Yesterday, Manhattan GMAT posted a 700 level GMAT question on our blog. Today, they have followed up with the answer:
If we want to distribute x biscuits among y patrons equally and with no split or leftover biscuits , then x must be divisible by y. Note that since both x and y count physical objects, both variables must be positive integers. The value of x is also constrained to be at least 2.
Since x must be divisible by y, we can also say that y must be a factor of x. Asking how many values of y satisfy the conditions is equivalent to asking how many factors x has.
(1) SUFFICIENT. If we can write the prime factorization of x as a^{2}b^{3}, where a and b are different prime numbers, then we can in fact count the factors of x – even though we do not know the values of x, a, or b. The reason is that we can construct every factor of x uniquely out of powers of a and powers of b. No factor of x can contain any primes other than a and b. Moreover, in any factor of x, the power of a cannot be larger than 2 (since x = a^{2}b^{3}, and if the factor had a higher power of a, then when we divide x by the factor, we would be left with uncanceled a’s in the denominator). By the same reasoning, the power of b in the factor cannot be larger than 3. Finally, both powers must be nonnegative integers (0 or positive integers. Thus, we can construct a table to see all the possibilities. Simply multiply together the row and column labels to get each entry:

a^{0} = 1 
a^{1} = a 
a^{2} 
b^{0} = 1 
1 
a 
a^{2} 
b^{1} = b 
b 
ab 
a^{2}b 
b^{2} 
b^{2} 
ab^{2} 
a^{2}b^{2}

b^{3} 
b^{3} 
ab^{3} 
a^{2}b^{3} 
Thus, there are 12 unique factors of x. In fact, we do not have to enumerate the factors. A shortcut is to add 1 to each prime’s power in the factorization (to account for the possibility of a^{0} or b^{0}) and then multiply the results together. In this case, since x = a^{2}b^{3}, we write (2 + 1)(3 + 1) = (3)(4) = 12.
(2) INSUFFICIENT. By itself, the statement does not refer to x or y, so it cannot be sufficient to answer the given question.
Incidentally, one trap in this problem is that if you put the statements together, you can actually figure out the values of a and b, and therefore the value of x. Since b = a + 1, we can conclude that a = 2 and b = 3. The only primes that differ by exactly 1 are 2 and 3. Since all primes greater than 2 are odd, the minimum difference between all other pairs of primes is 2.
The correct answer is (A).