What are the divisors of 5203?

1, 11, 43, 121, 473, 5203

There is a total of 6 positive divisors.
The sum of these divisors is 5852.
The arithmetic mean is 975.33333333333.

6 odd divisors

1, 11, 43, 121, 473, 5203

How to compute the divisors of 5203?

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 5203 by each of the numbers from 1 to 5203 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)

5203 / 1 = 5203 (the remainder is 0, so 1 is a divisor of 5203)
5203 / 2 = 2601.5 (the remainder is 1, so 2 is not a divisor of 5203)
5203 / 3 = 1734.3333333333 (the remainder is 1, so 3 is not a divisor of 5203)
...
5203 / 5202 = 1.0001922337562 (the remainder is 1, so 5202 is not a divisor of 5203)
5203 / 5203 = 1 (the remainder is 0, so 5203 is a divisor of 5203)

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 5203 (i.e. 72.131823767322). 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 $\frac{N}{d} = D$ )

5203 / 1 = 5203 (the remainder is 0, so 1 and 5203 are divisors of 5203)
5203 / 2 = 2601.5 (the remainder is 1, so 2 is not a divisor of 5203)
5203 / 3 = 1734.3333333333 (the remainder is 1, so 3 is not a divisor of 5203)
...
5203 / 71 = 73.281690140845 (the remainder is 20, so 71 is not a divisor of 5203)
5203 / 72 = 72.263888888889 (the remainder is 19, so 72 is not a divisor of 5203)