Two numbers and Alex and Bill

Here's a solution that doesn't require plenty of trial and error. The only downside of this question is to assume that the setup of this question is plausible.

So we have have to find distinct non-zero digits $A,B,C,D,E$ and $F$ satisfying the cryptogram below.

$\begin{array}{ccccc} & & & A & B&C\\ - & & & D & E &F \\ \hline & & & 2 & 8 & 8\\ \hline \end{array} \tag{1}$

Now we know that $1\leq A,B,C,D,E,F\leq 9$ , and we are given that $(A+B+C)+(D+E+F) = 12+12=24$ , so we have used up exactly 6 digits from the list $\{1,2,3,4,\ldots,9\}$ , or equivalently, we we won't be using $9-6=3$ digits from the same list of numbers. This tells us that we want to find 3 distinct digits from that list that adds up to $(1+2+3+\cdots+9) - (12+12) = 21$ , and the only possible cases are:

Case 1 : We won't be using the digits $4,8,9$ (because $4+8+9=21$ ).
Or equivalently, we will only be using the digits 1, 2, 3, 5, 6 and 7 for $A,B,C,D,E$ and $F$ .

Case 2 : We won't be using the digits $5,7,9$ (because $5+7+9=21$ ).
Or equivalently, we will only be using the digits 1, 2, 3, 4, 6 and 8 for $A,B,C,D,E$ and $F$ .

Case 3 : We won't be using the digits $6,7,8$ (because $6+7+8=21$ ).
Or equivalently, we will only be using the digits 1, 2, 3, 4, 5 and 9 for $A,B,C,D,E$ and $F$ .

By looking at the cryptogram, we know that $C - F \equiv 8 \pmod {10} \qquad \qquad \tag{2}$

Suppose Case 1 is true, then the possible values of $(C,F)$ are $(1,3), (3,5), (5,7)$ . So the possible value of the last digit of Alex and Bill's numbers can be 4, 8 or 2 only.

Suppose Case 2 is true, then the possible values of $(C,F)$ are $(1,3), (2,4), (4,6), (6,8)$ . So the possible value of the last digit of Alex and Bill's numbers can be 4, 6 or 0 only.

Suppose Case 3 is true, then the possible values of $(C,F)$ are $(1,3), (2,4), (3,5)$ . So the possible value of the last digit of Alex and Bill's numbers can be 4, 6 or 8 only.

Notice that
it is still possible to have the last digit of 4 if Case 1, Case 2 or Case 3 is true.
it is still possible to have the last digit of 6 if Case 2 or Case 3 is true.
it is still possible to have the last digit of 8 if Case 1 or Case 3 is true.

From the context of the information given in the question, we know that the last digit of the sum must be unique (that is, it only appears once), so the last digit cannot be 4, 6 or 8. Hence, we are able to reduce the number of cases to the following two scenarios: $(C,F) = (5,7)$ or $(C,F) = (4,6)$ only.

If $(C,F) = (5,7)$ , then we must find distinct values ${1,2,3,6}$ that represents $A,B,D,E$ in the cryptogram below.

$\begin{array}{ccccc} & & & A & B&5\\ + & & & D & E &7 \\ \hline & & & 2 & 8 & 8\\ \hline \end{array}$

Because $\overline{ABC} > 288$ , then $A> 3\implies A = 6$ only. With a little bit of patience, by trial and error we can show that $\overline{ABC} =615$ and $\overline{DEF}=327$ .

Since I've found a solution (and I assumed that your setup is plausible), then I assert that the only possible values of Alex's and Bill's numbers are 615 and 327, so our answer is $615+327=\boxed{942}$ .

First consider the possible digits in the larger number, we must have three distinct digits which sum to 12. The only options are

• (9, 2, 1)
• (8, 3, 1)
• (7, 4, 1)
• (7, 3, 2)
• (6, 5, 1)
• (6, 4, 2)
• (5, 4, 3)

Each of these corresponds to 6 possible values of the larger number (eg (9, 2, 1) could be 921, 912, 291, 219, 192, 129. Going through all 42 possibilities for the larger number (remembering that it must be greater than 388=288+100), we get that there are just 3 values A for which A-288 also has a digit sum of 12 and where all 6 digits in A and A-288 are distinct:

• 615
• 561
• 426

Of these, the possible sums are 942, 834, 564. So for Bill to have worked out the answer given the final digit, Alex must have written down 2, and the two numbers were 615 and 327, with a sum of 942 .