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---
id: 5900f3b61000cf542c50fec8
title: 'Problem 73: Counting fractions in a range'
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challengeType: 5
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forumTopicId: 302186
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dashedName: problem-73-counting-fractions-in-a-range
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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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$$\frac{1}{8}, \frac{1}{7}, \frac{1}{6}, \frac{1}{5}, \frac{1}{4}, \frac{2}{7}, \frac{1}{3}, \mathbf{\frac{3}{8}, \frac{2}{5}, \frac{3}{7}}, \frac{1}{2}, \frac{4}{7}, \frac{3}{5}, \frac{5}{8}, \frac{2}{3}, \frac{5}{7}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \frac{7}{8}$$
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It can be seen that there are `3` fractions between $\frac{1}{3}$ and $\frac{1}{2}$.
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How many fractions lie between $\frac{1}{3}$ and $\frac{1}{2}$ in the sorted set of reduced proper fractions for `d` ≤ `limit` ?
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# --hints--
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`countingFractionsInARange(8)` should return a number.
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```js
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assert(typeof countingFractionsInARange(8) === 'number');
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```
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`countingFractionsInARange(8)` should return `3` .
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```js
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assert.strictEqual(countingFractionsInARange(8), 3);
```
`countingFractionsInARange(1000)` should return `50695` .
```js
assert.strictEqual(countingFractionsInARange(1000), 50695);
```
`countingFractionsInARange(6000)` should return `1823861` .
```js
assert.strictEqual(countingFractionsInARange(6000), 1823861);
```
`countingFractionsInARange(12000)` should return `7295372` .
```js
assert.strictEqual(countingFractionsInARange(12000), 7295372);
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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 countingFractionsInARange(limit) {
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return true;
}
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countingFractionsInARange(8);
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```
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# --solutions--
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```js
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function countingFractionsInARange(limit) {
let result = 0;
const stack = [[3, 2]];
while (stack.length > 0) {
const [startDenominator, endDenominator] = stack.pop();
const curDenominator = startDenominator + endDenominator;
if (curDenominator < = limit) {
result++;
stack.push([startDenominator, curDenominator]);
stack.push([curDenominator, endDenominator]);
}
}
return result;
}
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```