SAT/ACT problem of the week, October 06, 2016 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.

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 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.