Don't Try From 99 down

Two digit numbers are 10 to 99. We might get the largest possible remainder from 99 down rather than 10 up.

2017 \mod{99} = 37 . When 2017 is divided by any of {2,...,38} , the remainder will be less than 38 . So the answer is in {39, ..., 99} .

2017 \mod{98} = 57 . Similarly the answer is in {59, ..., 98} .

2017 \mod{97} = 77 . Now we only need to look in {79, ..., 96}

Similarly,

2017 \mod{96} = 1 .

2017 \mod{95} = 22

2017 \mod{94} = 43

2017 \mod{93} = 64

2017 \mod{92} = 85 , Great we only have to look upto 87 now. Because less than 87 the remainder is anyway less than or equal to 85

2017 \mod{91, 90, 89, 88, 87} = {15, 37, 59, 81, 16}

Note This is not the elegant solution. But it might give some idea how to generalize it. I can't think of better way right now.