Difference between revisions of "2018 AMC 8 Problems/Problem 18"
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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>(1+6)*(1+1)*(1+2)=7*2*3=\boxed{42}, \textbf{(E)}</math> -shreyasb |
==See Also== | ==See Also== | ||
Revision as of 12:11, 24 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 | ||
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