What are the divisors of 3840?

1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 32, 40, 48, 60, 64, 80, 96, 120, 128, 160, 192, 240, 256, 320, 384, 480, 640, 768, 960, 1280, 1920, 3840

32 even divisors

2, 4, 6, 8, 10, 12, 16, 20, 24, 30, 32, 40, 48, 60, 64, 80, 96, 120, 128, 160, 192, 240, 256, 320, 384, 480, 640, 768, 960, 1280, 1920, 3840

4 odd divisors

1, 3, 5, 15

How to compute the divisors of 3840?

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 3840 by each of the numbers from 1 to 3840 to determine which ones have a remainder equal to 0.

Remainder = N ( M × N M )

(where N M is the integer part of the quotient)

  • 3840 / 1 = 3840 (the remainder is 0, so 1 is a divisor of 3840)
  • 3840 / 2 = 1920 (the remainder is 0, so 2 is a divisor of 3840)
  • 3840 / 3 = 1280 (the remainder is 0, so 3 is a divisor of 3840)
  • ...
  • 3840 / 3839 = 1.0002604845012 (the remainder is 1, so 3839 is not a divisor of 3840)
  • 3840 / 3840 = 1 (the remainder is 0, so 3840 is a divisor of 3840)

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 3840 (i.e. 61.967733539319). 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 × d = N

(thus, if N D = d , then N d = D )

  • 3840 / 1 = 3840 (the remainder is 0, so 1 and 3840 are divisors of 3840)
  • 3840 / 2 = 1920 (the remainder is 0, so 2 and 1920 are divisors of 3840)
  • 3840 / 3 = 1280 (the remainder is 0, so 3 and 1280 are divisors of 3840)
  • ...
  • 3840 / 60 = 64 (the remainder is 0, so 60 and 64 are divisors of 3840)
  • 3840 / 61 = 62.950819672131 (the remainder is 58, so 61 is not a divisor of 3840)