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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