Select Page

# 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.

# 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
4/10
##### Suggested Book
Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics

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.