problem129.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.
# 
# The least value of n for which A(n) first exceeds ten is 17.
# 
# Find the least value of n for which A(n) first exceeds one-million.

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 = 10**6
current = max + 1 #it is obvious that A(n) <= n
while true
	temp = a(current)
	puts "#{current}, #{temp}"
	break if temp != nil and temp > max
	current += 2
end