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>
1.9 KiB
1.9 KiB
id, title, challengeType, forumTopicId, dashedName
| id | title | challengeType | forumTopicId | dashedName |
|---|---|---|---|---|
| 5900f3b11000cf542c50fec4 | Problem 69: Totient maximum | 5 | 302181 | problem-69-totient-maximum |
--description--
Euler's Totient function, φ(n) [sometimes called the phi function], is used to determine the number of numbers less than 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.
| n | Relatively Prime | φ(n) | n/φ(n) |
|---|---|---|---|
| 2 | 1 | 1 | 2 |
| 3 | 1,2 | 2 | 1.5 |
| 4 | 1,3 | 2 | 2 |
| 5 | 1,2,3,4 | 4 | 1.25 |
| 6 | 1,5 | 2 | 3 |
| 7 | 1,2,3,4,5,6 | 6 | 1.1666... |
| 8 | 1,3,5,7 | 4 | 2 |
| 9 | 1,2,4,5,7,8 | 6 | 1.5 |
| 10 | 1,3,7,9 | 4 | 2.5 |
It can be seen that n=6 produces a maximum n/φ(n) for n ≤ 10.
Find the value of n ≤ 1,000,000 for which n/φ(n) is a maximum.
--hints--
totientMaximum() should return a number.
assert(typeof totientMaximum() === 'number');
totientMaximum() should return 510510.
assert.strictEqual(totientMaximum(), 510510);
--seed--
--seed-contents--
function totientMaximum() {
return true;
}
totientMaximum();
--solutions--
// solution required