The number of 100 digit number having the sum of their digit equal to 3 is :

Open in App

Solution

We have the few possibilities.

A 1 with two other ones somewhere. $(\frac{99}{2})$

A 1 with a two somewhere else $(\frac{99}{1})$

A 2 with a one somewhere else $(\frac{99}{1})$

A 3 with 99 zeroes $(\frac{1}{1})$

$(\frac{99}{2}) + (\frac{99}{1}) + (\frac{99}{1}) + (\frac{1}{1})$

$= 4851 + 99 + 99 + 1 = 5050$
This isthe required solution.

Suggest Corrections

Similar questions

The number of numbers from 1 to 10 ¹⁰⁰ having the sum of their digits equal to 3 is

Q. A number of two-digit numbers having the property that they are perfectly divided by the sum of their digits with quotient equal to 7, is:

Q. How many 3 digits numbers have the sum of their digits as 15?

The digit of positive integers having 3-digit number are in AP and their sum is 15 . Number obtained by reversing the digit is 594 less than the original number. Find the number.

Q. The number of six-digit numbers which have sum of their digits as an odd integer, is

Join BYJU'S Learning Program

The number of 100 digit number having the sum of their digit equal to 3 is :

We have the few possibilities.A 1 with two other ones somewhere. (992) A 1 with a two somewhere else (991)A 2 with a one somewhere else (991)A 3 with 99 zeroes (11)(992)+(991)+(991)+(11)=4851+99+99+1=5050This isthe required solution.