What are the divisors of 5925?

1, 3, 5, 15, 25, 75, 79, 237, 395, 1185, 1975, 5925

There is a total of 12 positive divisors.
The sum of these divisors is 9920.
The arithmetic mean is 826.66666666667.

12 odd divisors

1, 3, 5, 15, 25, 75, 79, 237, 395, 1185, 1975, 5925

How to compute the divisors of 5925?

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 5925 by each of the numbers from 1 to 5925 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)

5925 / 1 = 5925 (the remainder is 0, so 1 is a divisor of 5925)
5925 / 2 = 2962.5 (the remainder is 1, so 2 is not a divisor of 5925)
5925 / 3 = 1975 (the remainder is 0, so 3 is a divisor of 5925)
...
5925 / 5924 = 1.0001688048616 (the remainder is 1, so 5924 is not a divisor of 5925)
5925 / 5925 = 1 (the remainder is 0, so 5925 is a divisor of 5925)

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 5925 (i.e. 76.974021591703). 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 $\frac{N}{d} = D$ )

5925 / 1 = 5925 (the remainder is 0, so 1 and 5925 are divisors of 5925)
5925 / 2 = 2962.5 (the remainder is 1, so 2 is not a divisor of 5925)
5925 / 3 = 1975 (the remainder is 0, so 3 and 1975 are divisors of 5925)
...
5925 / 75 = 79 (the remainder is 0, so 75 and 79 are divisors of 5925)
5925 / 76 = 77.960526315789 (the remainder is 73, so 76 is not a divisor of 5925)