What are the divisors of 3102?

1, 2, 3, 6, 11, 22, 33, 47, 66, 94, 141, 282, 517, 1034, 1551, 3102

There is a total of 16 positive divisors.
The sum of these divisors is 6912.
The arithmetic mean is 432.

8 even divisors

2, 6, 22, 66, 94, 282, 1034, 3102

8 odd divisors

1, 3, 11, 33, 47, 141, 517, 1551

How to compute the divisors of 3102?

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 3102 by each of the numbers from 1 to 3102 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)

3102 / 1 = 3102 (the remainder is 0, so 1 is a divisor of 3102)
3102 / 2 = 1551 (the remainder is 0, so 2 is a divisor of 3102)
3102 / 3 = 1034 (the remainder is 0, so 3 is a divisor of 3102)
...
3102 / 3101 = 1.0003224766204 (the remainder is 1, so 3101 is not a divisor of 3102)
3102 / 3102 = 1 (the remainder is 0, so 3102 is a divisor of 3102)

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 3102 (i.e. 55.695601262577). 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 $\frac{N}{d} = D$ )

3102 / 1 = 3102 (the remainder is 0, so 1 and 3102 are divisors of 3102)
3102 / 2 = 1551 (the remainder is 0, so 2 and 1551 are divisors of 3102)
3102 / 3 = 1034 (the remainder is 0, so 3 and 1034 are divisors of 3102)
...
3102 / 54 = 57.444444444444 (the remainder is 24, so 54 is not a divisor of 3102)
3102 / 55 = 56.4 (the remainder is 22, so 55 is not a divisor of 3102)