What are the divisors of 5712?

1, 2, 3, 4, 6, 7, 8, 12, 14, 16, 17, 21, 24, 28, 34, 42, 48, 51, 56, 68, 84, 102, 112, 119, 136, 168, 204, 238, 272, 336, 357, 408, 476, 714, 816, 952, 1428, 1904, 2856, 5712

32 even divisors

2, 4, 6, 8, 12, 14, 16, 24, 28, 34, 42, 48, 56, 68, 84, 102, 112, 136, 168, 204, 238, 272, 336, 408, 476, 714, 816, 952, 1428, 1904, 2856, 5712

8 odd divisors

1, 3, 7, 17, 21, 51, 119, 357

How to compute the divisors of 5712?

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 5712 by each of the numbers from 1 to 5712 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)

  • 5712 / 1 = 5712 (the remainder is 0, so 1 is a divisor of 5712)
  • 5712 / 2 = 2856 (the remainder is 0, so 2 is a divisor of 5712)
  • 5712 / 3 = 1904 (the remainder is 0, so 3 is a divisor of 5712)
  • ...
  • 5712 / 5711 = 1.0001751006829 (the remainder is 1, so 5711 is not a divisor of 5712)
  • 5712 / 5712 = 1 (the remainder is 0, so 5712 is a divisor of 5712)

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 5712 (i.e. 75.577774510765). 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 )

  • 5712 / 1 = 5712 (the remainder is 0, so 1 and 5712 are divisors of 5712)
  • 5712 / 2 = 2856 (the remainder is 0, so 2 and 2856 are divisors of 5712)
  • 5712 / 3 = 1904 (the remainder is 0, so 3 and 1904 are divisors of 5712)
  • ...
  • 5712 / 74 = 77.189189189189 (the remainder is 14, so 74 is not a divisor of 5712)
  • 5712 / 75 = 76.16 (the remainder is 12, so 75 is not a divisor of 5712)