2016 AMC 10A Problem 22

Question

For some positive integer

n,

the number

110n^3

has

110

positive integer divisors, including

1

and the number

110n^3.

How many positive integer divisors does the number

81n^4

have?

110

191

261

325

425

The divisor count

110=2\cdot5\cdot11

forces exponents

1,4,10

Those exponents are all

1\pmod3

, so they can come from

110n^3

The prime factorization is

110=2\cdot5\cdot11

. If

110n^3

has

110=2\cdot5\cdot11

divisors, then it has exactly three prime factors, with exponents

1,4,10

in some order.

For each of the primes

2,5,11

, its exponent in

110n^3

is

1

more than a multiple of

3

. The exponents

1,4,10

all have this form, so the corresponding exponents in

n

are

0,1,3

in some order.

In

81n^4=3^4n^4

, the exponents from

n^4

are therefore

0,4,12

, in some order, along with the exponent

4

on prime

3

. Hence the divisor count is

\begin{aligned} &(0+1)\cdot(4+1)\cdot(12+1) \\ &\quad {}\cdot(4+1)=325. \end{aligned}

Thus, the correct answer is D.

Problem 22 in other years