What are the divisors of 2958?

1, 2, 3, 6, 17, 29, 34, 51, 58, 87, 102, 174, 493, 986, 1479, 2958

There is a total of 16 positive divisors.
The sum of these divisors is 6480.
The arithmetic mean is 405.

8 even divisors

2, 6, 34, 58, 102, 174, 986, 2958

8 odd divisors

1, 3, 17, 29, 51, 87, 493, 1479

How to compute the divisors of 2958?

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 2958 by each of the numbers from 1 to 2958 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)

2958 / 1 = 2958 (the remainder is 0, so 1 is a divisor of 2958)
2958 / 2 = 1479 (the remainder is 0, so 2 is a divisor of 2958)
2958 / 3 = 986 (the remainder is 0, so 3 is a divisor of 2958)
...
2958 / 2957 = 1.0003381805884 (the remainder is 1, so 2957 is not a divisor of 2958)
2958 / 2958 = 1 (the remainder is 0, so 2958 is a divisor of 2958)

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 2958 (i.e. 54.387498563549). 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 $\frac{N}{d} = D$ )

2958 / 1 = 2958 (the remainder is 0, so 1 and 2958 are divisors of 2958)
2958 / 2 = 1479 (the remainder is 0, so 2 and 1479 are divisors of 2958)
2958 / 3 = 986 (the remainder is 0, so 3 and 986 are divisors of 2958)
...
2958 / 53 = 55.811320754717 (the remainder is 43, so 53 is not a divisor of 2958)
2958 / 54 = 54.777777777778 (the remainder is 42, so 54 is not a divisor of 2958)