$n=\frac{k(k+1)}{2}\ ,\ m=\frac{(2^k)!}{2^{\frac{k(k+1)}{2}}}$

first we see that $m=n=k=1 $ is a solution so we have to consider that the LHS its undependet in any variablas of RHS so clearly we will have the solution in the form $(2^{k})!=t\frac{(2^{k})!}{t}$ from here you discus only some cases and the final result is like roza2010

But, why should $m$ be odd. I think $n \le v_2((2^k)!)=2^k-1$ i.e. $n=2^k-1-x,m=\frac{(2^k)!}{2^{2^k-1-x}}$ Edit: Sorry, actually got confused seeing the above post.

I agree with dibyo_95 but $ v_{2}((2^{k})!)=2^{k}-1 $

I'm sorry, am I missing something?! I might be wrong... but I knew there was a theorem (Lagrange's theorem or something like this) which stated that $v_2(n!)=\left[ \dfrac{n}{2} \right]+\left[ \dfrac{n}{2^2} \right]+\left[ \dfrac{n}{2^3} \right]+....$. If I'm not wrong $\left[ \dfrac{2^k}{2} \right]+\left[ \dfrac{2^k}{2^2} \right]+...+\left[ \dfrac{2^k}{2^k} \right]=1+2+...+2^{k-1}=2^k-1$ and not $\dfrac{k(k+1)}{2}$. And then, as dibyo_99 said, $n=2^k-1-x,\ m=\dfrac{(2^k)!}{2^{2^k-1-x}}$

Last post is the one correct, of course. A simple computation on one fingers would show that for $k=3$ the power of $2$ dividing $8!$ is $7$, not $\dfrac {3(3+1)} {2} = 6$ (and in effect this of course agrees with $2^3 - 1 = 7$). But scarcely anybody cares to verify his hasty answer on small cases ...

Sorry to bring life to an old post but seeing that the above commenter said something about the solution and the fact that I got a different one from the guys above, I decided to post my solution. Notice that $\displaystyle 2^k!$ has $\displaystyle \sum_{i=0}^{k-1}2^i$ twos (we notice this by either intuition or just trying out different values of k). This a geometric series so: $2^n=2^{\sum_{i=0}^{k-1}2^i} \Leftrightarrow n=2^{k-1}-1$ This leaves $m=\frac{2^k!}{2^n} \Leftrightarrow m=\frac{2^k!}{2^{k-1}-1}$ Not sure if this solution is correct as (SPOILER ALERT: Pretty sad fact incoming) this was the very first time I used the geometric sum formula

Aiscrim wrote: I'm sorry, am I missing something?! I might be wrong... but I knew there was a theorem (Lagrange's theorem or something like this) which stated that $v_2(n!)=\left[ \dfrac{n}{2} \right]+\left[ \dfrac{n}{2^2} \right]+\left[ \dfrac{n}{2^3} \right]+....$. Legendre theorem. @Mr.Chem-Mathy The solution is correct, but I don't like your explanation Mr.Chem-Mathy wrote: Notice that $\displaystyle 2^k!$ has $\displaystyle \sum_{i=0}^{k-1}2^i$ twos (we notice this by either intuition or just trying out different values of k) That's all.

WolfusA wrote: @Mr.Chem-Mathy The solution is correct, but I don't like your explanation Mr.Chem-Mathy wrote: Notice that $\displaystyle 2^k!$ has $\displaystyle \sum_{i=0}^{k-1}2^i$ twos (we notice this by either intuition or just trying out different values of k) That's all. Alright, thanks for the advice.

Mr.Chem-Mathy wrote: Sorry to bring life to an old post but seeing that the above commenter said something about the solution and the fact that I got a different one from the guys above, I decided to post my solution. Notice that $\displaystyle 2^k!$ has $\displaystyle \sum_{i=0}^{k-1}2^i$ twos (we notice this by either intuition or just trying out different values of k). This a geometric series so: $2^n=2^{\sum_{i=0}^{k-1}2^i} \Leftrightarrow n=2^{k-1}-1$ This leaves $m=\frac{2^k!}{2^n} \Leftrightarrow m=\frac{2^k!}{2^{k-1}-1}$ Not sure if this solution is correct as (SPOILER ALERT: Pretty sad fact incoming) this was the very first time I used the geometric sum formula Another fault which I see is, that n is not necessarily the largest power of 2 that divides LHS, hence comes into play the $x$ here Aiscrim wrote: I'm sorry, am I missing something?! I might be wrong... but I knew there was a theorem (Lagrange's theorem or something like this) which stated that $v_2(n!)=\left[ \dfrac{n}{2} \right]+\left[ \dfrac{n}{2^2} \right]+\left[ \dfrac{n}{2^3} \right]+....$. If I'm not wrong $\left[ \dfrac{2^k}{2} \right]+\left[ \dfrac{2^k}{2^2} \right]+...+\left[ \dfrac{2^k}{2^k} \right]=1+2+...+2^{k-1}=2^k-1$ and not $\dfrac{k(k+1)}{2}$. And then, as dibyo_99 said, $n=2^k-1-x,\ m=\dfrac{(2^k)!}{2^{2^k-1-x}}$

i think roza2010 is right

jiangyucheng wrote: i think roza2010 is right No,he ain't

( n, m ) = (n, $ \frac{(2^k)!}{2^n} $) for every n= 1, 2, 3, .... $2^k - 1$

I'm not sure if I understood the question, but wouldn't the max value of $n$ be equal to $2^k -1$ making $m$ equal to $\frac{(2^k)!}{2^n}$, with other couples of the form $(n_{i}, m_{i}) = (\frac{n}{2^i}, 2^i m)$ for a total of $2^k -1$ couples of $(n, m)$.

First of all, notice that by taking $\nu_2$ we obtain $\nu_2 (2^k!)=\nu_2 (2^n\cdot m)$ However notice that $\nu_2 (2^k!)=2^k-S_2(2^k)=2^k-1$ by Legendre, and $\nu_2 (2^n\cdot m)\ge n\Longrightarrow \nu_2(2^n\cdot m)=n+t\text{ for some }t\in\mathbb{Z}_{\ge0}$ Thus $2^k-1=n+t\Longrightarrow n=2^k-1-t$, furthermore plugging this into our original equation we obtain $2^k!=2^{2^k-1-t}\cdot m\Longrightarrow m=\frac{2^k!}{2^{2^k-1-t}}$ So, to sum up $\boxed{(n,m)=\left(2^k-1-t,\frac{2^k!}{2^{2^k-1-t}}\right)\text{ for some }t\in\mathbb{Z}_{\ge0}}$ $\blacksquare$.

A weird question! Surely the factorization of $2^{k}$ exists and only one (m,n) suits the function. Why he ask you to solve all the solutions .

Sol sketch: Evaluate v_2((2^k)!), consider maximal n, find all other solutions.

Is the answer infinitely many?