What are the divisors of 3750?

1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 125, 150, 250, 375, 625, 750, 1250, 1875, 3750

There is a total of 20 positive divisors.
The sum of these divisors is 9372.
The arithmetic mean is 468.6.

10 even divisors

2, 6, 10, 30, 50, 150, 250, 750, 1250, 3750

10 odd divisors

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

How to compute the divisors of 3750?

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 3750 by each of the numbers from 1 to 3750 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)

3750 / 1 = 3750 (the remainder is 0, so 1 is a divisor of 3750)
3750 / 2 = 1875 (the remainder is 0, so 2 is a divisor of 3750)
3750 / 3 = 1250 (the remainder is 0, so 3 is a divisor of 3750)
...
3750 / 3749 = 1.0002667377967 (the remainder is 1, so 3749 is not a divisor of 3750)
3750 / 3750 = 1 (the remainder is 0, so 3750 is a divisor of 3750)

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 3750 (i.e. 61.237243569579). 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 $\frac{N}{d} = D$ )

3750 / 1 = 3750 (the remainder is 0, so 1 and 3750 are divisors of 3750)
3750 / 2 = 1875 (the remainder is 0, so 2 and 1875 are divisors of 3750)
3750 / 3 = 1250 (the remainder is 0, so 3 and 1250 are divisors of 3750)
...
3750 / 60 = 62.5 (the remainder is 30, so 60 is not a divisor of 3750)
3750 / 61 = 61.475409836066 (the remainder is 29, so 61 is not a divisor of 3750)