What are the divisors of 4805?

1, 5, 31, 155, 961, 4805

There is a total of 6 positive divisors.
The sum of these divisors is 5958.
The arithmetic mean is 993.

6 odd divisors

1, 5, 31, 155, 961, 4805

How to compute the divisors of 4805?

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 4805 by each of the numbers from 1 to 4805 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)

4805 / 1 = 4805 (the remainder is 0, so 1 is a divisor of 4805)
4805 / 2 = 2402.5 (the remainder is 1, so 2 is not a divisor of 4805)
4805 / 3 = 1601.6666666667 (the remainder is 2, so 3 is not a divisor of 4805)
...
4805 / 4804 = 1.0002081598668 (the remainder is 1, so 4804 is not a divisor of 4805)
4805 / 4805 = 1 (the remainder is 0, so 4805 is a divisor of 4805)

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 4805 (i.e. 69.318107302493). 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 $\frac{N}{d} = D$ )

4805 / 1 = 4805 (the remainder is 0, so 1 and 4805 are divisors of 4805)
4805 / 2 = 2402.5 (the remainder is 1, so 2 is not a divisor of 4805)
4805 / 3 = 1601.6666666667 (the remainder is 2, so 3 is not a divisor of 4805)
...
4805 / 68 = 70.661764705882 (the remainder is 45, so 68 is not a divisor of 4805)
4805 / 69 = 69.63768115942 (the remainder is 44, so 69 is not a divisor of 4805)