What are the divisors of 8217?

1, 3, 9, 11, 33, 83, 99, 249, 747, 913, 2739, 8217

There is a total of 12 positive divisors.
The sum of these divisors is 13104.
The arithmetic mean is 1092.

12 odd divisors

1, 3, 9, 11, 33, 83, 99, 249, 747, 913, 2739, 8217

How to compute the divisors of 8217?

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 8217 by each of the numbers from 1 to 8217 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)

8217 / 1 = 8217 (the remainder is 0, so 1 is a divisor of 8217)
8217 / 2 = 4108.5 (the remainder is 1, so 2 is not a divisor of 8217)
8217 / 3 = 2739 (the remainder is 0, so 3 is a divisor of 8217)
...
8217 / 8216 = 1.0001217137293 (the remainder is 1, so 8216 is not a divisor of 8217)
8217 / 8217 = 1 (the remainder is 0, so 8217 is a divisor of 8217)

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 8217 (i.e. 90.647669578429). 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 $\frac{N}{d} = D$ )

8217 / 1 = 8217 (the remainder is 0, so 1 and 8217 are divisors of 8217)
8217 / 2 = 4108.5 (the remainder is 1, so 2 is not a divisor of 8217)
8217 / 3 = 2739 (the remainder is 0, so 3 and 2739 are divisors of 8217)
...
8217 / 89 = 92.325842696629 (the remainder is 29, so 89 is not a divisor of 8217)
8217 / 90 = 91.3 (the remainder is 27, so 90 is not a divisor of 8217)