What are the divisors of 8262?

1, 2, 3, 6, 9, 17, 18, 27, 34, 51, 54, 81, 102, 153, 162, 243, 306, 459, 486, 918, 1377, 2754, 4131, 8262

There is a total of 24 positive divisors.
The sum of these divisors is 19656.
The arithmetic mean is 819.

12 even divisors

2, 6, 18, 34, 54, 102, 162, 306, 486, 918, 2754, 8262

12 odd divisors

1, 3, 9, 17, 27, 51, 81, 153, 243, 459, 1377, 4131

How to compute the divisors of 8262?

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 8262 by each of the numbers from 1 to 8262 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)

8262 / 1 = 8262 (the remainder is 0, so 1 is a divisor of 8262)
8262 / 2 = 4131 (the remainder is 0, so 2 is a divisor of 8262)
8262 / 3 = 2754 (the remainder is 0, so 3 is a divisor of 8262)
...
8262 / 8261 = 1.0001210507203 (the remainder is 1, so 8261 is not a divisor of 8262)
8262 / 8262 = 1 (the remainder is 0, so 8262 is a divisor of 8262)

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 8262 (i.e. 90.895544445259). 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 $\frac{N}{d} = D$ )

8262 / 1 = 8262 (the remainder is 0, so 1 and 8262 are divisors of 8262)
8262 / 2 = 4131 (the remainder is 0, so 2 and 4131 are divisors of 8262)
8262 / 3 = 2754 (the remainder is 0, so 3 and 2754 are divisors of 8262)
...
8262 / 89 = 92.831460674157 (the remainder is 74, so 89 is not a divisor of 8262)
8262 / 90 = 91.8 (the remainder is 72, so 90 is not a divisor of 8262)