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I'd like help with this question :

What is the remainder when

$$2^{2} + 22^{2} + 222^{2}+ \ldots + \underbrace{2222...22^{2}}_{49 \text{ times}} $$

is divided by $9$

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In mod $9$, we have $$2^2+22^2+\cdots+22\ldots 2^2$$ $$\equiv 2^2+4^2+6^2+8^2+1^2+3^2+5^2+7^2+0^2+2^2+\cdots$$ $$\equiv 5(2^2+4^2+6^2+8^2+1^2+3^2+5^2+7^2+0^2)+2^2+4^2+6^2+8^2$$ $$\equiv 5\cdot\frac{8\cdot 9\cdot 17}{6}+4+7+0+1\equiv 5\cdot 12\cdot 17+3\equiv 5\cdot 3\cdot (-1)+3\equiv 6$$

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Since $10^k\equiv 1\pmod{9}$, your sum is

$$ 4\cdot\sum_{k=1}^{49} k^2 = \left.\frac{2n(2n+1)(2n+2)}{6}\right|_{n=49}=161700\equiv 1+6+1+7\equiv \color{red}{6}\pmod{9}.$$

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$22222=2222*10+2$, then with out the squares, modulo $9$, you have the sequence $$2,4,6,8,1,3,5,7,0,2,4...,$$ Then with the squares you have the sequence $$4,7,0,1,1,0,7,0,4,7,...$$ Can you take it from here?

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