Question

Let $N$ be the number of consecutive $0$ 's at the right end of the decimal representation of the product $1!\,2!\,3!\,4! \cdots 99!\,100!.$ Find the remainder when $N$ is divided by $1000.$

Answer

Answer: 124

Concepts:trailing zeros factorial summation

Difficulty rating: 2400

Solution:

Factors of $2$ are plentiful, so $N$ is the exponent of $5$ in the product. Each integer $j$ with $1 \le j \le 100$ appears as a factor in exactly $101 - j$ of the factorials, namely $j!, (j+1)!, \ldots, 100!.$

Every multiple of $5$ contributes one factor of $5$ per appearance, and every multiple of $25$ contributes one more. Over $j = 5, 10, \ldots, 100$ the appearances total $96 + 91 + \cdots + 1 = \frac{20 \cdot 97}{2} = 970,$ and over $j = 25, 50, 75, 100$ they total $76 + 51 + 26 + 1 = 154.$

Hence $N = 970 + 154 = 1124,$ and the remainder upon division by $1000$ is $124.$

2006 AIME I Problem 4

Answer: 124

Difficulty rating: 2400

Solution:

Problem 4 in Other Years