Divisibility Application

Remainder problems are tackled by writing the base as a multiple of the divisor plus a small adjustment and then expanding binomially, a powerful JEE Advanced application. Write 7 = 6 + 1, so 7^{83} = (6 + 1)^{83} = C(83,0)6^{83} + C(83,1)6^{82} + ... + C(83,82)6 + C(83,83). Every term except the last contains a factor of 6 and is therefore divisible by 6, leaving only the final term C(83,83) = 1 unaccounted for. Hence 7^{83} = 6k + 1 for some integer k, giving a remainder of 1 when divided by 6. Option 5 would correspond to a base congruent to -1 raised to an odd power, which is not the situation since 7 ≡ 1 (mod 6). Option 0 wrongly claims complete divisibility. Option 7 is not even a valid remainder modulo 6, since remainders must lie in 0 to 5. The reduction 7 ≡ 1 (mod 6) makes every power congruent to 1. Plausibility check: 7 ≡ 1 (mod 6) immediately gives 7^{83} ≡ 1^{83} = 1 (mod 6), confirming the binomial argument by a direct modular shortcut.