What are the divisors of 3410?

1, 2, 5, 10, 11, 22, 31, 55, 62, 110, 155, 310, 341, 682, 1705, 3410

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

8 even divisors

2, 10, 22, 62, 110, 310, 682, 3410

8 odd divisors

1, 5, 11, 31, 55, 155, 341, 1705

How to compute the divisors of 3410?

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 3410 by each of the numbers from 1 to 3410 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)

3410 / 1 = 3410 (the remainder is 0, so 1 is a divisor of 3410)
3410 / 2 = 1705 (the remainder is 0, so 2 is a divisor of 3410)
3410 / 3 = 1136.6666666667 (the remainder is 2, so 3 is not a divisor of 3410)
...
3410 / 3409 = 1.0002933411558 (the remainder is 1, so 3409 is not a divisor of 3410)
3410 / 3410 = 1 (the remainder is 0, so 3410 is a divisor of 3410)

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 3410 (i.e. 58.395205282626). 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 $\frac{N}{d} = D$ )

3410 / 1 = 3410 (the remainder is 0, so 1 and 3410 are divisors of 3410)
3410 / 2 = 1705 (the remainder is 0, so 2 and 1705 are divisors of 3410)
3410 / 3 = 1136.6666666667 (the remainder is 2, so 3 is not a divisor of 3410)
...
3410 / 57 = 59.824561403509 (the remainder is 47, so 57 is not a divisor of 3410)
3410 / 58 = 58.793103448276 (the remainder is 46, so 58 is not a divisor of 3410)