It's Raining Chocolates

$\begin{aligned} 2^{100} + 3^{100} & \equiv (\color{#3D99F6}{2^3})^{33+1} \text{(mod 7)} + (\color{#3D99F6}{3^3})^{33+1}\text{(mod 7)} \quad \quad \small \color{#3D99F6}{\text{We note that: } 2^3 = 8 \equiv 1 \text{(mod 7)} \text{ and } 3^3 = 27 \equiv -1 \text{(mod 7)}} \\ & \equiv 2(\color{#3D99F6}{1}) \text{(mod 7)} + 3(\color{#3D99F6}{-1}) \text{(mod 7)} \\ & \equiv 2 \text{(mod 7)} - 3 \text{(mod 7)} \\ & \equiv -1 \text{(mod 7)} \\ & \equiv \boxed{6} \text{(mod 7)}\end{aligned}$

$2^{10} (mod 7)= 2$

$3^{10} (mod 7)= 4$ ; thus,

$2^{100}=(2^{10})^{10}=2^{10} mod(7) =2 (mod 7)$ and

$3^{100}=(3^{10})^{10}=3^{10} mod (7)=4 (mod 7)$ ; therefore, $remainder=4+2=6$

We can see that $3^{1}$ is congurent to 3 (mod 7) and respected remainders are: $3,2,6,4,5,1$ for $3^{1} ,3^{2}....3^{6}$ now $3^{7}$ is congurent to $3^{1}$ and $3^{100}$ is congurent to $3^{4}$ so the remainder is $6$ for $3^{100}$ (congurent by module 7 of course). Similarily we find out that powers of 2 give remainders of $2,4,1$ respectively and $2^{100}$ is congurent to $2^{1}$ so the remainder is 2. Adding those two we get 6 and 6 divided by 7 gives us the remainder $6$