What are the divisors of 8253?

1, 3, 7, 9, 21, 63, 131, 393, 917, 1179, 2751, 8253

There is a total of 12 positive divisors.
The sum of these divisors is 13728.
The arithmetic mean is 1144.

12 odd divisors

1, 3, 7, 9, 21, 63, 131, 393, 917, 1179, 2751, 8253

How to compute the divisors of 8253?

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 8253 by each of the numbers from 1 to 8253 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)

8253 / 1 = 8253 (the remainder is 0, so 1 is a divisor of 8253)
8253 / 2 = 4126.5 (the remainder is 1, so 2 is not a divisor of 8253)
8253 / 3 = 2751 (the remainder is 0, so 3 is a divisor of 8253)
...
8253 / 8252 = 1.0001211827436 (the remainder is 1, so 8252 is not a divisor of 8253)
8253 / 8253 = 1 (the remainder is 0, so 8253 is a divisor of 8253)

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 8253 (i.e. 90.84602357836). 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 $\frac{N}{d} = D$ )

8253 / 1 = 8253 (the remainder is 0, so 1 and 8253 are divisors of 8253)
8253 / 2 = 4126.5 (the remainder is 1, so 2 is not a divisor of 8253)
8253 / 3 = 2751 (the remainder is 0, so 3 and 2751 are divisors of 8253)
...
8253 / 89 = 92.730337078652 (the remainder is 65, so 89 is not a divisor of 8253)
8253 / 90 = 91.7 (the remainder is 63, so 90 is not a divisor of 8253)