What are the divisors of 2905?

1, 5, 7, 35, 83, 415, 581, 2905

There is a total of 8 positive divisors.
The sum of these divisors is 4032.
The arithmetic mean is 504.

8 odd divisors

1, 5, 7, 35, 83, 415, 581, 2905

How to compute the divisors of 2905?

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 2905 by each of the numbers from 1 to 2905 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)

2905 / 1 = 2905 (the remainder is 0, so 1 is a divisor of 2905)
2905 / 2 = 1452.5 (the remainder is 1, so 2 is not a divisor of 2905)
2905 / 3 = 968.33333333333 (the remainder is 1, so 3 is not a divisor of 2905)
...
2905 / 2904 = 1.0003443526171 (the remainder is 1, so 2904 is not a divisor of 2905)
2905 / 2905 = 1 (the remainder is 0, so 2905 is a divisor of 2905)

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 2905 (i.e. 53.898051912847). 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 $\frac{N}{d} = D$ )

2905 / 1 = 2905 (the remainder is 0, so 1 and 2905 are divisors of 2905)
2905 / 2 = 1452.5 (the remainder is 1, so 2 is not a divisor of 2905)
2905 / 3 = 968.33333333333 (the remainder is 1, so 3 is not a divisor of 2905)
...
2905 / 52 = 55.865384615385 (the remainder is 45, so 52 is not a divisor of 2905)
2905 / 53 = 54.811320754717 (the remainder is 43, so 53 is not a divisor of 2905)