What are the divisors of 4938?

1, 2, 3, 6, 823, 1646, 2469, 4938

There is a total of 8 positive divisors.
The sum of these divisors is 9888.
The arithmetic mean is 1236.

4 even divisors

2, 6, 1646, 4938

4 odd divisors

1, 3, 823, 2469

How to compute the divisors of 4938?

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 4938 by each of the numbers from 1 to 4938 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)

4938 / 1 = 4938 (the remainder is 0, so 1 is a divisor of 4938)
4938 / 2 = 2469 (the remainder is 0, so 2 is a divisor of 4938)
4938 / 3 = 1646 (the remainder is 0, so 3 is a divisor of 4938)
...
4938 / 4937 = 1.0002025521572 (the remainder is 1, so 4937 is not a divisor of 4938)
4938 / 4938 = 1 (the remainder is 0, so 4938 is a divisor of 4938)

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 4938 (i.e. 70.270904363043). 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 $\frac{N}{d} = D$ )

4938 / 1 = 4938 (the remainder is 0, so 1 and 4938 are divisors of 4938)
4938 / 2 = 2469 (the remainder is 0, so 2 and 2469 are divisors of 4938)
4938 / 3 = 1646 (the remainder is 0, so 3 and 1646 are divisors of 4938)
...
4938 / 69 = 71.565217391304 (the remainder is 39, so 69 is not a divisor of 4938)
4938 / 70 = 70.542857142857 (the remainder is 38, so 70 is not a divisor of 4938)