What are the divisors of 3300?
1, 2, 3, 4, 5, 6, 10, 11, 12, 15, 20, 22, 25, 30, 33, 44, 50, 55, 60, 66, 75, 100, 110, 132, 150, 165, 220, 275, 300, 330, 550, 660, 825, 1100, 1650, 3300
- There is a total of 36 positive divisors.
- The sum of these divisors is 10416.
- The arithmetic mean is 289.33333333333.
24 even divisors
2, 4, 6, 10, 12, 20, 22, 30, 44, 50, 60, 66, 100, 110, 132, 150, 220, 300, 330, 550, 660, 1100, 1650, 3300
12 odd divisors
1, 3, 5, 11, 15, 25, 33, 55, 75, 165, 275, 825
How to compute the divisors of 3300?
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 3300 by each of the numbers from 1 to 3300 to determine which ones have a remainder equal to 0.
(where is the integer part of the quotient)
- 3300 / 1 = 3300 (the remainder is 0, so 1 is a divisor of 3300)
- 3300 / 2 = 1650 (the remainder is 0, so 2 is a divisor of 3300)
- 3300 / 3 = 1100 (the remainder is 0, so 3 is a divisor of 3300)
- ...
- 3300 / 3299 = 1.0003031221582 (the remainder is 1, so 3299 is not a divisor of 3300)
- 3300 / 3300 = 1 (the remainder is 0, so 3300 is a divisor of 3300)
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 3300 (i.e. 57.44562646538). 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 )
- 3300 / 1 = 3300 (the remainder is 0, so 1 and 3300 are divisors of 3300)
- 3300 / 2 = 1650 (the remainder is 0, so 2 and 1650 are divisors of 3300)
- 3300 / 3 = 1100 (the remainder is 0, so 3 and 1100 are divisors of 3300)
- ...
- 3300 / 56 = 58.928571428571 (the remainder is 52, so 56 is not a divisor of 3300)
- 3300 / 57 = 57.894736842105 (the remainder is 51, so 57 is not a divisor of 3300)