# Problem 318: 2011 nines

![graphic](img318.gif)

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