
I have a big doubt on your approach. 2^4 mod 5=1 and 2^198 mod 199=1 doesn't mean that 2^396 mod 1990=1. It can even be checked by the fact that 2^396 and 1990 are both even and hence the remainder should be even. My solution is like this 2^1990 mod 1990 = 2^1989/(199*5)using chinese remainder theorum,2^1989 mod 199(A) = 114(r1)2^1989 mod 5(B)= 2(r2)for 199x+5y=1, x=1 and y=40hence remainder = 199*2*1+114*5*40=22402or, 22402995*22=512since the numerator and denominator were both divided by 2 hence the actual remainder is 512*2=1024.
can u pls expln me parag kumar 