What are the divisors of 3933?

1, 3, 9, 19, 23, 57, 69, 171, 207, 437, 1311, 3933

There is a total of 12 positive divisors.
The sum of these divisors is 6240.
The arithmetic mean is 520.

12 odd divisors

1, 3, 9, 19, 23, 57, 69, 171, 207, 437, 1311, 3933

How to compute the divisors of 3933?

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 3933 by each of the numbers from 1 to 3933 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)

3933 / 1 = 3933 (the remainder is 0, so 1 is a divisor of 3933)
3933 / 2 = 1966.5 (the remainder is 1, so 2 is not a divisor of 3933)
3933 / 3 = 1311 (the remainder is 0, so 3 is a divisor of 3933)
...
3933 / 3932 = 1.0002543234995 (the remainder is 1, so 3932 is not a divisor of 3933)
3933 / 3933 = 1 (the remainder is 0, so 3933 is a divisor of 3933)

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 3933 (i.e. 62.713634881101). 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 $\frac{N}{d} = D$ )

3933 / 1 = 3933 (the remainder is 0, so 1 and 3933 are divisors of 3933)
3933 / 2 = 1966.5 (the remainder is 1, so 2 is not a divisor of 3933)
3933 / 3 = 1311 (the remainder is 0, so 3 and 1311 are divisors of 3933)
...
3933 / 61 = 64.475409836066 (the remainder is 29, so 61 is not a divisor of 3933)
3933 / 62 = 63.435483870968 (the remainder is 27, so 62 is not a divisor of 3933)