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. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC10 box|year=2016|ab=A|num-b=22|num-a=24}} | ||
| + | {{MAA Notice}} | ||
Revision as of 21:39, 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.
See Also
| 2016 AMC 10A (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
