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

(Solution 2)
(Solution 2)
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<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 the product cannot contain any <math>1</math>'s. 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>.
+
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 the product cannot contain any <math>1</math>'s. 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>.

Revision as of 21:42, 23 July 2017

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 the product cannot contain any $1$'s. 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 $5\cdot5\cdot13=\boxed{\textbf{(D) }325}$

See Also

2016 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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