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=p13b1​​p23b2​​⋯pn3bn​​ where the pi​ are distinct primes, then its divisors are all the numbers of the form p1a1​​p2a2​​⋯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.