What are the divisors of 5342?

1, 2, 2671, 5342

There is a total of 4 positive divisors.
The sum of these divisors is 8016.
The arithmetic mean is 2004.

2 even divisors

2, 5342

2 odd divisors

1, 2671

How to compute the divisors of 5342?

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 5342 by each of the numbers from 1 to 5342 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)

5342 / 1 = 5342 (the remainder is 0, so 1 is a divisor of 5342)
5342 / 2 = 2671 (the remainder is 0, so 2 is a divisor of 5342)
5342 / 3 = 1780.6666666667 (the remainder is 2, so 3 is not a divisor of 5342)
...
5342 / 5341 = 1.0001872308556 (the remainder is 1, so 5341 is not a divisor of 5342)
5342 / 5342 = 1 (the remainder is 0, so 5342 is a divisor of 5342)

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 5342 (i.e. 73.088986858486). 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 $\frac{N}{d} = D$ )

5342 / 1 = 5342 (the remainder is 0, so 1 and 5342 are divisors of 5342)
5342 / 2 = 2671 (the remainder is 0, so 2 and 2671 are divisors of 5342)
5342 / 3 = 1780.6666666667 (the remainder is 2, so 3 is not a divisor of 5342)
...
5342 / 72 = 74.194444444444 (the remainder is 14, so 72 is not a divisor of 5342)
5342 / 73 = 73.178082191781 (the remainder is 13, so 73 is not a divisor of 5342)