2013 AMC 8 Problems/Problem 15

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

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.

Page

Toolbox

Search