Child Labor #1

It's called p-adic valuation of a natural number $n$ , with $p$ prime, the maximum $k$ such that $p^k$ divides $n$ : $v_p (n)=max\{k \in \mathbb{N}: \; p^k \mid n\}$

Two theorems are useful in this problem: product theorem and Legendre-de Polignac's formula .

• Product theorem states that $v_p(m\cdot n)=v_p(m)+v_p(n)$

• Legendre-de Polignac's formula states that v_p(n!)= \sum_{i=1}^{\infty} \Big{\lfloor} \frac{n}{p^i} \Big{\rfloor}

Their proofs aren't particularly difficult; you can try yourself to find them out if you want.

( $1!\cdot 2! \cdot ... \cdot n!$ is called superfactorial and sometimes represented as $n\$$ , so I will do the same here.)

Now, the number of trailing zeros of $n\$$ is obviously equal to $v_5(n\$)$ . For product theorem and Legendre-de Polignac formula:

v_5(n\$)=v_5(1! \cdot 2! \cdot ... \cdot n!)=v_5(1!)+v_5(2!)+...+v_5(n!)=\displaystyle \sum_{i=1}^{\infty} \Big{\lfloor} \frac{1}{5^i} \Big{\rfloor}+ \displaystyle \sum_{i=1}^{\infty} \Big{\lfloor} \frac{2}{5^i} \Big{\rfloor}+...+\displaystyle \sum_{i=1}^{\infty} \Big{\lfloor} \frac{n}{5^i} \Big{\rfloor}\;\;\;\;\;[A]

We notice also that $v_5(n!)=v_5(m!)$ if and only if $\exists k \in \mathbb{N}: 5k \leq n,m < 5(k+1)\;\;\;\;\;[B]$ .

Now, the problem asks when $v_5(n\$)=n$ . If $n=1$ we have $v_5(1\$)=0$ , so $n>v_5(n\$)$ , but, for, let's say, $n=15$ we have $n<v_5(n\$)$ , since $v_5(15\$)=18$ , so, if $n=v_5(n\$)$ , it would happen before $15$ . This calculation are very easy to do, since $[A]$ and $[B]$ help us a lot. In fact, with very few attempts we find that $v_5(13\$)=13$ so $n=\boxed{13}$ .