What are the divisors of 2915?

1, 5, 11, 53, 55, 265, 583, 2915

There is a total of 8 positive divisors.
The sum of these divisors is 3888.
The arithmetic mean is 486.

8 odd divisors

1, 5, 11, 53, 55, 265, 583, 2915

How to compute the divisors of 2915?

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 2915 by each of the numbers from 1 to 2915 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)

2915 / 1 = 2915 (the remainder is 0, so 1 is a divisor of 2915)
2915 / 2 = 1457.5 (the remainder is 1, so 2 is not a divisor of 2915)
2915 / 3 = 971.66666666667 (the remainder is 2, so 3 is not a divisor of 2915)
...
2915 / 2914 = 1.0003431708991 (the remainder is 1, so 2914 is not a divisor of 2915)
2915 / 2915 = 1 (the remainder is 0, so 2915 is a divisor of 2915)

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 2915 (i.e. 53.990739946772). 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 $\frac{N}{d} = D$ )

2915 / 1 = 2915 (the remainder is 0, so 1 and 2915 are divisors of 2915)
2915 / 2 = 1457.5 (the remainder is 1, so 2 is not a divisor of 2915)
2915 / 3 = 971.66666666667 (the remainder is 2, so 3 is not a divisor of 2915)
...
2915 / 52 = 56.057692307692 (the remainder is 3, so 52 is not a divisor of 2915)
2915 / 53 = 55 (the remainder is 0, so 53 and 55 are divisors of 2915)