What are the divisors of 5607?

1, 3, 7, 9, 21, 63, 89, 267, 623, 801, 1869, 5607

There is a total of 12 positive divisors.
The sum of these divisors is 9360.
The arithmetic mean is 780.

12 odd divisors

1, 3, 7, 9, 21, 63, 89, 267, 623, 801, 1869, 5607

How to compute the divisors of 5607?

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 5607 by each of the numbers from 1 to 5607 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)

5607 / 1 = 5607 (the remainder is 0, so 1 is a divisor of 5607)
5607 / 2 = 2803.5 (the remainder is 1, so 2 is not a divisor of 5607)
5607 / 3 = 1869 (the remainder is 0, so 3 is a divisor of 5607)
...
5607 / 5606 = 1.0001783803068 (the remainder is 1, so 5606 is not a divisor of 5607)
5607 / 5607 = 1 (the remainder is 0, so 5607 is a divisor of 5607)

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 5607 (i.e. 74.879903846092). 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 $\frac{N}{d} = D$ )

5607 / 1 = 5607 (the remainder is 0, so 1 and 5607 are divisors of 5607)
5607 / 2 = 2803.5 (the remainder is 1, so 2 is not a divisor of 5607)
5607 / 3 = 1869 (the remainder is 0, so 3 and 1869 are divisors of 5607)
...
5607 / 73 = 76.808219178082 (the remainder is 59, so 73 is not a divisor of 5607)
5607 / 74 = 75.77027027027 (the remainder is 57, so 74 is not a divisor of 5607)