What are the divisors of 3940?

1, 2, 4, 5, 10, 20, 197, 394, 788, 985, 1970, 3940

There is a total of 12 positive divisors.
The sum of these divisors is 8316.
The arithmetic mean is 693.

8 even divisors

2, 4, 10, 20, 394, 788, 1970, 3940

4 odd divisors

1, 5, 197, 985

How to compute the divisors of 3940?

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 3940 by each of the numbers from 1 to 3940 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)

3940 / 1 = 3940 (the remainder is 0, so 1 is a divisor of 3940)
3940 / 2 = 1970 (the remainder is 0, so 2 is a divisor of 3940)
3940 / 3 = 1313.3333333333 (the remainder is 1, so 3 is not a divisor of 3940)
...
3940 / 3939 = 1.000253871541 (the remainder is 1, so 3939 is not a divisor of 3940)
3940 / 3940 = 1 (the remainder is 0, so 3940 is a divisor of 3940)

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 3940 (i.e. 62.769419305901). 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 $\frac{N}{d} = D$ )

3940 / 1 = 3940 (the remainder is 0, so 1 and 3940 are divisors of 3940)
3940 / 2 = 1970 (the remainder is 0, so 2 and 1970 are divisors of 3940)
3940 / 3 = 1313.3333333333 (the remainder is 1, so 3 is not a divisor of 3940)
...
3940 / 61 = 64.590163934426 (the remainder is 36, so 61 is not a divisor of 3940)
3940 / 62 = 63.548387096774 (the remainder is 34, so 62 is not a divisor of 3940)