What are the divisors of 2935?

1, 5, 587, 2935

There is a total of 4 positive divisors.
The sum of these divisors is 3528.
The arithmetic mean is 882.

4 odd divisors

1, 5, 587, 2935

How to compute the divisors of 2935?

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 2935 by each of the numbers from 1 to 2935 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)

2935 / 1 = 2935 (the remainder is 0, so 1 is a divisor of 2935)
2935 / 2 = 1467.5 (the remainder is 1, so 2 is not a divisor of 2935)
2935 / 3 = 978.33333333333 (the remainder is 1, so 3 is not a divisor of 2935)
...
2935 / 2934 = 1.0003408316292 (the remainder is 1, so 2934 is not a divisor of 2935)
2935 / 2935 = 1 (the remainder is 0, so 2935 is a divisor of 2935)

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 2935 (i.e. 54.175640282326). 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 $\frac{N}{d} = D$ )

2935 / 1 = 2935 (the remainder is 0, so 1 and 2935 are divisors of 2935)
2935 / 2 = 1467.5 (the remainder is 1, so 2 is not a divisor of 2935)
2935 / 3 = 978.33333333333 (the remainder is 1, so 3 is not a divisor of 2935)
...
2935 / 53 = 55.377358490566 (the remainder is 20, so 53 is not a divisor of 2935)
2935 / 54 = 54.351851851852 (the remainder is 19, so 54 is not a divisor of 2935)