What are the divisors of 3550?

1, 2, 5, 10, 25, 50, 71, 142, 355, 710, 1775, 3550

There is a total of 12 positive divisors.
The sum of these divisors is 6696.
The arithmetic mean is 558.

6 even divisors

2, 10, 50, 142, 710, 3550

6 odd divisors

1, 5, 25, 71, 355, 1775

How to compute the divisors of 3550?

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 3550 by each of the numbers from 1 to 3550 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)

3550 / 1 = 3550 (the remainder is 0, so 1 is a divisor of 3550)
3550 / 2 = 1775 (the remainder is 0, so 2 is a divisor of 3550)
3550 / 3 = 1183.3333333333 (the remainder is 1, so 3 is not a divisor of 3550)
...
3550 / 3549 = 1.0002817695125 (the remainder is 1, so 3549 is not a divisor of 3550)
3550 / 3550 = 1 (the remainder is 0, so 3550 is a divisor of 3550)

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 3550 (i.e. 59.581876439065). 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 $\frac{N}{d} = D$ )

3550 / 1 = 3550 (the remainder is 0, so 1 and 3550 are divisors of 3550)
3550 / 2 = 1775 (the remainder is 0, so 2 and 1775 are divisors of 3550)
3550 / 3 = 1183.3333333333 (the remainder is 1, so 3 is not a divisor of 3550)
...
3550 / 58 = 61.206896551724 (the remainder is 12, so 58 is not a divisor of 3550)
3550 / 59 = 60.169491525424 (the remainder is 10, so 59 is not a divisor of 3550)