What are the divisors of 2923?

1, 37, 79, 2923

There is a total of 4 positive divisors.
The sum of these divisors is 3040.
The arithmetic mean is 760.

4 odd divisors

1, 37, 79, 2923

How to compute the divisors of 2923?

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 2923 by each of the numbers from 1 to 2923 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)

2923 / 1 = 2923 (the remainder is 0, so 1 is a divisor of 2923)
2923 / 2 = 1461.5 (the remainder is 1, so 2 is not a divisor of 2923)
2923 / 3 = 974.33333333333 (the remainder is 1, so 3 is not a divisor of 2923)
...
2923 / 2922 = 1.0003422313484 (the remainder is 1, so 2922 is not a divisor of 2923)
2923 / 2923 = 1 (the remainder is 0, so 2923 is a divisor of 2923)

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 2923 (i.e. 54.064775963653). 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 $\frac{N}{d} = D$ )

2923 / 1 = 2923 (the remainder is 0, so 1 and 2923 are divisors of 2923)
2923 / 2 = 1461.5 (the remainder is 1, so 2 is not a divisor of 2923)
2923 / 3 = 974.33333333333 (the remainder is 1, so 3 is not a divisor of 2923)
...
2923 / 53 = 55.150943396226 (the remainder is 8, so 53 is not a divisor of 2923)
2923 / 54 = 54.12962962963 (the remainder is 7, so 54 is not a divisor of 2923)