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

Revision as of 12:35, 9 November 2020

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

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 $\boxed{\textbf{(E) }42}$ .

2018 AMC 8 (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 AJHSME/AMC 8 Problems and Solutions

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

@@ Line 10: / Line 10: @@
 We can first find the prime factorization of <math>23,232</math>, which is <math>2^6\cdot3^1\cdot11^2</math>. Now, we just add one to our powers and multiply. Therefore, the answer is <math>(6+1)\cdot(1+1)\cdot(2+1)=7\cdot2\cdot3=\boxed{\textbf{(E) }42}</math>
-==Solution 25678==
+==Solution 2==
 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>.

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