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Berkeley Math Circle: Monthly Contest 7

Combinatorics Difficulty 5.0 AIME Prove it United States

Problem:
Prove that there exist pairwise distinct positive integers $a_{0}, a_{1}, a_{2}, \ldots, a_{1000}$ such that
$a_{0}! = a_{1}!\, a_{2}!\, \ldots\, a_{1000}!$
Here $n! = 1 \times 2 \times \cdots \times n$ as usual.

Solution

Solution:
We proceed by induction on $n \geq 2$ . First, we can have $a_{1}=3$ , $a_{2}=5$ and $a_{0}=6$ . Now, given a working tuple $\left(a_{0}, a_{1}, \ldots, a_{n}\right)$ , note that the tuple
$\left(a_{0}!, a_{1}, \ldots, a_{n},\left(a_{0}-1\right)!\right)$
is a working tuple of length $n+1$ . This completes the proof.

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