Difference between revisions of "2018 AMC 8 Problems/Problem 18"
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==Solution 1== | ==Solution 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 | + | We can first find the prime factorization of <math>23,232</math>, which is <math>2^6\cdot3^1\cdot11^2</math>. Now, we add one to our powers and multiply. Therefore, the answer is <math>(6+1)\cdot(1+1)\cdot(2+1)=7\cdot2\cdot3=\boxed{\textbf{(E) }42}</math> |
Note: | Note: | ||
23232 is a large number, so we can look for shortcuts to factor it. | 23232 is a large number, so we can look for shortcuts to factor it. | ||
− | One way to factor it quickly is use 3 and 11 divisibility rules to observe that <math>23232 = 3 \cdot 7744 = 3 \cdot 11 \cdot 704 = 3 \cdot 11^2 \cdot 64 = 3^1 \cdot 11^2 \cdot 2^6</math>. | + | One way to factor it quickly is to use 3 and 11 divisibility rules to observe that <math>23232 = 3 \cdot 7744 = 3 \cdot 11 \cdot 704 = 3 \cdot 11^2 \cdot 64 = 3^1 \cdot 11^2 \cdot 2^6</math>. |
Another way is to spot the "32" and compute that <math>23232 = 32\cdot(101 + 10000/16) = 32\cdot (101+ 5^4) = 32\cdot 726 = 32 \cdot 11 \cdot 66</math>. | Another way is to spot the "32" and compute that <math>23232 = 32\cdot(101 + 10000/16) = 32\cdot (101+ 5^4) = 32\cdot 726 = 32 \cdot 11 \cdot 66</math>. | ||
− | ==Solution 2 (Using | + | ==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>. |
Latest revision as of 20:12, 18 January 2024
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 .
Solution 2 (Using 264^2)
Observe that =
, so this is
of
which is
, which has
factors. The answer is
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.