What are we learning ?

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

First look at the knowledge graph.

Next understand the 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 $s$?
Source of the problem
American Mathematical Contest 2013, AMC 8 Problem 15
Key Competency
Basic algebra and Powers of Numbers
Difficulty Level
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.

Connected Program at Cheenta

Amc 8 Master class

Cheenta AMC Training Camp consists of live group and one on one classes, 24/7 doubt clearing and continuous problem solving streams.

Similar Problems