}$ then $d(n)=\tau(n)=\prod_{i=1}^{r}(a_i+1)$ now
$(\prod_{i=1}^{r}(a_i+1))^3=4*\prod_{i=1}^{r}p_i^{a_i}\implies p_i\neq 3$ when $p_k\neq 2, 3|a_k\implies a_k=3*b_k$ when $p_k=2, a_k\equiv 1\mod 3\implies a_k=3*b_k+1$
Now let for all $p_i\neq 3,b_i\geq 1$ then $(3*b_i+1)<p_i^{3*b_i}$ and if $p_k=2$ then for all $b_k>3$ or $3*b_k+1>10$ then $(3*b_k+2)^3<2^{3b_k+1}$ so let
$n=2^{3b_1+1}*\prod_{i=2}^{r}p_i^{3*b_i}$
Clearly as mentioned above $b_1<3$ is only possible..
Case 1- $b_1=0$ we have
$2^3*\prod_{i=2}^{r}(3*b_i+1)=4*2*\prod_{i=2}^{r}p_i^{3*b_i}$ so clearly in this case $n=2$ is only solution.
Case 2- $b_1=1$ then we have
$5^3*\prod_{i=2}^{r}(3*b_i+1)=4*2^4*\prod_{i=2}^{r}p_i^{3*b_i}$ so clearly there must exist $p_k=5, a_k=3$ then
$5^3*4^3*\prod_{i=3}^{r}(3*b_i+1)=4*2^4*5^3*\prod_{i=3}^{r}p_i^{3*b_i}$ which $\prod_{i=3}^{r}(3*b_i+1)=\prod_{i=3}^{r}p_i^{3*b_i}$ for $p_i>5$ which is clear contradiction so $n$ has only 1 prime divisor $2$ . Hence only solution in this case is $n=2^4*5^3$
Case 3-) $b_i=2$ then $8^3*\prod_{i=2}^{r}(3*b_i+1)=4*2^7*\prod_{i=2}^{r}p_i^{3*b_i}$ we get
$\prod_{i=2}^{r}(3*b_i+1)=\prod_{i=2}^{r}p_i^{3*b_i}$ for $p_i>3$ this is clear contradiction so $n$ has only 1 prime divisor $2$ , Hence in this case only solution is $n=2^7$
Case 4- $b_i=3$ then $11^3*\prod_{i=2}^{r}(3*b_i+1)=4*2^{10}*\prod_{i=2}^{r}p_i^{3*b_i}$
But $5^3*\prod_{i=2}^{r}(3*b_i+1)< 4*2^4*\prod_{i=2}^{r}p_i^{3*b_i}$ so this case has no solution!!
Hence $n\in\{2, 2^4*5^3, 2^7\}$ $\blacksquare$

This post has been edited 4 times. Last edited by Maths_1729, Nov 20, 2020, 6:40 AM

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#15 h Nov 30, 2020, 7:36 AM • 3 Y

Y by centslordm, Mango247, Mango247

The answer is $\boxed{n = 2, 2^7, 2^45^3}$ .

From the given equation, we can express $n$ as $n = 2^{3\alpha + 1}p_1^{3\alpha_1}\cdots p_k^{3\alpha_k}.$
$d(n)^3 = 4n$ turns into $\left((3\alpha + 2)(3\alpha_1 + 1)\cdots (3\alpha_k + 1)\right)^3 = 2^{3\alpha+3} p_1^{3\alpha_1} \cdots p_k^{3\alpha_k} \qquad \qquad (1)$ or $(3\alpha + 2)(3\alpha_1 + 1)\cdots (3\alpha_k +1) = 2^{\alpha + 1}p_1^{\alpha_1}\cdots p_k^{\alpha_k}. \qquad \qquad (2)$
From $(1)$ , note that $3 \nmid n$ . In particular, $p_i \ne 3$ for any $i$ .

