## 2011 nines

### Problem 318

Published on Saturday, 1st January 2011, 04:00 pm; Solved by 483

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 pq, 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.