What are the divisors of 5775?

1, 3, 5, 7, 11, 15, 21, 25, 33, 35, 55, 75, 77, 105, 165, 175, 231, 275, 385, 525, 825, 1155, 1925, 5775

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

24 odd divisors

1, 3, 5, 7, 11, 15, 21, 25, 33, 35, 55, 75, 77, 105, 165, 175, 231, 275, 385, 525, 825, 1155, 1925, 5775

How to compute the divisors of 5775?

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 5775 by each of the numbers from 1 to 5775 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)

5775 / 1 = 5775 (the remainder is 0, so 1 is a divisor of 5775)
5775 / 2 = 2887.5 (the remainder is 1, so 2 is not a divisor of 5775)
5775 / 3 = 1925 (the remainder is 0, so 3 is a divisor of 5775)
...
5775 / 5774 = 1.0001731901628 (the remainder is 1, so 5774 is not a divisor of 5775)
5775 / 5775 = 1 (the remainder is 0, so 5775 is a divisor of 5775)

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 5775 (i.e. 75.993420767853). 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 $\frac{N}{d} = D$ )

5775 / 1 = 5775 (the remainder is 0, so 1 and 5775 are divisors of 5775)
5775 / 2 = 2887.5 (the remainder is 1, so 2 is not a divisor of 5775)
5775 / 3 = 1925 (the remainder is 0, so 3 and 1925 are divisors of 5775)
...
5775 / 74 = 78.040540540541 (the remainder is 3, so 74 is not a divisor of 5775)
5775 / 75 = 77 (the remainder is 0, so 75 and 77 are divisors of 5775)