Each letter from $A$ to $J$ represents a distinct number from 1 to 10. In the table below, each number written represents the product of the number in its row by the number in its column. For example, $A \cdot F=18$.

| $\times$ | $A$ | $B$ | $C$ | $D$ | $E$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $F$ | 18 |  |  |  |  |
| $G$ |  |  |  | 20 |  |
| $H$ |  | 42 |  |  |  |
| $I$ |  |  |  |  | 24 |
| $J$ |  |  | 10 |  |  |

Using the equation

$$
A+B+C+D+E=F+G+H+I
$$

and knowing the five values in the table, determine the value of each letter from $A$ to $J$.
Solution

Let's start with the number 42, the only way it can appear is with the product 6$\cdot$7. Thus, $\{B, H\}=\{6,7\}$. Next, let's analyze 18 and 24. Since 6 has already been used, we can conclude that $\{A, F\}=\{2,9\}$ and $\{E, I\}=\{3,8\}$. Looking at 10 and 20 and remembering that 2 has already been used, we have $\{C, J\}=\{1,10\}$ and $\{D, G\}=\{4,5\}$. The sum of all numbers from 1 to 10 is 55 and being $S=A+B+C+D+E$, we have $2 S+J=55$ implying that $J$ is odd. Since $\{C, J\}=\{1,10\}$, we have $J=1, C=10$ and $S=27$. Note that $A+B+D+E=17$. Each of these letters has only two possible values and the smallest possible sum is $2+6+4+3=15$. By replacing 3 with 8 or 2 with 9, we exceed the sum of 17, so the only way to reach 17 is to take $A=2, B=7, D=5$ and $E=3$. This allows us to find the values of the other letters $F=9, G=4, H=6$ and $I=8$.