Question

Consider a large number N = 1234567891011121314………979899100. What is the remainder when first 100 digits of N is divided by 9?
0
8
1
5

Answer 1

Answer:

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!

The question is "What is the remainder when first 100 digits of N is divided by 9?"

Hence the answer is "5".

Choice D is the correct answer.

Answer 2

Answer:

The correct answer is an option(d)5.

When the first 100 digits of N are divided by 9, the remainder is 5.

Step-by-step explanation:

Basically, N is a list of the 100 earliest natural numbers in ascending order!  

There are 189 digits in the first 99 natural numbers (9 + 90 * 2) =189 digits.  

Let's take a look at the first 49 numbers. Nine plus forty times two equals 89 digits  

There are 11 more digits to make 100. The 50s, 51s, 52s, 53s, 54s, and 55s are included  

Therefore, N's first 100 digits are 1234567891011121314.......52535455.

In order to determine if the above number is divisible by 9, we must find the sum of the above number.

There are 14 occurrences of each of the digits 1 to 4, six occurrences of digit 5, and five occurrences of each of the digits 6 to 9.

The sum of digits from 1 to 4 = 10  

The 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  

We get 5 as the remainder when 365 is divided by 9.

To know more about Natural numbers:

https://brainly.in/question/140221

To Know more about Remainders:

https://brainly.in/question/54935112

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