What are the divisors of 5650?

1, 2, 5, 10, 25, 50, 113, 226, 565, 1130, 2825, 5650

There is a total of 12 positive divisors.
The sum of these divisors is 10602.
The arithmetic mean is 883.5.

6 even divisors

2, 10, 50, 226, 1130, 5650

6 odd divisors

1, 5, 25, 113, 565, 2825

How to compute the divisors of 5650?

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 5650 by each of the numbers from 1 to 5650 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)

5650 / 1 = 5650 (the remainder is 0, so 1 is a divisor of 5650)
5650 / 2 = 2825 (the remainder is 0, so 2 is a divisor of 5650)
5650 / 3 = 1883.3333333333 (the remainder is 1, so 3 is not a divisor of 5650)
...
5650 / 5649 = 1.0001770224819 (the remainder is 1, so 5649 is not a divisor of 5650)
5650 / 5650 = 1 (the remainder is 0, so 5650 is a divisor of 5650)

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 5650 (i.e. 75.166481891865). 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 $\frac{N}{d} = D$ )

5650 / 1 = 5650 (the remainder is 0, so 1 and 5650 are divisors of 5650)
5650 / 2 = 2825 (the remainder is 0, so 2 and 2825 are divisors of 5650)
5650 / 3 = 1883.3333333333 (the remainder is 1, so 3 is not a divisor of 5650)
...
5650 / 74 = 76.351351351351 (the remainder is 26, so 74 is not a divisor of 5650)
5650 / 75 = 75.333333333333 (the remainder is 25, so 75 is not a divisor of 5650)