What are the divisors of 4902?

1, 2, 3, 6, 19, 38, 43, 57, 86, 114, 129, 258, 817, 1634, 2451, 4902

There is a total of 16 positive divisors.
The sum of these divisors is 10560.
The arithmetic mean is 660.

8 even divisors

2, 6, 38, 86, 114, 258, 1634, 4902

8 odd divisors

1, 3, 19, 43, 57, 129, 817, 2451

How to compute the divisors of 4902?

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 4902 by each of the numbers from 1 to 4902 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)

4902 / 1 = 4902 (the remainder is 0, so 1 is a divisor of 4902)
4902 / 2 = 2451 (the remainder is 0, so 2 is a divisor of 4902)
4902 / 3 = 1634 (the remainder is 0, so 3 is a divisor of 4902)
...
4902 / 4901 = 1.0002040399918 (the remainder is 1, so 4901 is not a divisor of 4902)
4902 / 4902 = 1 (the remainder is 0, so 4902 is a divisor of 4902)

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 4902 (i.e. 70.014284256857). 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 $\frac{N}{d} = D$ )

4902 / 1 = 4902 (the remainder is 0, so 1 and 4902 are divisors of 4902)
4902 / 2 = 2451 (the remainder is 0, so 2 and 2451 are divisors of 4902)
4902 / 3 = 1634 (the remainder is 0, so 3 and 1634 are divisors of 4902)
...
4902 / 69 = 71.04347826087 (the remainder is 3, so 69 is not a divisor of 4902)
4902 / 70 = 70.028571428571 (the remainder is 2, so 70 is not a divisor of 4902)