What are the divisors of 3969?

1, 3, 7, 9, 21, 27, 49, 63, 81, 147, 189, 441, 567, 1323, 3969

There is a total of 15 positive divisors.
The sum of these divisors is 6897.
The arithmetic mean is 459.8.

15 odd divisors

1, 3, 7, 9, 21, 27, 49, 63, 81, 147, 189, 441, 567, 1323, 3969

How to compute the divisors of 3969?

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 3969 by each of the numbers from 1 to 3969 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)

3969 / 1 = 3969 (the remainder is 0, so 1 is a divisor of 3969)
3969 / 2 = 1984.5 (the remainder is 1, so 2 is not a divisor of 3969)
3969 / 3 = 1323 (the remainder is 0, so 3 is a divisor of 3969)
...
3969 / 3968 = 1.000252016129 (the remainder is 1, so 3968 is not a divisor of 3969)
3969 / 3969 = 1 (the remainder is 0, so 3969 is a divisor of 3969)

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 3969 (i.e. 63). 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 $\frac{N}{d} = D$ )

3969 / 1 = 3969 (the remainder is 0, so 1 and 3969 are divisors of 3969)
3969 / 2 = 1984.5 (the remainder is 1, so 2 is not a divisor of 3969)
3969 / 3 = 1323 (the remainder is 0, so 3 and 1323 are divisors of 3969)
...
3969 / 62 = 64.016129032258 (the remainder is 1, so 62 is not a divisor of 3969)
3969 / 63 = 63 (the remainder is 0, so 63 and 63 are divisors of 3969)