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How to check whether number is perfect or not in PHP

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503 views
asked Feb 27, 2016 by avibootz

2 Answers

0 votes
function isPerfectNumber($n)
{
    $i = 1;
    $sum = 0;
     
    while ($i < $n)
    {
        if ($n % $i == 0)
        {
            $sum = $sum + $i;
        }
        $i++;
    }
    return $sum == $n;
}

$n = 496;
 
if (isPerfectNumber($n))
    echo $n . " is a Perfect Number";
else
    echo $n . " is Not a Perfect Number";

    
/*
run:

496 is a Perfect Number 
     
*/

 




answered Feb 27, 2016 by avibootz
0 votes
function isPerfectNumber($number)
{
    $sum = 0;
    for ($i = 2; $i <= sqrt($number); $i++)
    {
        if (!($number % $i))
        {
            $sum += $i;
            if ($i <> $number / $i)
                $sum += $number / $i;
        }
    }
    return ++$sum == $number;
}

$n = 496;
  
if (isPerfectNumber($n))
    echo $n . " is a Perfect Number";
else
    echo $n . " is Not a Perfect Number";    

 
/*
run: 
  
496 is a Perfect Number
  
*/ 

 




answered Sep 4 by avibootz
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