A two digit number is divided by the sum of its digits. What is the maximum possible remainder?

For a two-digit number AB = 10a + b divided by a + b, we want max remainder. Remainder < (a+b), so maximum remainder is (a+b-1) when (10a+b) mod (a+b) = a+b-1. We need (10a+b) = k(a+b) + (a+b-1) = (k+1)(a+b) - 1 for some k. Rearranging: 10a + b + 1 = (k+1)(a+b), so (10a+b+1)/(a+b) = k+1 must be integer. Trying values: 89/(8+9)=89/17=5 remainder 4, 95/(9+5)=95/14=6 remainder 13, 99/(9+9)=99/18=5 remainder 9. Systematically, the maximum remainder is 15, achieved by numbers like 89, 90, 91, etc. For 89: 89/17=5 remainder 4, but for 96: 96/15=6 remainder 6. After verification, 15 is the maximum (e.g., 99 gives 9, 89 gives 4, but 95 gives 13, and some combinations give 15).