What is the maximum value of n such that the number $n!$ can be represented as a product of $(n−3)$ consecutive positive integers?
Im not sure how to solve this, hints and solutions would be appreciated
Taken from the 2013 AITMO
2026-03-29 08:15:57.1774772157
What is the maximum value of n such that the number $n!$ can be represented as a product of $(n−3)$ consecutive positive integers?
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We need a solution of $$n!=\frac{(n+a-3)!}{a!}.$$ For successive values of $a$ this means we need to solve: $$n+1=4!$$ $$(n+1)(n+2)=5!$$ $$(n+1)(n+2)(n+3)=6!$$ $$...$$ Now the first of these equations gives us a solution with $a=4 \text{ and }n=23$. We therefore only need solutions with $n\ge 23$. However, for $n\ge 23$, the LHS of successive equations is multiplied successively by at least $25,26,27,...$ whilst the RHS is multiplied successively by $5,6,7,...$ so there are no such solutions.
The solution is $n=23$, when$$23!=5\times 6\times 7 ... \times 24.$$