Workbook Wednesdays: Answer

Pencils up! Here is the answer to yesterday’s Manhattan GMAT challenge problem.

Question

What is the greatest integer m for which the number

50!

10^m

is an integer?

(A) 5
(B) 8
(C) 10
(D) 11
(E) 12

Answer

A quotient of two integers will be an integer if the numerator is divisible by the denominator, so we need 50! to be divisible by 10^m. To check divisibility, we must compare the prime boxes of these two numbers (The prime box of a number is the collection of prime numbers that make up that number. The product of all the elements of a number’s prime box is the number itself. For example, the prime box of 12 contains the numbers 2,2,3).

Since 10 = 2 × 5, the prime box of 10^m is comprised of only 2’s and 5’s, namely m 2′s and m 5′s. That is becaues 10^m = (2 × 5)^m = (2^m) × (5^m). Now, some x is divisible by some y if x‘s prime box contains all the numbers in y‘s prime box. So in order for 50! to be divisible by 10^m, it has to have at least m 5′s and m 2′s in its prime box.

Let’s count how many 5′s 50! has in its prime box.
50! = 1 × 2 × 3 × … 50, so all we have to do is add the number of 5′s in the prime boxes of 1, 2, 3, …, 50. The only numbers that contribute 5′s are the multiples of 5, namely 5, 10, 15, 20, 25, 30, 35, 40, 45, 50. But don’t forget to notice that 25 and 50 are both divisible by 25, so they each contribute two 5’s.

That makes a total of 10 + 2 = twelve 5′s in the prime box of 50!.

As for 2′s, we have at least 25 (2, 4, 6, …, 50), so we shouldn’t waste time counting the exact number. The limiting factor for m is the number of 5′s, i.e. 12. Therefore, the greatest integer m that would work here is 12.

The correct answer is E.