The Quest for 700: Weekly GMAT Challenge (Answer)

The first task is to figure out a positive integer n for which the given condition is true: that all the unique positive divisors multiply up to n².

Let’s first take a number-testing approach.

If n = 1, the condition holds, but we can quickly see that all the answer choices are equal in this case.

If n = 2, the condition doesn’t hold. Take the unique positive factors: 1 and 2. The product is 2, not 2².

The same is true for n = 3 or for any other prime number.

If n = 4, the condition still doesn’t hold. The unique positive divisors are 1, 2, and 4, which multiply up to 8, not 16.

If n = 6, the condition holds (thankfully). The unique positive divisors are 1, 2, 3, and 6, which multiply up to 36, or 6².

Thus, we can solve the problem if we figure out the product of all the unique positive divisors of 36. We don’t need the digits; we just need to know what that number is as a power of 6 (our n). The fast, organized way is to take the factors in pairs:

1 × 36 = 36 (=6²)

2 × 18 = 36 (=6²)

3 × 12 = 36 (=6²)

4 × 9 = 36 (=6²)

And finally there’s a 6, of course, the square root of 36. We only count this once (the question specifies unique divisors).

The product of the four “6-squared’s” (6²) and the final 6 is 6² × 6² × 6² × 6² × 6 = 6⁹.

This gives us the correct answer of E.

Alternatively, we could solve this problem in a more abstract way. First, we can figure out that n must be the product of two distinct primes, say p and q (so that n = pq). The only unique positive factors of n would then be 1, p, q, and pq (that is, n itself). Then we could write n² as p²q². (By the way, if n were allowed to be a cube of a prime number (such that n = p³), then the product of all of n‘s factors would also be n², but this possibility is explicitly ruled out.)

In terms of p and q, then, the unique factors of n² would then be as follows:

1, p, p²

q, qp, qp²

q², q²p, q²p²

The product of all these factors is q⁹p⁹, or n⁹.