What are the divisors of 4335?

1, 3, 5, 15, 17, 51, 85, 255, 289, 867, 1445, 4335

There is a total of 12 positive divisors.
The sum of these divisors is 7368.
The arithmetic mean is 614.

12 odd divisors

1, 3, 5, 15, 17, 51, 85, 255, 289, 867, 1445, 4335

How to compute the divisors of 4335?

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 4335 by each of the numbers from 1 to 4335 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)

4335 / 1 = 4335 (the remainder is 0, so 1 is a divisor of 4335)
4335 / 2 = 2167.5 (the remainder is 1, so 2 is not a divisor of 4335)
4335 / 3 = 1445 (the remainder is 0, so 3 is a divisor of 4335)
...
4335 / 4334 = 1.0002307337333 (the remainder is 1, so 4334 is not a divisor of 4335)
4335 / 4335 = 1 (the remainder is 0, so 4335 is a divisor of 4335)

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 4335 (i.e. 65.840716885526). 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 $\frac{N}{d} = D$ )

4335 / 1 = 4335 (the remainder is 0, so 1 and 4335 are divisors of 4335)
4335 / 2 = 2167.5 (the remainder is 1, so 2 is not a divisor of 4335)
4335 / 3 = 1445 (the remainder is 0, so 3 and 1445 are divisors of 4335)
...
4335 / 64 = 67.734375 (the remainder is 47, so 64 is not a divisor of 4335)
4335 / 65 = 66.692307692308 (the remainder is 45, so 65 is not a divisor of 4335)