What are the divisors of 4950?

1, 2, 3, 5, 6, 9, 10, 11, 15, 18, 22, 25, 30, 33, 45, 50, 55, 66, 75, 90, 99, 110, 150, 165, 198, 225, 275, 330, 450, 495, 550, 825, 990, 1650, 2475, 4950

18 even divisors

2, 6, 10, 18, 22, 30, 50, 66, 90, 110, 150, 198, 330, 450, 550, 990, 1650, 4950

18 odd divisors

1, 3, 5, 9, 11, 15, 25, 33, 45, 55, 75, 99, 165, 225, 275, 495, 825, 2475

How to compute the divisors of 4950?

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 4950 by each of the numbers from 1 to 4950 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)

  • 4950 / 1 = 4950 (the remainder is 0, so 1 is a divisor of 4950)
  • 4950 / 2 = 2475 (the remainder is 0, so 2 is a divisor of 4950)
  • 4950 / 3 = 1650 (the remainder is 0, so 3 is a divisor of 4950)
  • ...
  • 4950 / 4949 = 1.0002020610224 (the remainder is 1, so 4949 is not a divisor of 4950)
  • 4950 / 4950 = 1 (the remainder is 0, so 4950 is a divisor of 4950)

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 4950 (i.e. 70.356236397351). 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 )

  • 4950 / 1 = 4950 (the remainder is 0, so 1 and 4950 are divisors of 4950)
  • 4950 / 2 = 2475 (the remainder is 0, so 2 and 2475 are divisors of 4950)
  • 4950 / 3 = 1650 (the remainder is 0, so 3 and 1650 are divisors of 4950)
  • ...
  • 4950 / 69 = 71.739130434783 (the remainder is 51, so 69 is not a divisor of 4950)
  • 4950 / 70 = 70.714285714286 (the remainder is 50, so 70 is not a divisor of 4950)