Problem:
If 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
Answer Choices:
A. 200
B. 201
C. 202
D. 203
E. 204
Solution:
The cubes
x=1,23,26,…,23k,…,(267)3
have
d=1,4,7,…,3k+1,…,202
divisors, respectively. In fact, for any prime p,(p67)3 has 202 divisors.
To show that, of the choices listed, d=202 is the only possible answer, we prove that for any perfect cube x>1,d must be of the form 3k+1:
If x=p13b1p23b2⋯pn3bn where the pi are distinct primes, then its divisors are all the numbers of the form p1a1p2a2⋯pnan, with 0≤ai≤3bi for i=1,2,…,n. Taking the product of the number of choices for each ai yields d=(3b1+1)(3b2+1)⋯(3bn+1)=3k+1 for some integer k.