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id	challengeType	title	forumTopicId
5900f3a51000cf542c50feb8	5	Problem 57: Square root convergents	302168

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?

Instructions

Tests

tests:
  - text: <code>squareRootConvergents()</code> should return a number.
    testString: assert(typeof squareRootConvergents() === 'number');
  - text: <code>squareRootConvergents()</code> should return 153.
    testString: assert.strictEqual(squareRootConvergents(), 153);

Challenge Seed

function squareRootConvergents() {

  return true;
}

squareRootConvergents();

Solution

// solution required

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