What are the divisors of 1875?

1, 3, 5, 15, 25, 75, 125, 375, 625, 1875

There is a total of 10 positive divisors.
The sum of these divisors is 3124.
The arithmetic mean is 312.4.

10 odd divisors

1, 3, 5, 15, 25, 75, 125, 375, 625, 1875

How to compute the divisors of 1875?

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 1875 by each of the numbers from 1 to 1875 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)

1875 / 1 = 1875 (the remainder is 0, so 1 is a divisor of 1875)
1875 / 2 = 937.5 (the remainder is 1, so 2 is not a divisor of 1875)
1875 / 3 = 625 (the remainder is 0, so 3 is a divisor of 1875)
...
1875 / 1874 = 1.0005336179296 (the remainder is 1, so 1874 is not a divisor of 1875)
1875 / 1875 = 1 (the remainder is 0, so 1875 is a divisor of 1875)

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 1875 (i.e. 43.301270189222). 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 $\frac{N}{d} = D$ )

1875 / 1 = 1875 (the remainder is 0, so 1 and 1875 are divisors of 1875)
1875 / 2 = 937.5 (the remainder is 1, so 2 is not a divisor of 1875)
1875 / 3 = 625 (the remainder is 0, so 3 and 625 are divisors of 1875)
...
1875 / 42 = 44.642857142857 (the remainder is 27, so 42 is not a divisor of 1875)
1875 / 43 = 43.604651162791 (the remainder is 26, so 43 is not a divisor of 1875)