What are the divisors of 2370?

1, 2, 3, 5, 6, 10, 15, 30, 79, 158, 237, 395, 474, 790, 1185, 2370

There is a total of 16 positive divisors.
The sum of these divisors is 5760.
The arithmetic mean is 360.

8 even divisors

2, 6, 10, 30, 158, 474, 790, 2370

8 odd divisors

1, 3, 5, 15, 79, 237, 395, 1185

How to compute the divisors of 2370?

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 2370 by each of the numbers from 1 to 2370 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)

2370 / 1 = 2370 (the remainder is 0, so 1 is a divisor of 2370)
2370 / 2 = 1185 (the remainder is 0, so 2 is a divisor of 2370)
2370 / 3 = 790 (the remainder is 0, so 3 is a divisor of 2370)
...
2370 / 2369 = 1.0004221190376 (the remainder is 1, so 2369 is not a divisor of 2370)
2370 / 2370 = 1 (the remainder is 0, so 2370 is a divisor of 2370)

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 2370 (i.e. 48.682645778552). 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 $\frac{N}{d} = D$ )

2370 / 1 = 2370 (the remainder is 0, so 1 and 2370 are divisors of 2370)
2370 / 2 = 1185 (the remainder is 0, so 2 and 1185 are divisors of 2370)
2370 / 3 = 790 (the remainder is 0, so 3 and 790 are divisors of 2370)
...
2370 / 47 = 50.425531914894 (the remainder is 20, so 47 is not a divisor of 2370)
2370 / 48 = 49.375 (the remainder is 18, so 48 is not a divisor of 2370)