What are the divisors of 1838?

1, 2, 919, 1838

There is a total of 4 positive divisors.
The sum of these divisors is 2760.
The arithmetic mean is 690.

2 even divisors

2, 1838

2 odd divisors

1, 919

How to compute the divisors of 1838?

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 1838 by each of the numbers from 1 to 1838 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)

1838 / 1 = 1838 (the remainder is 0, so 1 is a divisor of 1838)
1838 / 2 = 919 (the remainder is 0, so 2 is a divisor of 1838)
1838 / 3 = 612.66666666667 (the remainder is 2, so 3 is not a divisor of 1838)
...
1838 / 1837 = 1.0005443658138 (the remainder is 1, so 1837 is not a divisor of 1838)
1838 / 1838 = 1 (the remainder is 0, so 1838 is a divisor of 1838)

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 1838 (i.e. 42.871902220452). 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 $\frac{N}{d} = D$ )

1838 / 1 = 1838 (the remainder is 0, so 1 and 1838 are divisors of 1838)
1838 / 2 = 919 (the remainder is 0, so 2 and 919 are divisors of 1838)
1838 / 3 = 612.66666666667 (the remainder is 2, so 3 is not a divisor of 1838)
...
1838 / 41 = 44.829268292683 (the remainder is 34, so 41 is not a divisor of 1838)
1838 / 42 = 43.761904761905 (the remainder is 32, so 42 is not a divisor of 1838)