What are the divisors of 9100?

1, 2, 4, 5, 7, 10, 13, 14, 20, 25, 26, 28, 35, 50, 52, 65, 70, 91, 100, 130, 140, 175, 182, 260, 325, 350, 364, 455, 650, 700, 910, 1300, 1820, 2275, 4550, 9100

24 even divisors

2, 4, 10, 14, 20, 26, 28, 50, 52, 70, 100, 130, 140, 182, 260, 350, 364, 650, 700, 910, 1300, 1820, 4550, 9100

12 odd divisors

1, 5, 7, 13, 25, 35, 65, 91, 175, 325, 455, 2275

How to compute the divisors of 9100?

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 9100 by each of the numbers from 1 to 9100 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)

  • 9100 / 1 = 9100 (the remainder is 0, so 1 is a divisor of 9100)
  • 9100 / 2 = 4550 (the remainder is 0, so 2 is a divisor of 9100)
  • 9100 / 3 = 3033.3333333333 (the remainder is 1, so 3 is not a divisor of 9100)
  • ...
  • 9100 / 9099 = 1.0001099021871 (the remainder is 1, so 9099 is not a divisor of 9100)
  • 9100 / 9100 = 1 (the remainder is 0, so 9100 is a divisor of 9100)

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 9100 (i.e. 95.393920141695). 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 )

  • 9100 / 1 = 9100 (the remainder is 0, so 1 and 9100 are divisors of 9100)
  • 9100 / 2 = 4550 (the remainder is 0, so 2 and 4550 are divisors of 9100)
  • 9100 / 3 = 3033.3333333333 (the remainder is 1, so 3 is not a divisor of 9100)
  • ...
  • 9100 / 94 = 96.808510638298 (the remainder is 76, so 94 is not a divisor of 9100)
  • 9100 / 95 = 95.789473684211 (the remainder is 75, so 95 is not a divisor of 9100)