What are the divisors of 1733?

1, 1733

2 odd divisors

1, 1733

How to compute the divisors of 1733?

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 1733 by each of the numbers from 1 to 1733 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)

  • 1733 / 1 = 1733 (the remainder is 0, so 1 is a divisor of 1733)
  • 1733 / 2 = 866.5 (the remainder is 1, so 2 is not a divisor of 1733)
  • 1733 / 3 = 577.66666666667 (the remainder is 2, so 3 is not a divisor of 1733)
  • ...
  • 1733 / 1732 = 1.0005773672055 (the remainder is 1, so 1732 is not a divisor of 1733)
  • 1733 / 1733 = 1 (the remainder is 0, so 1733 is a divisor of 1733)

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 1733 (i.e. 41.629316592997). 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 )

  • 1733 / 1 = 1733 (the remainder is 0, so 1 and 1733 are divisors of 1733)
  • 1733 / 2 = 866.5 (the remainder is 1, so 2 is not a divisor of 1733)
  • 1733 / 3 = 577.66666666667 (the remainder is 2, so 3 is not a divisor of 1733)
  • ...
  • 1733 / 40 = 43.325 (the remainder is 13, so 40 is not a divisor of 1733)
  • 1733 / 41 = 42.268292682927 (the remainder is 11, so 41 is not a divisor of 1733)