What are the divisors of 3490?

1, 2, 5, 10, 349, 698, 1745, 3490

There is a total of 8 positive divisors.
The sum of these divisors is 6300.
The arithmetic mean is 787.5.

4 even divisors

2, 10, 698, 3490

4 odd divisors

1, 5, 349, 1745

How to compute the divisors of 3490?

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 3490 by each of the numbers from 1 to 3490 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)

3490 / 1 = 3490 (the remainder is 0, so 1 is a divisor of 3490)
3490 / 2 = 1745 (the remainder is 0, so 2 is a divisor of 3490)
3490 / 3 = 1163.3333333333 (the remainder is 1, so 3 is not a divisor of 3490)
...
3490 / 3489 = 1.000286615076 (the remainder is 1, so 3489 is not a divisor of 3490)
3490 / 3490 = 1 (the remainder is 0, so 3490 is a divisor of 3490)

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 3490 (i.e. 59.076221950968). 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 $\frac{N}{d} = D$ )

3490 / 1 = 3490 (the remainder is 0, so 1 and 3490 are divisors of 3490)
3490 / 2 = 1745 (the remainder is 0, so 2 and 1745 are divisors of 3490)
3490 / 3 = 1163.3333333333 (the remainder is 1, so 3 is not a divisor of 3490)
...
3490 / 58 = 60.172413793103 (the remainder is 10, so 58 is not a divisor of 3490)
3490 / 59 = 59.152542372881 (the remainder is 9, so 59 is not a divisor of 3490)