What are the divisors of 3042?

1, 2, 3, 6, 9, 13, 18, 26, 39, 78, 117, 169, 234, 338, 507, 1014, 1521, 3042

There is a total of 18 positive divisors.
The sum of these divisors is 7137.
The arithmetic mean is 396.5.

9 even divisors

2, 6, 18, 26, 78, 234, 338, 1014, 3042

9 odd divisors

1, 3, 9, 13, 39, 117, 169, 507, 1521

How to compute the divisors of 3042?

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 3042 by each of the numbers from 1 to 3042 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)

3042 / 1 = 3042 (the remainder is 0, so 1 is a divisor of 3042)
3042 / 2 = 1521 (the remainder is 0, so 2 is a divisor of 3042)
3042 / 3 = 1014 (the remainder is 0, so 3 is a divisor of 3042)
...
3042 / 3041 = 1.0003288391976 (the remainder is 1, so 3041 is not a divisor of 3042)
3042 / 3042 = 1 (the remainder is 0, so 3042 is a divisor of 3042)

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 3042 (i.e. 55.154328932551). 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 $\frac{N}{d} = D$ )

3042 / 1 = 3042 (the remainder is 0, so 1 and 3042 are divisors of 3042)
3042 / 2 = 1521 (the remainder is 0, so 2 and 1521 are divisors of 3042)
3042 / 3 = 1014 (the remainder is 0, so 3 and 1014 are divisors of 3042)
...
3042 / 54 = 56.333333333333 (the remainder is 18, so 54 is not a divisor of 3042)
3042 / 55 = 55.309090909091 (the remainder is 17, so 55 is not a divisor of 3042)