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 2e1​3e2​5e3​⋯ 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.