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

## 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 (Problems • Answer Key • Resources) Preceded byProblem 17 Followed byProblem 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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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