What are the divisors of 3398?

1, 2, 1699, 3398

There is a total of 4 positive divisors.
The sum of these divisors is 5100.
The arithmetic mean is 1275.

2 even divisors

2, 3398

2 odd divisors

1, 1699

How to compute the divisors of 3398?

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 3398 by each of the numbers from 1 to 3398 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)

3398 / 1 = 3398 (the remainder is 0, so 1 is a divisor of 3398)
3398 / 2 = 1699 (the remainder is 0, so 2 is a divisor of 3398)
3398 / 3 = 1132.6666666667 (the remainder is 2, so 3 is not a divisor of 3398)
...
3398 / 3397 = 1.0002943773918 (the remainder is 1, so 3397 is not a divisor of 3398)
3398 / 3398 = 1 (the remainder is 0, so 3398 is a divisor of 3398)

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 3398 (i.e. 58.292366567159). 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 $\frac{N}{d} = D$ )

3398 / 1 = 3398 (the remainder is 0, so 1 and 3398 are divisors of 3398)
3398 / 2 = 1699 (the remainder is 0, so 2 and 1699 are divisors of 3398)
3398 / 3 = 1132.6666666667 (the remainder is 2, so 3 is not a divisor of 3398)
...
3398 / 57 = 59.614035087719 (the remainder is 35, so 57 is not a divisor of 3398)
3398 / 58 = 58.586206896552 (the remainder is 34, so 58 is not a divisor of 3398)