What are the divisors of 2823?

1, 3, 941, 2823

4 odd divisors

1, 3, 941, 2823

How to compute the divisors of 2823?

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 2823 by each of the numbers from 1 to 2823 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)

  • 2823 / 1 = 2823 (the remainder is 0, so 1 is a divisor of 2823)
  • 2823 / 2 = 1411.5 (the remainder is 1, so 2 is not a divisor of 2823)
  • 2823 / 3 = 941 (the remainder is 0, so 3 is a divisor of 2823)
  • ...
  • 2823 / 2822 = 1.0003543586109 (the remainder is 1, so 2822 is not a divisor of 2823)
  • 2823 / 2823 = 1 (the remainder is 0, so 2823 is a divisor of 2823)

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 2823 (i.e. 53.131911315141). 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 )

  • 2823 / 1 = 2823 (the remainder is 0, so 1 and 2823 are divisors of 2823)
  • 2823 / 2 = 1411.5 (the remainder is 1, so 2 is not a divisor of 2823)
  • 2823 / 3 = 941 (the remainder is 0, so 3 and 941 are divisors of 2823)
  • ...
  • 2823 / 52 = 54.288461538462 (the remainder is 15, so 52 is not a divisor of 2823)
  • 2823 / 53 = 53.264150943396 (the remainder is 14, so 53 is not a divisor of 2823)