What are the divisors of 1120?

1, 2, 4, 5, 7, 8, 10, 14, 16, 20, 28, 32, 35, 40, 56, 70, 80, 112, 140, 160, 224, 280, 560, 1120

There is a total of 24 positive divisors.
The sum of these divisors is 3024.
The arithmetic mean is 126.

20 even divisors

2, 4, 8, 10, 14, 16, 20, 28, 32, 40, 56, 70, 80, 112, 140, 160, 224, 280, 560, 1120

4 odd divisors

1, 5, 7, 35

How to compute the divisors of 1120?

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 1120 by each of the numbers from 1 to 1120 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)

1120 / 1 = 1120 (the remainder is 0, so 1 is a divisor of 1120)
1120 / 2 = 560 (the remainder is 0, so 2 is a divisor of 1120)
1120 / 3 = 373.33333333333 (the remainder is 1, so 3 is not a divisor of 1120)
...
1120 / 1119 = 1.0008936550492 (the remainder is 1, so 1119 is not a divisor of 1120)
1120 / 1120 = 1 (the remainder is 0, so 1120 is a divisor of 1120)

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 1120 (i.e. 33.466401061363). 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 $\frac{N}{d} = D$ )

1120 / 1 = 1120 (the remainder is 0, so 1 and 1120 are divisors of 1120)
1120 / 2 = 560 (the remainder is 0, so 2 and 560 are divisors of 1120)
1120 / 3 = 373.33333333333 (the remainder is 1, so 3 is not a divisor of 1120)
...
1120 / 32 = 35 (the remainder is 0, so 32 and 35 are divisors of 1120)
1120 / 33 = 33.939393939394 (the remainder is 31, so 33 is not a divisor of 1120)