What are the divisors of 3402?

1, 2, 3, 6, 7, 9, 14, 18, 21, 27, 42, 54, 63, 81, 126, 162, 189, 243, 378, 486, 567, 1134, 1701, 3402

There is a total of 24 positive divisors.
The sum of these divisors is 8736.
The arithmetic mean is 364.

12 even divisors

2, 6, 14, 18, 42, 54, 126, 162, 378, 486, 1134, 3402

12 odd divisors

1, 3, 7, 9, 21, 27, 63, 81, 189, 243, 567, 1701

How to compute the divisors of 3402?

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 3402 by each of the numbers from 1 to 3402 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)

3402 / 1 = 3402 (the remainder is 0, so 1 is a divisor of 3402)
3402 / 2 = 1701 (the remainder is 0, so 2 is a divisor of 3402)
3402 / 3 = 1134 (the remainder is 0, so 3 is a divisor of 3402)
...
3402 / 3401 = 1.0002940311673 (the remainder is 1, so 3401 is not a divisor of 3402)
3402 / 3402 = 1 (the remainder is 0, so 3402 is a divisor of 3402)

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 3402 (i.e. 58.326666285671). 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 $\frac{N}{d} = D$ )

3402 / 1 = 3402 (the remainder is 0, so 1 and 3402 are divisors of 3402)
3402 / 2 = 1701 (the remainder is 0, so 2 and 1701 are divisors of 3402)
3402 / 3 = 1134 (the remainder is 0, so 3 and 1134 are divisors of 3402)
...
3402 / 57 = 59.684210526316 (the remainder is 39, so 57 is not a divisor of 3402)
3402 / 58 = 58.655172413793 (the remainder is 38, so 58 is not a divisor of 3402)