Sum of Digits in a Number2 months ago

The program must accept an integer N as the input. The program must print the sum of all the digits in N as the output.

Boundary Condition(s):
1 <= N <= 10^8

Input Format:
The first line contains N.

Output Format:
The first line contains the sum of all the digits in N.

Example Input/Output 1:
Input:
12345

Output:
15

Explanation:
The sum of all the digits in 12345 is 15.
Hence the output is 15

Example Input/Output 2:
Input:
5000

Output:
5

The below video explains the logic and the implementation in C programming language to find the sum of digits in a number.

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int N;
    scanf("%d", &N);
    int sum = 0;
    while(N != 0)
    {
        sum = sum + N%10;
        N = N/10;
    }
    printf("%d", sum);
    return 0;
}

 

#include <iostream>

using namespace std;

int main()
{
    int N;
    cin >> N;
    int sum = 0;
    while(N != 0)
    {
        sum = sum + N%10;
        N = N/10;
    }
    cout << sum;
}

 

import java.util.Scanner;

public class Hello {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int sum = 0;
        while (N != 0) {
            sum = sum + N % 10;
            N = N / 10;
        }
        System.out.print(sum);
    }

}

 

N = int(input())
sumOfDigits = 0
while N != 0:
    sumOfDigits += N%10
    N //= 10
print(sumOfDigits)