# Pandigital multiples

# Problem 38

# Take the number 192 and multiply it by each of 1, 2, and 3:

# 192 × 1 = 192

# 192 × 2 = 384

# 192 × 3 = 576

# By concatenating each product we get the 1 to 9 pandigital, 192384576.

# We will call 192384576 the concatenated product of 192 and (1,2,3)

# The same can be achieved by starting with 9 and multiplying by

# 1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is the

# concatenated product of 9 and (1,2,3,4,5).

# What is the largest 1 to 9 pandigital 9-digit number that can be formed as

# the concatenated product of an integer with (1,2, ... , n) where n > 1?

# Solution by Sam Gerber

def pandigital_multiples

max_integer = 9_999

biggest_product = 0

integer = 1

concatenated = "1"

while integer <= max_integer

concatenated = integer.to_s

(2..9).each do |n|

concatenated += (integer * n).to_s

length = concatenated.length

if length >= 9

if concatenated.split("").uniq.count == length &&

!concatenated.include?("0") &&

concatenated.to_i > biggest_product

biggest_product = concatenated.to_i

end

break

end

end

integer += 1

end

biggest_product

end

puts pandigital_multiples