2006 AIME A Problems/Problem 4

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.

2006 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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2006 AIME A Problems/Problem 4

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