Difference between revisions of "2016 AMC 10A Problems/Problem 22"
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<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== | + | ==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 | + | 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 11</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. | ||
− | + | The number of factors of <math>p_1^{n_1}\cdots p_k^{n_k}</math> is <math>(n_1+1)\cdots(n_k+1)</math> when the <math>p</math>'s are distinct primes. This tells us that none of these factors can be 1. The number of factors is given as 110. The only way to write 110 as a product of at least three factors without <math>1</math>s is <math>2\cdot 5\cdot 11</math>. | |
We conclude that <math>110n^3</math> 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 <math>n</math> all of the form <math>n=p_1\cdot p_2^3</math>. | We conclude that <math>110n^3</math> 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 <math>n</math> all of the form <math>n=p_1\cdot p_2^3</math>. | ||
<math>81n^4</math> thus has prime factorization <math>81n^4=3^4\cdot p_1^4\cdot p_2^{12}</math> and a factor count of <math>5\cdot5\cdot13=\boxed{\textbf{(D) }325}</math> | <math>81n^4</math> thus has prime factorization <math>81n^4=3^4\cdot p_1^4\cdot p_2^{12}</math> and a factor count of <math>5\cdot5\cdot13=\boxed{\textbf{(D) }325}</math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/hAEsBh64VSY | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/ZhAZ1oPe5Ds?t=3694 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=A|num-b=21|num-a=23}} | {{AMC10 box|year=2016|ab=A|num-b=21|num-a=23}} | ||
+ | {{AMC12 box|year=2016|ab=A|num-b=17|num-a=19}} | ||
+ | |||
+ | [[Category:Intermediate Number Theory Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:27, 17 January 2021
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 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.
The number of factors of is when the 's are distinct primes. This tells us that none of these factors can be 1. The number of factors is given as 110. The only way to write 110 as a product of at least three factors without s is .
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
Video Solution
~savannahsolver
Video Solution
https://youtu.be/ZhAZ1oPe5Ds?t=3694
~ pi_is_3.14
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 |
2016 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 12 Problems and Solutions |
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