What are the divisors of 3723?

1, 3, 17, 51, 73, 219, 1241, 3723

There is a total of 8 positive divisors.
The sum of these divisors is 5328.
The arithmetic mean is 666.

8 odd divisors

1, 3, 17, 51, 73, 219, 1241, 3723

How to compute the divisors of 3723?

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 3723 by each of the numbers from 1 to 3723 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)

3723 / 1 = 3723 (the remainder is 0, so 1 is a divisor of 3723)
3723 / 2 = 1861.5 (the remainder is 1, so 2 is not a divisor of 3723)
3723 / 3 = 1241 (the remainder is 0, so 3 is a divisor of 3723)
...
3723 / 3722 = 1.0002686727566 (the remainder is 1, so 3722 is not a divisor of 3723)
3723 / 3723 = 1 (the remainder is 0, so 3723 is a divisor of 3723)

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 3723 (i.e. 61.016391240387). 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 $\frac{N}{d} = D$ )

3723 / 1 = 3723 (the remainder is 0, so 1 and 3723 are divisors of 3723)
3723 / 2 = 1861.5 (the remainder is 1, so 2 is not a divisor of 3723)
3723 / 3 = 1241 (the remainder is 0, so 3 and 1241 are divisors of 3723)
...
3723 / 60 = 62.05 (the remainder is 3, so 60 is not a divisor of 3723)
3723 / 61 = 61.032786885246 (the remainder is 2, so 61 is not a divisor of 3723)