---
id: 5900f4ab1000cf542c50ffbd
challengeType: 5
title: 'Problem 318: 2011 nines'
forumTopicId: 301974
---

## Description
<section id='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.
</section>

## Instructions
<section id='instructions'>

</section>

## Tests
<section id='tests'>

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

```

</section>

## Challenge Seed
<section id='challengeSeed'>

<div id='js-seed'>

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

euler318();
```

</div>



</section>

## Solution
<section id='solution'>

```js
// solution required
```

</section>