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Difference between revisions of "1990 AIME Problems/Problem 11"

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(last part of solution is weak, but I think it works)
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== Problem ==
 
== Problem ==
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Someone observed that <math>6! = 8 \cdot 9 \cdot 10</math>.  Find the largest positive integer <math>n^{}_{}</math> for which <math>n^{}_{}!</math> can be expressed as the product of <math>n - 3_{}^{}</math> consecutive positive integers.
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Someone observed that <math>6! = 8 \cdot 9 \cdot 10</math>.  Find the largest [[positive]] [[integer]] <math>n^{}_{}</math> for which <math>n^{}_{}!</math> can be expressed as the [[product]] of <math>n - 3_{}^{}</math> [[consecutive]] positive integers.
  
 
== Solution ==
 
== Solution ==
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{{solution}}
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The product of <math>n - 3</math> consecutive integers can be written as <math>\frac{(n - 3 + a)!}{a!}</math> for some integer <math>a</math>. Thus, <math>n! = \frac{(n - 3 + a)!}{a!}</math>, from which it  becomes evident that <math>a \ge 3</math>. Since <math>\displaystyle (n - 3 + a)! > n!</math>, we can rewrite this as <math>\frac{n!(n+1)(n+2) \ldots (n-3+a)}{a!} = n! \Longrightarrow (n+1)(n+2) \ldots (n-3+a) = a!</math>. For <math>a = 4</math>, we get <math>n + 1 = 4!</math> so <math>n = 23</math>. For greater values of <math>a</math>, we need to find the product of <math>a-3</math> consecutive integers that equals <math>a!</math>. <math>n</math> can be approximated as <math>\displaystyle ^{a-3}\sqrt{a!}</math>, which decreases as <math>a</math> increases. Thus, <math>n = 23</math> is the greatest possible value to satisfy the given conditions.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1990|num-b=10|num-a=12}}
 
{{AIME box|year=1990|num-b=10|num-a=12}}

Revision as of 11:21, 3 March 2007

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.

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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