What are the divisors of 5903?

1, 5903

There is a total of 2 positive divisors.
The sum of these divisors is 5904.
The arithmetic mean is 2952.

2 odd divisors

1, 5903

How to compute the divisors of 5903?

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 5903 by each of the numbers from 1 to 5903 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)

5903 / 1 = 5903 (the remainder is 0, so 1 is a divisor of 5903)
5903 / 2 = 2951.5 (the remainder is 1, so 2 is not a divisor of 5903)
5903 / 3 = 1967.6666666667 (the remainder is 2, so 3 is not a divisor of 5903)
...
5903 / 5902 = 1.0001694340901 (the remainder is 1, so 5902 is not a divisor of 5903)
5903 / 5903 = 1 (the remainder is 0, so 5903 is a divisor of 5903)

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 5903 (i.e. 76.830983333548). 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 $\frac{N}{d} = D$ )

5903 / 1 = 5903 (the remainder is 0, so 1 and 5903 are divisors of 5903)
5903 / 2 = 2951.5 (the remainder is 1, so 2 is not a divisor of 5903)
5903 / 3 = 1967.6666666667 (the remainder is 2, so 3 is not a divisor of 5903)
...
5903 / 75 = 78.706666666667 (the remainder is 53, so 75 is not a divisor of 5903)
5903 / 76 = 77.671052631579 (the remainder is 51, so 76 is not a divisor of 5903)