What are the divisors of 3518?

1, 2, 1759, 3518

There is a total of 4 positive divisors.
The sum of these divisors is 5280.
The arithmetic mean is 1320.

2 even divisors

2, 3518

2 odd divisors

1, 1759

How to compute the divisors of 3518?

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 3518 by each of the numbers from 1 to 3518 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)

3518 / 1 = 3518 (the remainder is 0, so 1 is a divisor of 3518)
3518 / 2 = 1759 (the remainder is 0, so 2 is a divisor of 3518)
3518 / 3 = 1172.6666666667 (the remainder is 2, so 3 is not a divisor of 3518)
...
3518 / 3517 = 1.0002843332386 (the remainder is 1, so 3517 is not a divisor of 3518)
3518 / 3518 = 1 (the remainder is 0, so 3518 is a divisor of 3518)

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 3518 (i.e. 59.312730505348). 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 $\frac{N}{d} = D$ )

3518 / 1 = 3518 (the remainder is 0, so 1 and 3518 are divisors of 3518)
3518 / 2 = 1759 (the remainder is 0, so 2 and 1759 are divisors of 3518)
3518 / 3 = 1172.6666666667 (the remainder is 2, so 3 is not a divisor of 3518)
...
3518 / 58 = 60.655172413793 (the remainder is 38, so 58 is not a divisor of 3518)
3518 / 59 = 59.627118644068 (the remainder is 37, so 59 is not a divisor of 3518)