Files
gikf 7907f62337 fix(curriculum): clean-up Project Euler 121-140 (#42731)
* fix: clean-up Project Euler 121-140

* fix: corrections from review

Co-authored-by: Sem Bauke <46919888+Sembauke@users.noreply.github.com>

* fix: missing backticks

Co-authored-by: Kristofer Koishigawa <scissorsneedfoodtoo@gmail.com>

* fix: corrections from review

Co-authored-by: Tom <20648924+moT01@users.noreply.github.com>

* fix: missing delimiter

Co-authored-by: Sem Bauke <46919888+Sembauke@users.noreply.github.com>
Co-authored-by: Kristofer Koishigawa <scissorsneedfoodtoo@gmail.com>
Co-authored-by: Tom <20648924+moT01@users.noreply.github.com>
2021-07-16 21:38:37 +02:00

1.0 KiB

id, title, challengeType, forumTopicId, dashedName
id title challengeType forumTopicId dashedName
5900f3f21000cf542c50ff05 Problem 134: Prime pair connection 5 301762 problem-134-prime-pair-connection

--description--

Consider the consecutive primes p_1 = 19 and p_2 = 23. It can be verified that 1219 is the smallest number such that the last digits are formed by p_1 whilst also being divisible by p_2.

In fact, with the exception of p_1 = 3 and p_2 = 5, for every pair of consecutive primes, p_2 > p_1, there exist values of n for which the last digits are formed by p_1 and n is divisible by p_2. Let S be the smallest of these values of n.

Find \sum{S} for every pair of consecutive primes with 5 ≤ p_1 ≤ 1000000.

--hints--

primePairConnection() should return 18613426663617120.

assert.strictEqual(primePairConnection(), 18613426663617120);

--seed--

--seed-contents--

function primePairConnection() {

  return true;
}

primePairConnection();

--solutions--

// solution required