What are the divisors of 4712?

1, 2, 4, 8, 19, 31, 38, 62, 76, 124, 152, 248, 589, 1178, 2356, 4712

There is a total of 16 positive divisors.
The sum of these divisors is 9600.
The arithmetic mean is 600.

12 even divisors

2, 4, 8, 38, 62, 76, 124, 152, 248, 1178, 2356, 4712

4 odd divisors

1, 19, 31, 589

How to compute the divisors of 4712?

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 4712 by each of the numbers from 1 to 4712 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)

4712 / 1 = 4712 (the remainder is 0, so 1 is a divisor of 4712)
4712 / 2 = 2356 (the remainder is 0, so 2 is a divisor of 4712)
4712 / 3 = 1570.6666666667 (the remainder is 2, so 3 is not a divisor of 4712)
...
4712 / 4711 = 1.0002122691573 (the remainder is 1, so 4711 is not a divisor of 4712)
4712 / 4712 = 1 (the remainder is 0, so 4712 is a divisor of 4712)

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 4712 (i.e. 68.644009206922). 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 $\frac{N}{d} = D$ )

4712 / 1 = 4712 (the remainder is 0, so 1 and 4712 are divisors of 4712)
4712 / 2 = 2356 (the remainder is 0, so 2 and 2356 are divisors of 4712)
4712 / 3 = 1570.6666666667 (the remainder is 2, so 3 is not a divisor of 4712)
...
4712 / 67 = 70.328358208955 (the remainder is 22, so 67 is not a divisor of 4712)
4712 / 68 = 69.294117647059 (the remainder is 20, so 68 is not a divisor of 4712)