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?

$110$

$191$

$261$

$325$

$425$

Solution:

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 $(0+1)(4+1)(12+1)(4+1)=325.$

Thus, the correct answer is D.

2016 AMC 10A Problem 22

Solution:

Problem 22 in Other Years