N is nothing but first 100 natural numbers written in ascending order!
The first 99 natural numbers will give us 9 + (90×2) =189 digits.
So let us consider first 49 numbers. Number of digits = 9 + (40 * 2) = 89
11 more is needed to make 100 digits. So we can include 50, 51, 52, 53, 54 and 5
Hence the first 100 digits of N goes like this 1234567891011121314.......5253545
We need to find the sum of the above number to check for divisibility by 9.
From 1 to 50, each of the digits 1 to 4 occurs 14 times, digit 5 occurs 6 times and each of 6 to 9 occurs 5 times
Sum of digits from 1 to 4 = 10
Sum of digits from 6 to 9 = 30
Therefore, sum of digits from 1 to 50 = 10 * 15 + 5 * 6 + 30 * 5 = 330
Sum of digits from 51 to 54 plus a 5 = 5 * 4 + ( 1 + 2 + 3 + 4) + 5 = 20 + 15 = 35
Total sum = 330 + 35 = 365
When 365 is divided by 9, we get 5 as the remainder!
Correct Answer: 5
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