What are the divisors of 2952?

1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 41, 72, 82, 123, 164, 246, 328, 369, 492, 738, 984, 1476, 2952

There is a total of 24 positive divisors.
The sum of these divisors is 8190.
The arithmetic mean is 341.25.

18 even divisors

2, 4, 6, 8, 12, 18, 24, 36, 72, 82, 164, 246, 328, 492, 738, 984, 1476, 2952

6 odd divisors

1, 3, 9, 41, 123, 369

How to compute the divisors of 2952?

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 2952 by each of the numbers from 1 to 2952 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)

2952 / 1 = 2952 (the remainder is 0, so 1 is a divisor of 2952)
2952 / 2 = 1476 (the remainder is 0, so 2 is a divisor of 2952)
2952 / 3 = 984 (the remainder is 0, so 3 is a divisor of 2952)
...
2952 / 2951 = 1.0003388681803 (the remainder is 1, so 2951 is not a divisor of 2952)
2952 / 2952 = 1 (the remainder is 0, so 2952 is a divisor of 2952)

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 2952 (i.e. 54.332310828824). 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 $\frac{N}{d} = D$ )

2952 / 1 = 2952 (the remainder is 0, so 1 and 2952 are divisors of 2952)
2952 / 2 = 1476 (the remainder is 0, so 2 and 1476 are divisors of 2952)
2952 / 3 = 984 (the remainder is 0, so 3 and 984 are divisors of 2952)
...
2952 / 53 = 55.698113207547 (the remainder is 37, so 53 is not a divisor of 2952)
2952 / 54 = 54.666666666667 (the remainder is 36, so 54 is not a divisor of 2952)