problem130.rb

# A number consisting entirely of ones is called a repunit. We shall define R(k) to be a repunit of length k; for example, R(6) = 111111.
# 
# Given that n is a positive integer and GCD(n, 10) = 1, it can be shown that there always exists a value, k, for which R(k) is divisible by n, and let A(n) be the least such value of k; for example, A(7) = 6 and A(41) = 5.
# 
# You are given that for all primes, p > 5, that p − 1 is divisible by A(p). For example, when p = 41, A(41) = 5, and 40 is divisible by 5.
# 
# However, there are rare composite values for which this is also true; the first five examples being 91, 259, 451, 481, and 703.
# 
# Find the sum of the first twenty-five composite values of n for which
# GCD(n, 10) = 1 and n − 1 is divisible by A(n).
require 'miller-rabin'
require 'common'

def a(n)
	return nil if n % 2 == 0 or n % 5 == 0
	count = 1
	current = 1
	while current != 0
		count += 1
		current = (current * 10 + 1) % n
	end
	return count
end

max = 25
found = []
current = 2
while found.length < max
	puts "#{current}, #{found.length}"
	temp = a(current) unless current.prime?
	found << current if temp != nil and (current - 1) % temp == 0
	current += 1
end

puts found.inspect
puts found.sum