What are the divisors of 5360?

1, 2, 4, 5, 8, 10, 16, 20, 40, 67, 80, 134, 268, 335, 536, 670, 1072, 1340, 2680, 5360

There is a total of 20 positive divisors.
The sum of these divisors is 12648.
The arithmetic mean is 632.4.

16 even divisors

2, 4, 8, 10, 16, 20, 40, 80, 134, 268, 536, 670, 1072, 1340, 2680, 5360

4 odd divisors

1, 5, 67, 335

How to compute the divisors of 5360?

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 5360 by each of the numbers from 1 to 5360 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)

5360 / 1 = 5360 (the remainder is 0, so 1 is a divisor of 5360)
5360 / 2 = 2680 (the remainder is 0, so 2 is a divisor of 5360)
5360 / 3 = 1786.6666666667 (the remainder is 2, so 3 is not a divisor of 5360)
...
5360 / 5359 = 1.000186601978 (the remainder is 1, so 5359 is not a divisor of 5360)
5360 / 5360 = 1 (the remainder is 0, so 5360 is a divisor of 5360)

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 5360 (i.e. 73.212020870893). 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 $\frac{N}{d} = D$ )

5360 / 1 = 5360 (the remainder is 0, so 1 and 5360 are divisors of 5360)
5360 / 2 = 2680 (the remainder is 0, so 2 and 2680 are divisors of 5360)
5360 / 3 = 1786.6666666667 (the remainder is 2, so 3 is not a divisor of 5360)
...
5360 / 72 = 74.444444444444 (the remainder is 32, so 72 is not a divisor of 5360)
5360 / 73 = 73.424657534247 (the remainder is 31, so 73 is not a divisor of 5360)