What are the divisors of 2745?

1, 3, 5, 9, 15, 45, 61, 183, 305, 549, 915, 2745

There is a total of 12 positive divisors.
The sum of these divisors is 4836.
The arithmetic mean is 403.

12 odd divisors

1, 3, 5, 9, 15, 45, 61, 183, 305, 549, 915, 2745

How to compute the divisors of 2745?

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 2745 by each of the numbers from 1 to 2745 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)

2745 / 1 = 2745 (the remainder is 0, so 1 is a divisor of 2745)
2745 / 2 = 1372.5 (the remainder is 1, so 2 is not a divisor of 2745)
2745 / 3 = 915 (the remainder is 0, so 3 is a divisor of 2745)
...
2745 / 2744 = 1.0003644314869 (the remainder is 1, so 2744 is not a divisor of 2745)
2745 / 2745 = 1 (the remainder is 0, so 2745 is a divisor of 2745)

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 2745 (i.e. 52.392747589719). 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 $\frac{N}{d} = D$ )

2745 / 1 = 2745 (the remainder is 0, so 1 and 2745 are divisors of 2745)
2745 / 2 = 1372.5 (the remainder is 1, so 2 is not a divisor of 2745)
2745 / 3 = 915 (the remainder is 0, so 3 and 915 are divisors of 2745)
...
2745 / 51 = 53.823529411765 (the remainder is 42, so 51 is not a divisor of 2745)
2745 / 52 = 52.788461538462 (the remainder is 41, so 52 is not a divisor of 2745)