What are the divisors of 2924?

1, 2, 4, 17, 34, 43, 68, 86, 172, 731, 1462, 2924

There is a total of 12 positive divisors.
The sum of these divisors is 5544.
The arithmetic mean is 462.

8 even divisors

2, 4, 34, 68, 86, 172, 1462, 2924

4 odd divisors

1, 17, 43, 731

How to compute the divisors of 2924?

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 2924 by each of the numbers from 1 to 2924 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)

2924 / 1 = 2924 (the remainder is 0, so 1 is a divisor of 2924)
2924 / 2 = 1462 (the remainder is 0, so 2 is a divisor of 2924)
2924 / 3 = 974.66666666667 (the remainder is 2, so 3 is not a divisor of 2924)
...
2924 / 2923 = 1.0003421142662 (the remainder is 1, so 2923 is not a divisor of 2924)
2924 / 2924 = 1 (the remainder is 0, so 2924 is a divisor of 2924)

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 2924 (i.e. 54.074023338383). 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 $\frac{N}{d} = D$ )

2924 / 1 = 2924 (the remainder is 0, so 1 and 2924 are divisors of 2924)
2924 / 2 = 1462 (the remainder is 0, so 2 and 1462 are divisors of 2924)
2924 / 3 = 974.66666666667 (the remainder is 2, so 3 is not a divisor of 2924)
...
2924 / 53 = 55.169811320755 (the remainder is 9, so 53 is not a divisor of 2924)
2924 / 54 = 54.148148148148 (the remainder is 8, so 54 is not a divisor of 2924)