What are the divisors of 612?

1, 2, 3, 4, 6, 9, 12, 17, 18, 34, 36, 51, 68, 102, 153, 204, 306, 612

There is a total of 18 positive divisors.
The sum of these divisors is 1638.
The arithmetic mean is 91.

12 even divisors

2, 4, 6, 12, 18, 34, 36, 68, 102, 204, 306, 612

6 odd divisors

1, 3, 9, 17, 51, 153

How to compute the divisors of 612?

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 612 by each of the numbers from 1 to 612 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)

612 / 1 = 612 (the remainder is 0, so 1 is a divisor of 612)
612 / 2 = 306 (the remainder is 0, so 2 is a divisor of 612)
612 / 3 = 204 (the remainder is 0, so 3 is a divisor of 612)
...
612 / 611 = 1.0016366612111 (the remainder is 1, so 611 is not a divisor of 612)
612 / 612 = 1 (the remainder is 0, so 612 is a divisor of 612)

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 612 (i.e. 24.738633753706). 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 $\frac{N}{d} = D$ )

612 / 1 = 612 (the remainder is 0, so 1 and 612 are divisors of 612)
612 / 2 = 306 (the remainder is 0, so 2 and 306 are divisors of 612)
612 / 3 = 204 (the remainder is 0, so 3 and 204 are divisors of 612)
...
612 / 23 = 26.608695652174 (the remainder is 14, so 23 is not a divisor of 612)
612 / 24 = 25.5 (the remainder is 12, so 24 is not a divisor of 612)