Problem 318

2011 nines

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.

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