2020/21 AHSMC Part 1 Problem 8

Problem 8 of 16

The number of positive integer divisors of $10! = 10 \cdot 9 \cdot 8 \cdots 2 \cdot 1$ that have a remainder of $1$ when divided by $3$ is

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Solution

Since $10! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7$ the divisors of $10!$ which have remainder $1$ when divided by $3$ have the format $2^a 5^b 7^c$ , where $0 \le a \le 8$ , $0 \le b \le 2$ , $0 \le c \le 1$ and $a + b$ is even. There are $2(5 \cdot 2 + 4 \cdot 1) = 28$ solutions. The answer is (E).

University of Alberta

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