AIME I 2006 P4

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Problem Statement

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 $1000$.

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Solution

$$\text{Let } N = \text{ # of 0s at the right end of the decimal representation of } 1!2!3!4!\cdots99!100!$$ $$N = \sum_{i=1}^{100} \lfloor \dfrac{i}{5} \rfloor + \sum_{i=1}^{100} \lfloor \dfrac{i}{25} \rfloor$$ $$\sum_{i=1}^{100} \lfloor \dfrac{i}{5} \rfloor = (0+0+0+0+0) + (1+1+1+1+1) + \cdots + (19+19+19+19+19) + 20$$ $$\sum_{i=1}^{100} \lfloor \dfrac{i}{5} \rfloor = 5(1+2+ \cdots + 19) + 20 = 970$$ $$\sum_{i=1}^{100} \lfloor \dfrac{i}{25} \rfloor = 25(1+2+3) + 4 = 154$$ $$\therefore N = 970 + 154 = 1124 \implies N \equiv 124 \mod 1000$$