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Saudi Arabian Mathematical Competitions, 2012

Algebra Difficulty 4.6 AIME Prove it Saudi Arabia

Show that for every positive integers $n \ge 3$ there are distinct positive integers $a_1, a_2, \dots, a_n$ with
$a_1! a_2! \dots a_{n-1}! = a_n!$

Solution

For $n = 3$ we have $3! \cdot 5! = 6!$ .

Assume that $a_1! a_2! \dots a_{k-1}! = a_k!$ . Since $a_k!(a_k! - 1)! = (a_k!)!$ , it follows that
$a_1! a_2! \dots a_{k-1}! (a_k! - 1)! = a_k! (a_k! - 1)! = (a_k!)!$
and obviously $a_k! > a_k! - 1 > a_{k-1}!$ . We are done by induction.

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