What are the divisors of 2920?

1, 2, 4, 5, 8, 10, 20, 40, 73, 146, 292, 365, 584, 730, 1460, 2920

There is a total of 16 positive divisors.
The sum of these divisors is 6660.
The arithmetic mean is 416.25.

12 even divisors

2, 4, 8, 10, 20, 40, 146, 292, 584, 730, 1460, 2920

4 odd divisors

1, 5, 73, 365

How to compute the divisors of 2920?

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 2920 by each of the numbers from 1 to 2920 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)

2920 / 1 = 2920 (the remainder is 0, so 1 is a divisor of 2920)
2920 / 2 = 1460 (the remainder is 0, so 2 is a divisor of 2920)
2920 / 3 = 973.33333333333 (the remainder is 1, so 3 is not a divisor of 2920)
...
2920 / 2919 = 1.0003425830764 (the remainder is 1, so 2919 is not a divisor of 2920)
2920 / 2920 = 1 (the remainder is 0, so 2920 is a divisor of 2920)

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 2920 (i.e. 54.037024344425). 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 $\frac{N}{d} = D$ )

2920 / 1 = 2920 (the remainder is 0, so 1 and 2920 are divisors of 2920)
2920 / 2 = 1460 (the remainder is 0, so 2 and 1460 are divisors of 2920)
2920 / 3 = 973.33333333333 (the remainder is 1, so 3 is not a divisor of 2920)
...
2920 / 53 = 55.094339622642 (the remainder is 5, so 53 is not a divisor of 2920)
2920 / 54 = 54.074074074074 (the remainder is 4, so 54 is not a divisor of 2920)