IRL_ABooklet

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 $H=D$ or $H=B$. For a solution $EFGH$ to be bigger than $6840$ one of the numbers needs to be greater than $86$ because otherwise the product would not exceed $79\cdot 86=6794$. If we assume $AB<CD$ then $CD$ needs to be one of the following numbers $87,89,92,93,94,95,96,97,98$.

For each of these we can find the possible values of $AB$ which would make the product greater than $6840$, i.e. we work with the inequality $$CD>AB\ge \lfloor \frac{6840}{CD}\rfloor .$$ For $C=8$ we need to have $AB<80$. If $CD=87$ we even need to have $AB<70$ to avoid repeated digits. But $70\cdot 87<70\cdot 90=6300<6840$. If $CD=89$, we get $\lfloor \frac{6840}{CD}\rfloor =77$ and none of $77,78,79$ is possible for $AB$ because of repeated digits.

Note that when $AB<70$ we have $AB\times CD<70\cdot 98<7000$ and $E=6$ whenever $AB\times CD>6840$. Hence, we need to have $AB\ge 72$ to avoid repeated digits. Suppose now that $CD\ge 92$, then $A=7$ or $A=8$.

If $A=7$, we have $AB\times CD<80\cdot 98<8000$ which implies that $E=6$ whenever we have a solution with $AB\times CD>6840$. This implies $$72\le AB\le \lfloor \frac{7000}{CD}\rfloor .$$ We now list all possibilities in a table where we have excluded some by keeping in mind $E=6$ and by checking the last digits.

CD	92	93	94	95	98
7000/CD	76	75	74	73	71
AB	75	74	73	72
AB	74		72

For example, when $CD=92$ we excluded $AB=73$ because the last digit of the product would be $6$. We excluded $CD=96$ and $CD=97$ because $E=6$ and $A=7$. We now check that none of these possibilities leads to a solution. Indeed, $75\cdot 92=6900$ $74\cdot 92=6808$ $74\cdot 93=6882$ $73\cdot 94=6862$ $72\cdot 94=6768$ do not give solutions, except of course for $72\cdot 95=6840$.

If $A=8$, we have $AB\times CD<90\cdot 98<9000$ and $AB\times CD\ge 82\cdot 92=7544$ which implies that $E$ must be $7$. As $F$ can now be at most $6$, we actually have $AB\times CD<7700$, hence $$82\le AB\le \lfloor \frac{7700}{CD}\rfloor .$$ Because $94\cdot 82=7708$, the only values left for $CD$ are $92$ and $93$. If $CD=93$, $7700/93<83$ and the only possibility is $AB=82$. But $82\cdot 93=7626$, no solution. If $CD=92$, $AB\ne 82$ and $7700/92<84$, hence $AB$ could only be $83$. But $83\cdot 92=7636$. Thus $6840$ is the largest possible $EFGH$.