3*(a + b) or 3a + 3b
5a plus 2b
3(a+b)+a
3(b5)
3a+b
3*(a + b) or 3a + 3b
5a plus 2b
3(a+b)+a
3(b5)
The sum of b and 8 would be written b+8 or 8+b(commutative property of addition)
3a+b
int mul (int a, int b) { int sum= 0; for (; b>0; --b) sum -= -a; for (; b<0; ++b) sum -= a; return sum; }
You remember that 'sum' means addition so that's b+11
3(a + b) + a = 3a + 3b + a = 4a + 3b
(a+b)/6
4(2a+b) + 2(6a+2b) Working this out it is: 8a + 4b + 12a + 4b = 20a + 8b or 4(5a+2b).
//sum and product of 3 nos #include #include void main() { int a,b,c; printf("enter the 3 nos"); scanf("%d%d%d",&a,&b,&c); printf("sum of 3 nos",a+b+c); printf("product of 3 nos",a*b*c); getch(); }