What are the divisors of 3500?

1, 2, 4, 5, 7, 10, 14, 20, 25, 28, 35, 50, 70, 100, 125, 140, 175, 250, 350, 500, 700, 875, 1750, 3500

There is a total of 24 positive divisors.
The sum of these divisors is 8736.
The arithmetic mean is 364.

16 even divisors

2, 4, 10, 14, 20, 28, 50, 70, 100, 140, 250, 350, 500, 700, 1750, 3500

8 odd divisors

1, 5, 7, 25, 35, 125, 175, 875

How to compute the divisors of 3500?

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 3500 by each of the numbers from 1 to 3500 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)

3500 / 1 = 3500 (the remainder is 0, so 1 is a divisor of 3500)
3500 / 2 = 1750 (the remainder is 0, so 2 is a divisor of 3500)
3500 / 3 = 1166.6666666667 (the remainder is 2, so 3 is not a divisor of 3500)
...
3500 / 3499 = 1.0002857959417 (the remainder is 1, so 3499 is not a divisor of 3500)
3500 / 3500 = 1 (the remainder is 0, so 3500 is a divisor of 3500)

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 3500 (i.e. 59.160797830996). 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 $\frac{N}{d} = D$ )

3500 / 1 = 3500 (the remainder is 0, so 1 and 3500 are divisors of 3500)
3500 / 2 = 1750 (the remainder is 0, so 2 and 1750 are divisors of 3500)
3500 / 3 = 1166.6666666667 (the remainder is 2, so 3 is not a divisor of 3500)
...
3500 / 58 = 60.344827586207 (the remainder is 20, so 58 is not a divisor of 3500)
3500 / 59 = 59.322033898305 (the remainder is 19, so 59 is not a divisor of 3500)