20. In the addition sum shown, $J, K$ and $L$ stand for different digits. What is the value of $J+K+L$ ?
A 6
B 8
C 9
D 10
E 11
$$
\begin{array}{r}
J K L \\
J L L \\
+J K L \\
\hline 479
\end{array}
$$
SolUtion
$\mathbf{E}$

When we consider the units column, we have $3 L=9$ or $3 L=19$. Since only 9 is divisible by $3, L=3$. Then when we consider the tens column, we have $2 K+3=7$ or $2 K+3=17$. Therefore $K=2$ or $K=7$. However, if $K=2$, the hundreds column would tell us that $3 J=4$, which is not possible. Hence $K=7$. Now, when we consider the hundreds column, we have $3 J+1=4$, which has solution $J=1$. Therefore the value of $J+K+L$ is $1+7+3=11$.