What are the divisors of 4920?

1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 41, 60, 82, 120, 123, 164, 205, 246, 328, 410, 492, 615, 820, 984, 1230, 1640, 2460, 4920

24 even divisors

2, 4, 6, 8, 10, 12, 20, 24, 30, 40, 60, 82, 120, 164, 246, 328, 410, 492, 820, 984, 1230, 1640, 2460, 4920

8 odd divisors

1, 3, 5, 15, 41, 123, 205, 615

How to compute the divisors of 4920?

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 4920 by each of the numbers from 1 to 4920 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)

  • 4920 / 1 = 4920 (the remainder is 0, so 1 is a divisor of 4920)
  • 4920 / 2 = 2460 (the remainder is 0, so 2 is a divisor of 4920)
  • 4920 / 3 = 1640 (the remainder is 0, so 3 is a divisor of 4920)
  • ...
  • 4920 / 4919 = 1.0002032933523 (the remainder is 1, so 4919 is not a divisor of 4920)
  • 4920 / 4920 = 1 (the remainder is 0, so 4920 is a divisor of 4920)

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 4920 (i.e. 70.142711667001). 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 )

  • 4920 / 1 = 4920 (the remainder is 0, so 1 and 4920 are divisors of 4920)
  • 4920 / 2 = 2460 (the remainder is 0, so 2 and 2460 are divisors of 4920)
  • 4920 / 3 = 1640 (the remainder is 0, so 3 and 1640 are divisors of 4920)
  • ...
  • 4920 / 69 = 71.304347826087 (the remainder is 21, so 69 is not a divisor of 4920)
  • 4920 / 70 = 70.285714285714 (the remainder is 20, so 70 is not a divisor of 4920)