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# GMAT Tip: Divisibility Mother Lode

Today’s GMAT challenge question comes from ManhattanGMAT.  To help you with your GMAT studying, try to solve the problem on your own, and then read on for the explanation of its solution:

Problem

If m is the square of integer n and m is divisible by 98, m must also be divisible by:

I.   28

II.  196

III. 343

A) I only

B) II only

C) I & II only

D) II & III only

E) I, II, and III

Solution

First, break 98 into primes.  98 = 49 × 2 = 7 × 7 × 2.  The “ingredients” of 98 are 7, 7, and 2.

m is divisible by 98, so m must contain the “ingredients” that make a 98: 7, 7, and 2.  However, m could be a very large number with many more factors than just those – it’s simply divisible by 98, and so it contains, at minimum, the primes that make up 98.

However, we also know that m is a perfect square (because it is the square of an integer, n).  All perfect squares have primes that come in pairs.  For instance, 9 contains 3 and 3.  81 contains 3, 3, 3, and 3.  100 contains 2, 2, 5, and 5.

Since m is a perfect square, its primes must also come in pairs.   So, rather than just 7, 7, and 2, it must contain 7, 7, 2, AND ANOTHER 2.

If m contains, at minimum, 7, 7, 2, and 2, we know that its factors include all the possible combinations of those primes (plus the number 1):

1

2

7

2 x 2 = 4

2 x 7 = 14

7 x 7 = 49

2 x 2 x 7 = 28

2 x 7 x 7 = 98

2 x 2 x 7 x 7 = 196

Therefore, m must be divisible by 28 and 196, Roman numerals I and II.