Program to print the sum of digits without using modulus

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robaa

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2022-11-09

146

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

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robaa

2022-11-09

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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: