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If the number of factor of $(n+1)!$ is double than the number of factor of $n!$, then find the reminder if $n!$ is divided by $(n+1)$?

I'm not sure if the question mean factor = divisor. However in both cases I cant find a way to start.
Source: BdMO 2015 Noril regional Higher Secondary Question no. 10.

Rezwan Arefin
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  • "Factor" and "divisor" are synonyms, yes. – Arthur Jan 08 '17 at 22:41
  • A sufficient condition is that $n+1$ is prime, in which case Wilson's theorem tells you the remainder of $n!$ modulo $n+1$. So it seems that the main part of the problem is to show that $\tau((n+1)!) = 2\tau(n!)$ cannot occur in any other way. – Erick Wong Jan 08 '17 at 22:43
  • @Arthur I think "number of divisors" and "number of factors" are not same. Example: $10$ has $4$ divisors. And $n! = 1\times 2 \times 3 \times \cdots \times n$ has $n$ factors. So both are not same. – Rezwan Arefin Jan 08 '17 at 22:43
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    @ErickWong $\tau(4!) = 2\tau(3!)$ – Daniel Fischer Jan 08 '17 at 22:49
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    @ErickWong I think I got my answer... We can prove the opposite also. i.e. If $\tau((n+1)!) = 2\tau(n!)$ then $n+1$ is prime. The rest as easy. – Rezwan Arefin Jan 08 '17 at 22:49
  • @RezwanArefin I think in this context the words are the same, although what you say is not entirely without merit. Also, $n+1$ doesn't have to be prime as the above example of $n = 3$ illustrates. – Arthur Jan 08 '17 at 22:50
  • @Arthur :( Yep.. Then how to solve this? :( – Rezwan Arefin Jan 08 '17 at 22:51
  • For some value $x$, what is $\tau (x)$? I never knew there existed a function $\tau$, but only thought that $\tau$ is a value equal to $2\pi$. – Mr Pie Jan 15 '18 at 00:45

1 Answers1

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According to the below theorem, $n$ as in the problem statement is either one less than a prime, or $n=3$. In the former case, Wilson's theorem gives us $n!\equiv n\pmod {n+1}$; in the latter case $3!\equiv 2\pmod 4$.

Theorem. For a natural number $n$, we have $\tau(n!)=2\tau((n-1)!)$ iff $n$ is prime or $n=4$.

Proof. One direction is straightforward: We have $\tau(4!)=8$, $\tau(6)=4$, and if $n$ is prime, the divisors of $n!$ are precisely the numbers $d$ and $nd$ where $d$ is a divisor of $(n-1)!$.

For the other direction, let $n=p_1^{a_1}p_2^{a_2}\cdots p_m^{a_m}$ be the prime factorization of $n$ (with $m\ge1$ and all $a_i\ge 1$). Write $n!=p_1^{b_1}p_2^{b_2}\cdots p_m^{b_m}r$ with $r$ only divisible by other primes (i.e., co-prime to $n$). So $(n-1)!=p_1^{b_1-a_1}p_2^{b_2-a_2}\cdots p_m^{b_m-a_m}r$. Then by multiplicativity of the divisor counting function, $$\tau(n!)=(b_1+1)(b_2+1)\cdots (b_m+1)\tau(r)$$ and $$ \tau((n-1)!)=(b_1-a_1+1)(b_2-a_2+1)\cdots (b_m-a_m+1)\tau(r).$$ Hence for $\tau(n!)=2\tau((n-1)!)$ we need $$ 2\cdot (b_1-a_1+1)(b_2-a_2+1)\cdots (b_m-a_m+1)= (b_1+1)(b_2+1)\cdots (b_m+1)$$ or equivalently $$\tag1 \prod_{i=1}^m\left(1-\frac{a_i}{b_i+1}\right)=\frac12.$$ It is well known that the exponent $b_i$ of $p_i$ in $n!$ is given by $$\tag2 b_i=\left\lfloor \frac {n}{p_i}\right\rfloor+\left\lfloor \frac {n}{p_i^2}\right\rfloor+\left\lfloor \frac {n}{p_i^3}\right\rfloor+\ldots.$$ Let $Q_i=\frac{n}{p_i^{a_i}}$. Then for $1\le k\le a_i$, the $k$th summand in $(2)$ equals $p_i^{a_i-k}Q_i$, which gives us the simple estimates $$\tag3b_i\ge a_iQ_i$$ and $$\tag4b_i\ge (a_i-1)p_iQ_i+Q_i.$$ As $Q_i$ has $m-1$ distinct prime divisors, we conclude $Q_i\ge m!$. Using this, $(1)$, $(3)$, and Bernoulli's inequality, we obtain $$\tag5\begin{align}\frac12&=\prod_{i=1}^m\left(1-\frac{a_i}{b_i+1}\right)\\ & \ge \prod_{i=1}^m\left(1-\frac{a_i}{Q_ia_i+1}\right)\\ &>\prod_{i=1}^m\left(1-\frac{1}{Q_i}\right) \\ & \ge \left(1-\frac1{m!}\right)^m\\ &\ge 1-\frac 1{(m-1)!}.\end{align}$$ This leads to $(m-1)!<2$, i.e., $m\le2$.

Assume $m=2$. Then from $(5)$, we extract
$$\frac12 > \left(1-\frac1{Q_1}\right)\left(1-\frac1{Q_2}\right).$$ If none of the $Q_i$ equals $2$, this leads to the absurdity $\frac12>\frac23\cdot\frac34$ because $Q_1\ne Q_2$. Therefore, wlog. $Q_2=2$ and consequently $p_1=2$, $a_1=1$, $n=2p_2^{a_2}$ with $p_2\ge 3$, and $b_1\ge p_2^{a_2}+\lfloor p_2^{a_2}/2\rfloor \ge 4$. Also, from $(4)$, we get $b_2\ge 6a_2-4$. Plugging this into $(1)$, we arrive at $$ \frac12 = \left(1-\frac1{b_1+1}\right)\left(1-\frac{a_2}{b_2+1}\right)\ge \frac 45\cdot\left(1-\frac{a_2}{6a_2-3}\right)\ge\frac45\cdot\frac 23,$$ contradiction.

Therefore, $m=1$, $Q_1=1$, and as $(1)$ demands $2a_1=b_1+1$, we have from $(4)$ that $$2(a_1-1) = b_1-1 \ge (a_1-1)p_1$$ and conclude that $a_1=1$ (i.e., $n$ is prime) or $p_1=2$. In the latter case, $b_1=2^{a_1}-1$, and $2a_1=b_1+1$ leads to $a_1\le 2$, i.e., we get only $n=4$ as additional possibility. $\square$

Hagen von Eitzen
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