What are the divisors of 2352?

1, 2, 3, 4, 6, 7, 8, 12, 14, 16, 21, 24, 28, 42, 48, 49, 56, 84, 98, 112, 147, 168, 196, 294, 336, 392, 588, 784, 1176, 2352

There is a total of 30 positive divisors.
The sum of these divisors is 7068.
The arithmetic mean is 235.6.

24 even divisors

2, 4, 6, 8, 12, 14, 16, 24, 28, 42, 48, 56, 84, 98, 112, 168, 196, 294, 336, 392, 588, 784, 1176, 2352

6 odd divisors

1, 3, 7, 21, 49, 147

How to compute the divisors of 2352?

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 2352 by each of the numbers from 1 to 2352 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)

2352 / 1 = 2352 (the remainder is 0, so 1 is a divisor of 2352)
2352 / 2 = 1176 (the remainder is 0, so 2 is a divisor of 2352)
2352 / 3 = 784 (the remainder is 0, so 3 is a divisor of 2352)
...
2352 / 2351 = 1.0004253509145 (the remainder is 1, so 2351 is not a divisor of 2352)
2352 / 2352 = 1 (the remainder is 0, so 2352 is a divisor of 2352)

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 2352 (i.e. 48.497422611929). 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 $\frac{N}{d} = D$ )

2352 / 1 = 2352 (the remainder is 0, so 1 and 2352 are divisors of 2352)
2352 / 2 = 1176 (the remainder is 0, so 2 and 1176 are divisors of 2352)
2352 / 3 = 784 (the remainder is 0, so 3 and 784 are divisors of 2352)
...
2352 / 47 = 50.042553191489 (the remainder is 2, so 47 is not a divisor of 2352)
2352 / 48 = 49 (the remainder is 0, so 48 and 49 are divisors of 2352)