What are the divisors of 3021?

1, 3, 19, 53, 57, 159, 1007, 3021

There is a total of 8 positive divisors.
The sum of these divisors is 4320.
The arithmetic mean is 540.

8 odd divisors

1, 3, 19, 53, 57, 159, 1007, 3021

How to compute the divisors of 3021?

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 3021 by each of the numbers from 1 to 3021 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)

3021 / 1 = 3021 (the remainder is 0, so 1 is a divisor of 3021)
3021 / 2 = 1510.5 (the remainder is 1, so 2 is not a divisor of 3021)
3021 / 3 = 1007 (the remainder is 0, so 3 is a divisor of 3021)
...
3021 / 3020 = 1.0003311258278 (the remainder is 1, so 3020 is not a divisor of 3021)
3021 / 3021 = 1 (the remainder is 0, so 3021 is a divisor of 3021)

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 3021 (i.e. 54.963624334645). 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 $\frac{N}{d} = D$ )

3021 / 1 = 3021 (the remainder is 0, so 1 and 3021 are divisors of 3021)
3021 / 2 = 1510.5 (the remainder is 1, so 2 is not a divisor of 3021)
3021 / 3 = 1007 (the remainder is 0, so 3 and 1007 are divisors of 3021)
...
3021 / 53 = 57 (the remainder is 0, so 53 and 57 are divisors of 3021)
3021 / 54 = 55.944444444444 (the remainder is 51, so 54 is not a divisor of 3021)