What are the divisors of 4758?

1, 2, 3, 6, 13, 26, 39, 61, 78, 122, 183, 366, 793, 1586, 2379, 4758

There is a total of 16 positive divisors.
The sum of these divisors is 10416.
The arithmetic mean is 651.

8 even divisors

2, 6, 26, 78, 122, 366, 1586, 4758

8 odd divisors

1, 3, 13, 39, 61, 183, 793, 2379

How to compute the divisors of 4758?

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 4758 by each of the numbers from 1 to 4758 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)

4758 / 1 = 4758 (the remainder is 0, so 1 is a divisor of 4758)
4758 / 2 = 2379 (the remainder is 0, so 2 is a divisor of 4758)
4758 / 3 = 1586 (the remainder is 0, so 3 is a divisor of 4758)
...
4758 / 4757 = 1.000210216523 (the remainder is 1, so 4757 is not a divisor of 4758)
4758 / 4758 = 1 (the remainder is 0, so 4758 is a divisor of 4758)

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 4758 (i.e. 68.978257443922). 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 $\frac{N}{d} = D$ )

4758 / 1 = 4758 (the remainder is 0, so 1 and 4758 are divisors of 4758)
4758 / 2 = 2379 (the remainder is 0, so 2 and 2379 are divisors of 4758)
4758 / 3 = 1586 (the remainder is 0, so 3 and 1586 are divisors of 4758)
...
4758 / 67 = 71.014925373134 (the remainder is 1, so 67 is not a divisor of 4758)
4758 / 68 = 69.970588235294 (the remainder is 66, so 68 is not a divisor of 4758)