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> | ||
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==Solution== | ==Solution== | ||
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>7*2*3=\boxed{42}, \textbf{(E)}</math> -shreyasb | 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>7*2*3=\boxed{42}, \textbf{(E)}</math> -shreyasb | ||
| − | + | ==See Also== | |
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{{AMC8 box|year=2018|num-b=17|num-a=19}} | {{AMC8 box|year=2018|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:42, 21 November 2018
Problem 18
How many positive factors does 
have?
Solution
We can first find the prime factorization of , which is 
. Now, we just add one to our powers and multiply. Therefore, the answer is 
-shreyasb
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. 
