2018 AMC 8 Problems/Problem 18

Revision as of 21:32, 12 August 2020 by Pi is 3.14 (talk | contribs) (Solution 1)

Problem 18

How many positive factors does 23,232 have?

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

Video Solution

https://youtu.be/6xNkyDgIhEE?t=1515

Solution 1

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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