Program to print the sum of digits without using modulus

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Program to print the sum of digits without using modulus

In this program, we have to add the digits of the entry number without the logic of using modulus(%).

Example:

149: the sum of its digits (1, 4, 9) is 14.

Go through the algorithm to code for this program:

Algorithm

STEP 1: START
STEP 2: ENTER n as a string
STEP 3: SET sum =0
STEP 4: SET i = 0
STEP 4: REPEAT STEP 5 and 6 UNTIL (i<length(n))
STEP 5: sum =sum + n[i]
STEP 6: sum = sum - 48
STEP 7: i = i+1
STEP 7: PRINT sum
STEP 8: END

Java Program

import java.util.*;
class SumOfDigits {
    public static void main(String[] args) {
   Scanner sc = new Scanner(System.in); 
  System.out.println("Enter the number?");
  String str=sc.nextLine();
   char[] n = str.toCharArray();
   int sum=0;
for(int i =0;i<n.length;i++)
{
   sum = sum + ((int)n[i]);
   sum = sum-48;
}
System.out.println("sum of digits: "+sum);
}   
}

Output:

Enter the number?
345
sum of digits: 12

C Program

#include <stdio.h>
int main()
{
   int c, sum, t;
   char n[1000];
   printf("Enter the number?");
   scanf("%s", n);
   sum = c = 0;
   while (n[c] != '\0') {
      t   = n[c] - '0'; 
      sum = sum + t;
      c++;
   }
   printf("sum of digits: %d",sum);
   return 0;
}

Output:

Enter the number? 45
sum of digits: 9

C# Program

using System;
public class Sum_Of_Digits
{
public static void Main()
{
	int sum=0,i;
	Console.WriteLine("Enter the number?");
string str = Console.ReadLine();
char[] c = str.ToCharArray();
for(i =0;i<c.Length;i++)
{
   sum = sum + ((int)c[i]);
   sum = sum-48;
}
Console.WriteLine("sum of digits: "+sum);
}
}

Output:

Enter the number? 75
sum of digits: 12

Python Program

sum=0
Str = input("Enter the number?");
for i in range(len(Str)):
    sum = sum + (int(Str[i]))

print ("sum of digits: %d"%(sum) )

Output:

	Enter the number? 175
	sum of digits: 13

PHP Program

<?php
	echo("Enter the number?");
	$n=readline();
	$sum = 0;
	$c = 0;
	for($c = 0;$c<strlen($n);$c++){
   	   $t   = $n[$c] - '0'; 
      	   $sum = $sum + $t;
 	  }
   echo("sum of digits:");
   echo($sum);
?>

Output: