What are the divisors of 4719?

1, 3, 11, 13, 33, 39, 121, 143, 363, 429, 1573, 4719

There is a total of 12 positive divisors.
The sum of these divisors is 7448.
The arithmetic mean is 620.66666666667.

12 odd divisors

1, 3, 11, 13, 33, 39, 121, 143, 363, 429, 1573, 4719

How to compute the divisors of 4719?

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 4719 by each of the numbers from 1 to 4719 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)

4719 / 1 = 4719 (the remainder is 0, so 1 is a divisor of 4719)
4719 / 2 = 2359.5 (the remainder is 1, so 2 is not a divisor of 4719)
4719 / 3 = 1573 (the remainder is 0, so 3 is a divisor of 4719)
...
4719 / 4718 = 1.0002119542179 (the remainder is 1, so 4718 is not a divisor of 4719)
4719 / 4719 = 1 (the remainder is 0, so 4719 is a divisor of 4719)

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 4719 (i.e. 68.694977982382). 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 $\frac{N}{d} = D$ )

4719 / 1 = 4719 (the remainder is 0, so 1 and 4719 are divisors of 4719)
4719 / 2 = 2359.5 (the remainder is 1, so 2 is not a divisor of 4719)
4719 / 3 = 1573 (the remainder is 0, so 3 and 1573 are divisors of 4719)
...
4719 / 67 = 70.432835820896 (the remainder is 29, so 67 is not a divisor of 4719)
4719 / 68 = 69.397058823529 (the remainder is 27, so 68 is not a divisor of 4719)