What are the divisors of 2340?
1, 2, 3, 4, 5, 6, 9, 10, 12, 13, 15, 18, 20, 26, 30, 36, 39, 45, 52, 60, 65, 78, 90, 117, 130, 156, 180, 195, 234, 260, 390, 468, 585, 780, 1170, 2340
- There is a total of 36 positive divisors.
- The sum of these divisors is 7644.
- The arithmetic mean is 212.33333333333.
24 even divisors
2, 4, 6, 10, 12, 18, 20, 26, 30, 36, 52, 60, 78, 90, 130, 156, 180, 234, 260, 390, 468, 780, 1170, 2340
12 odd divisors
1, 3, 5, 9, 13, 15, 39, 45, 65, 117, 195, 585
How to compute the divisors of 2340?
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.
Brute force algorithm
We could start by using a brute-force method which would involve dividing 2340 by each of the numbers from 1 to 2340 to determine which ones have a remainder equal to 0.
(where is the integer part of the quotient)
- 2340 / 1 = 2340 (the remainder is 0, so 1 is a divisor of 2340)
- 2340 / 2 = 1170 (the remainder is 0, so 2 is a divisor of 2340)
- 2340 / 3 = 780 (the remainder is 0, so 3 is a divisor of 2340)
- ...
- 2340 / 2339 = 1.0004275331338 (the remainder is 1, so 2339 is not a divisor of 2340)
- 2340 / 2340 = 1 (the remainder is 0, so 2340 is a divisor of 2340)
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 2340 (i.e. 48.373546489791). Indeed, if a number N has a divisor D greater than its square root, then there is necessarily a smaller divisor d such that:
(thus, if , then )
- 2340 / 1 = 2340 (the remainder is 0, so 1 and 2340 are divisors of 2340)
- 2340 / 2 = 1170 (the remainder is 0, so 2 and 1170 are divisors of 2340)
- 2340 / 3 = 780 (the remainder is 0, so 3 and 780 are divisors of 2340)
- ...
- 2340 / 47 = 49.787234042553 (the remainder is 37, so 47 is not a divisor of 2340)
- 2340 / 48 = 48.75 (the remainder is 36, so 48 is not a divisor of 2340)