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() }
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.
The number is 45. The sum of its digits i.e. 4+5=9 Five times the sum of its digits i.e. 5 times 9 which is 45
The sum of the digits in the number 65 is calculated by adding the individual digits together. In this case, the digits are 6 and 5. Adding 6 + 5 equals 11. Therefore, the sum of the digits in the number 65 is 11.
1 + 1 = 2 The sum of the digits is therefore 2.
Add the digits together. The sum of the digits of 23 is 5.
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.
By the sum of its digits: 10. By each of its individual digits: 11.
The sum of the individual digit is itself
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the sum of my digits is 6? answer=60
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
Add the digits together. The sum of the digits of 23 is 5.
The number is 45. The sum of its digits i.e. 4+5=9 Five times the sum of its digits i.e. 5 times 9 which is 45
To find the sum of the digits in the number 34353, we can add each individual digit together. 3 + 4 + 3 + 5 + 3 = 18. Therefore, the sum of the values of the digits in the number 34353 is 18.
The sum of the digits in the number 65 is calculated by adding the individual digits together. In this case, the digits are 6 and 5. Adding 6 + 5 equals 11. Therefore, the sum of the digits in the number 65 is 11.
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.