What are the divisors of 735?

1, 3, 5, 7, 15, 21, 35, 49, 105, 147, 245, 735

There is a total of 12 positive divisors.
The sum of these divisors is 1368.
The arithmetic mean is 114.

12 odd divisors

1, 3, 5, 7, 15, 21, 35, 49, 105, 147, 245, 735

How to compute the divisors of 735?

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 735 by each of the numbers from 1 to 735 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)

735 / 1 = 735 (the remainder is 0, so 1 is a divisor of 735)
735 / 2 = 367.5 (the remainder is 1, so 2 is not a divisor of 735)
735 / 3 = 245 (the remainder is 0, so 3 is a divisor of 735)
...
735 / 734 = 1.0013623978202 (the remainder is 1, so 734 is not a divisor of 735)
735 / 735 = 1 (the remainder is 0, so 735 is a divisor of 735)

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 735 (i.e. 27.110883423452). 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 $\frac{N}{d} = D$ )

735 / 1 = 735 (the remainder is 0, so 1 and 735 are divisors of 735)
735 / 2 = 367.5 (the remainder is 1, so 2 is not a divisor of 735)
735 / 3 = 245 (the remainder is 0, so 3 and 245 are divisors of 735)
...
735 / 26 = 28.269230769231 (the remainder is 7, so 26 is not a divisor of 735)
735 / 27 = 27.222222222222 (the remainder is 6, so 27 is not a divisor of 735)