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Difference between revisions of "2013 AMC 8 Problems/Problem 15"

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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>
  
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==Video Solution==
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==Video Solution by OmegaLearn==
 
https://youtu.be/7an5wU9Q5hk?t=301
 
https://youtu.be/7an5wU9Q5hk?t=301
 +
 +
~ pi_is_3.14
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 +
==Video Solution 2==
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https://youtu.be/ew7QnjAAHcw ~savannahsolver
  
 
==Solution==
 
==Solution==
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===Solution 2: Process of Elimination===
 
===Solution 2: Process of Elimination===
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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>.
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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

Problem

If $3^p + 3^4 = 90$, $2^r + 44 = 76$, and $5^3 + 6^s = 1421$, what is the product of $p$, $r$, and $s$?

$\textbf{(A)}\ 27 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 70 \qquad \textbf{(E)}\ 90$

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 $p$. Start with $3^p+3^4=90$. Then, change $3^4$ to $81$. Subtract $81$ from both sides to get $3^p=9$ and see that $p$ is $2$. Now, solve for $r$. Since $2^r+44=76$, $2^r$ must equal $32$, so $r=5$. Now, solve for $s$. $5^3+6^s=1421$ can be simplified to $125+6^s=1421$ which simplifies further to $6^s=1296$. Therefore, $s=4$. $prs$ equals $2*5*4$ which equals $40$. So, the answer is $\boxed{\textbf{(B)}\ 40}$.

Solution 2: Process of Elimination

First, we solve for $s$. As Solution 1 perfectly states, $5^3+6^s=1421$ can be simplified to $125+6^s=1421$ which simplifies further to $6^s=1296$. Therefore, $s=4$. We know that you cannot take a root of any of the numbers raised to $p$, $r$, or $s$ and get a rational answer, and none of the answer choices are irrational, so that rules out the possibility that $p$, $r$, or $s$ is a fraction. The only answer choice that is divisible by $4$ is $\boxed{\textbf{(B)}\ 40}$.

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. AMC logo.png

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