What are the divisors of 2720?

1, 2, 4, 5, 8, 10, 16, 17, 20, 32, 34, 40, 68, 80, 85, 136, 160, 170, 272, 340, 544, 680, 1360, 2720

There is a total of 24 positive divisors.
The sum of these divisors is 6804.
The arithmetic mean is 283.5.

20 even divisors

2, 4, 8, 10, 16, 20, 32, 34, 40, 68, 80, 136, 160, 170, 272, 340, 544, 680, 1360, 2720

4 odd divisors

1, 5, 17, 85

How to compute the divisors of 2720?

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 2720 by each of the numbers from 1 to 2720 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)

2720 / 1 = 2720 (the remainder is 0, so 1 is a divisor of 2720)
2720 / 2 = 1360 (the remainder is 0, so 2 is a divisor of 2720)
2720 / 3 = 906.66666666667 (the remainder is 2, so 3 is not a divisor of 2720)
...
2720 / 2719 = 1.0003677822729 (the remainder is 1, so 2719 is not a divisor of 2720)
2720 / 2720 = 1 (the remainder is 0, so 2720 is a divisor of 2720)

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 2720 (i.e. 52.153619241621). 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 $\frac{N}{d} = D$ )

2720 / 1 = 2720 (the remainder is 0, so 1 and 2720 are divisors of 2720)
2720 / 2 = 1360 (the remainder is 0, so 2 and 1360 are divisors of 2720)
2720 / 3 = 906.66666666667 (the remainder is 2, so 3 is not a divisor of 2720)
...
2720 / 51 = 53.333333333333 (the remainder is 17, so 51 is not a divisor of 2720)
2720 / 52 = 52.307692307692 (the remainder is 16, so 52 is not a divisor of 2720)