What are the divisors of 940?

1, 2, 4, 5, 10, 20, 47, 94, 188, 235, 470, 940

There is a total of 12 positive divisors.
The sum of these divisors is 2016.
The arithmetic mean is 168.

8 even divisors

2, 4, 10, 20, 94, 188, 470, 940

4 odd divisors

1, 5, 47, 235

How to compute the divisors of 940?

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 940 by each of the numbers from 1 to 940 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)

940 / 1 = 940 (the remainder is 0, so 1 is a divisor of 940)
940 / 2 = 470 (the remainder is 0, so 2 is a divisor of 940)
940 / 3 = 313.33333333333 (the remainder is 1, so 3 is not a divisor of 940)
...
940 / 939 = 1.0010649627263 (the remainder is 1, so 939 is not a divisor of 940)
940 / 940 = 1 (the remainder is 0, so 940 is a divisor of 940)

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 940 (i.e. 30.659419433512). 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 $\frac{N}{d} = D$ )

940 / 1 = 940 (the remainder is 0, so 1 and 940 are divisors of 940)
940 / 2 = 470 (the remainder is 0, so 2 and 470 are divisors of 940)
940 / 3 = 313.33333333333 (the remainder is 1, so 3 is not a divisor of 940)
...
940 / 29 = 32.413793103448 (the remainder is 12, so 29 is not a divisor of 940)
940 / 30 = 31.333333333333 (the remainder is 10, so 30 is not a divisor of 940)