Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 2 (Using 69))
(→‎Solution 1)
 
Line 12: Line 12:
  
 
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>.
 +
 +
A third way to factor it is to observe <math>23232 = 24000 - 768</math>. Factoring out the 3 gives us <math>3(8000 - 256)</math>. Since <math>8000 = 2^6 \cdot 5^3</math> and <math>256 = 2^8</math>, we have <math>2^6 \cdot 3 (5^3 - 2^2) = 2^6 \cdot 3 (125-4) = 2^6 \cdot 3 \cdot 121 = 2^6 \cdot 3 \cdot 11^2</math>.
  
 
==Solution 2 (Using 264^2)==
 
==Solution 2 (Using 264^2)==

Latest revision as of 13:44, 16 January 2025

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 $23232 = 32\cdot(101 + 10000/16) = 32\cdot (101+ 5^4) = 32\cdot 726 = 32 \cdot 11 \cdot 66$.

A third way to factor it is to observe $23232 = 24000 - 768$. Factoring out the 3 gives us $3(8000 - 256)$. Since $8000 = 2^6 \cdot 5^3$ and $256 = 2^8$, we have $2^6 \cdot 3 (5^3 - 2^2) = 2^6 \cdot 3 (125-4) = 2^6 \cdot 3 \cdot 121 = 2^6 \cdot 3 \cdot 11^2$.

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_18&oldid=240259"