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

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==Solution==
 
==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.
 
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.
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==See Also==
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{{AMC10 box|year=2016|ab=A|num-b=22|num-a=24}}
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{{MAA Notice}}

Revision as of 21:39, 3 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

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.

See Also

2016 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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

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