What are the divisors of 3958?

1, 2, 1979, 3958

There is a total of 4 positive divisors.
The sum of these divisors is 5940.
The arithmetic mean is 1485.

2 even divisors

2, 3958

2 odd divisors

1, 1979

How to compute the divisors of 3958?

A number N is said to be divisible by a number M (with M non-zero) if, when we divide N by M, the remainder of the division is zero.

$N mod M = 0$

Brute force algorithm

We could start by using a brute-force method which would involve dividing 3958 by each of the numbers from 1 to 3958 to determine which ones have a remainder equal to 0.

$Remainder = N - (M \times ⌊ \frac{N}{M} ⌋)$

(where $⌊ \frac{N}{M} ⌋$ is the integer part of the quotient)

3958 / 1 = 3958 (the remainder is 0, so 1 is a divisor of 3958)
3958 / 2 = 1979 (the remainder is 0, so 2 is a divisor of 3958)
3958 / 3 = 1319.3333333333 (the remainder is 1, so 3 is not a divisor of 3958)
...
3958 / 3957 = 1.0002527167046 (the remainder is 1, so 3957 is not a divisor of 3958)
3958 / 3958 = 1 (the remainder is 0, so 3958 is a divisor of 3958)

Improved algorithm using square-root

However, there is another slightly better approach that reduces the number of iterations by testing only integers less than or equal to the square root of 3958 (i.e. 62.912637840103). Indeed, if a number N has a divisor D greater than its square root, then there is necessarily a smaller divisor d such that:

$D \times d = N$

(thus, if $\frac{N}{D} = d$ , then $\frac{N}{d} = D$ )

3958 / 1 = 3958 (the remainder is 0, so 1 and 3958 are divisors of 3958)
3958 / 2 = 1979 (the remainder is 0, so 2 and 1979 are divisors of 3958)
3958 / 3 = 1319.3333333333 (the remainder is 1, so 3 is not a divisor of 3958)
...
3958 / 61 = 64.885245901639 (the remainder is 54, so 61 is not a divisor of 3958)
3958 / 62 = 63.838709677419 (the remainder is 52, so 62 is not a divisor of 3958)