What are the divisors of 3038?

1, 2, 7, 14, 31, 49, 62, 98, 217, 434, 1519, 3038

There is a total of 12 positive divisors.
The sum of these divisors is 5472.
The arithmetic mean is 456.

6 even divisors

2, 14, 62, 98, 434, 3038

6 odd divisors

1, 7, 31, 49, 217, 1519

How to compute the divisors of 3038?

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 3038 by each of the numbers from 1 to 3038 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)

3038 / 1 = 3038 (the remainder is 0, so 1 is a divisor of 3038)
3038 / 2 = 1519 (the remainder is 0, so 2 is a divisor of 3038)
3038 / 3 = 1012.6666666667 (the remainder is 2, so 3 is not a divisor of 3038)
...
3038 / 3037 = 1.0003292723082 (the remainder is 1, so 3037 is not a divisor of 3038)
3038 / 3038 = 1 (the remainder is 0, so 3038 is a divisor of 3038)

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 3038 (i.e. 55.118055118083). 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 $\frac{N}{d} = D$ )

3038 / 1 = 3038 (the remainder is 0, so 1 and 3038 are divisors of 3038)
3038 / 2 = 1519 (the remainder is 0, so 2 and 1519 are divisors of 3038)
3038 / 3 = 1012.6666666667 (the remainder is 2, so 3 is not a divisor of 3038)
...
3038 / 54 = 56.259259259259 (the remainder is 14, so 54 is not a divisor of 3038)
3038 / 55 = 55.236363636364 (the remainder is 13, so 55 is not a divisor of 3038)