if 1995 is the given number

you write the code to add like this
1+9+9+5=24
2+4=6

final result is 6;

write code without for loop and don't take unique digit numbers like 9999 ....

solve this programming puzzle...

• void main()
  {
  int i=1994,sum=0,j;
  while(i>0)
  {
  j=i%10;
  sum=j+sum;
  i=i/10;
  }
  printf("%d",sum);
  }
• 7 years agoHelpfull: Yes(3) No(5)
• void main()
  {
  int n;
  int plus(int);
  printf("nEnter a number :");
  scanf("%d", &n);
  printf("nThe sum is %d", plus(n));
  }
  int plus(int n)
  {
  static int rem,sum;
  if(n>0)
  {
  rem = n % 10;
  sum += rem;
  plus(num/10);
  }
  return sum;
  }
  
  
• 7 years agoHelpfull: Yes(1) No(3)
• #include
  #include
  int main()
  {
  int n;
  clrscr();
  printf("enter number");
  scanf("%d",&n);
  printf("the sum of individual digits:%d",add(n));
  return 0;
  }
  int add(int a)
  {
  int r,sum=0;
  if(a>0)
  {
  r=a%10;
  sum+=r;
  a=a/10;
  add(a);
  }
  if(sum>10)
  {
  add(sum);
  }
  return sum;
  }
  
• 7 years agoHelpfull: Yes(1) No(2)
• main()
  {
  int n,sum=0,rem;
  printf("enter n value");
  scanf(%d,&n);
  while(n
• 7 years agoHelpfull: Yes(1) No(2)
• #include
  int add(int, int);
  int main()
  {
  int n,t;
  
  printf("enter number");
  scanf("%d",&n);
  t=n;
  printf("the sum of individual digits:%d",add(n,t));
  return 0;
  }
  int add(int n, int t)
  {
  int r,sum=0;
  while (t != 0)
  {
  r = t % 10;
  sum = sum + r;
  t = t / 10;
  }
  
  printf("Sum of digits of %d = %dn", n, sum);
  if(sum>10)
  {
  n=t=sum;
  add(sum,t);
  }
  
  return sum;
  }
• 6 years agoHelpfull: Yes(1) No(0)
• I have taken 'i' as 1995 . chnage the value of 'i' and get the respective output.
  
  #include
  void main()
  {
  int i=1995,j,sum=0,k,l,total=0;
  while(i>0)
  {
  j=i%10;
  sum=sum+j;
  i=i/10;
  }
  if(sum
• 6 years agoHelpfull: Yes(1) No(0)
• take i value as u wish and get the output
  
  #include
  void main()
  {
  int i=19957,j,sum=0,k,l,total=0;
  while(i>0)
  {
  j=i%10;
  sum=sum+j;
  i=i/10;
  }
  if(sum
• 6 years agoHelpfull: Yes(1) No(0)
• main()
  {
  int n;
  scanf("%d",&n);
  sum(n);
  }
  void sum(int n)
  {
  int rem;
  rem=x%9;
  if(rem9)
  sum(rem);
  else
  {
  printf("sum is 9");
  exit(0);
  }
  }
• 7 years agoHelpfull: Yes(0) No(2)
• int digitalRoot(int n){
  if(n%9==0)
  return 9;
  else return n%9;
  }
• 7 years agoHelpfull: Yes(0) No(0)
• using System;
  using System.Collections.Generic;
  using System.Linq;
  using System.Text;
  
  namespace sumofdigits
  {
  class Program
  {
  static void Main(string[] args)
  {
  int n = 1995;
  int i = 1;
  int sum=0;
  int r=0;
  while (i
• 7 years agoHelpfull: Yes(0) No(1)
• first printf:0
  second printf:garbage value
• 6 years agoHelpfull: Yes(0) No(1)
• #include
  int func(int ,int );
  int func(int n,int sum)
  {
  if(n>0)
  {
  sum=sum+(n%10);
  sum=func(n/10,sum);
  }
  else if(sum/10!=0)
  {
  sum=func(sum,0);
  }
  else
  return sum;
  }
  int main(void) {
  int n=1995;
  int sum=0;
  sum=func(n,sum);
  printf("%d",sum);
  return 0;
  }
• 6 years agoHelpfull: Yes(0) No(1)
• #include
  void main()
  {
  int i,j,sum=0;
  scanf("%d",&i); //input data
  while(i>0)
  {
  j=i%10;
  sum+=j;
  i=i/10;
  }
  printf("%d",sum); //output display
  }
• 6 years agoHelpfull: Yes(0) No(2)
• #include
  void main()
  {
  int i=1411997,j,sum=0,k,l,total=0;
  while(i>0)
  {
  j=i%10;
  sum=sum+j;
  i=i/10;
  }
  if(sum
• 6 years agoHelpfull: Yes(0) No(0)