What are the divisors of 3540?

1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 59, 60, 118, 177, 236, 295, 354, 590, 708, 885, 1180, 1770, 3540

There is a total of 24 positive divisors.
The sum of these divisors is 10080.
The arithmetic mean is 420.

16 even divisors

2, 4, 6, 10, 12, 20, 30, 60, 118, 236, 354, 590, 708, 1180, 1770, 3540

8 odd divisors

1, 3, 5, 15, 59, 177, 295, 885

How to compute the divisors of 3540?

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 3540 by each of the numbers from 1 to 3540 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)

3540 / 1 = 3540 (the remainder is 0, so 1 is a divisor of 3540)
3540 / 2 = 1770 (the remainder is 0, so 2 is a divisor of 3540)
3540 / 3 = 1180 (the remainder is 0, so 3 is a divisor of 3540)
...
3540 / 3539 = 1.0002825656965 (the remainder is 1, so 3539 is not a divisor of 3540)
3540 / 3540 = 1 (the remainder is 0, so 3540 is a divisor of 3540)

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 3540 (i.e. 59.497899122574). 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 $\frac{N}{d} = D$ )

3540 / 1 = 3540 (the remainder is 0, so 1 and 3540 are divisors of 3540)
3540 / 2 = 1770 (the remainder is 0, so 2 and 1770 are divisors of 3540)
3540 / 3 = 1180 (the remainder is 0, so 3 and 1180 are divisors of 3540)
...
3540 / 58 = 61.034482758621 (the remainder is 2, so 58 is not a divisor of 3540)
3540 / 59 = 60 (the remainder is 0, so 59 and 60 are divisors of 3540)