What are the divisors of 3403?

1, 41, 83, 3403

There is a total of 4 positive divisors.
The sum of these divisors is 3528.
The arithmetic mean is 882.

4 odd divisors

1, 41, 83, 3403

How to compute the divisors of 3403?

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 3403 by each of the numbers from 1 to 3403 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)

3403 / 1 = 3403 (the remainder is 0, so 1 is a divisor of 3403)
3403 / 2 = 1701.5 (the remainder is 1, so 2 is not a divisor of 3403)
3403 / 3 = 1134.3333333333 (the remainder is 1, so 3 is not a divisor of 3403)
...
3403 / 3402 = 1.0002939447384 (the remainder is 1, so 3402 is not a divisor of 3403)
3403 / 3403 = 1 (the remainder is 0, so 3403 is a divisor of 3403)

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 3403 (i.e. 58.335238064141). 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 $\frac{N}{d} = D$ )

3403 / 1 = 3403 (the remainder is 0, so 1 and 3403 are divisors of 3403)
3403 / 2 = 1701.5 (the remainder is 1, so 2 is not a divisor of 3403)
3403 / 3 = 1134.3333333333 (the remainder is 1, so 3 is not a divisor of 3403)
...
3403 / 57 = 59.701754385965 (the remainder is 40, so 57 is not a divisor of 3403)
3403 / 58 = 58.672413793103 (the remainder is 39, so 58 is not a divisor of 3403)