What are the divisors of 5403?

1, 3, 1801, 5403

There is a total of 4 positive divisors.
The sum of these divisors is 7208.
The arithmetic mean is 1802.

4 odd divisors

1, 3, 1801, 5403

How to compute the divisors of 5403?

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 5403 by each of the numbers from 1 to 5403 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)

5403 / 1 = 5403 (the remainder is 0, so 1 is a divisor of 5403)
5403 / 2 = 2701.5 (the remainder is 1, so 2 is not a divisor of 5403)
5403 / 3 = 1801 (the remainder is 0, so 3 is a divisor of 5403)
...
5403 / 5402 = 1.0001851166235 (the remainder is 1, so 5402 is not a divisor of 5403)
5403 / 5403 = 1 (the remainder is 0, so 5403 is a divisor of 5403)

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 5403 (i.e. 73.505101863748). 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 $\frac{N}{d} = D$ )

5403 / 1 = 5403 (the remainder is 0, so 1 and 5403 are divisors of 5403)
5403 / 2 = 2701.5 (the remainder is 1, so 2 is not a divisor of 5403)
5403 / 3 = 1801 (the remainder is 0, so 3 and 1801 are divisors of 5403)
...
5403 / 72 = 75.041666666667 (the remainder is 3, so 72 is not a divisor of 5403)
5403 / 73 = 74.013698630137 (the remainder is 1, so 73 is not a divisor of 5403)