Devide by $6$

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samiul_samin
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Devide by $6$

Unread post by samiul_samin » 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 $6$?

NABILA
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Re: Devide by $6$

Unread post by NABILA » Mon Jan 14, 2019 6:40 pm

zero(0), Z-E-R-O.
Wãlkîñg, lõvǐñg, $mīlïñg @nd lìvíñg thě Lîfè

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samiul_samin
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Re: Devide by $6$

Unread post by samiul_samin » Thu Jan 17, 2019 4:21 pm

NABILA wrote:
Mon Jan 14, 2019 6:40 pm
zero(0), Z-E-R-O.
Correct 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. :D

NABILA
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Re: Devide by $6$

Unread post by NABILA » Fri Jan 18, 2019 5:48 pm

$VETO$.Now I can see $4$.
Wãlkîñg, lõvǐñg, $mīlïñg @nd lìvíñg thě Lîfè

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samiul_samin
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Re: Devide by $6$

Unread post by samiul_samin » Fri Jan 18, 2019 6:54 pm

NABILA wrote:
Fri Jan 18, 2019 5:48 pm
$VETO$.Now I can see $4$.
I didn't understand this post.

NABILA
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Re: Devide by $6$

Unread post by NABILA » 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. :oops:
Wãlkîñg, lõvǐñg, $mīlïñg @nd lìvíñg thě Lîfè

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SINAN EXPERT
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Re: Devide by $6$

Unread post by SINAN EXPERT » 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 $4$ :?:.
Last edited by SINAN EXPERT on Sun Jan 20, 2019 2:09 pm, edited 1 time in total.

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samiul_samin
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Re: Devide by $6$

Unread post by samiul_samin » Sat Jan 19, 2019 11:54 pm

samiul_samin wrote:
Thu Jan 17, 2019 4:21 pm
NABILA wrote:
Mon Jan 14, 2019 6:40 pm
zero(0), Z-E-R-O.
Correct 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.

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Tasnood
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Re: Devide by $6$

Unread post by Tasnood » 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 $(6)$ :D

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

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