freeCodeCamp/problem-318-2011-nines.english.md at 81a0c0c8e07a6c3a4f99228a8af10967fa0c53d9

Valeriy 79d9012432 fix(curriculum): quotes in tests (#18828 )

* fix(curriculum): tests quotes

* fix(curriculum): fill seed-teardown

* fix(curriculum): fix tests and remove unneeded seed-teardown

2018-10-20 23:32:47 +05:30

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id, challengeType, title

id	challengeType	title
5900f4ab1000cf542c50ffbd	5	Problem 318: 2011 nines

Description

Consider the real number √2+√3. When we calculate the even powers of √2+√3 we get: (√2+√3)2 = 9.898979485566356... (√2+√3)4 = 97.98979485566356... (√2+√3)6 = 969.998969071069263... (√2+√3)8 = 9601.99989585502907... (√2+√3)10 = 95049.999989479221... (√2+√3)12 = 940897.9999989371855... (√2+√3)14 = 9313929.99999989263... (√2+√3)16 = 92198401.99999998915...

It looks like that the number of consecutive nines at the beginning of the fractional part of these powers is non-decreasing. In fact it can be proven that the fractional part of (√2+√3)2n approaches 1 for large n.

Consider all real numbers of the form √p+√q with p and q positive integers and p<q, such that the fractional part of (√p+√q)2n approaches 1 for large n.

Let C(p,q,n) be the number of consecutive nines at the beginning of the fractional part of (√p+√q)2n.

Let N(p,q) be the minimal value of n such that C(p,q,n) ≥ 2011.

Find ∑N(p,q) for p+q ≤ 2011.

Instructions

Tests

tests:
  - text: <code>euler318()</code> should return 709313889.
    testString: assert.strictEqual(euler318(), 709313889, '<code>euler318()</code> should return 709313889.');

Challenge Seed

function euler318() {
  // Good luck!
  return true;
}

euler318();

Solution

// solution required

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