Given, N = 98765432109876543210 ..... up to 1000 digits, find the smallest natural number n such that N + n is divisible by 11.

A. 2

B. 3

C. 4

D. 5

Answer: Option D

Solution(By Examveda Team)

For a no. to be divisible by 11,
Sum(odd digit nos) - Sum(even digit nos) = 0 or divisible by 11
If we look at 9876543210, the difference we get is 5
i.e. [(9 + 7 + 5 + 3 + 1) - (8 + 6 + 4 + 2 + 0) = 5]
The series is up to 1000 digit,
That means, we get $$\frac{{1000}}{{10}}$$  = 100 time 5,
then the difference will be 5 × 100 = 500
In order for the difference to be divisible by 11, we need to add 5 and the no will become 505
505 is divisible by 11