IRL ABooklet

Solution 1. Consider factorisations of numbers $EFGH$ with all digits different, and identify all factorisations consisting of two 2-digit numbers. Start with the smallest possible number $1023$ and stop when a solution is found.
$$\begin{align*} 1023 &= 3 \cdot 11 \cdot 31 = 11 \cdot 93 = 31 \cdot 33 \\ 1024 &= 2^{10} = 16 \cdot 64 = 32 \cdot 32 \\ 1025 &= 5^2 \cdot 41 = 25 \cdot 41 \\ 1026 &= 2 \cdot 3^3 \cdot 19 = 18 \cdot 57 = 19 \cdot 54 = 27 \cdot 38 \\ 1027 &= 13 \cdot 79 \\ 1028 &= 2^2 \cdot 257 \\ 1029 &= 3 \cdot 7^3 = 21 \cdot 49 \\ 1032 &= 2^3 \cdot 3 \cdot 43 = 12 \cdot 86 = 24 \cdot 43 \\ 1034 &= 2 \cdot 517 \\ 1035 &= 3^2 \cdot 5 \cdot 23 = 15 \cdot 69 = 23 \cdot 45 \\ 1036 &= 2^2 \cdot 7 \cdot 37 = 14 \cdot 74 = 28 \cdot 37 \\ 1037 &= 17 \cdot 61 \\ 1038 &= 2 \cdot 3 \cdot 173 \\ 1039 & \quad \text{prime} \end{align*}$$

$$\begin{align*} 1042 &= 2 \cdot 521 \\ 1043 &= 7 \cdot 149 \\ 1045 &= 5 \cdot 11 \cdot 19 = 11 \cdot 95 = 19 \cdot 55 \\ 1046 &= 2 \cdot 523 \\ 1047 &= 3 \cdot 349 \\ 1048 &= 2^3 \cdot 131 \\ 1049 &= \text{prime} \\ 1052 &= 2^2 \cdot 263 \\ 1053 &= 3^4 \cdot 13 = 13 \cdot 81 = 27 \cdot 39 \\ 1054 &= 2 \cdot 17 \cdot 31 = 17 \cdot 62 = 31 \cdot 34 \\ 1056 &= 2^5 \cdot 3 \cdot 11 = 11 \cdot 96 = 12 \cdot 88 = 16 \cdot 66 = 22 \cdot 48 = 24 \cdot 44 = 32 \cdot 33 \\ 1057 &= 7 \cdot 151 \\ 1058 &= 2 \cdot 23^2 = 23 \cdot 46 \quad \text{Eureka!} \end{align*}$$

Solution 2. Some simple observations help in reducing cases. Without loss of generality, we will assume throughout $AB < CD$. We cannot have $B = 0$ or $D = 0$, as this would imply $H = 0$. Similarly, we cannot have $B = 1$ or $D = 1$, as this would imply $H = B$ or $H = D$. We cannot have $A = 1$, as this would imply $E = 1$, since $AB \times CD < 20 \cdot 98 = 1960 < 2000$.
If $AB = 21$ then $AB \times CD \le 21 \cdot 98 = 2058$ and $E \in \{1, 2\}$ which is impossible. Thus the smallest possible value of $AB$ is $23$.
If $AB = 23$, the smallest possible value of $CD$ is $45$. But $23 \cdot 45 = 1035$, however $23 \cdot 46 = 1058$ yields a solution.
We need to show that there is no solution with a smaller value of $AB \times CD$. We only need to consider these possibilities: $AB = 24, 25, 26, 27, 28, 29, 32$, because $32^2 = 1024 < 1058 < 1089 = 33^2$. In each case we keep in mind that we wish to achieve $1000 < AB \times CD < 1058$.
For $AB = 24$ there are no possibilities for $CD$, since $CD$ cannot contain the digit $4$, and $24 \cdot 39 < 1000$ while $24 \cdot 50 > 1058$. Therefore no solutions exist in this case.
For $AB = 25$, the possibilities for $CD$ are $40$, $41$, $42$. None of these work.
For $AB = 26$, the possibilities for $CD$ are $39$, $40$. Neither works.
For $AB = 27$, the possibilities for $CD$ are $38$, $39$. Neither works.
For $AB = 28$, the possibilities for $CD$ are $36$, $37$. Neither works.
For $AB = 29$, the possibilities for $CD$ are $35$, $36$. Neither works.
For $AB = 32$, the possibilities for $CD$ are $32$, $33$. Neither works.
Hence $23 \cdot 46 = 1058$ is the smallest solution.