Question

Remainder when 6^83+8^83 is divided by 49 not by binomial

Answer 1

(7-1)^83+(7+1)^32

in the binomial expansion of both series except last two terms every term is divisible by 49

remaining term values are 83*7+32*7+1-1

=  805=49*16+21

hence answer is 21

Alternate method :

6^83mod49 =( 216^27×36) mod49 = 36 (20^27) mod49.

Reason- 216mod49 = 20.

6^83 mod49 = 36 (8000^9) mod49 = 36 (13^9) mod49
= (36×169^4×13) mod49
= (36×13×22^4) mod49
= (9×52×484^2) mod49
= [9×3×(-6)^2] mod49
= (27×36) mod49
= (54×18) mod49
= (5×18) mod49
= 90mod49
= 41.

8^32mod49 = 64^16mod49
= 15^16mod49
= 3375^5*15 mod49
= (-6)^5*15 mod49
= (-216)*36*15 mod49
= (-20)*540mod49
= (-20)*10*54mod49
= 29*10*5 mod49
= 29*50 mod49
= 29*1 mod49
= 29mod49
= 29.

So, 
(6^83+8^32) mod49 
= (41+29) mod49
= 70mod49
= 21.

Hope This Helps :)