Difference between revisions of "2018 AMC 8 Problems/Problem 18"
(→Problem 18) |
m (→Solution) |
||
Line 5: | Line 5: | ||
==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>(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 just add one to our powers and multiply. Therefore, the answer is <math>(6+1)*(1+1)*(2+1)=7*2*3=\boxed{42}, \textbf{(E)}</math> |
+ | |||
==Solution 2== | ==Solution 2== | ||
Revision as of 20:43, 20 June 2019
Contents
Problem 18
How many positive factors does 23,232 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
Solution 2
Observe that , so this is
of
which is
, which has
factors. The answer is
.
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.