What's remaining?!

Thanks, I have just learnt Fermat's theorem on Wikipedia. Its general from states that for $m\equiv n \mod{(p-1)}$ then for every integer $a$ we have $a^m\equiv a^n\mod{p}$ .

Therefore, for $m=100$ and $p = 7$ then, $100 \equiv n \mod {6} \Rightarrow n = 4$ .

$\Rightarrow 2^{100} + 3^{100} + 4^{100} + 5^{100} \equiv (2^4 + 3^4 + 4^4 + 5^4) \mod {7}$

$\equiv (16 + 81 + 256 + 625) \mod {7}$

$\equiv (2 + 4 + 4 + 2) \mod {7}$

$\equiv \boxed{5} \mod {7}$

$2\equiv-5(mod7)$ $\Longrightarrow2^{100}\equiv5^{100}mod7)$ $Same\ thing\ with\ 3\ and\ we\ are\ down\ to\ finding\ [2*5^{100}+2*4^{100}](mod7).$ $But\ 5^{2}\equiv4(mod7)\ \Longrightarrow\ we\ have\ to\ find:$ $[2*4^{50}+2*4^{100}](mod7)$ $Which\ is\ equal\ to\ finding:$ $[2*4^{50}(4^{50}+1)](mod7).$ $But\ 4^{3}\equiv1(mod7)$ $\Longrightarrow\ 4^{48}\equiv1(mod7)$ $\Longrightarrow\ 4^{50}\equiv2(mod7)$ $Putting\ in\ the\ values\ we\ get:$ $[2*2(2+1)]$ $Which\ is\ congruent\ to\ 5\ mod7.$

using the formula ( p + 1 )(q + 1)(r + 1 ) then (100+1)x4 = 404 divide by 7 the remainder is 5