What are the divisors of 2925?

1, 3, 5, 9, 13, 15, 25, 39, 45, 65, 75, 117, 195, 225, 325, 585, 975, 2925

There is a total of 18 positive divisors.
The sum of these divisors is 5642.
The arithmetic mean is 313.44444444444.

18 odd divisors

1, 3, 5, 9, 13, 15, 25, 39, 45, 65, 75, 117, 195, 225, 325, 585, 975, 2925

How to compute the divisors of 2925?

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 2925 by each of the numbers from 1 to 2925 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)

2925 / 1 = 2925 (the remainder is 0, so 1 is a divisor of 2925)
2925 / 2 = 1462.5 (the remainder is 1, so 2 is not a divisor of 2925)
2925 / 3 = 975 (the remainder is 0, so 3 is a divisor of 2925)
...
2925 / 2924 = 1.000341997264 (the remainder is 1, so 2924 is not a divisor of 2925)
2925 / 2925 = 1 (the remainder is 0, so 2925 is a divisor of 2925)

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 2925 (i.e. 54.08326913196). 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 $\frac{N}{d} = D$ )

2925 / 1 = 2925 (the remainder is 0, so 1 and 2925 are divisors of 2925)
2925 / 2 = 1462.5 (the remainder is 1, so 2 is not a divisor of 2925)
2925 / 3 = 975 (the remainder is 0, so 3 and 975 are divisors of 2925)
...
2925 / 53 = 55.188679245283 (the remainder is 10, so 53 is not a divisor of 2925)
2925 / 54 = 54.166666666667 (the remainder is 9, so 54 is not a divisor of 2925)