What are the divisors of 4940?

1, 2, 4, 5, 10, 13, 19, 20, 26, 38, 52, 65, 76, 95, 130, 190, 247, 260, 380, 494, 988, 1235, 2470, 4940

There is a total of 24 positive divisors.
The sum of these divisors is 11760.
The arithmetic mean is 490.

16 even divisors

2, 4, 10, 20, 26, 38, 52, 76, 130, 190, 260, 380, 494, 988, 2470, 4940

8 odd divisors

1, 5, 13, 19, 65, 95, 247, 1235

How to compute the divisors of 4940?

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 4940 by each of the numbers from 1 to 4940 to determine which ones have a remainder equal to 0.

$Remainder = N - (M \times ⌊ \frac{N}{M} ⌋)$

(where $⌊ \frac{N}{M} ⌋$ is the integer part of the quotient)

4940 / 1 = 4940 (the remainder is 0, so 1 is a divisor of 4940)
4940 / 2 = 2470 (the remainder is 0, so 2 is a divisor of 4940)
4940 / 3 = 1646.6666666667 (the remainder is 2, so 3 is not a divisor of 4940)
...
4940 / 4939 = 1.0002024701357 (the remainder is 1, so 4939 is not a divisor of 4940)
4940 / 4940 = 1 (the remainder is 0, so 4940 is a divisor of 4940)

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 4940 (i.e. 70.285133563222). Indeed, if a number N has a divisor D greater than its square root, then there is necessarily a smaller divisor d such that:

$D \times d = N$

(thus, if $\frac{N}{D} = d$ , then $\frac{N}{d} = D$ )

4940 / 1 = 4940 (the remainder is 0, so 1 and 4940 are divisors of 4940)
4940 / 2 = 2470 (the remainder is 0, so 2 and 2470 are divisors of 4940)
4940 / 3 = 1646.6666666667 (the remainder is 2, so 3 is not a divisor of 4940)
...
4940 / 69 = 71.594202898551 (the remainder is 41, so 69 is not a divisor of 4940)
4940 / 70 = 70.571428571429 (the remainder is 40, so 70 is not a divisor of 4940)