96 lines
2.3 KiB
Markdown
96 lines
2.3 KiB
Markdown
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---
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id: 5900f3b31000cf542c50fec6
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title: 'Problem 71: Ordered fractions'
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challengeType: 5
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forumTopicId: 302184
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dashedName: problem-71-ordered-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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$$\frac{1}{8}, \frac{1}{7}, \frac{1}{6}, \frac{1}{5}, \frac{1}{4}, \frac{2}{7}, \frac{1}{3}, \frac{3}{8}, \frac{\textbf2}{\textbf5}, \frac{3}{7}, \frac{1}{2}, \frac{4}{7}, \frac{3}{5}, \frac{5}{8}, \frac{2}{3}, \frac{5}{7}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \frac{7}{8}$$
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It can be seen that $\frac{2}{5}$ is the fraction immediately to the left of $\frac{3}{7}$.
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By listing the set of reduced proper fractions for `d` ≤ `limit` in ascending order of size, find the numerator of the fraction immediately to the left of $\frac{3}{7}$.
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# --hints--
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`orderedFractions(8)` should return a number.
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```js
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assert(typeof orderedFractions(8) === 'number');
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```
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`orderedFractions(8)` should return `2`.
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```js
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assert.strictEqual(orderedFractions(8), 2);
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```
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`orderedFractions(10)` should return `2`.
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```js
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assert.strictEqual(orderedFractions(10), 2);
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```
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`orderedFractions(9994)` should return `4283`.
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```js
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assert.strictEqual(orderedFractions(9994), 4283);
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```
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`orderedFractions(500000)` should return `214283`.
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```js
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assert.strictEqual(orderedFractions(500000), 214283);
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```
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`orderedFractions(1000000)` should return `428570`.
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```js
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assert.strictEqual(orderedFractions(1000000), 428570);
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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 orderedFractions(limit) {
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return true;
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}
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orderedFractions(8);
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```
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# --solutions--
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```js
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function orderedFractions(limit) {
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const fractions = [];
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const fractionValues = {};
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const highBoundary = 3 / 7;
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let lowBoundary = 2 / 7;
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for (let denominator = limit; denominator > 2; denominator--) {
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let numerator = Math.floor((3 * denominator - 1) / 7);
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let value = numerator / denominator;
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if (value > highBoundary || value < lowBoundary) {
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continue;
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}
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fractionValues[value] = [numerator, denominator];
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fractions.push(value);
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lowBoundary = value;
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}
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fractions.sort();
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return fractionValues[fractions[fractions.length - 1]][0];
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}
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```
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