Difference between revisions of "2018 AMC 8 Problems/Problem 18"
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<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 30== |
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> | 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> | ||
Revision as of 14:46, 2 January 2024
Contents
Problem
How many positive factors does have?
Solution 30
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 69)
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.