What are the divisors of 5325?

1, 3, 5, 15, 25, 71, 75, 213, 355, 1065, 1775, 5325

There is a total of 12 positive divisors.
The sum of these divisors is 8928.
The arithmetic mean is 744.

12 odd divisors

1, 3, 5, 15, 25, 71, 75, 213, 355, 1065, 1775, 5325

How to compute the divisors of 5325?

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 5325 by each of the numbers from 1 to 5325 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)

5325 / 1 = 5325 (the remainder is 0, so 1 is a divisor of 5325)
5325 / 2 = 2662.5 (the remainder is 1, so 2 is not a divisor of 5325)
5325 / 3 = 1775 (the remainder is 0, so 3 is a divisor of 5325)
...
5325 / 5324 = 1.0001878287002 (the remainder is 1, so 5324 is not a divisor of 5325)
5325 / 5325 = 1 (the remainder is 0, so 5325 is a divisor of 5325)

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 5325 (i.e. 72.972597596632). 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 $\frac{N}{d} = D$ )

5325 / 1 = 5325 (the remainder is 0, so 1 and 5325 are divisors of 5325)
5325 / 2 = 2662.5 (the remainder is 1, so 2 is not a divisor of 5325)
5325 / 3 = 1775 (the remainder is 0, so 3 and 1775 are divisors of 5325)
...
5325 / 71 = 75 (the remainder is 0, so 71 and 75 are divisors of 5325)
5325 / 72 = 73.958333333333 (the remainder is 69, so 72 is not a divisor of 5325)