Yesterday, Manhattan GMAT posted a GMAT question on our blog. Today, they have followed up with the 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 – 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.
The answer is C.