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---
id: 5900f3a51000cf542c50feb8
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title: 'Problem 57: Square root convergents'
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challengeType: 5
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forumTopicId: 302168
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dashedName: problem-57-square-root-convergents
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---
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# --description--
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It is possible to show that the square root of two can be expressed as an infinite continued fraction.
< div style = 'text-align: center;' > $\sqrt 2 =1+ \frac 1 {2+ \frac 1 {2 +\frac 1 {2+ \dots}}}$</ div >
By expanding this for the first four iterations, we get:
$1 + \\frac 1 2 = \\frac 32 = 1.5$
$1 + \\frac 1 {2 + \\frac 1 2} = \\frac 7 5 = 1.4$
$1 + \\frac 1 {2 + \\frac 1 {2+\\frac 1 2}} = \\frac {17}{12} = 1.41666 \\dots$
$1 + \\frac 1 {2 + \\frac 1 {2+\\frac 1 {2+\\frac 1 2}}} = \\frac {41}{29} = 1.41379 \\dots$
The next three expansions are $\\frac {99}{70}$, $\\frac {239}{169}$, and $\\frac {577}{408}$, but the eighth expansion, $\\frac {1393}{985}$, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator.
In the first one-thousand expansions, how many fractions contain a numerator with more digits than denominator?
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# --hints--
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`squareRootConvergents()` should return a number.
```js
assert(typeof squareRootConvergents() === 'number');
```
`squareRootConvergents()` should return 153.
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```js
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assert.strictEqual(squareRootConvergents(), 153);
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```
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# --seed--
## --seed-contents--
```js
function squareRootConvergents() {
return true;
}
squareRootConvergents();
```
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# --solutions--
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```js
// solution required
```