Digital Root

Easy Numbers

Repeatedly sum the digits of a number until a single digit remains.

Sample input

493193

Sample output

Solution

n = 493193
while n >= 10:
    n = sum(int(d) for d in str(n))
print(n)

let n = 493193;
while (n >= 10) {
  let sum = 0;
  while (n > 0) {
    sum += n % 10;
    n = Math.floor(n / 10);
  }
  n = sum;
}
console.log(n);

public class Main {
    public static void main(String[] args) {
        int n = 493193;
        while (n >= 10) {
            int sum = 0;
            while (n > 0) {
                sum += n % 10;
                n /= 10;
            }
            n = sum;
        }
        System.out.println(n);
    }
}

fun main() {
    var n = 493193
    while (n >= 10) {
        var sum = 0
        while (n > 0) {
            sum += n % 10
            n /= 10
        }
        n = sum
    }
    println(n)
}

var n = 493193
while n >= 10 {
    var sum = 0
    while n > 0 {
        sum += n % 10
        n /= 10
    }
    n = sum
}
print(n)

void main() {
  int n = 493193;
  while (n >= 10) {
    int sum = 0;
    while (n > 0) {
      sum += n % 10;
      n ~/= 10;
    }
    n = sum;
  }
  print(n);
}

#include <iostream>
using namespace std;

int main() {
    int n = 493193;
    while (n >= 10) {
        int sum = 0;
        while (n > 0) {
            sum += n % 10;
            n /= 10;
        }
        n = sum;
    }
    cout << n << endl;
    return 0;
}

#include <stdio.h>

int main() {
    int n = 493193;
    while (n >= 10) {
        int sum = 0;
        while (n > 0) {
            sum += n % 10;
            n /= 10;
        }
        n = sum;
    }
    printf("%d\n", n);
    return 0;
}