What are the divisors of 3483?

1, 3, 9, 27, 43, 81, 129, 387, 1161, 3483

There is a total of 10 positive divisors.
The sum of these divisors is 5324.
The arithmetic mean is 532.4.

10 odd divisors

1, 3, 9, 27, 43, 81, 129, 387, 1161, 3483

How to compute the divisors of 3483?

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 3483 by each of the numbers from 1 to 3483 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)

3483 / 1 = 3483 (the remainder is 0, so 1 is a divisor of 3483)
3483 / 2 = 1741.5 (the remainder is 1, so 2 is not a divisor of 3483)
3483 / 3 = 1161 (the remainder is 0, so 3 is a divisor of 3483)
...
3483 / 3482 = 1.0002871912694 (the remainder is 1, so 3482 is not a divisor of 3483)
3483 / 3483 = 1 (the remainder is 0, so 3483 is a divisor of 3483)

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 3483 (i.e. 59.016946718718). 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 $\frac{N}{d} = D$ )

3483 / 1 = 3483 (the remainder is 0, so 1 and 3483 are divisors of 3483)
3483 / 2 = 1741.5 (the remainder is 1, so 2 is not a divisor of 3483)
3483 / 3 = 1161 (the remainder is 0, so 3 and 1161 are divisors of 3483)
...
3483 / 58 = 60.051724137931 (the remainder is 3, so 58 is not a divisor of 3483)
3483 / 59 = 59.033898305085 (the remainder is 2, so 59 is not a divisor of 3483)