What are the divisors of 4918?

1, 2, 2459, 4918

There is a total of 4 positive divisors.
The sum of these divisors is 7380.
The arithmetic mean is 1845.

2 even divisors

2, 4918

2 odd divisors

1, 2459

How to compute the divisors of 4918?

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 4918 by each of the numbers from 1 to 4918 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)

4918 / 1 = 4918 (the remainder is 0, so 1 is a divisor of 4918)
4918 / 2 = 2459 (the remainder is 0, so 2 is a divisor of 4918)
4918 / 3 = 1639.3333333333 (the remainder is 1, so 3 is not a divisor of 4918)
...
4918 / 4917 = 1.0002033760423 (the remainder is 1, so 4917 is not a divisor of 4918)
4918 / 4918 = 1 (the remainder is 0, so 4918 is a divisor of 4918)

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 4918 (i.e. 70.128453569147). 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 $\frac{N}{d} = D$ )

4918 / 1 = 4918 (the remainder is 0, so 1 and 4918 are divisors of 4918)
4918 / 2 = 2459 (the remainder is 0, so 2 and 2459 are divisors of 4918)
4918 / 3 = 1639.3333333333 (the remainder is 1, so 3 is not a divisor of 4918)
...
4918 / 69 = 71.275362318841 (the remainder is 19, so 69 is not a divisor of 4918)
4918 / 70 = 70.257142857143 (the remainder is 18, so 70 is not a divisor of 4918)