What are the divisors of 4743?

1, 3, 9, 17, 31, 51, 93, 153, 279, 527, 1581, 4743

There is a total of 12 positive divisors.
The sum of these divisors is 7488.
The arithmetic mean is 624.

12 odd divisors

1, 3, 9, 17, 31, 51, 93, 153, 279, 527, 1581, 4743

How to compute the divisors of 4743?

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 4743 by each of the numbers from 1 to 4743 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)

4743 / 1 = 4743 (the remainder is 0, so 1 is a divisor of 4743)
4743 / 2 = 2371.5 (the remainder is 1, so 2 is not a divisor of 4743)
4743 / 3 = 1581 (the remainder is 0, so 3 is a divisor of 4743)
...
4743 / 4742 = 1.0002108814846 (the remainder is 1, so 4742 is not a divisor of 4743)
4743 / 4743 = 1 (the remainder is 0, so 4743 is a divisor of 4743)

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 4743 (i.e. 68.869441699494). 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 $\frac{N}{d} = D$ )

4743 / 1 = 4743 (the remainder is 0, so 1 and 4743 are divisors of 4743)
4743 / 2 = 2371.5 (the remainder is 1, so 2 is not a divisor of 4743)
4743 / 3 = 1581 (the remainder is 0, so 3 and 1581 are divisors of 4743)
...
4743 / 67 = 70.791044776119 (the remainder is 53, so 67 is not a divisor of 4743)
4743 / 68 = 69.75 (the remainder is 51, so 68 is not a divisor of 4743)