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 6-divisor number has two possible prime shapes, and only the 2N condition tells you which one you hold. Since 6 = 6 or 6 = 2 × 3, N is either p⁵ or p²q for distinct primes p and q. Now use the second condition. Since 9 = 3 × 3, the number 2N must be two distinct primes each squared, and one of those primes must be 2 because 2 divides 2N. So 2N = 2²q², and halving gives N = 2q² for an odd prime q.
Check the first condition: (1 + 1)(2 + 1) = 6 divisors, as required. The same condition kills the other branch: if N were p⁵, then 2N would have 12 divisors, or 7 if p = 2, never 9, which is why choice B's 11 is unreachable. Now square: N² = 2²q⁴, with (2 + 1)(4 + 1) = 15 divisors.
The tempting moves are the fake rules, doubling the count (6 × 2 = 12, choice C) or squaring it (6² = 36, choice E); neither survives even the smallest example, since 2 has two divisors but 4 has three. Choice A merely repeats the 9 from the stem. Divisor counts come from exponents, and the exponents say 15: the answer is choice D.