What are the divisors of 5312?

1, 2, 4, 8, 16, 32, 64, 83, 166, 332, 664, 1328, 2656, 5312

There is a total of 14 positive divisors.
The sum of these divisors is 10668.
The arithmetic mean is 762.

12 even divisors

2, 4, 8, 16, 32, 64, 166, 332, 664, 1328, 2656, 5312

2 odd divisors

1, 83

How to compute the divisors of 5312?

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 5312 by each of the numbers from 1 to 5312 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)

5312 / 1 = 5312 (the remainder is 0, so 1 is a divisor of 5312)
5312 / 2 = 2656 (the remainder is 0, so 2 is a divisor of 5312)
5312 / 3 = 1770.6666666667 (the remainder is 2, so 3 is not a divisor of 5312)
...
5312 / 5311 = 1.0001882884579 (the remainder is 1, so 5311 is not a divisor of 5312)
5312 / 5312 = 1 (the remainder is 0, so 5312 is a divisor of 5312)

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 5312 (i.e. 72.883468633154). 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 $\frac{N}{d} = D$ )

5312 / 1 = 5312 (the remainder is 0, so 1 and 5312 are divisors of 5312)
5312 / 2 = 2656 (the remainder is 0, so 2 and 2656 are divisors of 5312)
5312 / 3 = 1770.6666666667 (the remainder is 2, so 3 is not a divisor of 5312)
...
5312 / 71 = 74.816901408451 (the remainder is 58, so 71 is not a divisor of 5312)
5312 / 72 = 73.777777777778 (the remainder is 56, so 72 is not a divisor of 5312)