Problem 29: Distinct powers
Solutions
Solution 1 (Click to Show/Hide)
function distinctPowers(n) {
let powers = {};
for (let i = 2; i <= n; i++) {
for (let j = 2; j <= n; j++) {
powers[Math.pow(i, j)] = true; // Repeated instances will not add additional key
}
}
return Object.keys(powers).length;
}