Difference between revisions of "2013 AMC 8 Problems/Problem 15"
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===Solution 2: Process of Elimination=== | ===Solution 2: Process of Elimination=== | ||
− | First, we solve for <math>s</math>. As Solution 1 perfectly states, <math>5^3+6^s=1421</math> can be simplified to <math>125+6^s=1421</math> which simplifies further to <math>6^s=1296</math>. Therefore, <math>s=4</math>. The only answer choice that is divisible by <math>4</math> is <math>\boxed{\textbf{(B)}\ 40}</math>. | + | First, we solve for <math>s</math>. As Solution 1 perfectly states, <math>5^3+6^s=1421</math> can be simplified to <math>125+6^s=1421</math> which simplifies further to <math>6^s=1296</math>. Therefore, <math>s=4</math>. We know that you cannot take a root of any of the numbers raised to <math>p</math>, <math>r</math>, or <math>s</math> and get a rational answer, and none of the answer choices are irrational, so that rules out the possibility that <math>p</math>, <math>r</math>, or <math>s</math> is a fraction. The only answer choice that is divisible by <math>4</math> is <math>\boxed{\textbf{(B)}\ 40}</math>. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2013|num-b=14|num-a=16}} | {{AMC8 box|year=2013|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 17:21, 25 September 2021
Contents
Problem
If , , and , what is the product of , , and ?
Video Solution
https://youtu.be/7an5wU9Q5hk?t=301
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 |
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