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

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 $\displaystyle (n - 3 + a)! > n!$ , 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 $\displaystyle ^{a-3}\sqrt{a!}$ , which decreases as $a$ increases. Thus, $n = 23$ is the greatest possible value to satisfy the given conditions.

1990 AIME (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1990 AIME Problems/Problem 11

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