What are the divisors of 9153?

1, 3, 9, 27, 81, 113, 339, 1017, 3051, 9153

There is a total of 10 positive divisors.
The sum of these divisors is 13794.
The arithmetic mean is 1379.4.

10 odd divisors

1, 3, 9, 27, 81, 113, 339, 1017, 3051, 9153

How to compute the divisors of 9153?

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 9153 by each of the numbers from 1 to 9153 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)

9153 / 1 = 9153 (the remainder is 0, so 1 is a divisor of 9153)
9153 / 2 = 4576.5 (the remainder is 1, so 2 is not a divisor of 9153)
9153 / 3 = 3051 (the remainder is 0, so 3 is a divisor of 9153)
...
9153 / 9152 = 1.0001092657343 (the remainder is 1, so 9152 is not a divisor of 9153)
9153 / 9153 = 1 (the remainder is 0, so 9153 is a divisor of 9153)

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 9153 (i.e. 95.671312314612). 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 $\frac{N}{d} = D$ )

9153 / 1 = 9153 (the remainder is 0, so 1 and 9153 are divisors of 9153)
9153 / 2 = 4576.5 (the remainder is 1, so 2 is not a divisor of 9153)
9153 / 3 = 3051 (the remainder is 0, so 3 and 3051 are divisors of 9153)
...
9153 / 94 = 97.372340425532 (the remainder is 35, so 94 is not a divisor of 9153)
9153 / 95 = 96.347368421053 (the remainder is 33, so 95 is not a divisor of 9153)