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

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+ | ==Problem== | ||

For some positive integer <math>n</math>, the number <math>110n^3</math> has <math>110</math> positive integer divisors, including <math>1</math> and the number <math>110n^3</math>. How many positive integer divisors does the number <math>81n^4</math> have? | For some positive integer <math>n</math>, the number <math>110n^3</math> has <math>110</math> positive integer divisors, including <math>1</math> and the number <math>110n^3</math>. How many positive integer divisors does the number <math>81n^4</math> have? | ||

<math>\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425</math> | <math>\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425</math> | ||

+ | |||

+ | ==Solution== | ||

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

## Revision as of 19:24, 3 February 2016

## Problem

For some positive integer , the number has positive integer divisors, including and the number . How many positive integer divisors does the number have?

## Solution

Since the prime factorization of is , we have that the number is equal to . This has factors when . This needs a multiple of 11 factors, which we can achieve by setting , so we have has factors. To achieve the desired factors, we need the number of factors to also be divisible by , so we can set , so has factors. Therefore, . In order to find the number of factors of , we raise this to the fourth power and multiply it by , and find the factors of that number. We have , and this has factors.