What are the divisors of 4720?

1, 2, 4, 5, 8, 10, 16, 20, 40, 59, 80, 118, 236, 295, 472, 590, 944, 1180, 2360, 4720

There is a total of 20 positive divisors.
The sum of these divisors is 11160.
The arithmetic mean is 558.

16 even divisors

2, 4, 8, 10, 16, 20, 40, 80, 118, 236, 472, 590, 944, 1180, 2360, 4720

4 odd divisors

1, 5, 59, 295

How to compute the divisors of 4720?

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 4720 by each of the numbers from 1 to 4720 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)

4720 / 1 = 4720 (the remainder is 0, so 1 is a divisor of 4720)
4720 / 2 = 2360 (the remainder is 0, so 2 is a divisor of 4720)
4720 / 3 = 1573.3333333333 (the remainder is 1, so 3 is not a divisor of 4720)
...
4720 / 4719 = 1.0002119093028 (the remainder is 1, so 4719 is not a divisor of 4720)
4720 / 4720 = 1 (the remainder is 0, so 4720 is a divisor of 4720)

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 4720 (i.e. 68.702256149271). 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 $\frac{N}{d} = D$ )

4720 / 1 = 4720 (the remainder is 0, so 1 and 4720 are divisors of 4720)
4720 / 2 = 2360 (the remainder is 0, so 2 and 2360 are divisors of 4720)
4720 / 3 = 1573.3333333333 (the remainder is 1, so 3 is not a divisor of 4720)
...
4720 / 67 = 70.44776119403 (the remainder is 30, so 67 is not a divisor of 4720)
4720 / 68 = 69.411764705882 (the remainder is 28, so 68 is not a divisor of 4720)