class even
{
public static viod main(String args[])
{
System.out.println("<------even number from 1 to 1000------>");
for(int i=1;i<=1000;i++)
if(i%2==0)
System.out.print(i+" ");
}
}
Here is another way:
for (int i = 2; i <= 1000; i += 2)
System.out.print(i + " ");
You can use int i; for (i = 10; i <= 50; i += 2) {//print i} as a program to print even numbers between 10 and 50.
In BASIC: 10 INPUT X: IF X = 999 THEN STOP ELSE PRINT X; 20 IF X/2 = INT(X/2) THEN PRINT "EVEN" ELSE PRINT "ODD" 30 GOTO 10
Reference:http:cprogramming-bd.com/c_page2.aspx# strange number
find even number in array
printf ("100+...+200=%d\n", ((100+200)/2)*(202-100)/2);
To write a C program to determine if something is odd or even you need to be a programmer. To write a program in C is complicate and only done by programmers.
for (int i = 2; i < 10; i ++) printf("%d\n", i); You did say even and odd numbers between 1 and 10. That's allnumbers between 1 and 10.
# include<stdio.h> void main() { int i = 1; while(i<11) { if(i%2==1) { printf("\n%d",i); } i++; } }
Code Below: <?php $j = 100; // Set limit upper limit echo "Even Numbers are: <br/>"; for($i=1;$i<=$j;$i++) { if($i%2) { continue; } else { echo $i."<br/>"; } } ?>
Start print "the sum of all even numbers is infinite" end
10
This program checks whether a number is odd or even. NOTE: This site removes formatting from answers. Replace (tab) with a tab or four spaces. #!/usr/bin/python print("Type a number.") n = input(": ") l = len(n)-1 even = ("02468") if n[l] in even: (tab)print(n, "is even.") if n[l] not in even: (tab)print(n, "is odd.")