What are the divisors of 5252?

1, 2, 4, 13, 26, 52, 101, 202, 404, 1313, 2626, 5252

There is a total of 12 positive divisors.
The sum of these divisors is 9996.
The arithmetic mean is 833.

8 even divisors

2, 4, 26, 52, 202, 404, 2626, 5252

4 odd divisors

1, 13, 101, 1313

How to compute the divisors of 5252?

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 5252 by each of the numbers from 1 to 5252 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)

5252 / 1 = 5252 (the remainder is 0, so 1 is a divisor of 5252)
5252 / 2 = 2626 (the remainder is 0, so 2 is a divisor of 5252)
5252 / 3 = 1750.6666666667 (the remainder is 2, so 3 is not a divisor of 5252)
...
5252 / 5251 = 1.0001904399162 (the remainder is 1, so 5251 is not a divisor of 5252)
5252 / 5252 = 1 (the remainder is 0, so 5252 is a divisor of 5252)

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 5252 (i.e. 72.470683727974). 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 $\frac{N}{d} = D$ )

5252 / 1 = 5252 (the remainder is 0, so 1 and 5252 are divisors of 5252)
5252 / 2 = 2626 (the remainder is 0, so 2 and 2626 are divisors of 5252)
5252 / 3 = 1750.6666666667 (the remainder is 2, so 3 is not a divisor of 5252)
...
5252 / 71 = 73.971830985915 (the remainder is 69, so 71 is not a divisor of 5252)
5252 / 72 = 72.944444444444 (the remainder is 68, so 72 is not a divisor of 5252)