What are the divisors of 3470?

1, 2, 5, 10, 347, 694, 1735, 3470

There is a total of 8 positive divisors.
The sum of these divisors is 6264.
The arithmetic mean is 783.

4 even divisors

2, 10, 694, 3470

4 odd divisors

1, 5, 347, 1735

How to compute the divisors of 3470?

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 3470 by each of the numbers from 1 to 3470 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)

3470 / 1 = 3470 (the remainder is 0, so 1 is a divisor of 3470)
3470 / 2 = 1735 (the remainder is 0, so 2 is a divisor of 3470)
3470 / 3 = 1156.6666666667 (the remainder is 2, so 3 is not a divisor of 3470)
...
3470 / 3469 = 1.0002882675123 (the remainder is 1, so 3469 is not a divisor of 3470)
3470 / 3470 = 1 (the remainder is 0, so 3470 is a divisor of 3470)

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 3470 (i.e. 58.906705900093). 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 $\frac{N}{d} = D$ )

3470 / 1 = 3470 (the remainder is 0, so 1 and 3470 are divisors of 3470)
3470 / 2 = 1735 (the remainder is 0, so 2 and 1735 are divisors of 3470)
3470 / 3 = 1156.6666666667 (the remainder is 2, so 3 is not a divisor of 3470)
...
3470 / 57 = 60.877192982456 (the remainder is 50, so 57 is not a divisor of 3470)
3470 / 58 = 59.827586206897 (the remainder is 48, so 58 is not a divisor of 3470)