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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>
43 lines
1.0 KiB
Markdown
43 lines
1.0 KiB
Markdown
---
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id: 5900f3f21000cf542c50ff05
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title: 'Problem 134: Prime pair connection'
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challengeType: 5
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forumTopicId: 301762
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dashedName: problem-134-prime-pair-connection
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---
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# --description--
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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 $p_2$.
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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 $n$.
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Find $\sum{S}$ for every pair of consecutive primes with $5 ≤ p_1 ≤ 1000000$.
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# --hints--
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`primePairConnection()` should return `18613426663617120`.
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```js
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assert.strictEqual(primePairConnection(), 18613426663617120);
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```
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# --seed--
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## --seed-contents--
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```js
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function primePairConnection() {
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return true;
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}
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primePairConnection();
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```
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# --solutions--
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```js
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// solution required
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```
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