What are the divisors of 2900?

1, 2, 4, 5, 10, 20, 25, 29, 50, 58, 100, 116, 145, 290, 580, 725, 1450, 2900

There is a total of 18 positive divisors.
The sum of these divisors is 6510.
The arithmetic mean is 361.66666666667.

12 even divisors

2, 4, 10, 20, 50, 58, 100, 116, 290, 580, 1450, 2900

6 odd divisors

1, 5, 25, 29, 145, 725

How to compute the divisors of 2900?

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 2900 by each of the numbers from 1 to 2900 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)

2900 / 1 = 2900 (the remainder is 0, so 1 is a divisor of 2900)
2900 / 2 = 1450 (the remainder is 0, so 2 is a divisor of 2900)
2900 / 3 = 966.66666666667 (the remainder is 2, so 3 is not a divisor of 2900)
...
2900 / 2899 = 1.0003449465333 (the remainder is 1, so 2899 is not a divisor of 2900)
2900 / 2900 = 1 (the remainder is 0, so 2900 is a divisor of 2900)

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 2900 (i.e. 53.851648071345). 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 $\frac{N}{d} = D$ )

2900 / 1 = 2900 (the remainder is 0, so 1 and 2900 are divisors of 2900)
2900 / 2 = 1450 (the remainder is 0, so 2 and 1450 are divisors of 2900)
2900 / 3 = 966.66666666667 (the remainder is 2, so 3 is not a divisor of 2900)
...
2900 / 52 = 55.769230769231 (the remainder is 40, so 52 is not a divisor of 2900)
2900 / 53 = 54.716981132075 (the remainder is 38, so 53 is not a divisor of 2900)