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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 ...

#NumberTheory

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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Comments

Although 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

Kay Xspre - 5 years, 5 months ago

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?

Calvin Lin Staff - 5 years, 5 months ago

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

Kay Xspre - 5 years, 4 months ago

@Kay Xspre – I still believe that the equation is wrong. Notice that the RHS is summing up a constant, and doesn't depend on $n$ . Is this intentional?

Calvin Lin Staff - 5 years, 4 months ago

@Calvin Lin – I suppose to write it with $n$ by not substituting 503. This is then fixed.

Kay Xspre - 5 years, 4 months ago

@Kay Xspre – Looks good now :) Thanks!

Calvin Lin Staff - 5 years, 4 months ago

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.

Anshul Sanghi - 5 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$