What are the divisors of 5415?

1, 3, 5, 15, 19, 57, 95, 285, 361, 1083, 1805, 5415

There is a total of 12 positive divisors.
The sum of these divisors is 9144.
The arithmetic mean is 762.

12 odd divisors

1, 3, 5, 15, 19, 57, 95, 285, 361, 1083, 1805, 5415

How to compute the divisors of 5415?

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 5415 by each of the numbers from 1 to 5415 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)

5415 / 1 = 5415 (the remainder is 0, so 1 is a divisor of 5415)
5415 / 2 = 2707.5 (the remainder is 1, so 2 is not a divisor of 5415)
5415 / 3 = 1805 (the remainder is 0, so 3 is a divisor of 5415)
...
5415 / 5414 = 1.000184706317 (the remainder is 1, so 5414 is not a divisor of 5415)
5415 / 5415 = 1 (the remainder is 0, so 5415 is a divisor of 5415)

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 5415 (i.e. 73.586683577941). 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 $\frac{N}{d} = D$ )

5415 / 1 = 5415 (the remainder is 0, so 1 and 5415 are divisors of 5415)
5415 / 2 = 2707.5 (the remainder is 1, so 2 is not a divisor of 5415)
5415 / 3 = 1805 (the remainder is 0, so 3 and 1805 are divisors of 5415)
...
5415 / 72 = 75.208333333333 (the remainder is 15, so 72 is not a divisor of 5415)
5415 / 73 = 74.178082191781 (the remainder is 13, so 73 is not a divisor of 5415)