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* 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>
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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