Given, N = 98765432109876543210 ..... up to 1000 digits, find the smallest natural number n such that N + n is divisible by 11.
Options:
A .  2
B .  3
C .  4
D .  5
Answer: Option D 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
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