Difference between revisions of "2018 AMC 8 Problems/Problem 18"
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==Solution 2 (Using 264^2)== | ==Solution 2 (Using 264^2)== | ||
− | Observe that <math>69696</math> = <math>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>. | + | Observe that <math>69696</math> = <math>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>. Very true |
==Video Solution == | ==Video Solution == |
Latest revision as of 13:50, 14 August 2025
Contents
Problem
How many positive factors does have?
Solution 1
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 .
A third way to factor it is to observe . Factoring out the 3 gives us
. Since
and
, we have
.
Solution 2 (Using 264^2)
Observe that =
, so this is
of
which is
, which has
factors. The answer is
. Very true
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.