Break down a number into its prime factors

Is there a way to identify a number within the range (from 1 to a specified number) that, when broken down into prime numbers, has the highest quantity?

For Example:

Input: 9

Output: 8

Explanation:

Breaking down 8 gives us 2 * 2 * 2 (3 primes)

7 = 7 * 1 (1 prime)

6 = 3 * 2 (2 primes)

and so on... Ultimately, we find that 8 has the most primes in its decomposition.

Requirement:

If multiple numbers have the same number of primes in their decomposition, return the largest of those numbers.

Answer №1

Understood your request clearly.

Below is a straightforward script that accomplishes what you asked for:

//credit to: https://jsfiddle.net/JamesOR/RC7SY/

function getAllFactorsFor(remainder) {
    var factors = [], i;

    for (i = 2; i <= remainder; i++) {
        while ((remainder % i) === 0) {
            factors.push(i);
            remainder /= i;
        }
    }

    return factors;
}

function calculate(x) {

    lastFactorCount = 0;
    highestNumber = 0;

    while (x) {
        currentCount = getAllFactorsFor(x).length;

        if (currentCount > lastFactorCount) {
            lastFactorCount = currentCount;
            highestNumber = x;
        }

        x--;
    }

    return highestNumber;
}

console.log(calculate(7)); //output: 6
console.log(calculate(11)) //output: 8

This code works for the given test cases provided. I used the getAllFactorsFor() function from an existing jsfiddle as there's no need to start from scratch ;)

The calculate() function takes an input number, iterates through each number from x down to 0, counts its factors, and stores the number with the most factors count while reducing x in each iteration.

In the end, it returns the number with the highest factor count. Simple as that.

Hope this solution proves beneficial!

Answer №2

It is important to note that after the numbers 2 and 3, the next prime number is 5, which is obviously larger than 2 multiplied by 2. This means that using 2 multiplied by 2 will always result in a better amount of prime factors than using any higher prime number. The formula for finding the number with the highest amount of 2 as prime factors, while still being lower or equal to the given number, can be represented as 2 ** Math.floor(Math.log2(num)). The only consideration we need to make is whether replacing the last prime factor 2 with a 3 would exceed the number, potentially resulting in a larger number. Keep in mind that using more than one 3 (such as 3 multiplied by 3) would lead to a solution that surpasses 8 (which equals 2 multiplied by 2 multiplied by 2). Therefore, the optimal solution is determined by the following function:

const f = num => {
  let twoEnd = 2 ** Math.floor(Math.log2(num));
  let threeEnd = twoEnd / 2 * 3;
  return threeEnd <= num ? threeEnd : twoEnd;
}

In certain circumstances, additional precautions may need to be taken for numbers smaller than 2.

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