PK 2016 Q18

Since the remainder of a division is always less than the divisor used, we can begin our search for the largest possible remainder by looking at the largest possible divisor, that is the largest possible sum of the digits of a 2-digit number.
The largest possible sum of digits is \(9 + 9 = 18\). And \(99 \div 18\) has remainder 9.
The next largest is 17, which could come from 89 or 98. Doing the division, we see \(89 \div 17\) has remainder 4 and \(98 \div 17\) has remainder 13.
The next largest sum of digits is 16, which could come from 88, 97, 79. And division shows that \(88 \div 16\) has remainder 8; \(97 \div 16\) has remainder 1, and \(79 \div 16\) has remainder 15 (the largest remainder so far).
Any sum of digits below 16 will have remainder below 15, so the remainder of 15 that we have achieved must be the largest possible.