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Competency in Focus: Powers of Numbers This problem from American Mathematics contest (AMC 8, 2013) is based on basic  algebra and Powers of Numbers.

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If $3^p + 3^4 = 90$$2^r + 44 = 76$, and $5^3 + 6^s = 1421$, what is the product of $p$$r$, and $s$?
Source of the problem
American Mathematical Contest 2013, AMC 8 Problem 15
Key Competency
Basic algebra and Powers of Numbers
Difficulty Level
4/10
Suggested Book
Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics 

Start with hints 

Do you really need a hint? Try it first!
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 $3^p=9$ .Now we can write 9 as \(3^2\) .So, from here we can say that p=2.
Now, solve for $r$. Since $2^r+44=76$$2^r$ must equal $32$,  and 32 can be written as \( 2^5 \) .So from here we have r=5.
Similarly now, solve for $s$$5^3+6^s=1421$ can be simplified to $125+6^s=1421$ which simplifies further to $6^s=1296$=\(6^4\) , which gives s=4.
Lastly, $prs$ equals $2*5*4$ which equals $40$. So, the answer is 40.

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