?? +?? ——- ?98

99 + 99 = 198

Reason (short):

• Let the two numbers be AB and CD. The sum ends with 98, so B + D ≡ 8 (mod 10), hence B + D is 8 or 18.
• If B + D = 8, the tens digit would be (A + C) mod 10, which cannot be 9 given A, C ∈ 1..9 (it would be at most 8). Impossible.
• If B + D = 18, then sum = 10(A + C) + 18 = 100 + 10(A + C − 9) + 8. For a nonzero hundreds digit, A + C ≥ 9, and to get tens = 9 we need A + C = 18.
• Thus A = C = 9 and B = D = 9, i.e., 99 + 99 = 198.