What are the divisors of 6150?

1, 2, 3, 5, 6, 10, 15, 25, 30, 41, 50, 75, 82, 123, 150, 205, 246, 410, 615, 1025, 1230, 2050, 3075, 6150

There is a total of 24 positive divisors.
The sum of these divisors is 15624.
The arithmetic mean is 651.

12 even divisors

2, 6, 10, 30, 50, 82, 150, 246, 410, 1230, 2050, 6150

12 odd divisors

1, 3, 5, 15, 25, 41, 75, 123, 205, 615, 1025, 3075

How to compute the divisors of 6150?

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 6150 by each of the numbers from 1 to 6150 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)

6150 / 1 = 6150 (the remainder is 0, so 1 is a divisor of 6150)
6150 / 2 = 3075 (the remainder is 0, so 2 is a divisor of 6150)
6150 / 3 = 2050 (the remainder is 0, so 3 is a divisor of 6150)
...
6150 / 6149 = 1.0001626280696 (the remainder is 1, so 6149 is not a divisor of 6150)
6150 / 6150 = 1 (the remainder is 0, so 6150 is a divisor of 6150)

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 6150 (i.e. 78.421935706791). 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 $\frac{N}{d} = D$ )

6150 / 1 = 6150 (the remainder is 0, so 1 and 6150 are divisors of 6150)
6150 / 2 = 3075 (the remainder is 0, so 2 and 3075 are divisors of 6150)
6150 / 3 = 2050 (the remainder is 0, so 3 and 2050 are divisors of 6150)
...
6150 / 77 = 79.87012987013 (the remainder is 67, so 77 is not a divisor of 6150)
6150 / 78 = 78.846153846154 (the remainder is 66, so 78 is not a divisor of 6150)