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[SOLVED] Even Number display (count)?


shamuraq

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Hmm I think this should work.

 

$start_num = 2408;
$end_num = 2531;
$total = 0;
$even = 0;
$odd = 0;
while($start_num < $end_num) {
$start_num++;
if ($start_num%2 == 0) {
	$even ++;
}
else if($start_num%2 == 1) {
	$odd ++;
}
$total++;
}

echo 'total: '.$total;
echo '<br>';
echo 'even: '.$even;
echo '<br>';
echo 'odd: '.$odd;

 

returns

 

total: 123
even: 61
odd: 62

$start = 2408;
$end = 2531;

$num = $end - $start + 1;
$isEven = $num % 2 == 0;
$end_even = $num % 2 == 0;

if ($isEven) {
     $e = $num / 2;
     echo 'Total odd numbers between ' . $start . ' and ' . $end . ' is: ' . $e . '<br />Total even numbers between ' . $start . ' and ' . $end . ' is: ' . $e;
} else {
     $e = round($num / 2);
     echo 'Total odd numbers between ' . $start . ' and ' . $end . ' is: ';
     echo $end_even? $e : $e - 1;
     echo '<br />Total even numbers between ' . $start . ' and ' . $end . ' is: ';
     echo $end_even? $e - 1 : $e;
}

Hi Cosizzle...

i used your codes and inserted into my script; scary. my localhost took sometyme to respond and only part of the script echo out. Script goes like this:

 

<?

$a = rand(1000,9999);

$b= $a + rand(13,49);

$total = 0;

$even = 0;

 

echo "How many even numbers are there between $a and $b?";

 

while($a<$b || $a%2 ==0){

$even++;

}

$f1 = $even + rand(0,10);

$f2 = $even + rand(0,11);

$f3 = $even + rand(0,12);

?>

    <table width="36%" border="0" cellpadding="0" cellspacing="3">

      <tr>

        <td><? echo "A) $even";?></td>

        <td><? echo "B) $f1";?></td>

      </tr>

      <tr>

        <td><? echo "C) $f2";?></td>

        <td><? echo "D) $f3";?></td>

      </tr>

    </table>

    </td>

Try mine. :)

 

I'm trying to understand urs Ken...

Problem is both the values are randomised so there's no way to tell if the starting integer is odd or even. Whereas urs started by taking it as even. How do i manipulate your script to cater to randomised value?

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