My friend, reaminder may be 5 or 2

 

all powers of 13 have last digita as 3,9,7,1 and they repeat

 

In case of 19 it is 9,1

by using AP we get, that 99th term is 7 in case of 13 and 93 term is 1 in case of 19.

adding them we get 8

Now i have a confusion here.

Either we subtract num by 3 or we see the remainder left when we divide it by 3

Check it out

May be both answers are wrong !!!!!!

 

  

hiiiiiiii

this question will be done using binomial theorem

we know that 12=4*3 so 12⁴ will be divisible by 81 as 81=3⁴

18=9*2 so 18² will be divisible by 81

 

13⁹⁹+19⁹³=(1+12)⁹⁹+(1+18)⁹³

on expanding all the terms in first expression which have powers of 12 more than or equal to 4 will be divisible by 81 and in second expression,terms with power of 18 more than or equal to 2 will be divisible by 81

 

let me denote ⁿC_r as C(n,r)

so we are left with the terms:

C(99,0)+C(99,1)*12+C(99,2)*12*12+C(99,3)*12*12*12+C(93,0)+C(93,1)*18

on taking all the 3's together we can see that C(99,2)*12*12+C(99,3)*12*12*12 is divisible by 81

so we are left with:

1+99*12+1+93*18

solve and divide by 81 and find the remainder.