the sum of my digits is 6? answer=60
Add the digits together. The sum of the digits of 23 is 5.
Yes, yes, and no. 3- sum of digits must be multiple of 3. 6- sum of digits must be multiple of 3 and number must be even (multiple of 2). 9- sum of digits must be multiple of 9. (The sum of the digits here is 21.)
The sum is 22 times the sum of the three digits.
It isthe number must end in 0 or 5, andthe sum of the digits (and the sum of the digits of that sum and so on), must be 9.
By the sum of its digits: 10. By each of its individual digits: 11.
There are many shell programs that will find the sum of the square of individual digits of a number. A small example is: SD=3n=2, sum=2, and SD=2.
enter the number whose digits are to be added num is the given value num=0! k=num%10 sum=sum=k k=num/10 num=k print the sum of the digits
no thanks
the sum of my digits is 6? answer=60
Add the digits together. The sum of the digits of 23 is 5.
The sum of the digits is also a multiple of 9. And, if the sum of the digits is too large, you can sum those digits and keep going.
Add the digits 6+0+1+5 = 12 If the sum of the individual digits is divisibnle by 3 then the number is divisible by 3. In this case the answer is yes.
Add the digits together. The sum of the digits of 23 is 5.
void main() { int num,r,sum=0; clrscr(); printf("enter the number\n"); scanf("%d",&num); while(num!=0) { r=num%10; sum=sum+r; num=num/10 } printf("The sum of individual digit of given number is=%d",sum); getch() }
8 is the same as the sum of the digits of its cube, 512.
1 + 1 = 2 The sum of the digits is therefore 2.