Difference between revisions of "2006 AIME A Problems/Problem 4"

Revision as of 13:03, 25 September 2007

Problem

Let $N$ be the number of consecutive 0's at the right end of the decimal representation of the product $1!2!3!4!\cdots99!100!.$ Find the remainder when $N$ is divided by 1000.

Solution

Clearly, $a_6=1$ . Now, consider selecting $5$ of the remaining $11$ values. Sort these values in descending order, and sort the other $6$ values in ascending order. Now, let the $5$ selected values be $a_1$ through $a_5$ , and let the remaining $6$ be $a_7$ through ${a_{12}}$ . It is now clear that there is a bijection between the number of ways to select $5$ values from $11$ and ordered 12-tuples $(a_1,\ldots,a_{12})$ . Thus, there will be ${11 \choose 5}=462$ such ordered 12-tuples.

@@ Line 1: / Line 1: @@
 == Problem ==
-Let <math> (a_1,a_2,a_3,\ldots,a_{12}) </math> be a [[permutation]] of <math> (1,2,3,\ldots,12) </math> for which
+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.
-<div style="text-align:center;"><math> a_1>a_2>a_3>a_4>a_5>a_6 \mathrm{\  and \ } a_6<a_7<a_8<a_9<a_{10}<a_{11}<a_{12}. </math></div>
-An example of such a permutation is <math> (6,5,4,3,2,1,7,8,9,10,11,12). </math> Find the number of such permutations.
 == Solution ==

2006 AIME II (Problems • Answer Key • Resources)
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Difference between revisions of "2006 AIME A Problems/Problem 4"

Revision as of 13:03, 25 September 2007

Problem

Solution

See also