Mistake made Man

$\begin{aligned} S & = 2^2 + 4^2 + 8^2 + 16^2 + 32^2 + 64^2 + 128^2 + 256^2 \text{ (mod 100)} \\ & = 4^1 + 4^2 + 4^3 + 4^4 + 4^5 + 4^6 + 4^7 + 4^8 \text{ (mod 100)} \\ & = (4+4^2) + 4^2(4+4^2) + 4^4(4+4^2) + 4^6(4+4^2) \text{ (mod 100)} \\ & = 20(1+4^2 + 4^4 + 4^6) \text{ (mod 100)} \\ & = 20(1+{\color{#3D99F6}16} + {\color{#3D99F6}16^2} + {\color{#3D99F6}16^3}) \text{ (mod 100)} \quad \quad \small \color{#3D99F6} \text{Power of number ends with 6 always ends with 6} \\ & = 20(1+{\color{#3D99F6}6} + {\color{#3D99F6}6} + {\color{#3D99F6}6}) \text{ (mod 100)} \\ & = 20 + 20 + 20 + 20 \text{ (mod 100)} \\ & = 80 \text{ (mod 100)} \end{aligned}$

Therefore the second last digit of $S$ is $\boxed 8$ .

the sequence $\big(2^{k}\big)^2$ , for $k=1,2,\dots$ , modulo $25$ , has got a repetitive pattern. $4, 16, 14, 6, -1, -4, -16, -14, -6, 1, 4$ , with period $10$ .

Therefore, the summation, modulo $25$ , would mostly cancel out and we get

$2^2+\dots + 256^2 \equiv 5 \ mod \ 25$

Also, not hard to see

$2^2+\dots + 256^2 \equiv 0 \ mod \ 4$

Chinese remainder theorem can be used then

$2^2+\dots + 256^2 \equiv 80 \ mod \ 100$

https://drive.google.com/file/d/1Pg8csf_Nv3IPEj0KSYIESqr-0vluzpuO/view?usp=drivesdk