Sachin writes all the numbers from 1 to 1000 on a paper in order find the 2883 digit written by him

269308300None of these

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• How many digits are used in writing the numbers from 1 to 1000?
• How many digits are used to write numbers?
• How many digits is a thousand?

Answer : C

Solution : Before 1000 there are one digit, two digits and three digits numbers.
Number of times 3 appear in one digit number `= 20 xx 9`
Number of times 3 appear in two digit number `= 11 xx 9`
Number of times 3 appear in three digit number = 21
Hence total number of times the digit 3 appear while writing the integers from 1 to 1000
`= 180 + 99 + 21 = 300`

Given a number n, we need to find the sum of its digits such that: 

If n < 10 digSum(n) = n Else digSum(n) = Sum(digSum(n))

Examples : 

Input : 1234 Output : 1 Explanation : The sum of 1+2+3+4 = 10, digSum(x) == 10 Hence ans will be 1+0 = 1 Input : 5674 Output : 4

A brute force approach is to sum all the digits until the sum < 10. 
Flowchart: 

Below is the brute force program to find the sum. 

C++

#include

using namespace std;

int digSum(int n)

{

    int sum = 0;

    while(n > 0 || sum > 9)

    {

        if(n == 0)

        {

            n = sum;

            sum = 0;

        }

        sum += n % 10;

        n /= 10;

    }

    return sum;

}

int main()

{

    int n = 1234;

    cout << digSum(n);

    return 0;

}

Java

import java.util.*;

public class GfG {

    static int digSum(int n)

    {

        int sum = 0;

        while (n > 0 || sum > 9)

        {

            if (n == 0) {

                n = sum;

                sum = 0;

            }

            sum += n % 10;

            n /= 10;

        }

        return sum;

    }

    public static void main(String argc[])

    {

        int n = 1234;

        System.out.println(digSum(n));

    }

}

Python

import math

def digSum( n):

    sum = 0

    while(n > 0 or sum > 9):

        if(n == 0):

            n = sum

            sum = 0

        sum += n % 10

        n /= 10

    return sum

n = 1234

print (digSum(n))

C#

using System;

class GFG {

    static int digSum(int n)

    {

        int sum = 0;

        while (n > 0 || sum > 9)

        {

            if (n == 0)

            {

                n = sum;

                sum = 0;

            }

            sum += n % 10;

            n /= 10;

        }

        return sum;

    }

    public static void Main()

    {

        int n = 1234;

        Console.Write(digSum(n));

    }

}

PHP

function digSum( $n)

{

    $sum = 0;

    while($n > 0 || $sum > 9)

    {

        if($n == 0)

        {

            $n = $sum;

            $sum = 0;

        }

        $sum += $n % 10;

        $n = (int)$n / 10;

    }

    return $sum;

}

$n = 1234;

echo digSum($n);

?>

Javascript

    }

    document.write(getSum(n));

C

#include

int digSum(int n)

{

    int sum = 0;

    while(n > 0 || sum > 9)

    {

        if(n == 0)

        {

            n = sum;

            sum = 0;

        }

        sum += n % 10;

        n /= 10;

    }

    return sum;

}

int main()

{

    int n = 1234;

    printf("%d",digSum(n));

    return 0;

}

Output : 

Time Complexity: O(log(n)).
Auxiliary Space: O(1)

So, another challenge is “Could you do it without any loop/recursion in O(1) runtime?”

YES!!
There exists a simple and elegant O(1) solution for this too. The answer is given simply:- 

How does the above logic works? 

The logic behind this approach is :

To check if a number is divisible by 9, add the digits of the number and check if the sum is divisible by 9 or not. If yes, is the case,  the number is divisible by 9, otherwise, it’s not.

let’s take 27  i.e (2+7 = 9) hence divisible by 9.
If a number n is divisible by 9, then the sum of its digit until the sum becomes a single digit is always 9. For example, 
Let, n = 2880 
Sum of digits = 2 + 8 + 8 = 18: 18 = 1 + 8 = 9

Therefore,
A number can be of the form 9x or 9x + k. For the first case, the answer is always 9. For the second case, and is always k which is the remainder left.

The problem is widely known as the digit root problem.

You may find this Wikipedia article useful. -> https://en.wikipedia.org/wiki/Digital_root

Below is the implementation of the above idea : 

C++

#include

using namespace std;

int digSum(int n)

{

    if (n == 0)

       return 0;

    return (n % 9 == 0) ? 9 : (n % 9);

}

int main()

{

    int n = 9999;

    cout<

    return 0;

}

Java

import java.io.*;

class GFG {

    static int digSum(int n)

    {

        if (n == 0)

        return 0;

        return (n % 9 == 0) ? 9 : (n % 9);

    }

    public static void main (String[] args)

    {

        int n = 9999;

        System.out.println(digSum(n));

    }

}

Python3

def digSum(n):

    if (n == 0):

        return 0

    if (n % 9 == 0):

        return 9

    else:

       return (n % 9)

n = 9999

print(digSum(n))

C#

using System;

class GFG

{

    static int digSum(int n)

    {

        if (n == 0)

        return 0;

        return (n % 9 == 0) ? 9 : (n % 9);

    }

    public static void Main ()

    {

        int n = 9999;

        Console.Write(digSum(n));

    }

}

PHP

function digSum($n)

{

    if ($n == 0)

        return 0;

    return ($n % 9 == 0) ? 9 : ($n % 9);

}

$n = 9999;

echo digSum($n);

?>

Javascript

Output: 

Time Complexity: O(1)

Auxiliary Space: O(1)

Related Post : 
https://www.geeksforgeeks.org/digital-rootrepeated-digital-sum-given-integer/

This article is contributed by Ayush Khanduri. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to . See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

How many digits are used in writing the numbers from 1 to 1000?

How many digits are used to write numbers?

How many digits is a thousand?