What are the divisors of 5915?

1, 5, 7, 13, 35, 65, 91, 169, 455, 845, 1183, 5915

There is a total of 12 positive divisors.
The sum of these divisors is 8784.
The arithmetic mean is 732.

12 odd divisors

1, 5, 7, 13, 35, 65, 91, 169, 455, 845, 1183, 5915

How to compute the divisors of 5915?

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 5915 by each of the numbers from 1 to 5915 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)

5915 / 1 = 5915 (the remainder is 0, so 1 is a divisor of 5915)
5915 / 2 = 2957.5 (the remainder is 1, so 2 is not a divisor of 5915)
5915 / 3 = 1971.6666666667 (the remainder is 2, so 3 is not a divisor of 5915)
...
5915 / 5914 = 1.0001690902942 (the remainder is 1, so 5914 is not a divisor of 5915)
5915 / 5915 = 1 (the remainder is 0, so 5915 is a divisor of 5915)

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 5915 (i.e. 76.909037180295). 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 $\frac{N}{d} = D$ )

5915 / 1 = 5915 (the remainder is 0, so 1 and 5915 are divisors of 5915)
5915 / 2 = 2957.5 (the remainder is 1, so 2 is not a divisor of 5915)
5915 / 3 = 1971.6666666667 (the remainder is 2, so 3 is not a divisor of 5915)
...
5915 / 75 = 78.866666666667 (the remainder is 65, so 75 is not a divisor of 5915)
5915 / 76 = 77.828947368421 (the remainder is 63, so 76 is not a divisor of 5915)