The product of three consecutive positive integers is $8$ times their sum. What is the sum of their [squares](https://artofproblemsolving.com/wiki/index.php/Perfect_square)?
$\mathrm{(A)}\ 50 \qquad\mathrm{(B)}\ 77 \qquad\mathrm{(C)}\ 110 \qquad\mathrm{(D)}\ 149 \qquad\mathrm{(E)}\ 194$
Let the three consecutive positive integers be $a-1$, $a$, and $a+1$. Since the mean is $a$, the sum of the integers is $3a$. So $8$ times the sum is just $24a$. With this, we now know that $a(a-1)(a+1)=24a\Rightarrow(a-1)(a+1)=24$.  $24=4\times6$, so $a=5$. Hence, the sum of the squares is $4^2+5^2+6^2=\boxed{\mathrm{ (B)}\ 77}$.