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

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==Problem==
 
==Problem==
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How many positive factors does 23,232 have?
+
How many positive factors does <math>23,232</math> have?
  
 
<math>\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }42</math>
 
<math>\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }42</math>
  
 
==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 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>
  
 
Note:
 
Note:
 
23232 is a large number, so we can look for shortcuts to factor it.  
 
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>.
+
One way to factor it quickly is to 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>.
  
 
Another way is to spot the "32" and compute that <math>23232 = 32\cdot(101 + 10000/16) = 32\cdot (101+ 5^4) = 32\cdot 726 = 32 \cdot 11 \cdot 66</math>.
 
Another way is to spot the "32" and compute that <math>23232 = 32\cdot(101 + 10000/16) = 32\cdot (101+ 5^4) = 32\cdot 726 = 32 \cdot 11 \cdot 66</math>.
  
−
==Solution 2==
+
==Solution 2 (Using 264^2)==
  
−
Observe (how???) 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>.
+
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>.
  
−
==Video Solution (CREATIVE ANALYSIS!!!)==
+
==Video Solution ==
 
https://youtu.be/44kIAfS5iJk
 
https://youtu.be/44kIAfS5iJk
  

Latest revision as of 19:12, 18 January 2024

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 add one to our powers and multiply. Therefore, the answer is $(6+1)\cdot(1+1)\cdot(2+1)=7\cdot2\cdot3=\boxed{\textbf{(E) }42}$

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

Another way is to spot the "32" and compute that $23232 = 32\cdot(101 + 10000/16) = 32\cdot (101+ 5^4) = 32\cdot 726 = 32 \cdot 11 \cdot 66$.

Solution 2 (Using 264^2)

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

https://youtu.be/44kIAfS5iJk

~Education, the Study of Everything

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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