## Devide by $6$

For students of class 9-10 (age 14-16)
samiul_samin
Posts: 1007
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 $6$?

NABILA
Posts: 35
Joined: Sat Dec 15, 2018 5:19 pm
Location: Munshigonj, Dhaka

### Re: Devide by $6$

zero(0), Z-E-R-O.
Wãlkîñg, lõvǐñg, $mīlïñg @nd lìvíñg thě Lîfè samiul_samin Posts: 1007 Joined: Sat Dec 09, 2017 1:32 pm ### Re: Devide by$6$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. NABILA Posts: 35 Joined: Sat Dec 15, 2018 5:19 pm Location: Munshigonj, Dhaka ### Re: Devide by$6VETO$.Now I can see$4$. Wãlkîñg, lõvǐñg,$mīlïñg @nd lìvíñg thě Lîfè

samiul_samin
Posts: 1007
Joined: Sat Dec 09, 2017 1:32 pm

### Re: Devide by $6$

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

NABILA
Posts: 35
Joined: Sat Dec 15, 2018 5:19 pm
Location: Munshigonj, Dhaka

### 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. 