What are the divisors of 3050?

1, 2, 5, 10, 25, 50, 61, 122, 305, 610, 1525, 3050

There is a total of 12 positive divisors.
The sum of these divisors is 5766.
The arithmetic mean is 480.5.

6 even divisors

2, 10, 50, 122, 610, 3050

6 odd divisors

1, 5, 25, 61, 305, 1525

How to compute the divisors of 3050?

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 3050 by each of the numbers from 1 to 3050 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)

3050 / 1 = 3050 (the remainder is 0, so 1 is a divisor of 3050)
3050 / 2 = 1525 (the remainder is 0, so 2 is a divisor of 3050)
3050 / 3 = 1016.6666666667 (the remainder is 2, so 3 is not a divisor of 3050)
...
3050 / 3049 = 1.0003279763857 (the remainder is 1, so 3049 is not a divisor of 3050)
3050 / 3050 = 1 (the remainder is 0, so 3050 is a divisor of 3050)

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 3050 (i.e. 55.226805085936). 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 $\frac{N}{d} = D$ )

3050 / 1 = 3050 (the remainder is 0, so 1 and 3050 are divisors of 3050)
3050 / 2 = 1525 (the remainder is 0, so 2 and 1525 are divisors of 3050)
3050 / 3 = 1016.6666666667 (the remainder is 2, so 3 is not a divisor of 3050)
...
3050 / 54 = 56.481481481481 (the remainder is 26, so 54 is not a divisor of 3050)
3050 / 55 = 55.454545454545 (the remainder is 25, so 55 is not a divisor of 3050)