A number theory problem by Sayantan Saha

By Fermat's little theorem ,

$\large 5^6 \equiv 1 \pmod{7} \Rightarrow 5^{100} = 5^{6\times 16 + 4} = 5^{6\times 16} \times 5^4 \equiv 1\times 2 \equiv \boxed{2} \pmod{7}$

that's a cyclical in form where 5^4 divide by seven give us the same as 5^100 divide by 7

5 = -2 (mod 7), then 5^100 = 2^100 (mod 7). It is known that , 1024 = 2 (mod 7), or 2^10 = 2 (mod 7), implies that, 2^100 = 2^10 (mod 7) , or, 2^100 = 2 (mod 7) , hence, 5^100 = 2 (mod 7). [take " = " as congruence sign.]