## 2011 nines

Published on Saturday, 1st January 2011, 04:00 pm; Solved by 763;
Difficulty rating: 50%

### Problem 318

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

It looks as if 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 $(\sqrt 2 + \sqrt 3)^{2 n}$ 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 $(\sqrt p + \sqrt q)^{ 2 n}$ approaches $1$ for large $n$.

Let $C(p,q,n)$ be the number of consecutive nines at the beginning of the fractional part of $(\sqrt p + \sqrt q)^{ 2 n}$.

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

Find $\displaystyle \sum N(p,q) \,\, \text{ for } p+q \le 2011$.