Difference between revisions of "2013 AMC 8 Problems/Problem 15"
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==Problem== | ==Problem== | ||
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If <math>3^p + 3^4 = 90</math>, <math>2^r + 44 = 76</math>, and <math>5^3 + 6^s = 1421</math>, what is the product of <math>p</math>, <math>r</math>, and <math>s</math>? | If <math>3^p + 3^4 = 90</math>, <math>2^r + 44 = 76</math>, and <math>5^3 + 6^s = 1421</math>, what is the product of <math>p</math>, <math>r</math>, and <math>s</math>? | ||
<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> | ||
− | ==Solution== | + | ==Video Solution by OmegaLearn== |
+ | https://youtu.be/7an5wU9Q5hk?t=301 | ||
− | + | ~ pi_is_3.14 | |
− | + | ==Video Solution 2== | |
+ | https://youtu.be/ew7QnjAAHcw ~savannahsolver | ||
− | <math> | + | ==Solution== |
− | + | ===Solution 1: Solving=== | |
− | <math> | + | First, we're going to solve for <math>p</math>. Start with <math>3^p+3^4=90</math>. Then, change <math>3^4</math> to <math>81</math>. Subtract <math>81</math> from both sides to get <math>3^p=9</math> and see that <math>p</math> is <math>2</math>. Now, solve for <math>r</math>. Since <math>2^r+44=76</math>, <math>2^r</math> must equal <math>32</math>, so <math>r=5</math>. Now, solve for <math>s</math>. <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>. <math>prs</math> equals <math>2*5*4</math> which equals <math>40</math>. So, the answer is <math>\boxed{\textbf{(B)}\ 40}</math>. |
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− | <math>6 | ||
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− | <math>6 | ||
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− | <math> | ||
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− | Therefore the answer is <math> | + | ===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>. 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 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.