What are the divisors of 3438?

1, 2, 3, 6, 9, 18, 191, 382, 573, 1146, 1719, 3438

There is a total of 12 positive divisors.
The sum of these divisors is 7488.
The arithmetic mean is 624.

6 even divisors

2, 6, 18, 382, 1146, 3438

6 odd divisors

1, 3, 9, 191, 573, 1719

How to compute the divisors of 3438?

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 3438 by each of the numbers from 1 to 3438 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)

3438 / 1 = 3438 (the remainder is 0, so 1 is a divisor of 3438)
3438 / 2 = 1719 (the remainder is 0, so 2 is a divisor of 3438)
3438 / 3 = 1146 (the remainder is 0, so 3 is a divisor of 3438)
...
3438 / 3437 = 1.0002909514111 (the remainder is 1, so 3437 is not a divisor of 3438)
3438 / 3438 = 1 (the remainder is 0, so 3438 is a divisor of 3438)

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 3438 (i.e. 58.634460857076). 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 $\frac{N}{d} = D$ )

3438 / 1 = 3438 (the remainder is 0, so 1 and 3438 are divisors of 3438)
3438 / 2 = 1719 (the remainder is 0, so 2 and 1719 are divisors of 3438)
3438 / 3 = 1146 (the remainder is 0, so 3 and 1146 are divisors of 3438)
...
3438 / 57 = 60.315789473684 (the remainder is 18, so 57 is not a divisor of 3438)
3438 / 58 = 59.275862068966 (the remainder is 16, so 58 is not a divisor of 3438)