Difference between revisions of "2016 AMC 10A Problems/Problem 22"

Revision as of 12:20, 4 February 2016

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

Solution 1

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.

Solution 2

$110n^3$ clearly has at least three distinct prime factors, namely 2, 5, and 11.

Further, since the number of factors of $p_1^{n_1}\cdots p_k^{n_k}$ is $(n_1+1)\cdots(n_k+1)$ when the $p$ 's are distinct primes, we see that there can be at most three distinct prime factors for a number with 110 factors.

We conclude that $110n^3$ has only the three prime factors 2, 5, and 11 and that the multiplicities are 1, 4, and 10 in some order. I.e., there are six different possible values of $n$ all of the form $n=p_1\cdot p_2^3$ .

$81n^4$ thus has prime factorization $81n^4=3^4\cdot p_1^4\cdot p_2^{12}$ and a factor count of $5\cdot5\cdot13=\boxed{\textbf{(D) }325}$

2016 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2016 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 19: / Line 19: @@
 ==See Also==
 {{AMC10 box|year=2016|ab=A|num-b=21|num-a=23}}
+{{AMC12 box|year=2016|ab=A|num-b=17|num-a=19}}
 {{MAA Notice}}

Page

Toolbox

Search