Consider $\alpha = 0$ . From $(2)$ , we have $(3\alpha_1 + 1)\cdots (3\alpha_k + 1) = p_1^{\alpha_1}\cdots p_k^{\alpha_k}.$ Dividing by the LHS gives $1 = \frac{p_1^{\alpha_1}\cdots p_k^{\alpha_k}}{(3\alpha_1 + 1)\cdots (3\alpha_k + 1)}. \qquad \qquad (3)$
If $k = 0$ , the RHS is $1$ , giving $n=2$ .

If $k \geq 1$ , $p_i \geq 5$ , so $\frac{p_i^{\alpha_i}}{3\alpha_i + 1} \geq \frac{5}{3(1) + 1} = \frac{5}{4}. \qquad \qquad (4)$ This is indeed the minimum value since the numerator increases faster (exponentially) than the denominator (linear) as $\alpha_i$ increases. Clearly, the RHS of $(3)$ is then greater than $1$ , and thus there are no solutions for $k \geq 1$ .

Now, let $\alpha = 1$ in $(2)$ . This gives, after dividing, $\frac{5}{4} = \frac{p_1^{\alpha_1}\cdots p_k^{\alpha_k}}{(3\alpha_1 + 1)\cdots (3\alpha_k + 1)}.$ The minimum of $(4)$ only occurs at $p_i = 5, \alpha_i = 1$ . Thus, the equation above is forced to be $\frac{5}{4} = \frac{5^1}{3(1) + 1},$ giving $n = 2^45^3$ .

Lastly, let $\alpha \geq 2$ in $(2)$ . We find $1 \leq \frac{2^3}{3(2) + 2}\frac{p_1^{\alpha_1}\cdots p_k^{\alpha_k}}{(3\alpha_1 + 1)\cdots (3\alpha_k + 1)} = \frac{p_1^{\alpha_1}\cdots p_k^{\alpha_k}}{(3\alpha_1 + 1)\cdots (3\alpha_k + 1)},$ which is similar to $(3)$ with equality $k=0$ , giving $n = 2^7$ .

Having exhausted all cases, we are done. $\blacksquare$

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#16 h Jul 2, 2021, 7:35 PM • 2 Y

Y by RedFlame2112, centslordm

We claim that $n=2,128,2000$ are the answers. These all work, so we prove that these are the only ones. First, notice that $4n$ must be a perfect cube, so let $n=2k^3$ . Now the condition changes to $d(2k^3)=2k$ , and we would still like to solve for all positive integer $k$ .

Suppose that $x=v_2(k)$ , and let $l=\frac{k}{2^x}$ . Note that $l$ is odd. This means that the equation reduces to $d(2^{3x+1}l^3)=2\cdot 2^x\cdot l \Rightarrow (3x+2)d(l^3)=2^x\cdot l\Rightarrow \frac{d(l^3)}{l}=\frac{2^{x+1}}{3x+2}.$ Now if we let $l=p_1^{e_1}p_2^{e_2}\ldots p_q^{e_q}$ , for odd primes $p_i$ and positive integer exponenets $e_i$ , we get $\prod_{i=1}^q \frac{3e_i+1}{p_i^{e_i}}=\frac{2^{x+1}}{3x+2}.$ Notice that when $e_i>1$ , the expression $\frac{3e_i+1}{p_i^{e_i}}$ is always less than one, and when $e_i=1$ , $p_i=3$ is the only way to make the expression greater than one. Thus the maximum value of the LHS is $\frac{4}{3}$ . If $x>2$ , then $RHS>\frac{4}{3}$ , so we must have $x\le 2$ .

When $x=0$ or $x=2$ , the RHS is equal to $1$ . Then either $l=1$ , which works, or we must have $p_i=3, e_i=1$ , causing a term in the product to be $\frac{4}{3}$ . All other combinations are less than one. The only other term that does not cause the product to go below $1$ is $\frac{4}{5}$ when $p_i=5, e_i=1$ , and it doesn't work, so $l=1$ is the only solution in this case.

