# 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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