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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>
  
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==Solution==
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==Solution 1==
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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 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.
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==Solution 2==
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<math>110n^3</math> clearly has at least three distinct prime factors, namely 2, 5, and 11.
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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>.
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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>.
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<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>
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==Video Solution==
 +
https://youtu.be/hAEsBh64VSY
 +
 
 +
~savannahsolver
 +
 
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/ZhAZ1oPe5Ds?t=3694
 +
 
 +
~ pi_is_3.14
 +
 
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=IxIinK0p8o4&list=PLyhPcpM8aMvJewbDgg6W1h6WjWqP-v4Uy&index=2
 +
 
 +
- AMBRIGGS
  
 
==See Also==
 
==See Also==
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{{AMC10 box|year=2016|ab=A|num-b=22|num-a=24}}
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{{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 18:27, 30 December 2022

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 1

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 $2^{10} \cdot 5 \cdot 11$ 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.

Solution 2

$110n^3$ clearly has at least three distinct prime factors, namely 2, 5, and 11.

The number of factors of $p_1^{n_1}\cdots p_k^{n_k}$ is $(n_1+1)\cdots(n_k+1)$ when the $p$'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 $1$s is $2\cdot 5\cdot 11$.

We conclude that $110n^3$ 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 $n$ all of the form $n=p_1\cdot p_2^3$.

$81n^4$ thus has prime factorization $81n^4=3^4\cdot p_1^4\cdot p_2^{12}$ and a factor count of $5\cdot5\cdot13=\boxed{\textbf{(D) }325}$

Video Solution

https://youtu.be/hAEsBh64VSY

~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/ZhAZ1oPe5Ds?t=3694

~ pi_is_3.14

Video Solution

https://www.youtube.com/watch?v=IxIinK0p8o4&list=PLyhPcpM8aMvJewbDgg6W1h6WjWqP-v4Uy&index=2

- AMBRIGGS

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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