What are the divisors of 2541?

1, 3, 7, 11, 21, 33, 77, 121, 231, 363, 847, 2541

There is a total of 12 positive divisors.
The sum of these divisors is 4256.
The arithmetic mean is 354.66666666667.

12 odd divisors

1, 3, 7, 11, 21, 33, 77, 121, 231, 363, 847, 2541

How to compute the divisors of 2541?

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 2541 by each of the numbers from 1 to 2541 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)

2541 / 1 = 2541 (the remainder is 0, so 1 is a divisor of 2541)
2541 / 2 = 1270.5 (the remainder is 1, so 2 is not a divisor of 2541)
2541 / 3 = 847 (the remainder is 0, so 3 is a divisor of 2541)
...
2541 / 2540 = 1.0003937007874 (the remainder is 1, so 2540 is not a divisor of 2541)
2541 / 2541 = 1 (the remainder is 0, so 2541 is a divisor of 2541)

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 2541 (i.e. 50.408332644514). 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 $\frac{N}{d} = D$ )

2541 / 1 = 2541 (the remainder is 0, so 1 and 2541 are divisors of 2541)
2541 / 2 = 1270.5 (the remainder is 1, so 2 is not a divisor of 2541)
2541 / 3 = 847 (the remainder is 0, so 3 and 847 are divisors of 2541)
...
2541 / 49 = 51.857142857143 (the remainder is 42, so 49 is not a divisor of 2541)
2541 / 50 = 50.82 (the remainder is 41, so 50 is not a divisor of 2541)