1991 AMC 12 Problem 14

Attempt Problem 14 of the 1991 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1991 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

14.

x

is the cube of a positive integer and

d

is the number of positive integers that are divisors of

x,

then

d

could be

200

201

202

203

204

In the prime factorization of a cube, every exponent is divisible by

3

Try a cube having only one prime factor

For any prime

p,

take

x=p^{201}=(p^{67})^3.

This is a cube, and its positive divisors are

1,p,p^2,\ldots,p^{201},

a total of

202.

Thus

202

can occur.

Thus the correct answer is C.

Problem 14 in other years