1.Start 2. Input a,b,c 3. Sum = a+b+c 4. Average = sum/3 5. Output - Sum,Average 6. Stop
Sorry. No sum of even numbers can be an odd number.
What to whole numbers have a sum of 12 and quotient of 3?
The sum of two numbers is the result given when the two numbers are added together. For example, the sum of 2 and 3 is 5.
No, the sum of three odd numbers will always be odd.
1.Start 2. Input a,b,c 3. Sum = a+b+c 4. Average = sum/3 5. Output - Sum,Average 6. Stop
//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(); }
You add the numbers. Then you divide by the number of numbers (i.e., if you have three numbers, you divide the sum by 3).You add the numbers. Then you divide by the number of numbers (i.e., if you have three numbers, you divide the sum by 3).You add the numbers. Then you divide by the number of numbers (i.e., if you have three numbers, you divide the sum by 3).You add the numbers. Then you divide by the number of numbers (i.e., if you have three numbers, you divide the sum by 3).
Sorry. No sum of even numbers can be an odd number.
What to whole numbers have a sum of 12 and quotient of 3?
There are not three prime numbers that have the sum of 3. The smallest prime number is 2. If all three prime numbers were 2, the sum would 2 + 2 + 2 = 6, so that is the smallest number that is the sum of three prime numbers.
If the product of two numbers is 4.2 and their sum is 4.4. then the numbers are 3 and 1.4.
3 prime numbers which have the sum of 132: 2 + 3 + 127 2 + 17 + 113
The sum of two numbers is the result given when the two numbers are added together. For example, the sum of 2 and 3 is 5.
No, the sum of three odd numbers will always be odd.
The two (not tow) numbers with a sum of 12 and a quotient (not quotation) of 3 are 9 and 3.
The three numbers are 24, 25, 26. You can solve this as a system of 3 equations and 3 unknowns. Let A, B, & C be the 3 numbers, so: [A + B - C = 23] ; then two equations to describe they are consecutive: [B - A = 1] & [C - B = 1]. Solving gives A=24, B=25, & C=26