What are the divisors of 4340?

1, 2, 4, 5, 7, 10, 14, 20, 28, 31, 35, 62, 70, 124, 140, 155, 217, 310, 434, 620, 868, 1085, 2170, 4340

There is a total of 24 positive divisors.
The sum of these divisors is 10752.
The arithmetic mean is 448.

16 even divisors

2, 4, 10, 14, 20, 28, 62, 70, 124, 140, 310, 434, 620, 868, 2170, 4340

8 odd divisors

1, 5, 7, 31, 35, 155, 217, 1085

How to compute the divisors of 4340?

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 4340 by each of the numbers from 1 to 4340 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)

4340 / 1 = 4340 (the remainder is 0, so 1 is a divisor of 4340)
4340 / 2 = 2170 (the remainder is 0, so 2 is a divisor of 4340)
4340 / 3 = 1446.6666666667 (the remainder is 2, so 3 is not a divisor of 4340)
...
4340 / 4339 = 1.0002304678497 (the remainder is 1, so 4339 is not a divisor of 4340)
4340 / 4340 = 1 (the remainder is 0, so 4340 is a divisor of 4340)

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 4340 (i.e. 65.878676368002). 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 $\frac{N}{d} = D$ )

4340 / 1 = 4340 (the remainder is 0, so 1 and 4340 are divisors of 4340)
4340 / 2 = 2170 (the remainder is 0, so 2 and 2170 are divisors of 4340)
4340 / 3 = 1446.6666666667 (the remainder is 2, so 3 is not a divisor of 4340)
...
4340 / 64 = 67.8125 (the remainder is 52, so 64 is not a divisor of 4340)
4340 / 65 = 66.769230769231 (the remainder is 50, so 65 is not a divisor of 4340)