ee1e8abd87
* feat(tools): add seed/solution restore script * chore(curriculum): remove empty sections' markers * chore(curriculum): add seed + solution to Chinese * chore: remove old formatter * fix: update getChallenges parse translated challenges separately, without reference to the source * chore(curriculum): add dashedName to English * chore(curriculum): add dashedName to Chinese * refactor: remove unused challenge property 'name' * fix: relax dashedName requirement * fix: stray tag Remove stray `pre` tag from challenge file. Signed-off-by: nhcarrigan <nhcarrigan@gmail.com> Co-authored-by: nhcarrigan <nhcarrigan@gmail.com>
49 lines
1.2 KiB
Markdown
49 lines
1.2 KiB
Markdown
---
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id: 5900f3b21000cf542c50fec5
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title: 'Problem 70: Totient permutation'
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challengeType: 5
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forumTopicId: 302183
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dashedName: problem-70-totient-permutation
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---
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# --description--
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Euler's Totient function, φ(`n`) \[sometimes called the phi function], is used to determine the number of positive numbers less than or equal to `n` which are relatively prime to `n`. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, φ(9)=6. The number 1 is considered to be relatively prime to every positive number, so φ(1)=1.
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Interestingly, φ(87109)=79180, and it can be seen that 87109 is a permutation of 79180.
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Find the value of `n`, 1 < `n` < 10<sup>7</sup>, for which φ(`n`) is a permutation of `n` and the ratio `n`/φ(`n`) produces a minimum.
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# --hints--
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`totientPermutation()` should return a number.
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```js
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assert(typeof totientPermutation() === 'number');
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```
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`totientPermutation()` should return 8319823.
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```js
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assert.strictEqual(totientPermutation(), 8319823);
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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 totientPermutation() {
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return true;
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}
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totientPermutation();
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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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