What are the divisors of 5919?

1, 3, 1973, 5919

4 odd divisors

1, 3, 1973, 5919

How to compute the divisors of 5919?

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 5919 by each of the numbers from 1 to 5919 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)

  • 5919 / 1 = 5919 (the remainder is 0, so 1 is a divisor of 5919)
  • 5919 / 2 = 2959.5 (the remainder is 1, so 2 is not a divisor of 5919)
  • 5919 / 3 = 1973 (the remainder is 0, so 3 is a divisor of 5919)
  • ...
  • 5919 / 5918 = 1.0001689760054 (the remainder is 1, so 5918 is not a divisor of 5919)
  • 5919 / 5919 = 1 (the remainder is 0, so 5919 is a divisor of 5919)

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 5919 (i.e. 76.935037531673). 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 )

  • 5919 / 1 = 5919 (the remainder is 0, so 1 and 5919 are divisors of 5919)
  • 5919 / 2 = 2959.5 (the remainder is 1, so 2 is not a divisor of 5919)
  • 5919 / 3 = 1973 (the remainder is 0, so 3 and 1973 are divisors of 5919)
  • ...
  • 5919 / 75 = 78.92 (the remainder is 69, so 75 is not a divisor of 5919)
  • 5919 / 76 = 77.881578947368 (the remainder is 67, so 76 is not a divisor of 5919)