When $x=1$ , the RHS is equal to $\frac{4}{5}$ . Notice that $l=5$ only creates a term of $\frac{4}{5}$ and therefore works. As every other possible product term is less than $\frac{4}{5}$ except for $\frac{4}{3}$ , we must include $1$ factor of $3$ if we are to find any other solutions. The next two largest terms that don't involve the prime $3$ are $\frac{4}{5}$ from $p_i=5, e_i=1$ and $\frac{4}{7}$ for $p_i=7, e_i=1$ . However, combining these terms with $\frac{4}{3}$ gives us a product less than $\frac{4}{5}$ . Thus we can only include one more term, and checking the terms shows us that this is not possible.

Therefore we have three solutions:

1. When $l=1, x=0$ , we get $k=1$ , implying $n=2$ .
2. When $l=1, x=2$ we get $k=4$ , implying $n=128$ .
3. When $l=5, x=2$ we get $k=20$ , implying $n=2000$ .

The proof is complete.

Remarks: This is the first ISL problem I've seen that includes a nonobvious use of the year number into it.

I forgot that $n$ (and therefore $k$ and therefore $l$ ) cannot be a multiple of $3$ , it would have made the casework a lot quicker.

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oVlad

#17 h Oct 21, 2021, 8:48 AM

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To begin with, observe that $n=2k^3.$ Hence, our equation rewrites as $2k=\tau(2k^3).$

Let $k=2^a\cdot 3^b\cdot 5^c\cdot d$ and assume that $d\neq 1.$ Now, observe that for all $p\geq 7$ and $m\geq 1,$ $p^m\geq 7(3m+1)/4.$

Moreover, since $f(x)=2^{x+1}/(3x+2)$ and $g_i(x)=i^x/(3x+1), \ i\in\{3,5\}$ reach their minimums in $x=1,1$ and $0$ respectively, we have $\frac{2k}{\tau(2k^3)}=\frac{2^{a+1}}{3a+2}\cdot \frac{3^b}{3b+1}\cdot \frac{5^c}{3c+1}\cdot \frac{d}{\tau(d^3)}\geq \frac{4}{5}\cdot \frac{3}{4}\cdot 1\cdot \frac{7}{4}=\frac{21}{20}>1$ which is an obvious contradiction. Therefore, $k$ is not divisible by any prime number greater than $5.$ Let $k=2^a\cdot 3^b\cdot 5^c$ as before.

Then, we have $2^{a+1}\cdot 3^b\cdot 5^c = (3a+2)(3b+1)(3c+1)$ so clearly, $b=0.$ Thus, $2^{a+1}\cdot 5^c=(3a+2)(5c+1).$

Some more simple, crude, boundings will yield the solutions $n=2,128,2000.$

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mihaig

#18 h Oct 24, 2021, 4:00 AM

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Bravo tinere

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#19 h Feb 13, 2022, 9:34 PM

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ISL marabot solve

Since $n=1$ fails, assume $n>1$ . Let the prime factorization of $n=2^{3k+1}\cdot p_1^{3a_1}\cdot p_2^{3a_2}\cdots p_m^{3a_m}$ , where $p_i$ are distinct primes greater than $2$ .

The equation is $2^{k+1} \cdot p_1^{a_1}\cdot p_2^{a_2}\cdots p_m^{a_m}=(3k+2)(3a_1+1)(3a_2+1)\cdots(3a_m+1)$ .

Case 1: $k=0$ or $k=2$ .
Then $p_1^{a_1}\cdot p_2^{a_2}\cdots p_m^{a_m}=(3a_1+1)(3a_2+1)\cdots (3a_m+1)$ .

Note that $a_i\ne 1$ because the LHS is odd. For $a_i\ge 2$ , we get $p_i^{a_i}\ge 3^{a_i}>3a_i+1$ . This gives a contradiction unless $m=0$ . In this case, $n=2^1=\boxed{2}$ or $n=2^7=\boxed{128}$ . Both work.

