Manhattan GMAT Challenge Problem of the Week – 13 Oct 2010

by Manhattan Prep, Oct 13, 2010

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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

Answer

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 its 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.

The answer is C.

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