checking if a number is divisible by 6 PHP

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 ?

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;

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

$num += (6-$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.

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;

I see some of the other answers calling the modulo twice.

My preference is not to ask php to do the same thing more than once. For this reason, I cache the remainder.

Other devs may prefer to not generate the extra global variable or have other justifications for using modulo operator twice.

Code: (Demo)

$factor = 6;
for($x = 0; $x < 10; ++$x){  // battery of 10 tests
    $number = rand( 0 , 100 );
    echo "Number: $number Becomes: ";
    if( $remainder = $number % $factor ) {  // if not zero
        $number += $factor - $remainder;  // use cached $remainder instead of calculating again
    }
    echo "$number\n";
}

Possible Output:

Number: 80 Becomes: 84
Number: 57 Becomes: 60
Number: 94 Becomes: 96
Number: 48 Becomes: 48
Number: 80 Becomes: 84
Number: 36 Becomes: 36
Number: 17 Becomes: 18
Number: 41 Becomes: 42
Number: 3 Becomes: 6
Number: 64 Becomes: 66

checking if a number is divisible by 6 PHP

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