# 1990 AIME Problems/Problem 11

## Problem

Someone observed that $6! = 8 \cdot 9 \cdot 10$. Find the largest positive integer $n^{}_{}$ for which $n^{}_{}!$ can be expressed as the product of $n - 3_{}^{}$ consecutive positive integers.

## Solution 1

The product of $n - 3$ consecutive integers can be written as $\frac{(n - 3 + a)!}{a!}$ for some integer $a$. Thus, $n! = \frac{(n - 3 + a)!}{a!}$, from which it becomes evident that $a \ge 3$. Since $(n - 3 + a)! > n!$, we can rewrite this as $\frac{n!(n+1)(n+2) \ldots (n-3+a)}{a!} = n! \Longrightarrow (n+1)(n+2) \ldots (n-3+a) = a!$. For $a = 4$, we get $n + 1 = 4!$ so $n = 23$. For greater values of $a$, we need to find the product of $a-3$ consecutive integers that equals $a!$. $n$ can be approximated as $^{a-3}\sqrt{a!}$, which decreases as $a$ increases. Thus, $n = 23$ is the greatest possible value to satisfy the given conditions.

## Solution 2

Let the largest of the (n-3) consecutive positive integers be k. Clearly k cannot be less than or equal to n, else the product of (n-3) consecutive positive integers will be less than n!.

Key observation: Now for n to be maximum the smallest number (or starting number) of the (n-3) consecutive positive integers must be minimum, implying that k needs to be minimum. But the least k > n is (n+1).

So the (n-3) consecutive positive integers are (5, 6, 7…, n+1)

So we have (n+1)! /4! = n! => n+1 = 24 => n = 23

Kris17

## Generalization:

Largest positive integer n for which n! can be expressed as the product of (n-a) consecutive positive integers = (a+1)! – 1

For ex. largest n such that product of (n-6) consecutive positive integers is equal to n! is 7!-1 = 5039

Proof: Reasoning the same way as above, let the largest of the (n-a) consecutive positive integers be k. Clearly k cannot be less than or equal to n, else the product of (n-a) consecutive positive integers will be less than n!.

Now, observe that for n to be maximum the smallest number (or starting number) of the (n-a) consecutive positive integers must be minimum, implying that k needs to be minimum. But the least k > n is (n+1).

So the (n-a) consecutive positive integers are (a+2, a+3, … n+1)

So we have (n+1)! / (a+1)! = n! => n+1 = (a+1)! => n = (a+1)! -1

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 