when 6^83+8^83 is divided by 49 , tell the remainder left where ^ denotes raised to?

PLS EXPLAIN (Acco. to me its 0)......

I think answer is 2

procedure:

6^83+8^83/49 can also be written as (7-1)^83+(7+1)^83/49

now apply binomial theorem

 (x+1)^n=nC0+nC1(x)+nC2(x^2)...................

 (x-1)^n=nC0-nC1(x)+nC2(x^2)............

after applying we get 2(83C0)+2(83C2(7^2)+83C4(7^4)...........)/49

we can observe that every number in the second part is divisible by 49 where as first part cant.therefore i.e the remainder i.e 2

To get more clarity refer this eg:625+4/25 remainder? ans 4

ANS)1.84617489 × 1073

(7-1)⁸⁹ + (7+1)⁸⁹.

This will give 

= (⁸⁹C₀7⁸⁹-⁸⁹C₁7⁸⁸+⁸⁹C₂7⁸⁷+...-⁸⁹C₈₇7²+⁸⁹C₈₈7¹-⁸⁹C₈₉7⁰)+(⁸⁹C₀7⁸⁹+⁸⁹C₁7⁸⁸+⁸⁹C₂7⁸⁷+...-⁸⁹C₈₇7²+⁸⁹C₈₈7¹-⁸⁹C₈₉7⁰)

=2*(⁸⁹C₀7⁸⁹+⁸⁹C₂7⁸⁷+⁸⁹C₄7⁸⁵+...+⁸⁹C₈₆7³+⁸⁹C₈₈7¹)

⁸⁹C₀7⁸⁹+⁸⁹C₂7⁸⁷+⁸⁹C₄7⁸⁵+...+⁸⁹C₈₆7³ will all be divisible by 49 because they consist of 7² as one of their factor but not 2(⁸⁹C₈₈7¹)

Then 2(⁸⁹C₈₈7¹)%49 = 21 ---Remainder

sorry for my mistake, its (7-1)⁸³ + (7+1)⁸³ not (7-1)⁸⁹ + (7+1)⁸⁹.

In similar ways of my prev reply  the remainder will be  2(⁸³C₈₂7¹)%49 =35

What this %49 means ,I am not able to understand it ,please clarify it to me. Otherwise your answer is perfecg