What are the divisors of 4218?

1, 2, 3, 6, 19, 37, 38, 57, 74, 111, 114, 222, 703, 1406, 2109, 4218

There is a total of 16 positive divisors.
The sum of these divisors is 9120.
The arithmetic mean is 570.

8 even divisors

2, 6, 38, 74, 114, 222, 1406, 4218

8 odd divisors

1, 3, 19, 37, 57, 111, 703, 2109

How to compute the divisors of 4218?

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 4218 by each of the numbers from 1 to 4218 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)

4218 / 1 = 4218 (the remainder is 0, so 1 is a divisor of 4218)
4218 / 2 = 2109 (the remainder is 0, so 2 is a divisor of 4218)
4218 / 3 = 1406 (the remainder is 0, so 3 is a divisor of 4218)
...
4218 / 4217 = 1.0002371354043 (the remainder is 1, so 4217 is not a divisor of 4218)
4218 / 4218 = 1 (the remainder is 0, so 4218 is a divisor of 4218)

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 4218 (i.e. 64.946131524518). 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 $\frac{N}{d} = D$ )

4218 / 1 = 4218 (the remainder is 0, so 1 and 4218 are divisors of 4218)
4218 / 2 = 2109 (the remainder is 0, so 2 and 2109 are divisors of 4218)
4218 / 3 = 1406 (the remainder is 0, so 3 and 1406 are divisors of 4218)
...
4218 / 63 = 66.952380952381 (the remainder is 60, so 63 is not a divisor of 4218)
4218 / 64 = 65.90625 (the remainder is 58, so 64 is not a divisor of 4218)