The sum of is the total of everything being summed; the sum total. Thus the sum of a, b and c is therefore a + b + c.
int sum_digits (int num) { int sum; sum =0; while (num) { sum += num % 10; num /= 10; } return sum; } Example usage: int main (void) { int i; printf ("Input a 6-digit number: "); scanf ("%6d", &i); printf ("The sum of the digits in %d is: %d\n", i, sum_digits (i)); return 0; }
You add 2 fractions with the same denominator [c], so the sum is the sum of the numerators divided by the denominator: a/c + b/c = (a+b)/c
/*use "c-free" compiler*/ #include <stdio.h> main() { int a,b,c; printf("enter the value of a & b"); scanf("%d%d",&a,&b); c=a+b; printf("sum of the two numbers is a+b- %d",c); getch(); }
10 for t = 1 to 50 20 input a 30 c = c + a 40 next t 50 print c
triple the sum of eleven and six
The sum of is the total of everything being summed; the sum total. Thus the sum of a, b and c is therefore a + b + c.
By adding 5+1 to get the sum of six.
J C Sum was born in 1976.
The sum of the first six odd numbers is 36.
the sum of the digits one hundred and twenty six is 9!!
int sum_digits (int num) { int sum; sum =0; while (num) { sum += num % 10; num /= 10; } return sum; } Example usage: int main (void) { int i; printf ("Input a 6-digit number: "); scanf ("%6d", &i); printf ("The sum of the digits in %d is: %d\n", i, sum_digits (i)); return 0; }
The probability of not rolling a sum of six with two fair dice is 1 minus the probability of rolling a sum of six. There are 36 permutations of rolling two dice. Of these, five sum to six, 1+5, 2+4, 3+3, 4+2, and 5+1. The probability, then of rolling a sum of six is 5 in 36. The probability, then of not rolling a sum of six is 31 in 36, or about 0.8611.
The sum of six and seven is expressed as six plus seven. The full equation would be six plus seven equals thirteen.
The sum of the first six positive numbers (1 to 6) is 21.
The sum of the first six counting numbers (1-6) is 19.
c + n