What are the divisors of 3650?

1, 2, 5, 10, 25, 50, 73, 146, 365, 730, 1825, 3650

There is a total of 12 positive divisors.
The sum of these divisors is 6882.
The arithmetic mean is 573.5.

6 even divisors

2, 10, 50, 146, 730, 3650

6 odd divisors

1, 5, 25, 73, 365, 1825

How to compute the divisors of 3650?

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 3650 by each of the numbers from 1 to 3650 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)

3650 / 1 = 3650 (the remainder is 0, so 1 is a divisor of 3650)
3650 / 2 = 1825 (the remainder is 0, so 2 is a divisor of 3650)
3650 / 3 = 1216.6666666667 (the remainder is 2, so 3 is not a divisor of 3650)
...
3650 / 3649 = 1.0002740476843 (the remainder is 1, so 3649 is not a divisor of 3650)
3650 / 3650 = 1 (the remainder is 0, so 3650 is a divisor of 3650)

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 3650 (i.e. 60.415229867973). 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 $\frac{N}{d} = D$ )

3650 / 1 = 3650 (the remainder is 0, so 1 and 3650 are divisors of 3650)
3650 / 2 = 1825 (the remainder is 0, so 2 and 1825 are divisors of 3650)
3650 / 3 = 1216.6666666667 (the remainder is 2, so 3 is not a divisor of 3650)
...
3650 / 59 = 61.864406779661 (the remainder is 51, so 59 is not a divisor of 3650)
3650 / 60 = 60.833333333333 (the remainder is 50, so 60 is not a divisor of 3650)