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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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==Problem==
 
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Solve for <math>p</math>, <math>r</math>, and <math>s</math>.
 
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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>.
 
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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>
 
 
==Solution==
 
==Solution==
 
First, we're going to solve for <math>p</math>. Start with <math>3^p+3^4=90</math>. Then, change 3^4 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>.
 
First, we're going to solve for <math>p</math>. Start with <math>3^p+3^4=90</math>. Then, change 3^4 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>.

Revision as of 20:43, 11 November 2017

Problem $6\cdot6=36$

Solve for s, x, and r.

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

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

Solution

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 $\boxed{\textbf{(B)}\ 40}$.

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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