Case 2: $k=1$ .
Then we get $4p_1^{a_1}p_2^{a_2}\cdots p_m^{a_m}=5(3a_1+1)(3a_2+1)\cdots(3a_m+1)$ . Clearly, it's not divisible by $3$ and is divisible by $5$ , so $p_1=5$ .

Note that for $i>1$ , we have $p_i^{a_i}>5^{a_i}>3a_i+1$ for all positive integers $a_i$ . So we have $4\cdot 5^{a_1}\le 5(3a_1+1)$ . This implies $a_1=1$ and $m=1$ . So $n=16\cdot 5^3=\boxed{2000}$ , which works.

Case 3: $k\ge 3$ .
We get $2^{k+1}\ge \frac{16}{11}(3k+2)$ . Since both sides are not divisible by $3$ , we get $p_i^{a_i}>5^{a_i}>3a_i+1$ for all positive integers $a_i$ . This is a contradiction as it implies LHS>RHS.

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awesomeming327.

#20 h Mar 19, 2022, 3:55 AM

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Sus

$n=\frac{d(n)^3}{4},$ so if $k=\frac{d(n)}{2}$ then $n=2k^3$ where $k$ is an integer. Let $n=2^{3e_0+1}\cdot 3^{3e_1}\cdot 5^{3e_2}\cdot \ldots\cdot p_n^{3e_n}$ where $p_i$ denotes the $i$ th odd prime, $e_i$ are non-negative integers but $e_n>0$ , and $p_n$ is the largest odd prime in the prime factorization of $n$ .

Now, the original equation gives $(3e_0+2)(3e_1+1)(3e_2+1)\cdot \ldots\cdot (3e_n+1)=2^{e_0+1}\cdot 3^{e_1}\cdot 5^{e_2}\cdot\ldots\cdot p_n^{e_n}.$ Note that for $p_i\ge 5,$ $p_i^{e_i}>3e_i+1$ for all $e_i\ge 1.$ Thus, $2^{e_0+1}3^{e_1}\le (3e_0+2)(3e_1+0).$ It is not hard to see that the only solutions of this are:

$(e_0,e_1)=(0,0),(0,1),(1,0),(1,1),(2,0),(2,1).$ For each of those cases, with simple bounding, the answer is $2,128,2000$

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#21 h Jul 29, 2022, 3:26 AM

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The answers are $n=2$ and $n=128$ . (Proof is rushed)

First, note that $n=2k^3$ for some integer $k$ . Then note that every exponent $e_p$ of every prime $p$ in the prime factorization of $n$ satisfies $e_p\neq 2\pmod 3$ which means that $3\nmid d(n)$ so that $3\nmid n$ . Now suppose that
$k=2^{e_2}\cdot 5^{e_5}\cdot 7^{e_7}\cdots$ Then define $f(n)=\frac{d(n)^3}{n}$ and note that $f(2)=4$ . Suppose we start from $1$ and multiply by $p^{v_p(k)}$ until we reach $k$ , counting the value of $f(n)$ . It is important to note that $f(n)$ can't decrease on every single multiplication. For primes $p\ge 5$ note that multiplying by any power $p^{e_p}$ will decrease the value of $f$ , so any increase must come from the power of $2$ . Again, simple bounding gives that any increase must occur when $e_2=0$ or $e_2=2$ , with both making $f$ unchanged, meaning that we can't use any primes $p\ge 5$ . Our only solutions are $n=2$ and $n=128$ .

ACTUALLY, if $e_2=1$ , we get that we can't have $e_p>0$ for $p\ge 7$ , so a finite case check gives $e_5=1$ which implies $n=2000$ is a final solution.

