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Discussion on: Project Euler #4 - Largest Palindrome Product

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Massimo Artizzu

Sooo... there are so many good solutions, but they all kind of look the same, so I wanted a different approach 😁

What if, instead of starting with the factors and checking if their product is a palindrome, we start with palindromes and find two 3-digit factors?

We'd start with 999999, then 998899, then 997799... You get the point. So we can start with a 3-digit number (e.g. 356)... and get a palindrome out of it (e.g. 356653), and look for factors.

My solution in JavaScript:

const digits = 3;
const upper = 10 ** digits - 1;   // 999
const lower = 10 ** (digits - 1); // 100
function palindromize(num) {
  const padded = String(num).padStart(digits,'0');
  return +(padded + padded .split('').reverse().join(''));

let p, b;
out: for (let a = upper; a >= lower; a--) {
  p = palindromize(a);
  for (b = Math.floor(p ** .5); p/b <= upper; b--) {
    if (p % b === 0) break out;
console.log(p / b, b);

I've tested it with up to 7 digits and it's still basically instantaneous. Over that limit you go over Number.MAX_SAFE_INTEGER and it becomes unreliable. I didn't bother using BigInts, also because you can't get the square root out of them (you can approximate it, but it's a bother).

P.S.: yes, I've used a label in JavaScript. Sue me 😂 (also useful if you have to break out of more than one cycle... which is usually rare).