## 2011 nines

### Problem 318

Published on Saturday, 1st January 2011, 04:00 pm; Solved by 520
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.