What are the divisors of 5335?

1, 5, 11, 55, 97, 485, 1067, 5335

There is a total of 8 positive divisors.
The sum of these divisors is 7056.
The arithmetic mean is 882.

8 odd divisors

1, 5, 11, 55, 97, 485, 1067, 5335

How to compute the divisors of 5335?

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 5335 by each of the numbers from 1 to 5335 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)

5335 / 1 = 5335 (the remainder is 0, so 1 is a divisor of 5335)
5335 / 2 = 2667.5 (the remainder is 1, so 2 is not a divisor of 5335)
5335 / 3 = 1778.3333333333 (the remainder is 1, so 3 is not a divisor of 5335)
...
5335 / 5334 = 1.0001874765654 (the remainder is 1, so 5334 is not a divisor of 5335)
5335 / 5335 = 1 (the remainder is 0, so 5335 is a divisor of 5335)

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 5335 (i.e. 73.041084329301). 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 $\frac{N}{d} = D$ )

5335 / 1 = 5335 (the remainder is 0, so 1 and 5335 are divisors of 5335)
5335 / 2 = 2667.5 (the remainder is 1, so 2 is not a divisor of 5335)
5335 / 3 = 1778.3333333333 (the remainder is 1, so 3 is not a divisor of 5335)
...
5335 / 72 = 74.097222222222 (the remainder is 7, so 72 is not a divisor of 5335)
5335 / 73 = 73.082191780822 (the remainder is 6, so 73 is not a divisor of 5335)