#include
using std::cout;
using std::endl;
int main()
{
int arr[] = {2, 45, 34, 56, 56};//you define here your array of even and odd numbers
int arrSize = sizeof arr/sizeof arr[0];
int numEven = 0;
int numOdd = 0;
for (int i = 0; i < (arrSize - 1); i++)
{
if (arr[i] % 2 == 0)
{
numEven++;
}
else
{
numOdd++;
}
}
cout << endl << "Number of even numbers: " << numEven
<< endl << "Number of odd numbers: " << numOdd
<< endl;
system("PAUSE");
return 0;
}
-- First of all, if the number is a fraction or a decimal, then don't worry about it. Only whole numbers are considered odd or even. -- All multiples of 2 are even numbers, and that's easy to recognize. It doesn't matter how big the number is, or how many digits it has. You only have to look at the last digit (the one on the right-hand end). If that digit is 2, 4, 6, 8, or 0, then the whole big number is even. Otherwise it's odd.
Reference:http:cprogramming-bd.com/c_page2.aspx# strange number
Move the print out requesting the user to enter an integer outside of the for loop and it will only print once instead of each time around the loop. You'll need a way to save the even and odd numbers that you detect in the loop. One way is to have separate arrays to hold the even and the odd numbers as you go around the loop. Then at the end of the loop you can have more loops to print the contents of one array and then the contents of the other array. Another way is to concatenate the number onto separate Strings (even and odd) to be displayed after the data gathering loop.
import java.util.*; class Demo { public static void main(String args[]) { System.out.println("Even Numbers Are "); for (int i=1;i<=20;i++) { System.out.print(" "+(2*i)); } System.out.println("\n \n Odd Numbers Are "); for(int j=0;j<=20;j++) { System.out.print(" "+((2*j)+1)); } } }
Algorithm: sum_evenInput: an integer, n, such that n>0Output: the sum of all even integers within the open range (1:n)sum := 0val := 2while (val < n) do{sum := sum + valval := val + 2}return sumNote that you explicitly specified between 1 and n, which literally means both 1 and n should be excluded from the sum. 1 would be excluded anyway since it is not an even number, however if we wish to include n, then use the following algorithm instead:Algorithm: sum_evenInput: an integer, n, such that n>0Output: the sum of all even integers within the half-open range (1:n]sum := 0val := 2while (val
None of them. All counting numbers are either odd or even.
None. All counting numbers are even or odd.
1, 3, 5, 7, 9 counting by twos to 99 are the odd numbers. 2, 4, 6, 8, 10 counting by twos to 100 are the even numbers.
All numbers are either odd or even, none are both, so less than 30 there are 29 counting numbers that are either odd or even but not both. (Assuming you mean starting to count with 1.)
The sum of the first 500 odd counting numbers is 250,000.
irrational numbers odd numbers even numbers whole numbers prime numbers composite numbers counting numbers
Niether it wouldn't be counted in a group of odd and even numbers it would only be counted as a odd and even if you was counting in decimals
Only even numbers are divisible by two. If you have an odd number, for example 3, and divide by two, you will get 1 with a remander of 1.
No.
37
There is no such pattern because there are no even odd numbers. Odd numbers, by definition, are odd and therefore, not even.
Choose one odd and one even