# what is remainder when 123412341234...(400 digits) is divided by 909?

asked Nov 2, 2013 in TCS

+1 vote

Let the number = N = 12341234........
909 = 9 x 101.
So we first divide the given number by 9 and find the remainder.
Sum of the digits of 12341234 ......400 digits = (1+2+3+4) x 100 = 1000 = 1.
So the remainder when the given number is divided by 9 is 1. Therefore N = 9k + 1
Now we divide the given number by 101.  101 = $10^2 + 1$ So for any number which is in the format of $10^n + 1$ the divisibility rule is divide the given number into several groups of 2 digits. and put +ve and -ve signs from right hand side.
123412341234.................. =.................-12+34-12+34 -12+34
In this expansion, there are 100, 34's and 100, 12's So their sum = 34 x 100 - 12 x 100 = 2200
And 2200 when divided by 101 gives 79 as remainder. So N = 101 L + 79
N = 9k + 1 = 101L + 79
9k = 101 L + 78
k= $\displaystyle\frac{101L+78}{9}=11L+\frac{2L+6}{9}+8 = 11L+\displaystyle\frac{2(L+3)}{9}+8$
So for L = 6 the given equation gives intezer.
So The least number satisfies the condition = 101 x 6 + 79 = 685
And this is the required remainder.