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lrjr24

#22 h Aug 8, 2022, 3:53 AM

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We claim the answers are $n=2$ , $n=128$ , or $n=2000$ .
Note that $n=2k^3$ , so $d(2k^3)=2k$ . Let $k=2^{e_1}3^{e_2}5^{e_3} \cdots p_i^{e_i}$ . If there is a prime $p \ge 7$ dividing $k$ , then $\frac{4}{5} \cdot \frac{3}{4} \cdot \frac{p}{4} \le \frac{2^{e_1+1}}{3e_1+2}\frac{3^{e_2}}{3e_2+1} \cdots \frac{p_i^{e_i}}{3e_i+1} = 1,$ a contradiction. This means $k=2^a3^b5^c$ . Note that we need to have $\frac{2^{a+1}}{3a+2}\frac{3^b}{3b+1}\frac{5^b}{5b+1}=1$ . Note that the minimum possible value of this expression is $\frac{3}{5}$ . This means if one of the values is over $\frac{5}{3}$ we get a contradiction. We now track the values each fraction can achieve less than $\frac{5}{3}$ . We have that $\frac{2^{a+1}}{3a+2}$ can achieve $1,\frac{4}{5},1, \frac{16}{11}$ , $\frac{3^b}{3b+1}$ can achieve $1, \frac{3}{4}, \frac{9}{7}$ , and $\frac{5^c}{3c+1}$ can achieve $1, \frac{5}{4}$ . If $c=0$ , then the only way to achieve reciprocating fractions is if they are both $1$ , so we get $k=1$ and $k=4$ . If $c=1$ , the only way to get the product of $\frac{4}{5}$ is through $a=1$ and $b=0$ , so $k=10$ . This gives all the solutions and proves there is no more so we are done.

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#24 h Mar 24, 2023, 7:29 PM

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Sketch

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vsamc

#25 h Apr 9, 2023, 5:57 PM

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Solution

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Sagnik123Biswas

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Sagnik123Biswas

#26 h Jan 15, 2024, 6:01 PM

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This problem can be cracked by making a series of observations.

Consider $d(n)^3 = 4n$ . This means that at a structural level, $n = 2j^3$ for some positive integer $j$ . Thus we know that $d(n)^3 = 4n = 8j^3 \implies d(2j^3) = 8j^3 \implies d(2j^3) = 2j$ .

Now we can arbitrarily think of $j = 2^{e_2}3^{e_3}5^{e_5}7^{e_7} \dots$ . Then note that $d(2j^3) = d( 2^{3e_2+1}3^{3e_3}5^{3e_5}7^{3e_7} \dots) = (3e_2 + 2)(3e_3 + 1)(3e_5 + 1) \dots \equiv 2 \pmod{3}$ . So if $d(2j^3) \equiv 2 \pmod{3}$ for all $j$ it must be true that $2j \equiv 2 \pmod{3} \implies j \equiv 1 \pmod{3}$ .

$j = 1$ obeys $d(2) = 2$ . Now consider odd $j \geq 7$ . Its prime factorization assumes the form $5^{e_5}7^{e_7}11^{e_{11}} \dots$ . If we look at $d(2j^3) = 2(3e_5 + 1)(3e_7 + 1)(3e_{11}+1)\dots$ . This needs to equal $2j = 2 \times 5^{e_5}7^{e_7}11^{e_{11}} \dots$ implying that $(3e_5 + 1)(3e_7 + 1)(3e_{11}+1)\dots = 5^{e_5}7^{e_7}11^{e_{11}} \dots$ . But note that $p^m > 3m + 1$ for $m \geq 1$ and $p \geq 5$ . So this would not be possible.

Thus it must hold that if $j > 1$ , it is even. Now let us begin considering $v_2(j)$ . If $v_2(j) = 1$ , then we now that $j = 2 \times 3^{e_3}5^{e_5}7^{e_7} \dots$ . Then $d(2j^3) = 5(3e_5 + 1)(3e_7 + 1) \dots = 2j = 4 \times 5^{e_5}7^{e_7} \dots$ . This only holds if $e_5 = 1$ .

If $v_2(j) = 2$ , we see that the only valid option is no other primes dividing $j$ since $d(2j^3) = 8 \times (3e_5 + 1)(3e_7 + 1)(3e_{11} + 1) \dots = 2j = 8 \times 5^{e_5}7^{e_7}11^{e_{11}}$ . Since $p^m > 3m + 1$ for $m \geq 1$ and $p \geq 5$ , it must hold that $e_5 = e_7 = e_{11} = 0$ .

