What are the divisors of 5229?

1, 3, 7, 9, 21, 63, 83, 249, 581, 747, 1743, 5229

There is a total of 12 positive divisors.
The sum of these divisors is 8736.
The arithmetic mean is 728.

12 odd divisors

1, 3, 7, 9, 21, 63, 83, 249, 581, 747, 1743, 5229

How to compute the divisors of 5229?

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 5229 by each of the numbers from 1 to 5229 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)

5229 / 1 = 5229 (the remainder is 0, so 1 is a divisor of 5229)
5229 / 2 = 2614.5 (the remainder is 1, so 2 is not a divisor of 5229)
5229 / 3 = 1743 (the remainder is 0, so 3 is a divisor of 5229)
...
5229 / 5228 = 1.0001912777353 (the remainder is 1, so 5228 is not a divisor of 5229)
5229 / 5229 = 1 (the remainder is 0, so 5229 is a divisor of 5229)

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 5229 (i.e. 72.311824759164). 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 $\frac{N}{d} = D$ )

5229 / 1 = 5229 (the remainder is 0, so 1 and 5229 are divisors of 5229)
5229 / 2 = 2614.5 (the remainder is 1, so 2 is not a divisor of 5229)
5229 / 3 = 1743 (the remainder is 0, so 3 and 1743 are divisors of 5229)
...
5229 / 71 = 73.647887323944 (the remainder is 46, so 71 is not a divisor of 5229)
5229 / 72 = 72.625 (the remainder is 45, so 72 is not a divisor of 5229)