What are the divisors of 4941?

1, 3, 9, 27, 61, 81, 183, 549, 1647, 4941

There is a total of 10 positive divisors.
The sum of these divisors is 7502.
The arithmetic mean is 750.2.

10 odd divisors

1, 3, 9, 27, 61, 81, 183, 549, 1647, 4941

How to compute the divisors of 4941?

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 4941 by each of the numbers from 1 to 4941 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)

4941 / 1 = 4941 (the remainder is 0, so 1 is a divisor of 4941)
4941 / 2 = 2470.5 (the remainder is 1, so 2 is not a divisor of 4941)
4941 / 3 = 1647 (the remainder is 0, so 3 is a divisor of 4941)
...
4941 / 4940 = 1.0002024291498 (the remainder is 1, so 4940 is not a divisor of 4941)
4941 / 4941 = 1 (the remainder is 0, so 4941 is a divisor of 4941)

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 4941 (i.e. 70.29224708316). 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 $\frac{N}{d} = D$ )

4941 / 1 = 4941 (the remainder is 0, so 1 and 4941 are divisors of 4941)
4941 / 2 = 2470.5 (the remainder is 1, so 2 is not a divisor of 4941)
4941 / 3 = 1647 (the remainder is 0, so 3 and 1647 are divisors of 4941)
...
4941 / 69 = 71.608695652174 (the remainder is 42, so 69 is not a divisor of 4941)
4941 / 70 = 70.585714285714 (the remainder is 41, so 70 is not a divisor of 4941)