What are the divisors of 6050?

1, 2, 5, 10, 11, 22, 25, 50, 55, 110, 121, 242, 275, 550, 605, 1210, 3025, 6050

There is a total of 18 positive divisors.
The sum of these divisors is 12369.
The arithmetic mean is 687.16666666667.

9 even divisors

2, 10, 22, 50, 110, 242, 550, 1210, 6050

9 odd divisors

1, 5, 11, 25, 55, 121, 275, 605, 3025

How to compute the divisors of 6050?

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 6050 by each of the numbers from 1 to 6050 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)

6050 / 1 = 6050 (the remainder is 0, so 1 is a divisor of 6050)
6050 / 2 = 3025 (the remainder is 0, so 2 is a divisor of 6050)
6050 / 3 = 2016.6666666667 (the remainder is 2, so 3 is not a divisor of 6050)
...
6050 / 6049 = 1.0001653165813 (the remainder is 1, so 6049 is not a divisor of 6050)
6050 / 6050 = 1 (the remainder is 0, so 6050 is a divisor of 6050)

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 6050 (i.e. 77.78174593052). 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 $\frac{N}{d} = D$ )

6050 / 1 = 6050 (the remainder is 0, so 1 and 6050 are divisors of 6050)
6050 / 2 = 3025 (the remainder is 0, so 2 and 3025 are divisors of 6050)
6050 / 3 = 2016.6666666667 (the remainder is 2, so 3 is not a divisor of 6050)
...
6050 / 76 = 79.605263157895 (the remainder is 46, so 76 is not a divisor of 6050)
6050 / 77 = 78.571428571429 (the remainder is 44, so 77 is not a divisor of 6050)