curriculum/challenges/english/blocks/project-euler-problems-101-to-200/5900f3e91000cf542c50fefc.md
The palindromic number 595 is interesting because it can be written as the sum of consecutive squares: $6^2 + 7^2 + 8^2 + 9^2 + 10^2 + 11^2 + 12^2$.
There are exactly eleven palindromes below one-thousand that can be written as consecutive square sums, and the sum of these palindromes is 4164. Note that $1 = 0^2 + 1^2$ has not been included as this problem is concerned with the squares of positive integers.
Find the sum of all the numbers less than the limit that are both palindromic and can be written as the sum of consecutive squares.
palindromicSums(100000000) should return 2906969179.
assert.strictEqual(palindromicSums(100000000), 2906969179);
palindromicSums(100) should return 137.
assert.strictEqual(palindromicSums(100), 137);
palindromicSums(1000) should return 4164.
assert.strictEqual(palindromicSums(1000),4164);
function palindromicSums(limit) {
return true;
}
palindromicSums(100);
function isPalindrome(num) {
return num
.toString()
.split('')
.every((digit, i, arr) => digit === arr[arr.length - 1 - i]);
}
function palindromicSums(limit) {
let sumOfPalindromes = 0;
const sqrtLimit = Math.sqrt(limit);
const list = new Set();
for (let i = 1; i <= sqrtLimit; i++) {
let sumOfSquares = i * i;
for (let j = i + 1; j <= sqrtLimit; j++) {
sumOfSquares += j * j;
if (sumOfSquares > limit) break;
if (isPalindrome(sumOfSquares) && !list.has(sumOfSquares)) {
sumOfPalindromes += sumOfSquares;
list.add(sumOfSquares);
}
}
}
return sumOfPalindromes;
}