Project Euler 318: Project Euler 318: Counting the number of consecutive nines at the beginning of the fractional part of (√p+√q)^{2n}
Project Euler 318: 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 nondecreasing.
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.
Project Euler 318 Solution
Runs < 0.470 seconds in Python 2.7.#Project Euler Problem 318 Solution
from math import sqrt, log10, ceil
s, L = 0, 2011
for p in xrange(1, L):
for q in xrange(p+1, Lp+1):
c = p + q  2*sqrt(p*q)
if c < 1:
s += ceil(L / log10(c))
print "sum of N(p,q) for p+q <=", L, "is", int(s)
Use this link to get the Project Euler 318 Solution Python 2.7 source.Answer
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