Binomial theorem

\(6701^{83} = \sum \limits_{k=0}^{83} ~\dbinom{83}{k} (6000)^{k}(701)^{83-k}\\ \text{It should be clear that all terms but 1, }k=0 \text{ are divisible by }1000\\ \text{we are then left with }\\ (701)^{83} = \sum \limits_{k=0}^{83}\dbinom{83}{k}(700)^k\\ \text{here there are two terms that are not divisible by }1000,~k=0,1\\ (701)^{83}\pmod{1000} = \\ 1+83(700) \pmod{1000} = \\ 1 + (80+3)(700) \pmod{1000}= \\ 1+ 56000+ 2100 \pmod{1000} = \\ 2101 \pmod{1000} = 101\)

Binomial Expansion  approach

(6701)^83  =  

(6700 + 1)^83  =  6700^83  + C(83,1)*6700^82 +  ...+ C(83,81)*6700^2 + C(83,82)*6700 + 1

Note that every term in red will have at least 4 trailing zeros, so  each of these terms is divisible by 1000

So....we only need consider the last two terms

C(83, 82) * 6700  +  1     =

83 * (6700)  + 1  =

(80 + 3) * (6700)  + 1 =

80*6700 + 3*6700  + 1 

The first term  will have 3 trailing zeros, so....it is divisible  by 1000....and

3 * 6700 + 1 = 20100  + 1 =  20(1000) +  100  +  1

So  we have  left

 [ 100 + 1 ] / 1000   ⇒    remainder 101