1996 AMC 12 Problem 29

Question

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?

32

34

35

36

38

Write

n=2^a3^bq

, where

q

is relatively prime to

6

, and let

t

be the number of divisors of

q

Use the two divisor-count equations to restrict

t

to a divisor of both

28

and

30

Write

n=2^a3^bq

with

\gcd(q,6)=1,

and let

t=\tau(q).

Then

\begin{aligned} (a+2)(b+1)t&=28,\\ (a+1)(b+2)t&=30. \end{aligned}

Hence

t

is

1

or

2.

Checking the factor pairs of

14,15

gives no solution when

t=2.

Checking those of

28,30

when

t=1

gives the unique solution

a=5,b=3.

Therefore

\begin{aligned} \tau(6n)&=(a+2)(b+2)t\\ &=7\cdot5=35, \end{aligned}

so the correct answer is C.

Problem 29 in other years