What are the divisors of 117?

1, 3, 9, 13, 39, 117

There is a total of 6 positive divisors.
The sum of these divisors is 182.
The arithmetic mean is 30.333333333333.

6 odd divisors

1, 3, 9, 13, 39, 117

How to compute the divisors of 117?

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 117 by each of the numbers from 1 to 117 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)

117 / 1 = 117 (the remainder is 0, so 1 is a divisor of 117)
117 / 2 = 58.5 (the remainder is 1, so 2 is not a divisor of 117)
117 / 3 = 39 (the remainder is 0, so 3 is a divisor of 117)
...
117 / 116 = 1.0086206896552 (the remainder is 1, so 116 is not a divisor of 117)
117 / 117 = 1 (the remainder is 0, so 117 is a divisor of 117)

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 117 (i.e. 10.816653826392). 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 $\frac{N}{d} = D$ )

117 / 1 = 117 (the remainder is 0, so 1 and 117 are divisors of 117)
117 / 2 = 58.5 (the remainder is 1, so 2 is not a divisor of 117)
117 / 3 = 39 (the remainder is 0, so 3 and 39 are divisors of 117)
...
117 / 9 = 13 (the remainder is 0, so 9 and 13 are divisors of 117)
117 / 10 = 11.7 (the remainder is 7, so 10 is not a divisor of 117)