What are the divisors of 5333?

1, 5333

2 odd divisors

1, 5333

How to compute the divisors of 5333?

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 5333 by each of the numbers from 1 to 5333 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)

  • 5333 / 1 = 5333 (the remainder is 0, so 1 is a divisor of 5333)
  • 5333 / 2 = 2666.5 (the remainder is 1, so 2 is not a divisor of 5333)
  • 5333 / 3 = 1777.6666666667 (the remainder is 2, so 3 is not a divisor of 5333)
  • ...
  • 5333 / 5332 = 1.0001875468867 (the remainder is 1, so 5332 is not a divisor of 5333)
  • 5333 / 5333 = 1 (the remainder is 0, so 5333 is a divisor of 5333)

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 5333 (i.e. 73.027392121039). 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 )

  • 5333 / 1 = 5333 (the remainder is 0, so 1 and 5333 are divisors of 5333)
  • 5333 / 2 = 2666.5 (the remainder is 1, so 2 is not a divisor of 5333)
  • 5333 / 3 = 1777.6666666667 (the remainder is 2, so 3 is not a divisor of 5333)
  • ...
  • 5333 / 72 = 74.069444444444 (the remainder is 5, so 72 is not a divisor of 5333)
  • 5333 / 73 = 73.054794520548 (the remainder is 4, so 73 is not a divisor of 5333)