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id	title	challengeType	forumTopicId	dashedName
5900f3a51000cf542c50feb8	Problem 57: Square root convergents	5	302168	problem-57-square-root-convergents

--description--

It is possible to show that the square root of two can be expressed as an infinite continued fraction.

$\sqrt 2 =1+ \frac 1 {2+ \frac 1 {2 +\frac 1 {2+ \dots}}}$

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?

--hints--

squareRootConvergents() should return a number.

assert(typeof squareRootConvergents() === 'number');

squareRootConvergents() should return 153.

assert.strictEqual(squareRootConvergents(), 153);

--seed--

--seed-contents--

function squareRootConvergents() {

  return true;
}

squareRootConvergents();

--solutions--

// solution required

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