What are the divisors of 1879?

1, 1879

There is a total of 2 positive divisors.
The sum of these divisors is 1880.
The arithmetic mean is 940.

2 odd divisors

1, 1879

How to compute the divisors of 1879?

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 1879 by each of the numbers from 1 to 1879 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)

1879 / 1 = 1879 (the remainder is 0, so 1 is a divisor of 1879)
1879 / 2 = 939.5 (the remainder is 1, so 2 is not a divisor of 1879)
1879 / 3 = 626.33333333333 (the remainder is 1, so 3 is not a divisor of 1879)
...
1879 / 1878 = 1.0005324813632 (the remainder is 1, so 1878 is not a divisor of 1879)
1879 / 1879 = 1 (the remainder is 0, so 1879 is a divisor of 1879)

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 1879 (i.e. 43.347433603386). 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 $\frac{N}{d} = D$ )

1879 / 1 = 1879 (the remainder is 0, so 1 and 1879 are divisors of 1879)
1879 / 2 = 939.5 (the remainder is 1, so 2 is not a divisor of 1879)
1879 / 3 = 626.33333333333 (the remainder is 1, so 3 is not a divisor of 1879)
...
1879 / 42 = 44.738095238095 (the remainder is 31, so 42 is not a divisor of 1879)
1879 / 43 = 43.697674418605 (the remainder is 30, so 43 is not a divisor of 1879)