Solution: $\mathbf{C}$
Let the input factors be $p, q, r, s$ and $t$ along the top row, and $v, w, x$, $y$ and $z$ in the first column, as shown in the grid on the right.
We see that $w p=15$ and $x p=18$, so that $p$ is a divisor of both 15 and 18 , and hence is either 1 or 3 . However, if $p=1$, then it would follow that $w=15$. But this is not possible as $w r=40$. We deduce that $p=3$. It follows that $w=5$ and $x=6$. Since $w=5$ and $w r=40$, we have that $r=8$. Therefore, as $z r=56, z=7$. Also, as $x=6$ and $x s=60, s=10$. Since $v s=20$, it follows that $v=2$. Then, as $v q=10, q=5$. Hence, as $y q=20$, we have that $y=4$. Finally, as $y t=24$, we deduce that $t=6$.

This enables us to complete the table, as shown on the left. (Though, really, we need only to calculate the diagonal entries that we have shown in bold.)
Therefore, we have $\mathrm{A}+\mathrm{B}+\mathrm{C}+\mathrm{D}+\mathrm{E}=6+25+48+40+42=161$.
13. In the multiplication grid on the right, the input factors (in the first row and the first column) are all missing and only some of the products within the table have been given.
What is the value of $\mathrm{A}+\mathrm{B}+\mathrm{C}+\mathrm{D}+\mathrm{E}$ ?
A 132
B 145
C 161
D 178
E 193