Difference between revisions of "2018 AMC 8 Problems/Problem 18"
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==Solution 2== | ==Solution 2== | ||
| − | 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>\textbf{(E)}</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}}</math>. |
==See Also== | ==See Also== | ||
Revision as of 19:07, 25 November 2018
Contents
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 
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. 
