What are the divisors of 5412?

1, 2, 3, 4, 6, 11, 12, 22, 33, 41, 44, 66, 82, 123, 132, 164, 246, 451, 492, 902, 1353, 1804, 2706, 5412

There is a total of 24 positive divisors.
The sum of these divisors is 14112.
The arithmetic mean is 588.

16 even divisors

2, 4, 6, 12, 22, 44, 66, 82, 132, 164, 246, 492, 902, 1804, 2706, 5412

8 odd divisors

1, 3, 11, 33, 41, 123, 451, 1353

How to compute the divisors of 5412?

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 5412 by each of the numbers from 1 to 5412 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)

5412 / 1 = 5412 (the remainder is 0, so 1 is a divisor of 5412)
5412 / 2 = 2706 (the remainder is 0, so 2 is a divisor of 5412)
5412 / 3 = 1804 (the remainder is 0, so 3 is a divisor of 5412)
...
5412 / 5411 = 1.000184808723 (the remainder is 1, so 5411 is not a divisor of 5412)
5412 / 5412 = 1 (the remainder is 0, so 5412 is a divisor of 5412)

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 5412 (i.e. 73.566296630998). 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 $\frac{N}{d} = D$ )

5412 / 1 = 5412 (the remainder is 0, so 1 and 5412 are divisors of 5412)
5412 / 2 = 2706 (the remainder is 0, so 2 and 2706 are divisors of 5412)
5412 / 3 = 1804 (the remainder is 0, so 3 and 1804 are divisors of 5412)
...
5412 / 72 = 75.166666666667 (the remainder is 12, so 72 is not a divisor of 5412)
5412 / 73 = 74.13698630137 (the remainder is 10, so 73 is not a divisor of 5412)