Unknown number

Unknown number is divisible by exactly three different primes. When we compare these primes in ascending order, the following applies:
• Difference first and second prime number is half the difference between the third and second prime numbers.
• The product of the difference the second with the first primes
and difference the third and the second prime number is a multiple of 17.

Determine the smallest number that has all the above properties.

Result

n =  2014

Solution:

2014= 2 × 19 × 53

$primes = array(2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67);
for($a=0;$a<count($primes); $a++)
{
    for($b=$a+1;$b<count($primes); $b++)
    {
        for($c=$b+1;$c<count($primes); $c++)
        {
            $p1 = $primes[$a];
            $p2 = $primes[$b];
            $p3 = $primes[$c];

            if($p2-$p1 == 0.5*($p3-$p2) &&  mod(($p2-$p1)*($p3-$p2),17)==0)
            {
                $rv[$p1*$p2*$p3] = "$p1 $p2 $p3";
            }

        }
    }

}
ksort($rv);
print_r($rv);

$2014= 2 \cdot 19 \cdot 53$

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