## Devide by $6$

- samiul_samin
**Posts:**999**Joined:**Sat Dec 09, 2017 1:32 pm

### Devide by $6$

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

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

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

- samiul_samin
**Posts:**999**Joined:**Sat Dec 09, 2017 1:32 pm

- samiul_samin
**Posts:**999**Joined:**Sat Dec 09, 2017 1:32 pm

### Re: Devide by $6$

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

Wãlkîñg, lõvǐñg, $mīlïñg @nd lìvíñg thě Lîfè

- SINAN EXPERT
**Posts:**38**Joined:**Sat Jan 19, 2019 3:35 pm**Location:**Dhaka, Bangladesh-
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### Re: Devide by $6$

$****VETO****$

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

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 $4$ .

Last edited by SINAN EXPERT on Sun Jan 20, 2019 2:09 pm, edited 1 time in total.

- samiul_samin
**Posts:**999**Joined:**Sat Dec 09, 2017 1:32 pm

### Re: Devide by $6$

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!

Last edited by samiul_samin on Mon Jan 21, 2019 11:03 am, edited 1 time in total.

### Re: Devide by $6$

This answer is wrong.

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

**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: $(2n+1)=2 \times 2019 +1=4039$