What are the divisors of 9117?

1, 3, 9, 1013, 3039, 9117

There is a total of 6 positive divisors.
The sum of these divisors is 13182.
The arithmetic mean is 2197.

6 odd divisors

1, 3, 9, 1013, 3039, 9117

How to compute the divisors of 9117?

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 9117 by each of the numbers from 1 to 9117 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)

9117 / 1 = 9117 (the remainder is 0, so 1 is a divisor of 9117)
9117 / 2 = 4558.5 (the remainder is 1, so 2 is not a divisor of 9117)
9117 / 3 = 3039 (the remainder is 0, so 3 is a divisor of 9117)
...
9117 / 9116 = 1.0001096972356 (the remainder is 1, so 9116 is not a divisor of 9117)
9117 / 9117 = 1 (the remainder is 0, so 9117 is a divisor of 9117)

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 9117 (i.e. 95.482982777037). 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 $\frac{N}{d} = D$ )

9117 / 1 = 9117 (the remainder is 0, so 1 and 9117 are divisors of 9117)
9117 / 2 = 4558.5 (the remainder is 1, so 2 is not a divisor of 9117)
9117 / 3 = 3039 (the remainder is 0, so 3 and 3039 are divisors of 9117)
...
9117 / 94 = 96.989361702128 (the remainder is 93, so 94 is not a divisor of 9117)
9117 / 95 = 95.968421052632 (the remainder is 92, so 95 is not a divisor of 9117)