## Blog

### The Quest for 700: Weekly GMAT Challenge (Answer)

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 left-over 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 a2b3, 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 = a2b3, 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 non-negative 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:

 a0 = 1 a1 = a a2 b0 = 1 1 a a2 b1 = b b ab a2b b2 b2 ab2 a2b2 b3 b3 ab3 a2b3

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 a0 or b0) and then multiply the results together. In this case, since x = a2b3, 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.

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