What are the divisors of 4935?

1, 3, 5, 7, 15, 21, 35, 47, 105, 141, 235, 329, 705, 987, 1645, 4935

There is a total of 16 positive divisors.
The sum of these divisors is 9216.
The arithmetic mean is 576.

16 odd divisors

1, 3, 5, 7, 15, 21, 35, 47, 105, 141, 235, 329, 705, 987, 1645, 4935

How to compute the divisors of 4935?

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 4935 by each of the numbers from 1 to 4935 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)

4935 / 1 = 4935 (the remainder is 0, so 1 is a divisor of 4935)
4935 / 2 = 2467.5 (the remainder is 1, so 2 is not a divisor of 4935)
4935 / 3 = 1645 (the remainder is 0, so 3 is a divisor of 4935)
...
4935 / 4934 = 1.0002026753141 (the remainder is 1, so 4934 is not a divisor of 4935)
4935 / 4935 = 1 (the remainder is 0, so 4935 is a divisor of 4935)

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 4935 (i.e. 70.249555158734). 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 $\frac{N}{d} = D$ )

4935 / 1 = 4935 (the remainder is 0, so 1 and 4935 are divisors of 4935)
4935 / 2 = 2467.5 (the remainder is 1, so 2 is not a divisor of 4935)
4935 / 3 = 1645 (the remainder is 0, so 3 and 1645 are divisors of 4935)
...
4935 / 69 = 71.521739130435 (the remainder is 36, so 69 is not a divisor of 4935)
4935 / 70 = 70.5 (the remainder is 35, so 70 is not a divisor of 4935)