What are the divisors of 2600?

1, 2, 4, 5, 8, 10, 13, 20, 25, 26, 40, 50, 52, 65, 100, 104, 130, 200, 260, 325, 520, 650, 1300, 2600

There is a total of 24 positive divisors.
The sum of these divisors is 6510.
The arithmetic mean is 271.25.

18 even divisors

2, 4, 8, 10, 20, 26, 40, 50, 52, 100, 104, 130, 200, 260, 520, 650, 1300, 2600

6 odd divisors

1, 5, 13, 25, 65, 325

How to compute the divisors of 2600?

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 2600 by each of the numbers from 1 to 2600 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)

2600 / 1 = 2600 (the remainder is 0, so 1 is a divisor of 2600)
2600 / 2 = 1300 (the remainder is 0, so 2 is a divisor of 2600)
2600 / 3 = 866.66666666667 (the remainder is 2, so 3 is not a divisor of 2600)
...
2600 / 2599 = 1.0003847633705 (the remainder is 1, so 2599 is not a divisor of 2600)
2600 / 2600 = 1 (the remainder is 0, so 2600 is a divisor of 2600)

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 2600 (i.e. 50.990195135928). 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 $\frac{N}{d} = D$ )

2600 / 1 = 2600 (the remainder is 0, so 1 and 2600 are divisors of 2600)
2600 / 2 = 1300 (the remainder is 0, so 2 and 1300 are divisors of 2600)
2600 / 3 = 866.66666666667 (the remainder is 2, so 3 is not a divisor of 2600)
...
2600 / 49 = 53.061224489796 (the remainder is 3, so 49 is not a divisor of 2600)
2600 / 50 = 52 (the remainder is 0, so 50 and 52 are divisors of 2600)