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Find all positive integers $k$ such that for any positive integer $n$, $2^{(k-1)n+1}$ does not divide $\frac{(kn)!}{n!}$.

From olympiad problem

I'm curious So far no one to solve this problem,Maybe it is very difficult May be other reasons? But I really want to know how to solve this problem more easy?

Alex M.
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math110
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    Denote $s(x)$ as the sum of digits in the binary presentation of $x$. As @darkness have shown, all left is to prove that for $k=b\times2^a$, where $b>1$ is an odd number, we can always find $n$ so that $s(kn)=s(bn)<s(n)$. Here is a hint: we know that there exists $q$ so that $b|(2^q−1)$. Assume $2^{p−1}<b<2^p$, the claim is that $n=1+2^p(2^{pq}−1)/b$ does the job. – Wiley Jun 14 '16 at 14:38

2 Answers2

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The number of factors $2$ in a number $a!$ is: $$ \left\lfloor \frac a 2 \right\rfloor + \left\lfloor \frac a 4 \right\rfloor + \left\lfloor \frac a 8 \right\rfloor + \cdots $$

So we have to find all $k$ such that for every $n$, we have $$ kn - n + 1 > \sum_{x=1} \left\lfloor \frac{kn}{2^x} \right\rfloor - \sum_{x=1} \left\lfloor \frac{n}{2^x} \right\rfloor $$

Now, to save typing, I define $F(a) = \sum_{x=1} \left\lfloor \frac{a}{2^x} \right\rfloor$. So how much is $F(a)$ compared to $a$?

Lemma: $F(2^a) = 2^a-1$

Proof: We have $F(2^a) = 2^{a-1} + 2^{a-2} + \cdots + 1 = 2^a-1$. $\blacksquare$

Lemma: If $b < 2^a$, then $F(2^a + b) = F(2^a) + F(b)$.

Proof: For every $x < a$, $\frac{2^a}{2^x}$ is an integer and thus $$ \left\lfloor \frac{2^a + b}{2^x} \right\rfloor = \left\lfloor \frac{2^a}{2^x} \right\rfloor + \left\lfloor \frac{b}{2^x} \right\rfloor $$ For every $x \geq a$, we have $\left\lfloor \frac{b}{2^x} \right\rfloor = 0$, so the equation holds as well. $\blacksquare$

A consequence of this is that $a - F(a)$ is the number of ones in the binary representation of $a$.

So $kn - n + 1 > F(kn) - F(n)$ means that $kn - F(kn) \geq n - F(n)$, which means that the binary representation of $kn$ has at least as many ones as the one of $n$.

So it's obvious that $k = 2^a$ always works. I'll think of some reasons why other $k$s don't work shortly.

dankness
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It holds that for any number n the exponent $e_p(n)$ of a prime p in n! is exactly

$e_p(n) = \frac{n-d_p(n)}{p-1}$

where $d_p(n)$ denotes the sum of digits in base p. This leads to the following reformulation of the problem:

Find all positive integers $k$ such that for all positive integers n

$\frac{(k-1)n-d_2(kn)+d_2(n)}{1}<(k-1)n+1 \\ d_2(n)-d_2(kn) < 1 \\ d_2(n) \leq d_2(kn)$

We note for all $k$ not relatively prime to 2:

Let $p_2$ be the exponent of 2 in the prime factorization of $k$. We define $k^*$ as then $k^* = k/(2^{p_2})$ which is relatively prime to 2. Now $d_2(n) \leq d_2(k^*n) \Leftrightarrow d_2(n) \leq d_2(kn)$. Hence we only need to prove or disprove the statement for numbers relatively prime to 2.

As we can easily see, if k = 1 then $d_2(n) = d_2(kn)$ and the above holds.

Let $k$ be relatively prime to 2 and $2^l>k>1$. Consider the number $q := 2^{φ(k)-1}-1$, which is divisible by $k$, and let $s_k := d_2(\frac{q}{k}) \geq 1$.

Now look at $(\sum_{i=0}^{l}2^{i(φ(k)-1)}) \cdot (2^{φ(k)-1} -1) =: M = 2^{l(φ(k)-1)}-1$

We have $d_2(M+k) = d_2(2^{l(φ(k)-1)} + k-1)= 1+d_s(k-1)\lt 1+l \leq d_2(M/k+1)$ since

$d_2(1+M/k) = \\ d_2(1+\sum_{i=0}^{l}(2^{i(φ(k)-1)}\cdot \frac{q}{k}) = \\ d_2(1+\frac{q}{k})+d_2(\sum_{i=1}^{l}(2^{i(φ(k)-1)}\cdot \frac{q}{k})) =\\ d_2(1+\frac{q}{k}) + \sum_{i=1}^{l}d_2(2^{i(k-1)}\cdot \frac{q}{k}) = \\ d_2(1+\frac{q}{k}) + \sum_{i=1}^{l}d_2(\frac{q}{k}) \geq\\ 1+l \cdot s_k \geq 1+l$

Using the above inequalities and $n=1+M/k$, we can see that $d_2(kn) < 1+l \leq d_2(n)$, which can be found for all k > 1 relatively prime to 2. But as outlined above, for every number which is not a power of 2 can be reduced to such a $k$. Hence, only power of 2 fulfill the problem.

HeroicKatora
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  • As a reference for the first formula, see http://math.stackexchange.com/a/1098127/198914 – HeroicKatora Jun 20 '16 at 19:03
  • I guess your $(2^{p-1}+1)/k$ should be $2^{e_0+1}/k$? Also there seems to be no justification that one could find a number such that $2^{e_0}\equiv k-1\pmod k$, e.g., take k=15. – Wiley Jun 20 '16 at 19:56
  • I just realized that the existance of such an $e_0$ is not guaranteed. In fact, I conjecture that the existance is a necessary condition for such a k. I'm trying to prove that right now – HeroicKatora Jun 20 '16 at 19:57
  • That would be the wrong route unfortunately. The issue here is that your strategy is overly aggressive to reduce $d_2(kn)$ all the way down to 2. – Wiley Jun 20 '16 at 20:02
  • I think I have fixed the argument, and as you correctly noted, I thought too narrow – HeroicKatora Jun 20 '16 at 20:51
  • Yep, this is correct except one minor issue: $2^{k-1}\equiv1\pmod k$ only holds when k is prime. You would need to replace the exponent with the totient function $\varphi(k)$ – Wiley Jun 20 '16 at 21:06