What are the divisors of 4905?

1, 3, 5, 9, 15, 45, 109, 327, 545, 981, 1635, 4905

There is a total of 12 positive divisors.
The sum of these divisors is 8580.
The arithmetic mean is 715.

12 odd divisors

1, 3, 5, 9, 15, 45, 109, 327, 545, 981, 1635, 4905

How to compute the divisors of 4905?

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 4905 by each of the numbers from 1 to 4905 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)

4905 / 1 = 4905 (the remainder is 0, so 1 is a divisor of 4905)
4905 / 2 = 2452.5 (the remainder is 1, so 2 is not a divisor of 4905)
4905 / 3 = 1635 (the remainder is 0, so 3 is a divisor of 4905)
...
4905 / 4904 = 1.0002039151713 (the remainder is 1, so 4904 is not a divisor of 4905)
4905 / 4905 = 1 (the remainder is 0, so 4905 is a divisor of 4905)

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 4905 (i.e. 70.035705179573). 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 $\frac{N}{d} = D$ )

4905 / 1 = 4905 (the remainder is 0, so 1 and 4905 are divisors of 4905)
4905 / 2 = 2452.5 (the remainder is 1, so 2 is not a divisor of 4905)
4905 / 3 = 1635 (the remainder is 0, so 3 and 1635 are divisors of 4905)
...
4905 / 69 = 71.086956521739 (the remainder is 6, so 69 is not a divisor of 4905)
4905 / 70 = 70.071428571429 (the remainder is 5, so 70 is not a divisor of 4905)