<s>[INST] Remainder when divided by 9 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  $9$
 [/INST] 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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