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 $\frac{(n+1)!}{4!} = n!$ $\Longrightarrow n+1 = 24$ $\Longrightarrow n = 23$

Generalization

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

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 $\frac{(n+1)!}{(a+1)!} = n!$ $\Longrightarrow n+1 = (a+1)!$ $\Longrightarrow n = (a+1)! -1$

$Kris17$

Video Solution!!!

https://www.youtube.com/watch?v=H-ZmsYjF-xE

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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