Difference between revisions of "2006 AIME I Problems/Problem 4"
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== Problem == | == Problem == | ||
Let <math> N </math> be the number of consecutive 0's at the right end of the decimal representation of the product <math> 1!2!3!4!\cdots99!100!. </math> Find the remainder when <math> N </math> is divided by 1000. | Let <math> N </math> be the number of consecutive 0's at the right end of the decimal representation of the product <math> 1!2!3!4!\cdots99!100!. </math> Find the remainder when <math> N </math> is divided by 1000. | ||
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== Solution == | == Solution == | ||
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== See also == | == See also == | ||
| − | * [[2006 AIME I]] | + | * [[2006 AIME I Problems]] |
Revision as of 11:13, 30 June 2006
Problem
Let 
be the number of consecutive 0's at the right end of the decimal representation of the product 
Find the remainder when is divided by 1000.
