What are the divisors of 4080?

1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 34, 40, 48, 51, 60, 68, 80, 85, 102, 120, 136, 170, 204, 240, 255, 272, 340, 408, 510, 680, 816, 1020, 1360, 2040, 4080

32 even divisors

2, 4, 6, 8, 10, 12, 16, 20, 24, 30, 34, 40, 48, 60, 68, 80, 102, 120, 136, 170, 204, 240, 272, 340, 408, 510, 680, 816, 1020, 1360, 2040, 4080

8 odd divisors

1, 3, 5, 15, 17, 51, 85, 255

How to compute the divisors of 4080?

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 4080 by each of the numbers from 1 to 4080 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)

  • 4080 / 1 = 4080 (the remainder is 0, so 1 is a divisor of 4080)
  • 4080 / 2 = 2040 (the remainder is 0, so 2 is a divisor of 4080)
  • 4080 / 3 = 1360 (the remainder is 0, so 3 is a divisor of 4080)
  • ...
  • 4080 / 4079 = 1.000245158127 (the remainder is 1, so 4079 is not a divisor of 4080)
  • 4080 / 4080 = 1 (the remainder is 0, so 4080 is a divisor of 4080)

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 4080 (i.e. 63.874877690685). 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 )

  • 4080 / 1 = 4080 (the remainder is 0, so 1 and 4080 are divisors of 4080)
  • 4080 / 2 = 2040 (the remainder is 0, so 2 and 2040 are divisors of 4080)
  • 4080 / 3 = 1360 (the remainder is 0, so 3 and 1360 are divisors of 4080)
  • ...
  • 4080 / 62 = 65.806451612903 (the remainder is 50, so 62 is not a divisor of 4080)
  • 4080 / 63 = 64.761904761905 (the remainder is 48, so 63 is not a divisor of 4080)