What are the divisors of 5355?

1, 3, 5, 7, 9, 15, 17, 21, 35, 45, 51, 63, 85, 105, 119, 153, 255, 315, 357, 595, 765, 1071, 1785, 5355

There is a total of 24 positive divisors.
The sum of these divisors is 11232.
The arithmetic mean is 468.

24 odd divisors

1, 3, 5, 7, 9, 15, 17, 21, 35, 45, 51, 63, 85, 105, 119, 153, 255, 315, 357, 595, 765, 1071, 1785, 5355

How to compute the divisors of 5355?

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 5355 by each of the numbers from 1 to 5355 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)

5355 / 1 = 5355 (the remainder is 0, so 1 is a divisor of 5355)
5355 / 2 = 2677.5 (the remainder is 1, so 2 is not a divisor of 5355)
5355 / 3 = 1785 (the remainder is 0, so 3 is a divisor of 5355)
...
5355 / 5354 = 1.0001867762421 (the remainder is 1, so 5354 is not a divisor of 5355)
5355 / 5355 = 1 (the remainder is 0, so 5355 is a divisor of 5355)

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 5355 (i.e. 73.177865505903). 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 $\frac{N}{d} = D$ )

5355 / 1 = 5355 (the remainder is 0, so 1 and 5355 are divisors of 5355)
5355 / 2 = 2677.5 (the remainder is 1, so 2 is not a divisor of 5355)
5355 / 3 = 1785 (the remainder is 0, so 3 and 1785 are divisors of 5355)
...
5355 / 72 = 74.375 (the remainder is 27, so 72 is not a divisor of 5355)
5355 / 73 = 73.356164383562 (the remainder is 26, so 73 is not a divisor of 5355)