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when 6^83+8^83 is divided by 49 , tell the remainder left where ^ denotes raised to?PLS EXPLAIN (Acco. to me its 0)......

10 years ago

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
10 years ago

(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+...+89C867will 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

10 years ago

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