c0aaecba63
* fix: rework challenge to use argument in function * fix: add solution * fix: use MathJax in fractions and equations * fix: missing backticks
91 lines
2.0 KiB
Markdown
91 lines
2.0 KiB
Markdown
---
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id: 5900f3b41000cf542c50fec7
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title: 'Problem 72: Counting fractions'
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challengeType: 5
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forumTopicId: 302185
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dashedName: problem-72-counting-fractions
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---
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# --description--
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Consider the fraction, $\frac{n}{d}$, where `n` and `d` are positive integers. If `n` < `d` and highest common factor, ${HCF}(n, d) = 1$, it is called a reduced proper fraction.
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If we list the set of reduced proper fractions for `d` ≤ 8 in ascending order of size, we get:
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$$\frac{1}{8}, \frac{1}{7}, \frac{1}{6}, \frac{1}{5}, \frac{1}{4}, \frac{2}{7}, \frac{1}{3}, \frac{3}{8}, \frac{2}{5}, \frac{3}{7}, \frac{1}{2}, \frac{4}{7}, \frac{3}{5}, \frac{5}{8}, \frac{2}{3}, \frac{5}{7}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \frac{7}{8}$$
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It can be seen that there are `21` elements in this set.
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How many elements would be contained in the set of reduced proper fractions for `d` ≤ `limit`?
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# --hints--
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`countingFractions(8)` should return a number.
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```js
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assert(typeof countingFractions(8) === 'number');
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```
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`countingFractions(8)` should return `21`.
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```js
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assert.strictEqual(countingFractions(8), 21);
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```
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`countingFractions(20000)` should return `121590395`.
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```js
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assert.strictEqual(countingFractions(20000), 121590395);
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```
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`countingFractions(500000)` should return `75991039675`.
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```js
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assert.strictEqual(countingFractions(500000), 75991039675);
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```
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`countingFractions(1000000)` should return `303963552391`.
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```js
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assert.strictEqual(countingFractions(1000000), 303963552391);
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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 countingFractions(limit) {
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return true;
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}
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countingFractions(8);
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```
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# --solutions--
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```js
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function countingFractions(limit) {
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const phi = {};
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let count = 0;
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for (let i = 2; i <= limit; i++) {
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if (!phi[i]) {
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phi[i] = i;
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}
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if (phi[i] === i) {
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for (let j = i; j <= limit; j += i) {
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if (!phi[j]) {
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phi[j] = j;
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}
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phi[j] = (phi[j] / i) * (i - 1);
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}
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}
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count += phi[i];
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}
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return count;
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}
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```
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