Adding the last digits gives R + E+ E = E mod 10 which requires R + E = 10

R must = 1 because the maximum sum of two 4-digit and one 3-digit number is less than 20K if the leading digits of the larger numbers cannot both be 9’s. And that requires E = 9

B + R + carry must either be 10 or else be >= 12 since the 1 is already taken and the sum must be >= 10. But the max carry is 2 and the max B is 8 so B + R + carry <= 11. Which requires that B + R + carry = 10 and H = 0. And since A >= 2 (0 and 1 are in use), the previous column gives 18 + A >= 20 making the carry 2, which requires B = 7.

From the tens digits we have A + 3 (R + R + carry from the ones) = M mod 10, and neither A nor M can be in (0, 1, 7, 9) The allowed values of A, then, are: 2, 3, 5, and there's no carry from the tens into the hundreds.  But from the hundreds digits we have 18 + A = Y mod 10, and that excludes A = 2 or 3 since Y can’t be 0 or 1. So A = 5.

Since A = 5, that requires M = 8 and Y = 3 and that’s all the digits:

7951 + 1519 + 919 = 10389 is therefore the only solution.