What are the divisors of 2912?

1, 2, 4, 7, 8, 13, 14, 16, 26, 28, 32, 52, 56, 91, 104, 112, 182, 208, 224, 364, 416, 728, 1456, 2912

There is a total of 24 positive divisors.
The sum of these divisors is 7056.
The arithmetic mean is 294.

20 even divisors

2, 4, 8, 14, 16, 26, 28, 32, 52, 56, 104, 112, 182, 208, 224, 364, 416, 728, 1456, 2912

4 odd divisors

1, 7, 13, 91

How to compute the divisors of 2912?

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 2912 by each of the numbers from 1 to 2912 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)

2912 / 1 = 2912 (the remainder is 0, so 1 is a divisor of 2912)
2912 / 2 = 1456 (the remainder is 0, so 2 is a divisor of 2912)
2912 / 3 = 970.66666666667 (the remainder is 2, so 3 is not a divisor of 2912)
...
2912 / 2911 = 1.000343524562 (the remainder is 1, so 2911 is not a divisor of 2912)
2912 / 2912 = 1 (the remainder is 0, so 2912 is a divisor of 2912)

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 2912 (i.e. 53.962950252928). 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 $\frac{N}{d} = D$ )

2912 / 1 = 2912 (the remainder is 0, so 1 and 2912 are divisors of 2912)
2912 / 2 = 1456 (the remainder is 0, so 2 and 1456 are divisors of 2912)
2912 / 3 = 970.66666666667 (the remainder is 2, so 3 is not a divisor of 2912)
...
2912 / 52 = 56 (the remainder is 0, so 52 and 56 are divisors of 2912)
2912 / 53 = 54.943396226415 (the remainder is 50, so 53 is not a divisor of 2912)