A positive integer N has exactly 6 positive divisors, and 2N has exactly 9 positive divisors. How many positive divisors does N² have?
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Answer & Explanation
Correct answer
D
The whole problem turns on one fact: a number with exactly 6 divisors has two possible prime shapes, and only the second condition tells you which one N has. Because 6 = 6 or 6 = 2 × 3, N is either p⁵ or p²q for distinct primes p and q.
Now use the second condition, that 2N has exactly 9 divisors. Because 9 = 9 or 9 = 3 × 3, a number with exactly 9 divisors is either p⁸ or p²q² for distinct primes. The p⁸ shape is impossible here: since 2 divides 2N, that prime would have to be 2, making N = 2⁷, which has 8 divisors, not 6. So 2N must be two distinct primes, each squared, and one of those primes has to be 2 (since 2 divides 2N). So 2N = 2²q², and halving gives N = 2q² for some odd prime q. Check the first condition: N = 2¹q² has (1 + 1)(2 + 1) = 6 divisors, exactly as required.
The same condition rules out the other shape. If N were p⁵, then 2N would have 12 divisors (or 7 when p = 2), never 9, so that branch is impossible. This is why 11 is unreachable: 11 is the divisor count of (p⁵)² = p¹⁰, and N can never be p⁵.
To finish, square N: N² = 2²q⁴, which has (2 + 1)(4 + 1) = 15 divisors. Divisor counts come from the exponents, and the exponents give 15. The answer is 15, choice D.