What are the divisors of 3735?

1, 3, 5, 9, 15, 45, 83, 249, 415, 747, 1245, 3735

There is a total of 12 positive divisors.
The sum of these divisors is 6552.
The arithmetic mean is 546.

12 odd divisors

1, 3, 5, 9, 15, 45, 83, 249, 415, 747, 1245, 3735

How to compute the divisors of 3735?

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 3735 by each of the numbers from 1 to 3735 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)

3735 / 1 = 3735 (the remainder is 0, so 1 is a divisor of 3735)
3735 / 2 = 1867.5 (the remainder is 1, so 2 is not a divisor of 3735)
3735 / 3 = 1245 (the remainder is 0, so 3 is a divisor of 3735)
...
3735 / 3734 = 1.0002678093198 (the remainder is 1, so 3734 is not a divisor of 3735)
3735 / 3735 = 1 (the remainder is 0, so 3735 is a divisor of 3735)

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 3735 (i.e. 61.11464636239). 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 $\frac{N}{d} = D$ )

3735 / 1 = 3735 (the remainder is 0, so 1 and 3735 are divisors of 3735)
3735 / 2 = 1867.5 (the remainder is 1, so 2 is not a divisor of 3735)
3735 / 3 = 1245 (the remainder is 0, so 3 and 1245 are divisors of 3735)
...
3735 / 60 = 62.25 (the remainder is 15, so 60 is not a divisor of 3735)
3735 / 61 = 61.229508196721 (the remainder is 14, so 61 is not a divisor of 3735)