# 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);``` Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...): Be the first to comment! ## Next similar examples:

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