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

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==Solution 1==
 
==Solution 1==
 
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>
 
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>
 +
 +
23232 is a large number, so we can look for shortcuts to factor it.
 +
One way to factor it quickly is use 3 and 11 divisibility rules to observe that <math>23232 = 3 \cdot 7744 = 3 \cdot 11 \cdot 704 = 3 \cdot 11^2 \cdot 64 = 3^1  \cdot 11^2  \cdot  2^6</math>.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 13:36, 2 January 2023

Problem

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 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 $(6+1)\cdot(1+1)\cdot(2+1)=7\cdot2\cdot3=\boxed{\textbf{(E) }42}$

23232 is a large number, so we can look for shortcuts to factor it. One way to factor it quickly is use 3 and 11 divisibility rules to observe that $23232 = 3 \cdot 7744 = 3 \cdot 11 \cdot 704 = 3 \cdot 11^2 \cdot 64 = 3^1 \cdot 11^2 \cdot 2^6$.

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

Video Solution by OmegaLearn

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

~ pi_is_3.14

Video Solution

https://youtu.be/sC2-sdUjm40

~savannahsolver

See Also

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

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