What are the divisors of 3492?

1, 2, 3, 4, 6, 9, 12, 18, 36, 97, 194, 291, 388, 582, 873, 1164, 1746, 3492

There is a total of 18 positive divisors.
The sum of these divisors is 8918.
The arithmetic mean is 495.44444444444.

12 even divisors

2, 4, 6, 12, 18, 36, 194, 388, 582, 1164, 1746, 3492

6 odd divisors

1, 3, 9, 97, 291, 873

How to compute the divisors of 3492?

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 3492 by each of the numbers from 1 to 3492 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)

3492 / 1 = 3492 (the remainder is 0, so 1 is a divisor of 3492)
3492 / 2 = 1746 (the remainder is 0, so 2 is a divisor of 3492)
3492 / 3 = 1164 (the remainder is 0, so 3 is a divisor of 3492)
...
3492 / 3491 = 1.0002864508737 (the remainder is 1, so 3491 is not a divisor of 3492)
3492 / 3492 = 1 (the remainder is 0, so 3492 is a divisor of 3492)

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 3492 (i.e. 59.093146810777). 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 $\frac{N}{d} = D$ )

3492 / 1 = 3492 (the remainder is 0, so 1 and 3492 are divisors of 3492)
3492 / 2 = 1746 (the remainder is 0, so 2 and 1746 are divisors of 3492)
3492 / 3 = 1164 (the remainder is 0, so 3 and 1164 are divisors of 3492)
...
3492 / 58 = 60.206896551724 (the remainder is 12, so 58 is not a divisor of 3492)
3492 / 59 = 59.186440677966 (the remainder is 11, so 59 is not a divisor of 3492)