### Devide by $6$

Posted:

**Thu Jan 10, 2019 11:01 am**$C=1^2+2^2+3^2+...+2018^2+2019^2$

What is remainder if we divide $C$ by $6$?

What is remainder if we divide $C$ by $6$?

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Posted: **Thu Jan 10, 2019 11:01 am**

$C=1^2+2^2+3^2+...+2018^2+2019^2$

What is remainder if we divide $C$ by $6$?

What is remainder if we divide $C$ by $6$?

Posted: **Mon Jan 14, 2019 6:40 pm**

zero(0), Z-E-R-O.

Posted: **Thu Jan 17, 2019 4:21 pm**

Posted: **Fri Jan 18, 2019 5:48 pm**

$VETO$.Now I can see $4$.

Posted: **Fri Jan 18, 2019 6:54 pm**

Posted: **Sat Jan 19, 2019 5:40 pm**

I mean at first I was wrong. Then， I've tried to solve it and found $4$.. Now，I think I'm wrong.

Posted: **Sat Jan 19, 2019 8:21 pm**

$****VETO****$

$S_{n^{2}}=\dfrac {n\left( n+1\right) \left( 2n+1\right) }{6}$

So, $S_{2019^{2}}=\dfrac {2019\times 2020\times 4025}{6}$

$\Rightarrow S_{2019^{2}}=2735913250$

Now, we can easily find the remainder $4$ .

$S_{n^{2}}=\dfrac {n\left( n+1\right) \left( 2n+1\right) }{6}$

So, $S_{2019^{2}}=\dfrac {2019\times 2020\times 4025}{6}$

$\Rightarrow S_{2019^{2}}=2735913250$

Now, we can easily find the remainder $4$ .

Posted: **Sat Jan 19, 2019 11:54 pm**

samiul_samin wrote: ↑Thu Jan 17, 2019 4:21 pmCorrect answer is $2$

We can easily get the total value of $C$ by using the formula of series.

Then it is an easy Modular arithmatic. Answer is 2!

Posted: **Mon Jan 21, 2019 1:10 am**

This answer is wrong. **Nabila** was right.

We can write any number in this form: $(6n+x) : x \in {(1,2,3,4,5)}$

So, by squaring them, we get:

$(6n+1)^2 \equiv 1$

$(6n+1)^2 \equiv 4$

$(6n+3)^2 \equiv 3$

$(6n+4)^2 \equiv 4$

$(6n+5)^2 \equiv 1$

$(6n)^2 \equiv 0$ mod $(6)$

So the pattern of the remainder is: $1,4,3,4,1,0$

$\lfloor {\frac {2019}{6}} \rfloor = 336.$ There are $336$ such groups.

$(1+4+3+4+1+0) \times 336 \equiv 1 \times 0 \equiv 0$ mod $(6)$

$1^2+2^2+...+2016^2 \equiv 0$ mod $(6)$

$\blacktriangleright (1^2+2^2+...+2016^2)+2017^2+2018^2+2019^2 \equiv 0+1+4+3 \equiv 8 \equiv 2$ mod $(6)$

The problem was: $(2n+1)=2 \times 2019 +1=4039$

We can write any number in this form: $(6n+x) : x \in {(1,2,3,4,5)}$

So, by squaring them, we get:

$(6n+1)^2 \equiv 1$

$(6n+1)^2 \equiv 4$

$(6n+3)^2 \equiv 3$

$(6n+4)^2 \equiv 4$

$(6n+5)^2 \equiv 1$

$(6n)^2 \equiv 0$ mod $(6)$

So the pattern of the remainder is: $1,4,3,4,1,0$

$\lfloor {\frac {2019}{6}} \rfloor = 336.$ There are $336$ such groups.

$(1+4+3+4+1+0) \times 336 \equiv 1 \times 0 \equiv 0$ mod $(6)$

$1^2+2^2+...+2016^2 \equiv 0$ mod $(6)$

$\blacktriangleright (1^2+2^2+...+2016^2)+2017^2+2018^2+2019^2 \equiv 0+1+4+3 \equiv 8 \equiv 2$ mod $(6)$

The problem was: $(2n+1)=2 \times 2019 +1=4039$