What are the divisors of 4023?

1, 3, 9, 27, 149, 447, 1341, 4023

There is a total of 8 positive divisors.
The sum of these divisors is 6000.
The arithmetic mean is 750.

8 odd divisors

1, 3, 9, 27, 149, 447, 1341, 4023

How to compute the divisors of 4023?

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 4023 by each of the numbers from 1 to 4023 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)

4023 / 1 = 4023 (the remainder is 0, so 1 is a divisor of 4023)
4023 / 2 = 2011.5 (the remainder is 1, so 2 is not a divisor of 4023)
4023 / 3 = 1341 (the remainder is 0, so 3 is a divisor of 4023)
...
4023 / 4022 = 1.0002486325211 (the remainder is 1, so 4022 is not a divisor of 4023)
4023 / 4023 = 1 (the remainder is 0, so 4023 is a divisor of 4023)

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 4023 (i.e. 63.427123535598). 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 $\frac{N}{d} = D$ )

4023 / 1 = 4023 (the remainder is 0, so 1 and 4023 are divisors of 4023)
4023 / 2 = 2011.5 (the remainder is 1, so 2 is not a divisor of 4023)
4023 / 3 = 1341 (the remainder is 0, so 3 and 1341 are divisors of 4023)
...
4023 / 62 = 64.887096774194 (the remainder is 55, so 62 is not a divisor of 4023)
4023 / 63 = 63.857142857143 (the remainder is 54, so 63 is not a divisor of 4023)