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| − | == Problem ==
| + | #REDIRECT [[2006 AIME I Problems/Problem 4]] |
| − | 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 ==
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| − | Clearly, <math>a_6=1</math>. Now, consider selecting <math>5</math> of the remaining <math>11</math> values. Sort these values in descending order, and sort the other <math>6</math> values in ascending order. Now, let the <math>5</math> selected values be <math>a_1</math> through <math>a_5</math>, and let the remaining <math>6</math> be <math>a_7</math> through <math>{a_{12}}</math>. It is now clear that there is a [[bijection]] between the number of ways to select <math>5</math> values from <math>11</math> and ordered 12-tuples <math>(a_1,\ldots,a_{12})</math>. Thus, there will be <math>{11 \choose 5}=462</math> such ordered 12-tuples.
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| − | == See also ==
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| − | {{AIME box|year=2006|n=II|num-b=3|num-a=5}}
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| − | [[Category:Intermediate Combinatorics Problems]]
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