What are the divisors of 5902?

1, 2, 13, 26, 227, 454, 2951, 5902

There is a total of 8 positive divisors.
The sum of these divisors is 9576.
The arithmetic mean is 1197.

4 even divisors

2, 26, 454, 5902

4 odd divisors

1, 13, 227, 2951

How to compute the divisors of 5902?

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 5902 by each of the numbers from 1 to 5902 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)

5902 / 1 = 5902 (the remainder is 0, so 1 is a divisor of 5902)
5902 / 2 = 2951 (the remainder is 0, so 2 is a divisor of 5902)
5902 / 3 = 1967.3333333333 (the remainder is 1, so 3 is not a divisor of 5902)
...
5902 / 5901 = 1.0001694628029 (the remainder is 1, so 5901 is not a divisor of 5902)
5902 / 5902 = 1 (the remainder is 0, so 5902 is a divisor of 5902)

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 5902 (i.e. 76.824475266675). 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 $\frac{N}{d} = D$ )

5902 / 1 = 5902 (the remainder is 0, so 1 and 5902 are divisors of 5902)
5902 / 2 = 2951 (the remainder is 0, so 2 and 2951 are divisors of 5902)
5902 / 3 = 1967.3333333333 (the remainder is 1, so 3 is not a divisor of 5902)
...
5902 / 75 = 78.693333333333 (the remainder is 52, so 75 is not a divisor of 5902)
5902 / 76 = 77.657894736842 (the remainder is 50, so 76 is not a divisor of 5902)