freeCodeCamp/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-318-2011-nines.md

id, title, challengeType, forumTopicId, dashedName

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

--description--

Consider the real number \sqrt{2} + \sqrt{3}.

When we calculate the even powers of \sqrt{2} + \sqrt{3} we get:

$$\begin{align} & {(\sqrt{2} + \sqrt{3})}^2 = 9.898979485566356\ldots \\ & {(\sqrt{2} + \sqrt{3})}^4 = 97.98979485566356\ldots \\ & {(\sqrt{2} + \sqrt{3})}^6 = 969.998969071069263\ldots \\ & {(\sqrt{2} + \sqrt{3})}^8 = 9601.99989585502907\ldots \\ & {(\sqrt{2} + \sqrt{3})}^{10} = 95049.999989479221\ldots \\ & {(\sqrt{2} + \sqrt{3})}^{12} = 940897.9999989371855\ldots \\ & {(\sqrt{2} + \sqrt{3})}^{14} = 9313929.99999989263\ldots \\ & {(\sqrt{2} + \sqrt{3})}^{16} = 92198401.99999998915\ldots \\ \end{align}$$

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 {(\sqrt{2} + \sqrt{3})}^{2n} approaches 1 for large n.

Consider all real numbers of the form \sqrt{p} + \sqrt{q} with p and q positive integers and p < q, such that the fractional part of {(\sqrt{p} + \sqrt{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 {(\sqrt{p} + \sqrt{q})}^{2n}.

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

Find \sum N(p,q) for p + q ≤ 2011.

--hints--

twoThousandElevenNines() should return 709313889.

assert.strictEqual(twoThousandElevenNines(), 709313889);

--seed--

--seed-contents--

function twoThousandElevenNines() {

  return true;
}

twoThousandElevenNines();

--solutions--

// solution required