What are the divisors of 120?

1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120

There is a total of 16 positive divisors.
The sum of these divisors is 360.
The arithmetic mean is 22.5.

12 even divisors

2, 4, 6, 8, 10, 12, 20, 24, 30, 40, 60, 120

4 odd divisors

1, 3, 5, 15

How to compute the divisors of 120?

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 120 by each of the numbers from 1 to 120 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)

120 / 1 = 120 (the remainder is 0, so 1 is a divisor of 120)
120 / 2 = 60 (the remainder is 0, so 2 is a divisor of 120)
120 / 3 = 40 (the remainder is 0, so 3 is a divisor of 120)
...
120 / 119 = 1.0084033613445 (the remainder is 1, so 119 is not a divisor of 120)
120 / 120 = 1 (the remainder is 0, so 120 is a divisor of 120)

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 120 (i.e. 10.954451150103). 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 $\frac{N}{d} = D$ )

120 / 1 = 120 (the remainder is 0, so 1 and 120 are divisors of 120)
120 / 2 = 60 (the remainder is 0, so 2 and 60 are divisors of 120)
120 / 3 = 40 (the remainder is 0, so 3 and 40 are divisors of 120)
...
120 / 9 = 13.333333333333 (the remainder is 3, so 9 is not a divisor of 120)
120 / 10 = 12 (the remainder is 0, so 10 and 12 are divisors of 120)