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

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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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== Solution ==
 
== Solution ==
 
{{solution}}
 
{{solution}}
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== See also ==
 
== See also ==
* [[1990 AIME Problems/Problem 10 | Previous problem]]
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{{AIME box|year=1990|num-b=10|num-a=12}}
* [[1990 AIME Problems/Problem 12 | Next problem]]
 
* [[1990 AIME Problems]]
 

Revision as of 00:34, 2 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

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See also

1990 AIME (ProblemsAnswer KeyResources)
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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