patternjavascriptModerate

Getting the divisors count of an integer

Submitted by: @import:stackexchange-codereview·Mar 10, 2026·

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thegettingdivisorscountinteger

Problem

I am using mod and my code works. I am wondering if there is a more efficient way to write a method that returns the total number of divisors for an integer.

function getDivisorsCnt(n){
    var divisors = 0;
    mod = n;
    while (mod > 0){
      if(n % mod === 0){
        divisors++;
      }
      mod--;
    }
    return divisors;
}

Solution

Some general remarks:

Variables should be declared explicitly with var, otherwise you

create a global variable.

The indent amount is different between the first level and the

deeper levels.

The spacing is not consistent: while ( vs if(. Generally I would

insert a space at least after keywords, and around parentheses/braces etc.

Your implementation is correct but not efficient, as the number of loop iterations
and remainder operations is equal to the input number.

The count of divisors can be efficiently computed from the
prime number factorization: If
$$
n = p_1^{e_1} \, p_2^{e_2} \cdots p_k^{e_k}
$$
is the factorization of \$ n \$ into prime numbers \$ p_i \$
with exponents \$ e_i \$, then
$$
\sigma_0(n) = (e_1+1)(e_2+1) \cdots (e_k+1)
$$
is the number of divisors of \$ n \$, see for example
Wikipedia: Divisor function. Example:
$$
720 = 2^4 \cdot 3^2 \cdot 5^1 \Longrightarrow
\sigma_0(720) = (4+1)(2+1)(1+1) = 30 \, .
$$

An implementation in JavaScript would be

function getDivisorsCnt(n){

    var numDivisors = 1;
    var factor = 2; // Candidate for prime factor of `n`

    // If `n` is not a prime number then it must have one factor
    // which is  1) {
        numDivisors *= 2;
    }

    return numDivisors;
}

As an example, getDivisorsCnt(720) requires 720 remainder operations
in your algorithm, but only 8 remainder operations and 6 divisions in this algorithm.

Code Snippets

function getDivisorsCnt(n){

    var numDivisors = 1;
    var factor = 2; // Candidate for prime factor of `n`

    // If `n` is not a prime number then it must have one factor
    // which is <= `sqrt(n)`, so we try these first:
    while (factor * factor <= n) {
        if (n % factor === 0) {
            // `factor` is a prime factor of `n`, determine the exponent:
            var exponent = 0;
            do {
                n /= factor;
                exponent++;
            } while (n % factor === 0)
            // `factor^exponent` is one term in the prime factorization of n,
            // this contributes as factor `exponent + 1`:
            numDivisors *= exponent + 1;
        }
        // Next possible prime factor:
        factor = factor == 2 ? 3 : factor + 2
    }

    // Now `n` is either 1 or a prime number. In the latter case,
    // it contributes a factor 2:
    if (n > 1) {
        numDivisors *= 2;
    }

    return numDivisors;
}

Context

StackExchange Code Review Q#120642, answer score: 10

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