What are the divisors of 4650?

1, 2, 3, 5, 6, 10, 15, 25, 30, 31, 50, 62, 75, 93, 150, 155, 186, 310, 465, 775, 930, 1550, 2325, 4650

There is a total of 24 positive divisors.
The sum of these divisors is 11904.
The arithmetic mean is 496.

12 even divisors

2, 6, 10, 30, 50, 62, 150, 186, 310, 930, 1550, 4650

12 odd divisors

1, 3, 5, 15, 25, 31, 75, 93, 155, 465, 775, 2325

How to compute the divisors of 4650?

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 4650 by each of the numbers from 1 to 4650 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)

4650 / 1 = 4650 (the remainder is 0, so 1 is a divisor of 4650)
4650 / 2 = 2325 (the remainder is 0, so 2 is a divisor of 4650)
4650 / 3 = 1550 (the remainder is 0, so 3 is a divisor of 4650)
...
4650 / 4649 = 1.0002151000215 (the remainder is 1, so 4649 is not a divisor of 4650)
4650 / 4650 = 1 (the remainder is 0, so 4650 is a divisor of 4650)

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 4650 (i.e. 68.190908484929). 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 $\frac{N}{d} = D$ )

4650 / 1 = 4650 (the remainder is 0, so 1 and 4650 are divisors of 4650)
4650 / 2 = 2325 (the remainder is 0, so 2 and 2325 are divisors of 4650)
4650 / 3 = 1550 (the remainder is 0, so 3 and 1550 are divisors of 4650)
...
4650 / 67 = 69.402985074627 (the remainder is 27, so 67 is not a divisor of 4650)
4650 / 68 = 68.382352941176 (the remainder is 26, so 68 is not a divisor of 4650)