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.