The sum reads 6247663 + 6837633 = 13085296.
The sum is comprised of six letters, as you can see if you squint: E, X, M, R, K and H. Since we know that the sum uses all ten digits, that means all of the smudged parts of the sum must represent a different digit each.
It's going to take a long and steady stream of logic, but it'll make sense in the end. For example, we can immediately deduce that the first dash, representing the ten millions slot, must be 1, because the sum of two digits with carryover can’t be greater than 19.
We also know that E must be even, because adding two of the same number will always produce an even number (specifically, two Ks). We also know that E + E must be pretty big, because it creates carryover on the left side of the equation. So E is either 6 or 8, which means that K must be 3 or 4.
Let's imagine the scenario where K = 4 and E = 8. this works for the ones digit of the answer, but when we hit the millions, we run into problems: 8 + 8 = 16, which has a 6 in the ones place (the K value in the sum), and we know K isn't 6. Therefore, K is 3 and and E is 6: you can check everything and it's still numerically legal.
Now let's evaluate a bit more. In the tens slot, subbing K and E, we see 6 + 3 = 9. No carryover here. The sum in the hundreds slot is 6 + 6. That's a total of 12, so there is carryover of 1 for the thousands slot, and X must therefore represent 2: the last digit of 12.
So we can re-write the sum, with the info we have so far, as follows:
(1) (1)
6 2 M R 6 6 3
+ 6 H 3 R 6 3 3
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1 3 - H - 2 9 6So far, we know:
Now let's move on to the millions. Since we know X =2, H must either be 8 or 9, because we need carryover to make the 1 in our ten millions slot. But we know H can't be 9, because we've already used 9 in our tens slot, and with only 5 variables left to fill and 5 digits left to assign, it leaves no room for another 9. So H must therefore be 8.
Now, we have:
(1) (1)
6 2 M R 6 6 3
+ 6 8 3 R 6 3 3
__________________________
1 3 0 8 - 2 9 6Just a little further! As we can see, M + 3 = 8, or 1 + M + 3 = 8. So M is either 4 or 5. But using similar logic outlined when solving for K and E, M cannot be 5, therefore it must be 4. That leaves us with enough info to determine that H = 8, and R = 7. We just need to fill in the rest of the blanks, and we have our complete sum!