Manhattan GMAT Challenge Problem of the Week – 8 Feb 2011

by Manhattan Prep, Feb 8, 2011

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Question

If the product of all the unique positive divisors of n, a positive integer which is not a perfect cube, is [pmath]n^2[/pmath], then the product of all the unique positive divisors of [pmath]n^2[/pmath] is

(A) [pmath]n^3[/pmath]

(B) [pmath]n^4[/pmath]

(C) [pmath]n^6[/pmath]

(D) [pmath]n^8[/pmath]

(E) [pmath]n^9[/pmath]

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 [pmath]n^2[/pmath].

Lets 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 doesnt hold. Take the unique positive factors: 1 and 2. The product is 2, not [pmath]2^2[/pmath].

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

If n = 4, the condition still doesnt 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 [pmath]6^2[/pmath].

Thus, we can solve the problem if we figure out the product of all the unique positive divisors of 36. We dont 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 (=[pmath]6^2[/pmath])

2 18 = 36 (=[pmath]6^2[/pmath])

3 12 = 36 (=[pmath]6^2[/pmath])

4 9 = 36 (=[pmath]6^2[/pmath])

And finally theres 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-squareds ([pmath]6^2[/pmath]) and the final 6 is

[pmath]6^2[/pmath] [pmath]6^2[/pmath] [pmath]6^2[/pmath] [pmath]6^2[/pmath] 6 = [pmath]6^9[/pmath].

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 [pmath]n^2[/pmath] as [pmath]p^2q^2[/pmath]. (By the way, if n were allowed to be a cube of a prime number (such that n = [pmath]p^3[/pmath]), then the product of all of n's factors would also be [pmath]n^2[/pmath], but this possibility is explicitly ruled out.)

In terms of p and q, then, the unique factors of [pmath]n^2[/pmath] would then be as follows:

1, p,[pmath]p^2[/pmath]

q, qp, [pmath]qp^2[/pmath]

[pmath]q^2[/pmath], [pmath]q^2[/pmath]p, [pmath]q^2p^2[/pmath]

The product of all these factors is [pmath]q^9p^9[/pmath], or [pmath]n^9[/pmath].

The correct answer is E.

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