What are the divisors of 9207?

1, 3, 9, 11, 27, 31, 33, 93, 99, 279, 297, 341, 837, 1023, 3069, 9207

There is a total of 16 positive divisors.
The sum of these divisors is 15360.
The arithmetic mean is 960.

16 odd divisors

1, 3, 9, 11, 27, 31, 33, 93, 99, 279, 297, 341, 837, 1023, 3069, 9207

How to compute the divisors of 9207?

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 9207 by each of the numbers from 1 to 9207 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)

9207 / 1 = 9207 (the remainder is 0, so 1 is a divisor of 9207)
9207 / 2 = 4603.5 (the remainder is 1, so 2 is not a divisor of 9207)
9207 / 3 = 3069 (the remainder is 0, so 3 is a divisor of 9207)
...
9207 / 9206 = 1.0001086248099 (the remainder is 1, so 9206 is not a divisor of 9207)
9207 / 9207 = 1 (the remainder is 0, so 9207 is a divisor of 9207)

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 9207 (i.e. 95.953113550317). 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 $\frac{N}{d} = D$ )

9207 / 1 = 9207 (the remainder is 0, so 1 and 9207 are divisors of 9207)
9207 / 2 = 4603.5 (the remainder is 1, so 2 is not a divisor of 9207)
9207 / 3 = 3069 (the remainder is 0, so 3 and 3069 are divisors of 9207)
...
9207 / 94 = 97.946808510638 (the remainder is 89, so 94 is not a divisor of 9207)
9207 / 95 = 96.915789473684 (the remainder is 87, so 95 is not a divisor of 9207)