What are the divisors of 4923?

1, 3, 9, 547, 1641, 4923

There is a total of 6 positive divisors.
The sum of these divisors is 7124.
The arithmetic mean is 1187.3333333333.

6 odd divisors

1, 3, 9, 547, 1641, 4923

How to compute the divisors of 4923?

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 4923 by each of the numbers from 1 to 4923 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)

4923 / 1 = 4923 (the remainder is 0, so 1 is a divisor of 4923)
4923 / 2 = 2461.5 (the remainder is 1, so 2 is not a divisor of 4923)
4923 / 3 = 1641 (the remainder is 0, so 3 is a divisor of 4923)
...
4923 / 4922 = 1.0002031694433 (the remainder is 1, so 4922 is not a divisor of 4923)
4923 / 4923 = 1 (the remainder is 0, so 4923 is a divisor of 4923)

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 4923 (i.e. 70.164093381159). 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 $\frac{N}{d} = D$ )

4923 / 1 = 4923 (the remainder is 0, so 1 and 4923 are divisors of 4923)
4923 / 2 = 2461.5 (the remainder is 1, so 2 is not a divisor of 4923)
4923 / 3 = 1641 (the remainder is 0, so 3 and 1641 are divisors of 4923)
...
4923 / 69 = 71.347826086957 (the remainder is 24, so 69 is not a divisor of 4923)
4923 / 70 = 70.328571428571 (the remainder is 23, so 70 is not a divisor of 4923)