Difference between revisions of "2016 AMC 10A Problems/Problem 22"
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==Solution== | ==Solution== | ||
| + | ===Solution 1=== | ||
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. | ||
| − | ==Solution== | + | ===Solution 2=== |
<math>110n^3</math> clearly has at least three distinct prime factors, namely 2, 5, and 11. | <math>110n^3</math> clearly has at least three distinct prime factors, namely 2, 5, and 11. | ||
Revision as of 07:19, 4 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
Solution 1
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.
Solution 2
clearly has at least three distinct prime factors, namely 2, 5, and 11.
Further, since the number of factors of 
is 
when the 
'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 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 all of the form 
.
thus has prime factorization 
and a factor count of 
See Also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
