What are the divisors of 5303?

1, 5303

There is a total of 2 positive divisors.
The sum of these divisors is 5304.
The arithmetic mean is 2652.

2 odd divisors

1, 5303

How to compute the divisors of 5303?

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 5303 by each of the numbers from 1 to 5303 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)

5303 / 1 = 5303 (the remainder is 0, so 1 is a divisor of 5303)
5303 / 2 = 2651.5 (the remainder is 1, so 2 is not a divisor of 5303)
5303 / 3 = 1767.6666666667 (the remainder is 2, so 3 is not a divisor of 5303)
...
5303 / 5302 = 1.0001886080724 (the remainder is 1, so 5302 is not a divisor of 5303)
5303 / 5303 = 1 (the remainder is 0, so 5303 is a divisor of 5303)

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 5303 (i.e. 72.82170006255). 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 $\frac{N}{d} = D$ )

5303 / 1 = 5303 (the remainder is 0, so 1 and 5303 are divisors of 5303)
5303 / 2 = 2651.5 (the remainder is 1, so 2 is not a divisor of 5303)
5303 / 3 = 1767.6666666667 (the remainder is 2, so 3 is not a divisor of 5303)
...
5303 / 71 = 74.69014084507 (the remainder is 49, so 71 is not a divisor of 5303)
5303 / 72 = 73.652777777778 (the remainder is 47, so 72 is not a divisor of 5303)