What are the divisors of 2950?

1, 2, 5, 10, 25, 50, 59, 118, 295, 590, 1475, 2950

There is a total of 12 positive divisors.
The sum of these divisors is 5580.
The arithmetic mean is 465.

6 even divisors

2, 10, 50, 118, 590, 2950

6 odd divisors

1, 5, 25, 59, 295, 1475

How to compute the divisors of 2950?

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 2950 by each of the numbers from 1 to 2950 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)

2950 / 1 = 2950 (the remainder is 0, so 1 is a divisor of 2950)
2950 / 2 = 1475 (the remainder is 0, so 2 is a divisor of 2950)
2950 / 3 = 983.33333333333 (the remainder is 1, so 3 is not a divisor of 2950)
...
2950 / 2949 = 1.0003390979993 (the remainder is 1, so 2949 is not a divisor of 2950)
2950 / 2950 = 1 (the remainder is 0, so 2950 is a divisor of 2950)

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 2950 (i.e. 54.313902456001). 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 $\frac{N}{d} = D$ )

2950 / 1 = 2950 (the remainder is 0, so 1 and 2950 are divisors of 2950)
2950 / 2 = 1475 (the remainder is 0, so 2 and 1475 are divisors of 2950)
2950 / 3 = 983.33333333333 (the remainder is 1, so 3 is not a divisor of 2950)
...
2950 / 53 = 55.660377358491 (the remainder is 35, so 53 is not a divisor of 2950)
2950 / 54 = 54.62962962963 (the remainder is 34, so 54 is not a divisor of 2950)