It is actually quite easy to write a program in java to do this. The easiest way to do this that I can think of is to use the remainder operator (%) to test whether a number is odd or not. Here is a simple program that will print out all the odd numbers between 1 and 50.
public class OddNumbers {
public static void main(String[] args) {
int i=1;
while(i < 50) {
if(i%2 != 0) {
System.out.println(i);
}
i++;
}
}
}
In GE BASIC, you can print odd numbers using a FOR-NEXT loop by specifying a starting point and incrementing by 2. Here’s a simple example: FOR I = 1 TO 99 STEP 2 PRINT I NEXT I This will print all odd numbers from 1 to 99. The STEP 2 ensures the loop only increments by 2, thereby producing only odd numbers.
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.
#include<stdio.h> int main () { int odd=1; int count=0; while (count++<10) { printf (%d\n", odd); odd+=2; } return 0; }
seq 1 2 99
first we write start and then read number and after that check the number is totaly divide by 2 or not if number is totally divide by 2 then number is even else number is odd.
You can use the following JavaScript code to achieve this: for (let i = 0; i <= 15; i++) { let message = `${i} is ${i % 2 === 0 ? 'even' : 'odd'}`; console.log(message); } This loop iterates from 0 to 15, checks if the current number i is even or odd using the modulus operator, and logs the appropriate message to the console.
In GE BASIC, you can print odd numbers using a FOR-NEXT loop by specifying a starting point and incrementing by 2. Here’s a simple example: FOR I = 1 TO 99 STEP 2 PRINT I NEXT I This will print all odd numbers from 1 to 99. The STEP 2 ensures the loop only increments by 2, thereby producing only odd numbers.
odd loop means at least the loop execute once.
how to use even and odd number with for loop and if condition plz dont use "int"..
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.
Using those 4 digits 6451 is the largest odd number you can make.
The largest odd number that can be made with those digits is 9407 .
9825
87603
Get number Find modulo 2 of number If the result is zero, number is even Else number is odd
#include<stdio.h> int main () { int odd=1; int count=0; while (count++<10) { printf (%d\n", odd); odd+=2; } return 0; }
it's impossible because the sum of any two odd numbers equals an even number; odd# + odd# = even#. even# +even# = even#. odd# + odd# + odd# + odd# = even#.