What are the divisors of 102?

1, 2, 3, 6, 17, 34, 51, 102

There is a total of 8 positive divisors.
The sum of these divisors is 216.
The arithmetic mean is 27.

4 even divisors

2, 6, 34, 102

4 odd divisors

1, 3, 17, 51

How to compute the divisors of 102?

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 102 by each of the numbers from 1 to 102 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)

102 / 1 = 102 (the remainder is 0, so 1 is a divisor of 102)
102 / 2 = 51 (the remainder is 0, so 2 is a divisor of 102)
102 / 3 = 34 (the remainder is 0, so 3 is a divisor of 102)
...
102 / 101 = 1.009900990099 (the remainder is 1, so 101 is not a divisor of 102)
102 / 102 = 1 (the remainder is 0, so 102 is a divisor of 102)

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 102 (i.e. 10.099504938362). 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 $\frac{N}{d} = D$ )

102 / 1 = 102 (the remainder is 0, so 1 and 102 are divisors of 102)
102 / 2 = 51 (the remainder is 0, so 2 and 51 are divisors of 102)
102 / 3 = 34 (the remainder is 0, so 3 and 34 are divisors of 102)
...
102 / 9 = 11.333333333333 (the remainder is 3, so 9 is not a divisor of 102)
102 / 10 = 10.2 (the remainder is 2, so 10 is not a divisor of 102)