PHP - If number is divisible by 3 and 5 then echo

Question

I\'m new to PHP and trying to create the following whilst minimizing the amount of code needed. PHP should show a list of 100 then display if the number is / by 3, 5 or 3 and 5.

Answer 1

      <?php

for ($i = 1; $i <= 100; $i++) {

    if ($i % 15 == 0) echo "This Number is Divisible by 3 and 5<br>";
    else if ($i % 3 == 0) echo "This Number is Divisible by 3 only<br>";
    else if ($i % 5 == 0) echo "This number is Divisible by 5 only<br>";
    else{
        echo "$i<br>";
    }

}

   ?>

Answer 2

<?php

if($number % 5 == 0 && $number % 3 == 0)  {
    echo "BY3 AND 5";
} elseif ($number % 5 == 0) {
    echo "BY5";
} elseif ($number % 3 == 0) {
    echo "BY3";
} else{
    echo "NOT BY3 OR 5";
}   
?>

Answer 3

This is neater and completed to be run:

<?php

for ($i = 1; $i <= 100; $i++) { 
                if ($i % 15 == 0)
                {
                    echo"Divisible by 3 and 5</br>";
                }
                elseif ($i%3==0)
                {
                    echo"Divisible by 3</br>";
                }
                elseif ($i%5==0)
                {
                    echo"Divisible by 5</br>";
                }
                else
                {
                    echo $i,"</br>";
                }

}
?>

Answer 4

$num_count = 100;
    $div_3 = "Divisible by 3";
    $div_5 = "Divisible by 5";
    $div_both = "Divisible by 3 and 5";
    $not_div = "Not Divisible by 3 or 5";

    for($i=0;$i<=$num_count;$i++)
    {
        switch($i)
        {
            case ($i%15==0):
            echo $i." (".$div_both.")</br>";
            break;
            case ($i%3==0):
            echo $i." (".$div_3.")</br>";
            break;
            case ($i%5==0):
            echo $i." (".$div_5.")</br>";
            break;
            default:
            echo $i."</br>";
            break;
        }
    }

Answer 5

No need to do three if statements:

echo "<table border='1'>";

for ($i = 1; $i <= 100; $i++) {

    echo "<tr><td>{$i}</td><td>";

    if ($i % 3 == 0) echo "BY3 ";
    if ($i % 5 == 0) echo "BY5";

    echo "</td></tr>\n";
}
echo "</table>";

Answer 6

if($number % 15 == 0)  
{
     echo "Divisible by 3 and 5";
} 
elseif ($number % 5 == 0) 
{
    echo "Divisible by 5";
} 
elseif ($number % 3 == 0) 
{
 echo "Divisible by 3";
}