Difference between revisions of "2018 AMC 8 Problems/Problem 18"

(Solution 2)
(Solution 2)
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==Solution 2==
 
==Solution 2==
  
Observe that <math>69696% = </math>264^2<math>, so this is </math>\frac{1}{3}<math> of </math>264^2<math> which is </math>88 \cdot 264 = 11^2 \cdot 8^2 \cdot 3 = 11^2 \cdot 2^6 \cdot 3<math>, which has </math>3 \cdot 7 \cdot 2 = 42<math> factors. The answer is </math>\boxed{\textbf{(E) }42}$.
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Observe that <math>69696</math> = <math>264^2</math>, so this is <math>\frac{1}{3}</math> of <math>264^2</math> which is <math>88 \cdot 264 = 11^2 \cdot 8^2 \cdot 3 = 11^2 \cdot 2^6 \cdot 3</math>, which has <math>3 \cdot 7 \cdot 2 = 42</math> factors. The answer is <math>\boxed{\textbf{(E) }42}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 19:55, 8 November 2019

Problem 18

How many positive factors does 23,232 have?

$\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }42$

Solution

We can first find the prime factorization of $23,232$, which is $2^6\cdot3^1\cdot11^2$. Now, we just add one to our powers and multiply. Therefore, the answer is $(6+1)\cdot(1+1)\cdot(2+1)=7\cdot2\cdot3=\boxed{\textbf{(E) }42}$

Solution 2

Observe that $69696$ = $264^2$, so this is $\frac{1}{3}$ of $264^2$ which is $88 \cdot 264 = 11^2 \cdot 8^2 \cdot 3 = 11^2 \cdot 2^6 \cdot 3$, which has $3 \cdot 7 \cdot 2 = 42$ factors. The answer is $\boxed{\textbf{(E) }42}$.

See Also

2018 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AJHSME/AMC 8 Problems and Solutions

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