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. $$\begin{array}{rl}1023& =3\cdot 11\cdot 31=11\cdot 93=31\cdot 33\\ 1024& =2^{10}=16\cdot 64=32\cdot 32\\ 1025& =5^2\cdot 41=25\cdot 41\\ 1026& =2\cdot 3^3\cdot 19=18\cdot 57=19\cdot 54=27\cdot 38\\ 1027& =13\cdot 79\\ 1028& =2^2\cdot 257\\ 1029& =3\cdot 7^3=21\cdot 49\\ 1032& =2^3\cdot 3\cdot 43=12\cdot 86=24\cdot 43\\ 1034& =2\cdot 517\\ 1035& =3^2\cdot 5\cdot 23=15\cdot 69=23\cdot 45\\ 1036& =2^2\cdot 7\cdot 37=14\cdot 74=28\cdot 37\\ 1037& =17\cdot 61\\ 1038& =2\cdot 3\cdot 173\\ 1039& \text{prime}\end{array}$$

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

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.