Cryptotastic

If we can, we will want $A,C,E$ to be $1,2,3$ in some order, so without loss of generality let $(A,C,E) = (1,2,3)$ . Now of the digits remaining, the minimum sum of any three of them will be $4 + 5 + 6 = 15$ , so the best we can do is have $G = 7$ . Now for $(B,D,F)$ we can't use $(4,5,6)$ since that will make $H = 5$ . The next best options would be $(4,5,8)$ , but that would make $H = 7$ , which is already taken. We move on to $(4,5,9)$ which will make $H = 8$ and all the digits distinct. So the minimum desired value is $\overline{GH} = \boxed{78}$ .

(Note that $(B,D,F) = (4,6,8)$ would also yield $\overline{GH} = 78$ but this would involve a repeated digit.)

Edit: To be complete, the only other choice for $(A,C,E)$ to have a minimum less than $80$ would be if we had $(A,C,E) = (1,2,4)$ , but as the least digits remaining for $B,D,F$ are $3,5,6$ the final sum would be necessarily greater than $80$ .

Even without trying for the minimum value, I was surprised how few ways there are to fill in the digits at all. I only came up with five possible values of GH.

In keeping the sum under 100, it is hard to avoid repeating a digit.

To keep the number GH restricted to 2 digits, that is, to arrive at any solution that even satisfies the question is 70% of the work done.

Obviously, we'll want A, C, E, G to be as small as possible.

So we'll assume A, C, E to be 1, 2 and 3. Now we only have larger numbers left.

We'll first asume B, D ,F to be 4, 5 and 6, as that is the logical conclusion.

However, on adding the resultant digits, it appears that H turns out to be 5! We can't afford a repetition!

So we keep increasing the value of F, from 5 to 6 and beyond, while keeping B and D at the lowest possible values.

Finally we get something like this - 14 + 25 + 39 = 78 No repetition.

And using the smallest possible values.

P.S. - There are actually really few ways to solve this problem. If you have a good one, share it!!

Let's first try to find a lower bound for the value of $\overline{GH}$ .

Clearly, $1+2+3=6\leq A+C+E\leq G$

However, $G\neq6$ because if $A+C+E=6$ , then $B+D+F>4+5+6>9$ forcing $G>6$ .

Now we check by trial and error which (if any) values of $H$ solve the cryptogram when $G=7$ and take the smallest.

If $A+C+E=7$ , then $B+D+F>9$ forcing $G>7$ , hence $A+C+E=6$ . ( $1,2,3,7$ are reserved.)

If $H<7$ , we have $B+D+F\geq 4+5+8>16$ which either carries at least $2$ over to $G$ or forces $H>6$ .

If $H=7$ , we have $G\neq7$ .

If $H=8$ we have $\left(A,B,C,D,E,F,G,H\right)=\left(1,4,2,5,3,9,7,8\right)$ as a solution.

Since $G\geq7$ and if $G=7$ , $H\geq8$ and if $H=8$ , $\overline{GH}\geq78$ and since $\overline{GH}=78$ is possible, the minimum value of $\overline{GH}$ is $\boxed{78}$ .

I selected the minimum for A,C,E (=1,2,3). Then I selected the next minimum for B,D.F (=4,5,6). But F=6/7/8 doesn't work, so F=9. Sum: 14+25+39=78. However, this is a trial-and-error method and I was "lucky" to get to the solution fast! So, I would like to see a solution that is obtained directly based on an algorithm or other logic method, but w/o trying ...

The answer is 78 demonstrated as follows:

Best try for a minimum value is A, B, C as 1,2,3 in order to minimize the ten's place.

The order of the digits in each column have no effect on the sum. This is evident if the sum is expressed as GH = 10 (A+C+E) + (B+D+F)). So let A, C, E be 1,2,3 in any order. (1)

Any choice of digits for the one's places will have a carry of at least one since the minimum choice {4,5,6} sums to 15 (and a maximum of 2 with {7,8,9}), Therefore the minimum value of G is 7 and our choices for B,D,E are limited to {4,5,6,8,9).

The minimum choices for B.D,F in ascending order are {4,5,6} with H=5 that fails by duplicating 5, {4,5,8} with H = 7 that fails by duplicating G = 7, and {4,5,9} with H = 8 that solves the puzzle with GH = 78. The solution is minimal because all minimum possibilities were exhausted in the process of finding the solution, none lower can exist.

(1) I hope you find that clear and concise with the additional virtue of avoiding the dreaded "assume without loss of generality".

Lowest total is 14+25+36=75 but this ends up repeating the digit 5 and we we can't have that as no two letters are the same numbers. So you just increase the 36 to 37,38,39 and you quickly discover that 39 will increase the total to 78, now leaving you with all different digits and by definition the lowest total. Hope that makes sense!

A=1; B=2; C=3; D=4; E=5; F=6; G=7; H=8