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 $9$
[Math] Remainder when divided by 9
modular arithmeticnumber theory
modular arithmeticnumber theory
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 $9$
Best Answer
In mod $9$, we have $$2^2+22^2+\cdots+22\ldots 2^2$$ $$\equiv 2^2+4^2+6^2+8^2+1^2+3^2+5^2+7^2+0^2+2^2+\cdots$$ $$\equiv 5(2^2+4^2+6^2+8^2+1^2+3^2+5^2+7^2+0^2)+2^2+4^2+6^2+8^2$$ $$\equiv 5\cdot\frac{8\cdot 9\cdot 17}{6}+4+7+0+1\equiv 5\cdot 12\cdot 17+3\equiv 5\cdot 3\cdot (-1)+3\equiv 6$$