What are the divisors of 5517?

1, 3, 9, 613, 1839, 5517

There is a total of 6 positive divisors.
The sum of these divisors is 7982.
The arithmetic mean is 1330.3333333333.

6 odd divisors

1, 3, 9, 613, 1839, 5517

How to compute the divisors of 5517?

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 5517 by each of the numbers from 1 to 5517 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)

5517 / 1 = 5517 (the remainder is 0, so 1 is a divisor of 5517)
5517 / 2 = 2758.5 (the remainder is 1, so 2 is not a divisor of 5517)
5517 / 3 = 1839 (the remainder is 0, so 3 is a divisor of 5517)
...
5517 / 5516 = 1.0001812907904 (the remainder is 1, so 5516 is not a divisor of 5517)
5517 / 5517 = 1 (the remainder is 0, so 5517 is a divisor of 5517)

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 5517 (i.e. 74.27651041884). 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 $\frac{N}{d} = D$ )

5517 / 1 = 5517 (the remainder is 0, so 1 and 5517 are divisors of 5517)
5517 / 2 = 2758.5 (the remainder is 1, so 2 is not a divisor of 5517)
5517 / 3 = 1839 (the remainder is 0, so 3 and 1839 are divisors of 5517)
...
5517 / 73 = 75.575342465753 (the remainder is 42, so 73 is not a divisor of 5517)
5517 / 74 = 74.554054054054 (the remainder is 41, so 74 is not a divisor of 5517)