111 lines
3.0 KiB
Markdown
111 lines
3.0 KiB
Markdown
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---
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id: 5900f3b11000cf542c50fec4
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title: 'Problem 69: Totient maximum'
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challengeType: 5
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forumTopicId: 302181
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dashedName: problem-69-totient-maximum
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---
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# --description--
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Euler's Totient function, ${\phi}(n)$ (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, ${\phi}(9) = 6$.
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<div style='margin-left: 4em;'>
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| $n$ | $\text{Relatively Prime}$ | $\displaystyle{\phi}(n)$ | $\displaystyle\frac{n}{{\phi}(n)}$ |
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| --- | ------------------------- | ------------------------ | ---------------------------------- |
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| 2 | 1 | 1 | 2 |
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| 3 | 1,2 | 2 | 1.5 |
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| 4 | 1,3 | 2 | 2 |
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| 5 | 1,2,3,4 | 4 | 1.25 |
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| 6 | 1,5 | 2 | 3 |
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| 7 | 1,2,3,4,5,6 | 6 | 1.1666... |
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| 8 | 1,3,5,7 | 4 | 2 |
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| 9 | 1,2,4,5,7,8 | 6 | 1.5 |
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| 10 | 1,3,7,9 | 4 | 2.5 |
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</div>
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It can be seen that `n` = 6 produces a maximum $\displaystyle\frac{n}{{\phi}(n)}$ for `n` ≤ 10.
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Find the value of `n` ≤ `limit` for which $\displaystyle\frac{n}{{\phi(n)}}$ is a maximum.
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# --hints--
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`totientMaximum(10)` should return a number.
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```js
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assert(typeof totientMaximum(10) === 'number');
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```
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`totientMaximum(10)` should return `6`.
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```js
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assert.strictEqual(totientMaximum(10), 6);
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```
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`totientMaximum(10000)` should return `2310`.
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```js
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assert.strictEqual(totientMaximum(10000), 2310);
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```
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`totientMaximum(500000)` should return `30030`.
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```js
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assert.strictEqual(totientMaximum(500000), 30030);
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```
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`totientMaximum(1000000)` should return `510510`.
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```js
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assert.strictEqual(totientMaximum(1000000), 510510);
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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 totientMaximum(limit) {
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return true;
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}
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totientMaximum(10);
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```
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# --solutions--
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```js
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function totientMaximum(limit) {
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function getSievePrimes(max) {
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const primesMap = new Array(max).fill(true);
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primesMap[0] = false;
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primesMap[1] = false;
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const primes = [];
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for (let i = 2; i < max; i = i + 2) {
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if (primesMap[i]) {
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primes.push(i);
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for (let j = i * i; j < max; j = j + i) {
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primesMap[j] = false;
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}
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}
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if (i === 2) {
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i = 1;
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}
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}
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return primes;
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}
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const MAX_PRIME = 50;
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const primes = getSievePrimes(MAX_PRIME);
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let result = 1;
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for (let i = 0; result * primes[i] < limit; i++) {
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result *= primes[i];
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}
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return result;
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}
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```
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