What are the divisors of 3381?

1, 3, 7, 21, 23, 49, 69, 147, 161, 483, 1127, 3381

There is a total of 12 positive divisors.
The sum of these divisors is 5472.
The arithmetic mean is 456.

12 odd divisors

1, 3, 7, 21, 23, 49, 69, 147, 161, 483, 1127, 3381

How to compute the divisors of 3381?

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 3381 by each of the numbers from 1 to 3381 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)

3381 / 1 = 3381 (the remainder is 0, so 1 is a divisor of 3381)
3381 / 2 = 1690.5 (the remainder is 1, so 2 is not a divisor of 3381)
3381 / 3 = 1127 (the remainder is 0, so 3 is a divisor of 3381)
...
3381 / 3380 = 1.0002958579882 (the remainder is 1, so 3380 is not a divisor of 3381)
3381 / 3381 = 1 (the remainder is 0, so 3381 is a divisor of 3381)

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 3381 (i.e. 58.146367040427). 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 $\frac{N}{d} = D$ )

3381 / 1 = 3381 (the remainder is 0, so 1 and 3381 are divisors of 3381)
3381 / 2 = 1690.5 (the remainder is 1, so 2 is not a divisor of 3381)
3381 / 3 = 1127 (the remainder is 0, so 3 and 1127 are divisors of 3381)
...
3381 / 57 = 59.315789473684 (the remainder is 18, so 57 is not a divisor of 3381)
3381 / 58 = 58.293103448276 (the remainder is 17, so 58 is not a divisor of 3381)