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:

Write $110n^3=p_1^{r_1}p_2^{r_2}\cdots$ so that the number of divisors is $(r_1+1)(r_2+1)\cdots=110.$ Since $110=2\cdot5\cdot11,$ there are exactly three distinct primes, which must be $2,5,11,$ with exponents $1,4,10$ in some order.

Taking $r_1=1,$ $r_2=4,$ $r_3=10$ for the primes $2,5,11$ gives $n^3=\dfrac{2^1\cdot 5^4\cdot 11^{10}}{2\cdot5\cdot11}=5^3\cdot 11^9,\quad\text{so}\quad n=5\cdot 11^3.$

Then $81n^4=3^4\cdot 5^4\cdot 11^{12},$ and since $3,5,11$ are distinct primes, the number of divisors is $(4+1)(4+1)(12+1)=5\cdot5\cdot13=325.$

Thus, the correct answer is D.

2016 AMC 12A Problem 18

Solution:

Problem 18 in Other Years