If $3^p + 3^4 = 90$, $2^r + 44 = 76$, and $5^3 + 6^s = 1421$, what is the product of $p$, $r$, and $s$?
$\textbf{(A)}\ 27 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 70 \qquad \textbf{(E)}\ 90$
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 $\boxed{\textbf{(B)}\ 40}$.