If $S = 2^2 + 4^2 + 8^2 + ... ... + 256^2 + 512^2$, Then what is the second digit from the end of $S$?

Question

$S = 2^2 + 4^2 + 8^2 + ... ... + 256^2 + 512^2$, then what is the second digit from the end of $S$ ?

Answer 1

Your sum is $$ (2^1)^2 + (2^2)^2 + \cdots + (2^8)^2 + (2^9)^2 = 4^1 + 4^2 + \cdots + 4^8 + 4^9 $$ which is a finite geometric series that we can throw the standard formula at to get $$ 4\cdot \frac{4^9-1}{3} $$ To evaluate this modulo $100$, use that $3^{-1} \equiv 67 \pmod{100}$ and $4^9 = (2^9)^2 = 512^2 \equiv 12^2 \pmod{100}$, so you're looking for the tens digit of $$ 4\cdot(12^2-1)\cdot 67 $$

Answer 2

Hint:

Let us find $$R=\dfrac{4^9-1}3\pmod{\dfrac{100}4}$$

$4^9=2^{18},2^9=512\equiv12\pmod{25},2^{18}\equiv12^2\equiv-6$

$3^{-1}\equiv-8\pmod{25}$

$R\equiv(-6-1)(-8)\equiv6\pmod{25}$

$4R\equiv6\cdot4\pmod{25\cdot4}$