Start with $3^{}p+3^{4}=90$. Then, change $3^{4}$ to $81$. Subtract from $81$ 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 $2\times5\times4$ which equals $40$. So, the answer is $40$. 
If $3^{}p+3^{4}=90$, $2^{}r+44=76$, and $5^{3}+6^{}s=1421$, what is the product of $p$, $r$, and $s$? (2013 AMC 8 Problem, Question \#15) 

A. $$27$$ 
B. $$40$$ 
C. $$50$$ 
D. $$70$$ 
E. $$90$$ 