re: Project Euler #4 - Largest Palindrome Product VIEW POST

Here's my Ruby solution. There's probably a much more clever way to do this. Probably a more Ruby way to do it for that matter.

``````# Find the largest palindrome number that is the product of two three digit factors.

def check_equality(num)
num.to_s == num.to_s.reverse
end

def find_palindrome
r1 = (999..1)
r2 = r1

(r1.first).downto(r1.last).each do |i|
(r2.first).downto(r2.last).each do |j|
if check_equality( i * j )
puts "#{i} * #{j} = #{i*j}"
return
end
end
end
end

find_palindrome
``````

If I understand it correctly (correct me if I'm wrong, I don't know Ruby), this doesn't work in general, as it prints the first palindrome product you find. But you don't know if it's the largest.

Unless you can prove it is 🤷‍♂️ (I have no idea).

I'm working backwards through the range of numbers beginning with '999.' Hence the extra verbosity in the block with the call to the downto method. (999..1).each do doesn't work, and (1..999).each do really shouldn't work either because ranges are not defined by their content, just a beginning state and an end state. So counting backwards the first palindrome I find is the largest. And the outer block isn't necessary, but I include it just for the sake of being thorough I guess.

The problem here is that the products you're getting aren't ordered. Which means that if you get a palindrome, you cannot be sure it's the largest.

In fact, I run your code and I got `999 * 91 = 90909`, which is not correct. Even if you limit your range to 999 to 100 (we're looking for 3-digit factors after all), you'd get `995 * 583 = 580085`. But the largest palindrome is `993 * 913 = 906609`, which comes after the others.

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