Olympiad Maths Prep

Track / Stage 5 / 44 of 400 #644 of 2000

Problem 644

AIME late
Number theory Difficulty 5.1 Find the answer

9) If nn is a natural number with 6 positive integer divisors, how many positive integer divisors does n2n^{2} have? N.B.: among the divisors of a number, we also count 1 and the number itself.
(A) 11
(B) 12
(C) 15
(D) 36
(E) the answer depends on nn

Official solution

9. The answer is (E)(\mathbf{E}).

Consider the number n=25=32n=2^{5}=32 which has exactly 6 divisors: 1,2,4,8,16,321,2,4,8,16,32. Its square n2=210=1024n^{2}=2^{10}=1024 has exactly 11 divisors, namely all numbers of the form 2k2^{k}, with k=0,1,2,,10k=0,1,2, \ldots, 10. This example rules out answers (B), (C), and (D). On the other hand, if we consider n=12n=12, we see that it also has exactly 6 divisors: 1, 2, 3, 4, 6, 12; its square n2=144=2432n^{2}=144=2^{4} \cdot 3^{2} has as divisors all numbers of the form 2k3h2^{k} \cdot 3^{h} with k=0,1,2,3,4k=0,1,2,3,4 and h=0,1,2h=0,1,2, which are a total of 15. Therefore, answer (A) is not correct, and we see that the number of divisors of n2n^{2} in general depends on nn.

[Problem proposed by A. Bianchi]

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.