<section id="description">
<p>Consider the consecutive primes p1 = 19 and p2 = 23. It can be verified that 1219 is the smallest number such that the last digits are formed by p1 whilst also being divisible by p2.</p>
<p>In fact, with the exception of p1 = 3 and p2 = 5, for every pair of consecutive primes, p2 > p1, there exist values of n for which the last digits are formed by p1 and n is divisible by p2. Let S be the smallest of these values of n.</p>
<p>Find ∑ S for every pair of consecutive primes with 5 ≤ p1 ≤ 1000000.</p></section>

function euler134() {
  // Good luck!
  return true;
}

euler134();


index

Проблема 134: Соединение с главной парой

Problem 134: Prime pair connection

<code>euler134()</code> should return 18613426663617120.