Difference between revisions of "2018 AMC 8 Problems/Problem 18"
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| − | ==Problem | + | ==Problem== |
| − | 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== | + | ==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 | + | 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: |
| + | 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 <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>. | ||
| − | Observe that <math>69696 = 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 | + | 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 (Using 264^2)== | ||
| + | |||
| + | 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>. | ||
| + | |||
| + | ==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== | ==See Also== | ||
Latest revision as of 20:12, 18 January 2024
Contents
Problem
How many positive factors does 
have?
Solution 1
We can first find the prime factorization of , which is 
. Now, we add one to our powers and multiply. Therefore, the answer is 
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 
.
Another way is to spot the "32" and compute that 
.
Solution 2 (Using 264^2)
Observe that 
= 
, so this is 
of which is 
, which has 
factors. The answer is 
.
Video Solution
~Education, the Study of Everything
Video Solution by OmegaLearn
https://youtu.be/6xNkyDgIhEE?t=1515
~ pi_is_3.14
Video Solution
~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. 