If $v_2(j) = e_2 \geq 3$ , then $3e_2 + 2 < 2^{e_2 + 1}$ not allowing equality.

So $j = 1, 4, 10 \implies n = 2j^3 \implies n = 2, 128, 2000$

This post has been edited 1 time. Last edited by Sagnik123Biswas, Jan 16, 2024, 1:30 AM
Reason: Add some clarity

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pie854

#27 h Sep 29, 2024, 8:58 PM

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Not a very nice problem

Let $p_k$ be the $k$ th prime. Let $n=\prod_{i=1}^\infty p_i^{a_i}$ , where only a finite number of $a_i$ are nonzero. So we have $(a_1+1)^3(a_2+1)^3(a_3+1)^3(a_4+1)^3\cdots=2^{a_1+2}3^{a_2}5^{a_3}7^{a_4} \cdots.$ Note that this implies $3\mid a_1+2$ , $3\mid a_2$ , $3\mid a_3$ , et cetera. We now present a lemma.

Lemma: If $k\geq 0$ , $3\mid k$ and $p\geq 5$ then $p^k \geq (k+1)^3$ and equality holds when $k=0$ .
Proof. Easy. ////

Case 1: $a_2=0$ , $a_1=1$
We have $(a_3+1)^3(a_4+1)^3(a_5+1)^3\cdots =5^{a_3}7^{a_4}11^{a_5}\cdots$ For this to hold we must have $a_3=a_4=a_5=\dots=0$ due to the lemma. Thus it follows that $n=2$ .

Case 2: $a_2=0$ , $a_1=4$
We have $5^3(a_3+1)^3(a_4+1)^3 \cdots =2^6 5^{a_3}7^{a_4} \cdots.$ Note that this implies $a_3\geq 3$ . Then we have the inequality $2^6 5^{a_3} \geq 5^3 (a_3+1)^3.$ With this and the lemma it follows that $a_3=3$ , $a_4=a_5=\dots=0$ . So $n=2^4\cdot 5^3=2000$ .

Case 3: $a_2=0$ , $a_1\geq 7$
Then we have the inequality $2^{a_1+2}\geq (a_1+1)^3$ with equality when $a_1=7$ . And the lemma gives $a_3=a_4=\dots=0$ . So $n=2^7=128$ .

Case 4: $a_2=3$
We have $2^6(a_1+1)^3(a_3+1)^3 \cdots=2^{a_1+2}3^3 5^{a_3} 7^{a_4}\cdots \qquad (1)$ so that $a_1\geq 4$ .
If $a_1=4$ then we have $5^3(a_3+1)^3 \cdots=3^3 5^{a_3} 7^{a_4}\cdots \qquad (2)$ So $a_3\geq 3$ , but if $a_3=3$ then by mod $2$ there is a contradiction. So $a_3\geq 6$ in which case we must have $3^3 5^{a_3} > 5^3 (a_3+1)^3.$ This inequality and the lemma imply that $(2)$ cannot hold.
If $a_1=7$ then it's easy to get a contradiction mod $2$ in $(1)$ .
So, $a_1\geq 10$ . Then we have the inequality $2^{a_1+2}3^3>2^6(a_1+1)^3$ . So once again using the lemma it follows that $(1)$ can't hold.

Case 5: $a_2\geq 6$
The inequality $2^{a_1+2}3^{a_2}>(a_1+1)^3(a_2+1)^3$ holds and so, by the lemma, there's no solution in this case.

We've dealt with all possible cases, so we're done.

This post has been edited 1 time. Last edited by pie854, Sep 29, 2024, 9:15 PM

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ezpotd

#28 h Oct 15, 2024, 4:58 AM

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I claim the answers are only $2,128,2000$ . It is easy to see that all of these work.

