Find the largest palindrome made from the product of two 3-digit numbers

Question

package testing.project;

public class PalindromeThreeDigits {

    public static void main(String[] args) {
        int value = 0;
        for(int i = 100;i <=999;i+         


        

Answer 1

I believe there is a simpler approach: Examine palindromes descending from the largest product of two three digit numbers, selecting the first palindrome with two three digit factors.

Here is the Ruby code:

require './palindrome_range'
require './prime'

def get_3_digit_factors(n)
  prime_factors = Prime.factors(n)

  rf = [prime_factors.pop]
  rf << prime_factors.shift while rf.inject(:*) < 100 || prime_factors.inject(:*) > 999

  lf = prime_factors.inject(:*)
  rf = rf.inject(:*)

  lf < 100 || lf > 999 || rf < 100 || rf > 999 ? [] : [lf, rf]
end

def has_3_digit_factors(n)
  return !get_3_digit_factors(n).empty?
end

pr = PalindromeRange.new(0, 999 * 999)
n = pr.downto.find {|n| has_3_digit_factors(n)}
puts "Found #{n} - Factors #{get_3_digit_factors(n).inspect}, #{Prime.factors(n).inspect}"

prime.rb:

class Prime

  class<

palindrome_range.rb:

class PalindromeRange

  FIXNUM_MAX = (2**(0.size * 8 -2) -1)

  def initialize(min = 0, max = FIXNUM_MAX)
    @min = min
    @max = max
  end

  def downto
    return enum_for(:downto) unless block_given?

    n = @max
    while n >= @min
      yield n if is_palindrome(n)
      n -= 1
    end
    nil
  end

  def each
    return upto
  end

  def upto
    return enum_for(:downto) unless block_given?

    n = @min
    while n <= @max
      yield n if is_palindrome(n)
      n += 1
    end
    nil
  end

  private

  def is_palindrome(n)
    s = n.to_s
    i = 0
    j = s.length - 1
    while i <= j
      break if s[i] != s[j]
      i += 1
      j -= 1
    end
    i > j
  end

end