checking if a number is divisible by 6 PHP

Question

Hi,

I want to check if a number is divisible by 6 and if not i need to increase it until it becomes divisible.

how can i do that ?

Thanks

Answer 1

Use the Mod % (modulus) operator

if ($x % 6 == 0) return 1;


function nearest_multiple_of_6($x) {
    if ($x % 6 == 0) return $x;    

    return (($x / 6) + 1) * 6;
}

Answer 2

So you want the next multiple of 6, is that it?

You can divide your number by 6, then ceil it, and multiply it again:

$answer = ceil($foo / 6) * 6;

Answer 3

Simply run a while loop that will continue to loop (and increase the number) until the number is divisible by 6.

while ($number % 6 != 0) {
    $number++;
}

Answer 4

For micro-optimisation freaks:

if ($num % 6 != 0)
    $num += 6 - $num % 6;

More evaluations of %, but less branching/looping. :-P

Answer 5

$num += (6-$num%6)%6;

no need for a while loop! Modulo (%) returns the remainder of a division. IE 20%6 = 2. 6-2 = 4. 20+4 = 24. 24 is divisible by 6.

Answer 6

Why don't you use the Modulus Operator?

Try this:

while ($s % 6 != 0) $s++;

Or is this what you meant?

<?

 $s= <some_number>;
 $k= $s % 6;

 if($k !=0)    $s=$s+6-$k;
?>

Answer 7

Simple:

if($number %6 != 0) {
  $number += 6 - ($number % 6);
}

The modulus operator gives the remainder of the division, so $number % 6 is the amount left over when dividing by 6. This will be faster than doing a loop and continually rechecking.

If decreasing is acceptable then this is even faster:

$number -= $number % 6;

Answer 8

if ($variable % 6 == 0) {
    echo 'This number is divisible by 6.';
}:

Make divisible by 6:

$variable += (6 - ($variable % 6)) % 6; // faster than while for large divisors

Answer 9

result = initial number + (6 - initial number % 6)

Answer 10

Assuming $foo is an integer:

$answer = (int) (floor(($foo+5)/6)*6)