Observe the function $\frac{d(n)^3}{n}$ is multiplicative over prime powers. Clearly, $d(n)$ and $n$ are even. Analyzing the exponents of the primes in $n$ , we see they are all divisible by $3$ except the power of $2$ , which has exponent $1$ mod $3$ . By the number of divisors formula, this gives $d(n) \equiv 2 \mod 3$ , so $d(n)$ and $n$ both do not divide $3$ .

We now consider the power of $2$ in $n$ . Note that for all prime powers other than the powers of $2$ up to $2^10$ , some of the powers of $3$ (which don't matter, since $n$ can't divide $3$ anyways), and $5, 5^2,7$ , we have $\frac{d(n)^3}{n} < 1$ . Letting the power of $2$ in $n$ be $2^x$ , we see that $\frac{d(2^x)}{2^x} \ge \frac{4}{\frac 85 \frac 87}$ , since all the other prime powers decrease the value of the product. This then forces $0 < x < 9$ , we also know that $x \equiv 1 mod 3$ , so $x$ can be either $1,4,7$ . In both the first and third cases, we have $\frac{d(2^x)}{2^x} = 4$ , so we require the product of $\frac{d(p)^3}{p}$ for the other prime powers $p$ to be $1$ . Clearly, one prime power must be one of $5, 5^2, 7$ , since the total product is $1$ and no individual factor is $1$ . We see $5,7$ fail because we would require a power of $2$ in the denominator of the product to cancel out the $8$ on the numerator, impossible, likewise $5^2$ fails because we would require a power of $3$ in the denominator. Thus no set of odd prime powers gives the desired product, so we get the desired solutions of $2, 2^7$ . In the other case, we have $\frac{d(2^x)}{2^x} = \frac{125}{16}$ , so we must have a power of $5$ to cancel out the $125$ on the top, note that the power of $5$ must be a cube, so testing we see that $5^3$ just works (and we cannot add anymore prime powers, since no set of prime powers $p$ has the product of $\frac{d(p)^3}{p} = 1$ ), and any larger powers of $125$ fail due to size reasons.

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#29 h Apr 27, 2025, 12:23 AM

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bruh so much cases

Clearly $n=2k^3$ for some $k \in \mathbb Z^+.$ Let $k=2^x \prod p_i^{e_i},$ then it suffices to solve $2^{x+1} \prod p_i^{e_i} = (3x+2) \prod (3e_i+1).$ If $x \geq 3,$ $\frac{2^{x+1}}{3x+2} \geq \frac{16}{11}.$ If $k$ has at least two prime factors other than $2$ $\frac{\prod p_i^{e_i}}{\prod (3e_i+1)} \geq \frac34 \cdot \frac54.$ So then $1 = \frac{2^{x+1}}{3x+2} \cdot \frac{\prod p_i^{e_i}}{\prod (3e_i+1)} \geq \frac{15}{11},$ contradiction. But if $x \geq 3,$ and $k$ has a prime factor other than $2$ greater than $3$ we get a similar result. But checking when $k$ has a prime factor of $3,$ if the exponent is at least $2$ we get a size contradiction, and if $k$ is $2^x \cdot 3$ the equation becomes $2^{x+1} \cdot 3 = (3x+2) \cdot 4$ which can be manually checked to have no solution. Finally if $k=1$ we also have a size contradiction as $2^{x+1} > 3x+2.$

Thus $x \leq 2.$ If $x=0, 2$ then $2^{x+1}=3x+2,$ so $\prod p_i^{e_i} = \prod (3e_i+1).$ By similar reasoning to above $k=1$ which yields the solutions $n=2, 2^7.$ If $x=1,$ we have $4 \cdot \prod p_i^{e_i} = 5 \cdot \prod (3e_i+1),$ by similar arguments to above we can't have primes at least $7$ and exponent at least $2$ so we check $k=2 \cdot 3, k=2 \cdot 5$ and only $k=10$ ends up working. This yields $n=2000.$

Hence the solutions are $n=2,128, 2000,$ which all clearly work.

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