Question 382749
We have

(100D + 10N + A) + (100D + 10A + N) = 100A + 10N + D

200D + 11N + 11A = 100A + 10N + D

199D + N = 89A

Suppose we write the last expression modulo 89 to obtain

21D + N = 0 = 89 = 178 etc. (modulo 89)

21D + N cannot be 178 because D would become larger than 5 (which would give a 4-digit sum). Therefore 21D + N = 89 --> D = 4, N = 5.

Solving, we have

796 + 5 = 89A --> 801 = 89A --> A = 9

Therefore D + N + A = 18.