Problem:
If n is a positive integer such that 2n has 28 positive divisors and 3n has 30 positive divisors, then how many positive divisors does 6n have?
Answer Choices:
A. 32
B. 34
C. 35
D. 36
E. 38
Solution:
Let 2e13e25e3⋯ be the prime factorization of n. Then the number of positive divisors of n is (e1+1)(e2+1)(e3+1)⋯. In view of the given information, we have
28=(e1+2)(e2+1)P
and
30=(e1+1)(e2+2)P,
where P=(e3+1)(e4+1)⋯. Subtracting the first equation from the second, we obtain 2=(e1−e2)P, so either e1−e2=1 and P=2, or e1−e2=2 and P=1. The first case yields 14=(e1+2)e1 and (e1+1)2=15; since e1 is a nonnegative integer, this is impossible. In the second case, e2=e1−2 and 30=(e1+1)e1, from which we find e1=5 and e2=3. Thus n=2533, so 6n=2634 has (6+1)(4+1)=35 positive divisors.