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1990 AIME Problems/Problem 11

Revision as of 17:52, 15 March 2017 by Warrenwangtennis (talk | contribs) (→‎Generalization:)

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 $(n+1)! /4! = n!$ $\Longrightarrow n+1 = 24$ $\Longrightarrow n = 23$

Kris17

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 4n$such that product of$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$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)n-a$consecutive positive integers are$a+2, a+3, … n+1$So we have$\frac{(n+1)!}/{(a+1)!} = n!$$ (Error compiling LaTeX. Unknown error_msg)\Longrightarrow n+1 = (a+1)!$$ (Error compiling LaTeX. Unknown error_msg)\Longrightarrow n = (a+1)! -1$

Kris17

See also

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
All AIME Problems and Solutions

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