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I conjecture that:

$\begin{equation*} \dfrac{(7n)!}{7^n n!} \equiv \left\{ \begin{aligned} & 1 \text{ if } n \text{ is even} \\ & 6 \text{ if } n \text{ is odd} \end{aligned} \right. \pmod{7} \end{equation*}$

Based on hand observations.

  • I tried to reduce the factorials as binomial coefficients, that didn't yield to some problem reduction.
  • Observing it in $\mathbb{Z}/7\mathbb{Z}$ didn't give much help also.
  • Simplifying the $n!$ using the $(7n)!$ didn't help also.
  • We can see using p-adic valuations that this expression is never a multiple of 7, due to the fact that all powers of 7 vanish (using Legendre formula for instance).
  • Attempts to separating odd / even cases didn't help also.
  • Considering the equation $\pmod{7^{n + 1}}$ didn't help also.
Raito
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2 Answers2

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Hint:

$$\frac{(7n)!}{7^n\cdot n!}\equiv \frac{(6!)^n\cdot (7\times 14\times\cdots \times 7n)}{7^n\cdot n!}\equiv\frac{7^n\cdot n!\cdot (6!)^n}{7^n\cdot n!}\equiv (6!)^n\pmod 7$$

Now, use Wilson's Theorem.

This can further be generalized for all primes $p$. We have,

$$\frac{(pn)!}{p^n\cdot n!}\equiv (-1)^n\pmod p$$

Prasun Biswas
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  • But I don't understand how $(7 × 14 × … × 7n) = 7^n × n!$ – Raito Sep 27 '17 at 15:44
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    @Raito, $$7\times 14\times\cdots\times 7n=(7\times 1)\times (7\times 2)\times\cdots (7\times n)=7^n\times (1\times 2\times\cdots\times n)=7^n\times n!$$ – Prasun Biswas Sep 27 '17 at 15:47
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Induction. $$\frac{(7(n+1))!}{7^{n+1}(n+1)!}\equiv\frac{(7n)!(7n+1)\cdots(7n+6)(7n+7)}{7^nn!7(n+1)}\equiv\frac{(7n)!}{7^nn!}(1\cdot2\cdots6)\equiv-\frac{(7n)!}{7^nn!} \mod 7.$$

Teddy38
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