What are the divisors of 4032?

1, 2, 3, 4, 6, 7, 8, 9, 12, 14, 16, 18, 21, 24, 28, 32, 36, 42, 48, 56, 63, 64, 72, 84, 96, 112, 126, 144, 168, 192, 224, 252, 288, 336, 448, 504, 576, 672, 1008, 1344, 2016, 4032

36 even divisors

2, 4, 6, 8, 12, 14, 16, 18, 24, 28, 32, 36, 42, 48, 56, 64, 72, 84, 96, 112, 126, 144, 168, 192, 224, 252, 288, 336, 448, 504, 576, 672, 1008, 1344, 2016, 4032

6 odd divisors

1, 3, 7, 9, 21, 63

How to compute the divisors of 4032?

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 4032 by each of the numbers from 1 to 4032 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)

  • 4032 / 1 = 4032 (the remainder is 0, so 1 is a divisor of 4032)
  • 4032 / 2 = 2016 (the remainder is 0, so 2 is a divisor of 4032)
  • 4032 / 3 = 1344 (the remainder is 0, so 3 is a divisor of 4032)
  • ...
  • 4032 / 4031 = 1.0002480774001 (the remainder is 1, so 4031 is not a divisor of 4032)
  • 4032 / 4032 = 1 (the remainder is 0, so 4032 is a divisor of 4032)

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 4032 (i.e. 63.49803146555). 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 )

  • 4032 / 1 = 4032 (the remainder is 0, so 1 and 4032 are divisors of 4032)
  • 4032 / 2 = 2016 (the remainder is 0, so 2 and 2016 are divisors of 4032)
  • 4032 / 3 = 1344 (the remainder is 0, so 3 and 1344 are divisors of 4032)
  • ...
  • 4032 / 62 = 65.032258064516 (the remainder is 2, so 62 is not a divisor of 4032)
  • 4032 / 63 = 64 (the remainder is 0, so 63 and 64 are divisors of 4032)