Solution to problem 26 in Julia

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David Doblas Jiménez 2021-08-16 20:25:12 +02:00
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#=
Created on 16 Aug 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for Problem 26 of Project Euler
https://projecteuler.net/problem=26
=#
function Problem26()
#=
A unit fraction contains 1 in the numerator. The decimal representation
of the unit fractions with denominators 2 to 10 are given:
1/2 = 0.5
1/3 = 0.(3)
1/4 = 0.25
1/5 = 0.2
1/6 = 0.1(6)
1/7 = 0.(142857)
1/8 = 0.125
1/9 = 0.(1)
1/10 = 0.1
Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle.
It can be seen that 1/7 has a 6-digit recurring cycle.
Find the value of d < 1000 for which 1/d contains the longest recurring
cycle in its decimal fraction part.
=#
cycle_length = 0
number_d = 0
for number in 3:2:999
if number % 5 == 0
continue
end
p = 1
while (big(10)^p % number) != 1
p += 1
end
if p > cycle_length
cycle_length, number_d = p, number
end
end
return number_d
end
println("Time to evaluate Problem 26:")
@time Problem26()
println("")
println("Result for Problem 26: ", Problem26())