# SAT problem of the week, October 06, 2014 solution

Hints for problem of elimination: You might get an intuitive feeling that 1 and 3 are too small, leaving you with three choices. Even if you don’t know the full answer, you must guess at this point.

Read on for a full solution.

Let $P = 1 \times 2 \times 3 \times \cdots \times 100$ has as many zeros as it does multiples of 10. You might think the correct answer is to count multiples of ten, leading you to choice C. However, a number is a multiple of $10^n$ if and only if it is a multiple of both $2^n$ and $5^n$. For example, $1 \times 2 \times 3 \times 4 \times 5 = 120$. All multiples of 5 between 1 and 100 provide one power of 5, providing twenty 5s. The numbers 25, 50, 75, and 100 each a second 5, since they re equal to $1\times 5^2$, $2\times 5^2$, $3\times 5^2$, and $4\times 5^2$. Therefore, $P$ is divisible by $5^{20 + 4} = 5^{24}$. Similar arguments show $P$ is divisible by $2^{97}$, which is divisible by $2^{24}$. Therefore, $P$ is divisible by $10^{24}$, so $P$ ends in 24 zeros.