What are the divisors of 3945?

1, 3, 5, 15, 263, 789, 1315, 3945

There is a total of 8 positive divisors.
The sum of these divisors is 6336.
The arithmetic mean is 792.

8 odd divisors

1, 3, 5, 15, 263, 789, 1315, 3945

How to compute the divisors of 3945?

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 3945 by each of the numbers from 1 to 3945 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)

3945 / 1 = 3945 (the remainder is 0, so 1 is a divisor of 3945)
3945 / 2 = 1972.5 (the remainder is 1, so 2 is not a divisor of 3945)
3945 / 3 = 1315 (the remainder is 0, so 3 is a divisor of 3945)
...
3945 / 3944 = 1.0002535496957 (the remainder is 1, so 3944 is not a divisor of 3945)
3945 / 3945 = 1 (the remainder is 0, so 3945 is a divisor of 3945)

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 3945 (i.e. 62.809234989769). 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 $\frac{N}{d} = D$ )

3945 / 1 = 3945 (the remainder is 0, so 1 and 3945 are divisors of 3945)
3945 / 2 = 1972.5 (the remainder is 1, so 2 is not a divisor of 3945)
3945 / 3 = 1315 (the remainder is 0, so 3 and 1315 are divisors of 3945)
...
3945 / 61 = 64.672131147541 (the remainder is 41, so 61 is not a divisor of 3945)
3945 / 62 = 63.629032258065 (the remainder is 39, so 62 is not a divisor of 3945)