What are the divisors of 4300?

1, 2, 4, 5, 10, 20, 25, 43, 50, 86, 100, 172, 215, 430, 860, 1075, 2150, 4300

There is a total of 18 positive divisors.
The sum of these divisors is 9548.
The arithmetic mean is 530.44444444444.

12 even divisors

2, 4, 10, 20, 50, 86, 100, 172, 430, 860, 2150, 4300

6 odd divisors

1, 5, 25, 43, 215, 1075

How to compute the divisors of 4300?

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 4300 by each of the numbers from 1 to 4300 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)

4300 / 1 = 4300 (the remainder is 0, so 1 is a divisor of 4300)
4300 / 2 = 2150 (the remainder is 0, so 2 is a divisor of 4300)
4300 / 3 = 1433.3333333333 (the remainder is 1, so 3 is not a divisor of 4300)
...
4300 / 4299 = 1.0002326122354 (the remainder is 1, so 4299 is not a divisor of 4300)
4300 / 4300 = 1 (the remainder is 0, so 4300 is a divisor of 4300)

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 4300 (i.e. 65.57438524302). 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 $\frac{N}{d} = D$ )

4300 / 1 = 4300 (the remainder is 0, so 1 and 4300 are divisors of 4300)
4300 / 2 = 2150 (the remainder is 0, so 2 and 2150 are divisors of 4300)
4300 / 3 = 1433.3333333333 (the remainder is 1, so 3 is not a divisor of 4300)
...
4300 / 64 = 67.1875 (the remainder is 12, so 64 is not a divisor of 4300)
4300 / 65 = 66.153846153846 (the remainder is 10, so 65 is not a divisor of 4300)