How to know if a # is evenly divisible by another #?

[Czaries]

I need a simple function to tell if a number is evenly divisible by another number.

For instance:

function is_divisible_by($num,$num2) {
/
Would be code here to return TRUE is $num was evenly divisible by $num2, and FALSE if it was not
/
}

[LordShryku]

Here's one of a thousand ways to do this...

function is_divisible_by($num,$num2) {
   $num = number_format(($num/$num2),2);
   if(substr($num, -3) == ".00") {
      return true;
   }
   return false;
}

[pjleonhardt]

if i understand you correctly you can use the modulus function..

basically, it returns the remainder of division, so if the remainder is 0, it divisible by the number yes?

<?
$num1 = 10;
$num2 = 5;

if($num1%$num2 == 0)
{
 $num3 = $num1/$num2;
  echo $num1 . " is divisible by $num2, $num2 times $num3 = $num1"; 
}
else
{
  echo $num1 . " is NOT divisible by $num2":
}
?>

hth

[ahundiak]

Or use the modulus operator
if (($num % $num2) == 0) {
echo 'num exactly divisible by num2";
}
http://www.php.net/manual/en/language.operators.arithmetic.php

[Czaries]

Yeah, thanks guys... I remembered the modulus operation about 30 mins after posting this, but I want to thank you all for your replies and help.