Which, now that I think about it, could have been used to reduce the
total number of "nums" that need checked. nums:reverse(n-m)+til
m:div[;10]n:prd x#10; This doubles performance from original code.
Further improvements could be made by not creating...
Hi gberkeley, well spotted! You are totally correct in what you have
said here, the last 10% of "pals" will have been joined with "0" and
will therefore not be palindromes when cast back to a long. You're also
right that I don't check any of the odd-...
Good explaination of your approach here. I've found a more novel
approach where we can create palindromes and then check if it has
products, rather than finding all products and checking if they are
palindromes. This has the benefit of not needing to...