What are the divisors of 5347?

1, 5347

2 odd divisors

1, 5347

How to compute the divisors of 5347?

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 5347 by each of the numbers from 1 to 5347 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)

  • 5347 / 1 = 5347 (the remainder is 0, so 1 is a divisor of 5347)
  • 5347 / 2 = 2673.5 (the remainder is 1, so 2 is not a divisor of 5347)
  • 5347 / 3 = 1782.3333333333 (the remainder is 1, so 3 is not a divisor of 5347)
  • ...
  • 5347 / 5346 = 1.0001870557426 (the remainder is 1, so 5346 is not a divisor of 5347)
  • 5347 / 5347 = 1 (the remainder is 0, so 5347 is a divisor of 5347)

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 5347 (i.e. 73.123183738128). 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 )

  • 5347 / 1 = 5347 (the remainder is 0, so 1 and 5347 are divisors of 5347)
  • 5347 / 2 = 2673.5 (the remainder is 1, so 2 is not a divisor of 5347)
  • 5347 / 3 = 1782.3333333333 (the remainder is 1, so 3 is not a divisor of 5347)
  • ...
  • 5347 / 72 = 74.263888888889 (the remainder is 19, so 72 is not a divisor of 5347)
  • 5347 / 73 = 73.246575342466 (the remainder is 18, so 73 is not a divisor of 5347)