What are the divisors of 2025?

1, 3, 5, 9, 15, 25, 27, 45, 75, 81, 135, 225, 405, 675, 2025

There is a total of 15 positive divisors.
The sum of these divisors is 3751.
The arithmetic mean is 250.06666666667.

15 odd divisors

1, 3, 5, 9, 15, 25, 27, 45, 75, 81, 135, 225, 405, 675, 2025

How to compute the divisors of 2025?

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 2025 by each of the numbers from 1 to 2025 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)

2025 / 1 = 2025 (the remainder is 0, so 1 is a divisor of 2025)
2025 / 2 = 1012.5 (the remainder is 1, so 2 is not a divisor of 2025)
2025 / 3 = 675 (the remainder is 0, so 3 is a divisor of 2025)
...
2025 / 2024 = 1.0004940711462 (the remainder is 1, so 2024 is not a divisor of 2025)
2025 / 2025 = 1 (the remainder is 0, so 2025 is a divisor of 2025)

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 2025 (i.e. 45). 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 $\frac{N}{d} = D$ )

2025 / 1 = 2025 (the remainder is 0, so 1 and 2025 are divisors of 2025)
2025 / 2 = 1012.5 (the remainder is 1, so 2 is not a divisor of 2025)
2025 / 3 = 675 (the remainder is 0, so 3 and 675 are divisors of 2025)
...
2025 / 44 = 46.022727272727 (the remainder is 1, so 44 is not a divisor of 2025)
2025 / 45 = 45 (the remainder is 0, so 45 and 45 are divisors of 2025)