I didn’t write this one, but solving this one on-the-fly,
\frac{a}{b} = \frac{29}{8}
so possible values of a and b would be multiples of 29 and 8 respectively:
- a = 29, b = 8 → remainder = 5
- a = 58, b = 16 → remainder = 10
- a = 87, b = 24 → remainder = 15
- a = 116, b = 32 → remainder = 20
so the remainder has to be a multiple of 5 → options D and F are correct.
Another option would be to show that, for k being a positive integer,
a = 29 b + 5
and hence multiplying by k on both sides
ka = k(29b + 5)
ka = 29kb + 5k
→ remainder when ka is divided by kb is 5k.