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Problem 1072

AMC 12 late, AIME early

Number theory Difficulty 5.0 Prove it Berkeley Math Circle · United States

As usual, let $n!$ denote the product of the integers from $1$ to $n$ inclusive. Determine the largest integer $m$ such that $m!$ divides $100! + 99! + 98!$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution:

The answer is $m = 98$ . Set
$N = 98! + 99! + 100! = 98! (1 + 99 + 99 \cdot 100)$
Hence $N$ is divisible by $98!$ . But
$\frac{N}{98!} = 1 + 99 \cdot 101$
is not divisible by $99$ . Hence $N$ is not divisible by $99!$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.