What are the divisors of 2809?

1, 53, 2809

3 odd divisors

1, 53, 2809

How to compute the divisors of 2809?

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 2809 by each of the numbers from 1 to 2809 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)

  • 2809 / 1 = 2809 (the remainder is 0, so 1 is a divisor of 2809)
  • 2809 / 2 = 1404.5 (the remainder is 1, so 2 is not a divisor of 2809)
  • 2809 / 3 = 936.33333333333 (the remainder is 1, so 3 is not a divisor of 2809)
  • ...
  • 2809 / 2808 = 1.0003561253561 (the remainder is 1, so 2808 is not a divisor of 2809)
  • 2809 / 2809 = 1 (the remainder is 0, so 2809 is a divisor of 2809)

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 2809 (i.e. 53). 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 )

  • 2809 / 1 = 2809 (the remainder is 0, so 1 and 2809 are divisors of 2809)
  • 2809 / 2 = 1404.5 (the remainder is 1, so 2 is not a divisor of 2809)
  • 2809 / 3 = 936.33333333333 (the remainder is 1, so 3 is not a divisor of 2809)
  • ...
  • 2809 / 52 = 54.019230769231 (the remainder is 1, so 52 is not a divisor of 2809)
  • 2809 / 53 = 53 (the remainder is 0, so 53 and 53 are divisors of 2809)