Recursive Digit Sum

Problem Statement :

We define super digit of an integer x using the following rules:

Given an integer, we need to find the super digit of the integer.

If x has only 1 digit, then its super digit is x.
Otherwise, the super digit of x  is equal to the super digit of the sum of the digits of x.

For example, the super digit of9875  will be calculated as:

	super_digit(9875)   	9+8+7+5 = 29 
	super_digit(29) 	2 + 9 = 11
	super_digit(11)		1 + 1 = 2
	super_digit(2)		= 2  

Function Description

Complete the function superDigit in the editor below. It must return the calculated super digit as an integer.

superDigit has the following parameter(s):

string n: a string representation of an integer
int k: the times to concatenate n to make p


int: the super digit of n repeated k times

Input Format

The first line contains two space separated integers, n and k.


1   <=   n   <=   10^100000
1  <=   k   <=  10^5

Solution :


                            Solution in C :

In  C++ :

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int k=0,flag=1;
int sum(string s){
   int64_t a=0,b;
    for(int i=0;i<s.size();i++){
    return a;
void super(int64_t a){
      int64_t val=0;

int main() {
    string s;
    int64_t a;
    return 0;

In   Java  :

import java.util.*;

public class Solution {

    public static void main(String[] args) {
        Solution s = new Solution();
        Scanner sc = new Scanner(;
        String str_n =;
        int k = sc.nextInt();
        int pSum = Integer.parseInt(s.supdig(str_n));
        pSum *= k;
        String sup = Integer.toString(s.supdig(pSum));
    String supdig(String n) {
        if(n.length() == 1) return n;
        else {
            int np = 0;
            for(int i = 0; i < n.length(); i++) {
                np += Character.getNumericValue( n.charAt(i) );    
            return supdig(Integer.toString(np));
    int supdig(int n) {
        if(n / 10 == 0) return n;
        else {
            int r = 0;
            while(n > 0) {
                r += n % 10;
                n /= 10;
            return supdig(r);

In    C  :

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

int main() {
    long long sd=0,h;
    int k;
    int c;
        if(c != ' ')
            sd += c -'0'; 
    }while(c != ' ');
    sd *= k;
    while(sd > 10){
        while(sd > 0){
            h+= sd %10;
            sd = sd /10;
        sd =h;

    return 0;

In   Python3  :


def digits_sum(n):
    s = 0
    while n > 9:
        s += n % 10
        n //= 10
    s += n
    return s

def super_digit(n):
    d = digits_sum(n)
    if d < 10:
        return d
    return super_digit(d)

def super_digit_iter(n):
    d = digits_sum(n)
    while d > 9:
        d = digits_sum(d)
    return d

def main():
    n, k = map(int, input().strip().split(' '))
    print(super_digit_iter(digits_sum(n) * k))

if __name__ == "__main__":

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