(a) 12
3815 = (39 – 1)15 = 3915 – 15C1 3914 (1) + 15C2 (39)13 (1)2 – 15C3 (39)12 (1)3 ... + 15C14 (39)1 (1) – 15C15 (1)
Except - 1 all other terms are divisible by 13.
∴ When 1 is added to it the number is divisible by 13. So the remainder is 13 – 1 = 12.