2011 Nines

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 \cdots $
$(\sqrt 2 + \sqrt 3)^4 = 97.98979485566356 \cdots $
$(\sqrt 2 + \sqrt 3)^6 = 969.998969071069263 \cdots $
$(\sqrt 2 + \sqrt 3)^8 = 9601.99989585502907 \cdots $
$(\sqrt 2 + \sqrt 3)^{10} = 95049.999989479221 \cdots $
$(\sqrt 2 + \sqrt 3)^{12} = 940897.9999989371855 \cdots $
$(\sqrt 2 + \sqrt 3)^{14} = 9313929.99999989263 \cdots $
$(\sqrt 2 + \sqrt 3)^{16} = 92198401.99999998915 \cdots $

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

To solve this problem, we can use Python code to iterate through all possible values of p and q, calculate the powers of (√p + √q), and find the minimum value of n for which C(p,q,n) >= 2011. We can then calculate the sum of all these minimum values.

Here is the Python code to solve this problem:

“`python
import math

def get_fractional_part(x):
return x – int(x)

def count_consecutive_nines(x):
count = 0
fractional_part = get_fractional_part(x)
while int(fractional_part * 10) == 9:
count += 1
fractional_part = get_fractional_part(fractional_part * 10)
return count

def calculate_N(p, q):
n = 1
while True:
power = (math.sqrt(p) + math.sqrt(q))**(2*n)
consecutive_nines = count_consecutive_nines(power)
if consecutive_nines >= 2011:
return n
n += 1

# Main code
sum_N = 0
for p in range(1, 2011):
for q in range(p+1, 2011):
sum_N += calculate_N(p, q)

print(“Sum of N(p,q) for p+q <= 2011:”, sum_N)
“`

Note that this solution may take some time to execute since it involves a large number of calculations. You can run this code on your computer to find the desired sum.

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