What is the remainder if \[1^{2} + 3^{2} + 5^{2} + \cdots+ 2011^{2}\] is divided by 8?

There must be an easy way to solve this. But I couldn't remember that! Help me to find the correct path ...

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## Comments

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TopNewestAlthough I have provided a solution before, I discovered a much quicker way to do it. If you put them into pairs, it will be

\[(1^2+3^2)+(5^2+7^2)+(9^2+11^2)+\dots+(2009^2+2011^2)\]

This may be written into the form of

\[\sum_{n=1}^{503}((4n-3)^2+(4n-1)^2) = \sum_{n=1}^{503}(32n^2-32n+10)\]

Since \(8|32\), all it matters to find the reminder is

\[\sum_{n=1}^{503}10 = 5030 = 8(628)+6\]

Hence, the reminder is 6

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Nice pairing.

I'm not sure what your central equation is about, since neither of the expressions depend on \(n\). Can you edit it accordingly?

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Ouch, thanks Calvin. It supposed to be \(n\) but I am too familiar with \(x\) when writing the expression. I'll edit them now.

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If you apply remainder theorem then remainder of 1^2 , 3^2 , 5^2 , 7^2 .............2011^2 when divided by 8 seperately will all equal to 1. so their will be 1006 times 1 .which totals 1006. Then remainder of 1006/8 =6. So the remainder will be 6.

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