JavaScript Program to Find All Factors of a Natural Number

Finding all factors of a natural number is a common task in programming and mathematics. A factor of a number is an integer that can be multiplied by another integer to produce the original number. In this article, we will discuss various approaches to finding all factors of a natural number using JavaScript.

Find All Factors of a Natural Number using for Loop

The simplest way to find all factors of a number is to iterate from 1 to the number itself and check if the number is divisible by the current iteration index. If it is, then the index is a factor of the number.

Example:

Javascript

function findFactors(n) {
    const factors = [];
    for (let i = 1; i <= n; i++) {
        if (n % i === 0) {
            factors.push(i);
        }
    }
    return factors;
}

console.log(findFactors(12));

Output

[ 1, 2, 3, 4, 6, 12 ]

Find All Factors of a Natural Number using Optimized Approach

We can optimize the above approach by iterating only up to the square root of the number. This is because if n is a factor of the number, then num/n is also a factor. However, we need to be careful to avoid duplicate factors.

Example:

Javascript

function findFactors(num) {
    let factors = [];
    for (let i = 1; i <= Math.sqrt(num); i++) {
        if (num % i === 0) {
            factors.push(i);
            if (i !== num / i) {
                factors.push(num / i);
            }
        }
    }
    return factors.sort((a, b) => a - b);
}

console.log(findFactors(12));

Output

[ 1, 2, 3, 4, 6, 12 ]

Find All Factors of a Natural Number using Recursion

We can also use recursion to find the factors of a number. This approach is not the most efficient but can be useful in understanding recursive algorithms.

Javascript

function findFactors(num, current = 1, factors = []) {
    if (current > num) {
        return factors;
    }
    if (num % current === 0) {
        factors.push(current);
    }
    return findFactors(num, current + 1, factors);
}

console.log(findFactors(12));

Output

[ 1, 2, 3, 4, 6, 12 ]