The Xs in the left column produce a 1-digit sum, so X = 1, 2, 3, or 4. But looking at the right column, X is the sum of Y and Z, which are distinct numbers, so the smallest possible value of X is 1 + 2 = 3. So X must be either 3 or 4. That means (looking at the left column again) that Y must be at least 6. But Y and Z are summed in both the center and the right columns, producing different digits. So Y and Z must total at least 11, and that means that a 1 is being carried into the total X + X in the left column, producing a total for that column (and a value for Y) of either 3 + 3 + 1 = 7 or 4 + 4 + 1 = 9.

The fact that we’re carrying a 1 into the center column shows us that X + 1 = Z. We’ve established that X is 3 or 4, so Z must be 4 or 5. So:

X = 3 or 4
Y = 7 or 9
Z = 4 or 5

A bit of trial and error shows that X = 4, Y = 9, Z = 5.

08/18/2017 UPDATE: There’s a simple solution that I overlooked:

The tens place shows Z + Y = Z. So either Y is 0 or Y is 9 and we’ve carried a 1 from the ones place. And Y can’t be 0 because it’s the first digit of the solution. So Y is 9.

The hundreds place shows X + X = 9. That’s possible only if X is 4 and we’ve carried a 1 from the tens column.

And the ones place shows Y + Z = X; that is, 9 + Z = 4. So Z is 5.

Also, my argument that X can’t be 1 or 2 isn’t valid — the X in the first column could be the ones digit of some two-digit sum.

Thanks to everyone who wrote in about this; sorry I couldn’t reply to each of you individually.

06/01/2018 Yet another solution:

• From the first column we know that X < Y, since there is no 1 carried forward.
• This means the third column is effectively Y + Z = X + 10 (a 1 has to be carried forward, per the observation above).
• Now the second column gives this equation: Z + Y + 1 = Z + 10 (again a 1 must be carried forward, since Y can’t be zero).
• Hence the first column gives the equation: X + X + 1 = Y.

Solving the three equations gives the three variables. (Thanks, Tom.)