What are the divisors of 5841?

1, 3, 9, 11, 33, 59, 99, 177, 531, 649, 1947, 5841

There is a total of 12 positive divisors.
The sum of these divisors is 9360.
The arithmetic mean is 780.

12 odd divisors

1, 3, 9, 11, 33, 59, 99, 177, 531, 649, 1947, 5841

How to compute the divisors of 5841?

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 5841 by each of the numbers from 1 to 5841 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)

5841 / 1 = 5841 (the remainder is 0, so 1 is a divisor of 5841)
5841 / 2 = 2920.5 (the remainder is 1, so 2 is not a divisor of 5841)
5841 / 3 = 1947 (the remainder is 0, so 3 is a divisor of 5841)
...
5841 / 5840 = 1.0001712328767 (the remainder is 1, so 5840 is not a divisor of 5841)
5841 / 5841 = 1 (the remainder is 0, so 5841 is a divisor of 5841)

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 5841 (i.e. 76.426435217142). 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 $\frac{N}{d} = D$ )

5841 / 1 = 5841 (the remainder is 0, so 1 and 5841 are divisors of 5841)
5841 / 2 = 2920.5 (the remainder is 1, so 2 is not a divisor of 5841)
5841 / 3 = 1947 (the remainder is 0, so 3 and 1947 are divisors of 5841)
...
5841 / 75 = 77.88 (the remainder is 66, so 75 is not a divisor of 5841)
5841 / 76 = 76.855263157895 (the remainder is 65, so 76 is not a divisor of 5841)