someone who think he is best then try this........(on remainders)

please avoid giving such titles
\underbrace{\underbrace{111100}\ \underbrace{111100}...........}_{100 \ times }13 \\ \texttt{we can see if we divide 111100 by 13, we get remainder as 11,}\\\texttt{ so the same process continnues,and finally we get reminder as 8 }

GOOD TRY
BUT NOT CORRECT

answer is 6
thread closed

@Harsh Can u post ur method plzz

ok
see
first of all start finding remainders when 10,100,1000,....is divided by 13
see when you divide 10 u get 10 as remainder
when you divide 100 u get 9 as remainder
now for finding remainder of 1000 multiply remainder of 10 and 100
(u get 90)
now find remainder on 90 it comes 12 & this is also remainder of 1000

similarly remainder of 10000,100000,1000000 are 3,4,1 respectively

now remainder on thousand is 12
so 1000+1 must be divisible by 13
from here 1001 is divisible by 13
now also 100100,10010,1001 are divisible by 13
add these
100100+10010+1001 = 111111 is also divisible by 13
now remainder on 111100 is 2
above remainder u can also find by adding remainder of 100,1000,10000,100000
the sum of these remainders comes to be 28
this gives remainder 2
now the digit can be written as
(111100 x 10^594 + 111100 x 10^588 + 111100 x 10^582+.........+111100)x100 +13
remainder on 10^6 is 1
multiply 10^6 by 10^6
u get 10^12 the remainder of 10^12 is also 1 ( remainders can be multiplied)
so remainder of 10^(6n) is also 1 where n is positive integer
now the number
(111100 x 10^594 + 111100 x 10^588 + 111100 x 10^582+.........+111100)x100 +13
can be written as in the form of remainders (2+2+2+2+.......100terms )x100 + 0
= (200)x100
= 2 x 10000
now write this again in the form of remainders
=2x3 = 6
so 6 is the last remainder