**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.

**Answer: **E

**Solution: **This problem is extremely difficult, and very few people get it right, so don’t feel bad if you got it wrong.

Let $latex 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 $latex 10^n $ if and only if it is a multiple of both $latex 2^n $ and $latex 5^n $. For example, $latex 1 \times 2 \times 3 \times 4 \times 5 = 120$. 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 $latex 1\times 5^2 $, $latex 2\times 5^2 $, $latex 3\times 5^2 $, and $latex 4\times 5^2 $. Therefore, $latex P $ is divisible by $latex 5^{20 + 4} = 5^{24} $. Similar arguments show $latex P $ is divisible by $latex 2^{97} $, which is divisible by $latex 2^{24} $. Therefore, $latex P $ is divisible by $latex 10^{24} $, so $latex P $ ends in 24 zeros.