Track / Stage 4 / 271 of 340 #531 of 1964

Problem 531

AMC 12 late, AIME early

Number theory Difficulty 4.9 Find the answer

We notice that $6!=8 \cdot 9 \cdot 10$ . Try to find the largest positive integer $n$ such that $n!$ can be expressed as the product of $n-3$ consecutive natural numbers.

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

Solution: According to the requirements of the problem, write $n!$ as
$n!=n(n-1) \cdot \cdots \cdot 5 \cdot 24 \text {. }$

Therefore, $n+1 \leqslant 24, n \leqslant 23$ . Hence, the maximum value of $n$ is 23. At this point, we have
$23!=24 \times 23 \times \cdots \times 5 .$

The right-hand side of the above equation is the product of 20 consecutive natural numbers.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.