I was preparing a lesson example, and found
it's not as straightforward as i thought -

write a simple loop that outputs all the odd
numbers less than 50.

--use only while loop
--do not use the modulus(%) operator

Just want to see the different approaches
of fellow programmers.

With Java
class Bakenda
{
public static void main(String[] param)
{
int x=1;
while(x<50){
System.out.println( x );
x+=2;
}
}
}

or with Ruby

(1. .50).step(2) { |n| puts n}

Remove the space between the dots. .

Gooood.


I should av added a 3rd condition - use only a 1-step increment.

I Dont understand that condition, Does that apply to me??, Do you mean it has to be +1 always?

Fayimora:

I Dont understand that condition, Does that apply to me??, Do you mean it has to be +1 always?

Yeah. You may want to try it out with +1

You can do it in closed form, I think. So no reason to do any looping.

sum of the even numbers from 0 to 50 is equal to 2 times the sum of the numbers from 0 to 25. And there is a closed form expression for the latter (don't recall it off the top of my head, but it is on wikipedia.)

Oh wow. I skimmed the post and misread it, thought he wanted a sum too. My bad. .

Wow, pretty bad fail on my part. Whoops :x

bakenda:

Yeah. You may want to try it out with +1

Guess thers no point again, ma solution is more or less the same with @omo's python solution

^^^

Yeah. My solution is similar too, but I didn't arrive at it
"instantly" like that. I was working on an example on
how to use the while loop(to output odd numbers), I
thought it was a simple thing(it is, actually), my  initial
mistake was to write something like:

while($n<50)
  {
  echo (2*$n)+1 . "
";
$n++;
  }

Now i c why computer science is tied to mathematics @ omo to dun u too much