777/3 = 259. So the numbers are 258, 259 and 260.
//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(); }
3,5&7 3*5*7=105
Yes, the sum of any three consecutive counting numbers has a factor of 3.
It need not be. 2 and 3 are consecutive integers and their sum is 5, not 31.
The sum of 3 consecutive whole numbers is always equal to 3 times the middle number in that sequence.
The sum of any three consecutive even numbers must be divisible by 3. 32 is not, so there is no solution.
2 and 3.
You probably want to know "What are the 3 consecutive whole numbers whose sum is 72 ?" They are 23, 24, and 25.
Sorry. No sum of even numbers can be an odd number.
The sum of three consecutive integers starting with ( n ) can be expressed as ( n + (n + 1) + (n + 2) ). Simplifying this expression gives ( 3n + 3 ). Therefore, the sum of the three consecutive integers is ( 3n + 3 ) or ( 3(n + 1) ).
The sum is 3*(x-1) or 3x - 3
There are four consecutive 3-digit numbers with a sum of 2,494: 622, 623, 624 and 625.