44 lines
1.2 KiB
Python
44 lines
1.2 KiB
Python
from fractions import Fraction
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from lib_misc import proper_divisors
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from math import ceil
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def get_distinct_solutions(n):
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""" Get the number of distinct solutions for 1/x + 1/y = 1/n. The idea is
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that 1/n*2 + 1/n*2 = 1/n is always a solution. So what we do is
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brute-force all 1/x for x in [2, n * 2]. We have a solution if the
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difference between 1/n and 1/x has a numerator of 1. """
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n_inv = Fraction(1, n)
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n_distinct = 0
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for x in range(2, n * 2 + 1):
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x_inv = Fraction(1, x)
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y_inv = n_inv - x_inv
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if y_inv.numerator == 1:
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n_distinct += 1
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return n_distinct
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def get_distinct_solutions2(n):
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ds = proper_divisors(n * n)
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return ceil((len(ds) + 1) / 2)
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def euler_108():
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d_max, n_prev = 0, 0
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# I arrived at the starting values empirically by observing the deltas.
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for n in range(1260, 1000000, 420):
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d = get_distinct_solutions2(n)
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if d > d_max:
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# print("n={} d={} delta={}".format(n, d, n - n_prev))
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n_prev = n
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d_max = d
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if d > 1000:
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return n
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if __name__ == "__main__":
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solution = euler_108()
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print("e108.py: " + str(solution))
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assert(solution == 180180)
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