What are the divisors of 3118?

1, 2, 1559, 3118

There is a total of 4 positive divisors.
The sum of these divisors is 4680.
The arithmetic mean is 1170.

2 even divisors

2, 3118

2 odd divisors

1, 1559

How to compute the divisors of 3118?

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 3118 by each of the numbers from 1 to 3118 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)

3118 / 1 = 3118 (the remainder is 0, so 1 is a divisor of 3118)
3118 / 2 = 1559 (the remainder is 0, so 2 is a divisor of 3118)
3118 / 3 = 1039.3333333333 (the remainder is 1, so 3 is not a divisor of 3118)
...
3118 / 3117 = 1.0003208213025 (the remainder is 1, so 3117 is not a divisor of 3118)
3118 / 3118 = 1 (the remainder is 0, so 3118 is a divisor of 3118)

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 3118 (i.e. 55.839054433255). 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 $\frac{N}{d} = D$ )

3118 / 1 = 3118 (the remainder is 0, so 1 and 3118 are divisors of 3118)
3118 / 2 = 1559 (the remainder is 0, so 2 and 1559 are divisors of 3118)
3118 / 3 = 1039.3333333333 (the remainder is 1, so 3 is not a divisor of 3118)
...
3118 / 54 = 57.740740740741 (the remainder is 40, so 54 is not a divisor of 3118)
3118 / 55 = 56.690909090909 (the remainder is 38, so 55 is not a divisor of 3118)