2016 AMC 10A Problems/Problem 22

Problem

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?

$\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425$

Solution

Since the prime factorization of $110$ is $2 \cdot 5 \cdot 11$ , we have that the number is equal to $2 \cdot 5 \cdot 11 \cdot n^3$ . This has $2 \cdot 2 \cdot 2=8$ factors when $n=1$ . This needs a multiple of 11 factors, which we can achieve by setting $n=2^3$ , so we have $2^10 \cdot 5 \cdot 8$ has $44$ factors. To achieve the desired $110$ factors, we need the number of factors to also be divisible by $5$ , so we can set $n=2^3 \cdot 5$ , so $2^10 \cdot 5^4 \cdot 11$ has $110$ factors. Therefore, $n=2^3 \cdot 5$ . In order to find the number of factors of $81n^4$ , we raise this to the fourth power and multiply it by $81$ , and find the factors of that number. We have $3^4 \cdot 2^12 \cdot 5^4$ , and this has $5 \cdot 13 \cdot 5=\boxed{\textbf{(D) }325}$ factors.

2016 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2016 AMC 10A Problems/Problem 22

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