(In reply to

Puzzle Solution With Explanation by K Sengupta)

Since the tens digit of the addition result (A) is different from the tens digit (M) of the only two digit summand MA, it follows that, there is a carryover due to the said addition.

Now, there are only two summands, and accordingly, the sum of the units digits must be less than 20, and accordingly, the caryover due to the said addition can only be 1. This gives the identities: A = M+1 and 2A = 10 +M, so that:

2A = 10+A-1, giving A = 9, so that: M = 8

Thus, the required addition when completed is:

89 + 9 = 98