What are the divisors of 5244?

1, 2, 3, 4, 6, 12, 19, 23, 38, 46, 57, 69, 76, 92, 114, 138, 228, 276, 437, 874, 1311, 1748, 2622, 5244

There is a total of 24 positive divisors.
The sum of these divisors is 13440.
The arithmetic mean is 560.

16 even divisors

2, 4, 6, 12, 38, 46, 76, 92, 114, 138, 228, 276, 874, 1748, 2622, 5244

8 odd divisors

1, 3, 19, 23, 57, 69, 437, 1311

How to compute the divisors of 5244?

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 5244 by each of the numbers from 1 to 5244 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)

5244 / 1 = 5244 (the remainder is 0, so 1 is a divisor of 5244)
5244 / 2 = 2622 (the remainder is 0, so 2 is a divisor of 5244)
5244 / 3 = 1748 (the remainder is 0, so 3 is a divisor of 5244)
...
5244 / 5243 = 1.0001907304978 (the remainder is 1, so 5243 is not a divisor of 5244)
5244 / 5244 = 1 (the remainder is 0, so 5244 is a divisor of 5244)

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 5244 (i.e. 72.415467960927). 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 $\frac{N}{d} = D$ )

5244 / 1 = 5244 (the remainder is 0, so 1 and 5244 are divisors of 5244)
5244 / 2 = 2622 (the remainder is 0, so 2 and 2622 are divisors of 5244)
5244 / 3 = 1748 (the remainder is 0, so 3 and 1748 are divisors of 5244)
...
5244 / 71 = 73.859154929577 (the remainder is 61, so 71 is not a divisor of 5244)
5244 / 72 = 72.833333333333 (the remainder is 60, so 72 is not a divisor of 5244)