What are the divisors of 3900?

1, 2, 3, 4, 5, 6, 10, 12, 13, 15, 20, 25, 26, 30, 39, 50, 52, 60, 65, 75, 78, 100, 130, 150, 156, 195, 260, 300, 325, 390, 650, 780, 975, 1300, 1950, 3900

24 even divisors

2, 4, 6, 10, 12, 20, 26, 30, 50, 52, 60, 78, 100, 130, 150, 156, 260, 300, 390, 650, 780, 1300, 1950, 3900

12 odd divisors

1, 3, 5, 13, 15, 25, 39, 65, 75, 195, 325, 975

How to compute the divisors of 3900?

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 3900 by each of the numbers from 1 to 3900 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)

  • 3900 / 1 = 3900 (the remainder is 0, so 1 is a divisor of 3900)
  • 3900 / 2 = 1950 (the remainder is 0, so 2 is a divisor of 3900)
  • 3900 / 3 = 1300 (the remainder is 0, so 3 is a divisor of 3900)
  • ...
  • 3900 / 3899 = 1.0002564760195 (the remainder is 1, so 3899 is not a divisor of 3900)
  • 3900 / 3900 = 1 (the remainder is 0, so 3900 is a divisor of 3900)

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 3900 (i.e. 62.449979983984). 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 )

  • 3900 / 1 = 3900 (the remainder is 0, so 1 and 3900 are divisors of 3900)
  • 3900 / 2 = 1950 (the remainder is 0, so 2 and 1950 are divisors of 3900)
  • 3900 / 3 = 1300 (the remainder is 0, so 3 and 1300 are divisors of 3900)
  • ...
  • 3900 / 61 = 63.934426229508 (the remainder is 57, so 61 is not a divisor of 3900)
  • 3900 / 62 = 62.903225806452 (the remainder is 56, so 62 is not a divisor of 3900)