Ben / Nujel

def reverseNum(a)
    ret = 0
    while (a > 0) do
          ret = (ret * 10) + (a % 10);
          a = (a / 10).floor;
    end
    return ret
end

def palindromeP(a)
    a == reverseNum(a)
end

def startSearch
    ret = 0
    for a in 0..999 do
        for b in 0..999 do
            p = a * b
            if (palindromeP(p) && (p > ret)) then
               ret = p
            end
        end
    end
    return ret
end

ret = startSearch
puts "The biggest product of 2 3-digit numbers that is a palindrome is: #{ret}"