What are the divisors of 5913?

1, 3, 9, 27, 73, 81, 219, 657, 1971, 5913

There is a total of 10 positive divisors.
The sum of these divisors is 8954.
The arithmetic mean is 895.4.

10 odd divisors

1, 3, 9, 27, 73, 81, 219, 657, 1971, 5913

How to compute the divisors of 5913?

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 5913 by each of the numbers from 1 to 5913 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)

5913 / 1 = 5913 (the remainder is 0, so 1 is a divisor of 5913)
5913 / 2 = 2956.5 (the remainder is 1, so 2 is not a divisor of 5913)
5913 / 3 = 1971 (the remainder is 0, so 3 is a divisor of 5913)
...
5913 / 5912 = 1.0001691474966 (the remainder is 1, so 5912 is not a divisor of 5913)
5913 / 5913 = 1 (the remainder is 0, so 5913 is a divisor of 5913)

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 5913 (i.e. 76.896033707858). 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 $\frac{N}{d} = D$ )

5913 / 1 = 5913 (the remainder is 0, so 1 and 5913 are divisors of 5913)
5913 / 2 = 2956.5 (the remainder is 1, so 2 is not a divisor of 5913)
5913 / 3 = 1971 (the remainder is 0, so 3 and 1971 are divisors of 5913)
...
5913 / 75 = 78.84 (the remainder is 63, so 75 is not a divisor of 5913)
5913 / 76 = 77.802631578947 (the remainder is 61, so 76 is not a divisor of 5913)