Difference between revisions of "2013 AMC 8 Problems/Problem 15"
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Pi is 3.14 (talk | contribs) |
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<math>\textbf{(A)}\ 27 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 70 \qquad \textbf{(E)}\ 90</math> | <math>\textbf{(A)}\ 27 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 70 \qquad \textbf{(E)}\ 90</math> | ||
− | ==Video Solution== | + | ==Video Solution by OmegaLearn== |
https://youtu.be/7an5wU9Q5hk?t=301 | https://youtu.be/7an5wU9Q5hk?t=301 | ||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution 2== | ||
https://youtu.be/ew7QnjAAHcw ~savannahsolver | https://youtu.be/ew7QnjAAHcw ~savannahsolver | ||
Latest revision as of 02:58, 29 December 2022
Contents
Problem
If , , and , what is the product of , , and ?
Video Solution by OmegaLearn
https://youtu.be/7an5wU9Q5hk?t=301
~ pi_is_3.14
Video Solution 2
https://youtu.be/ew7QnjAAHcw ~savannahsolver
Solution
Solution 1: Solving
First, we're going to solve for . Start with . Then, change to . Subtract from both sides to get and see that is . Now, solve for . Since , must equal , so . Now, solve for . can be simplified to which simplifies further to . Therefore, . equals which equals . So, the answer is .
Solution 2: Process of Elimination
First, we solve for . As Solution 1 perfectly states, can be simplified to which simplifies further to . Therefore, . We know that you cannot take a root of any of the numbers raised to , , or and get a rational answer, and none of the answer choices are irrational, so that rules out the possibility that , , or is a fraction. The only answer choice that is divisible by is .
See Also
2013 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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.