What are the divisors of 2557?

1, 2557

There is a total of 2 positive divisors.
The sum of these divisors is 2558.
The arithmetic mean is 1279.

2 odd divisors

1, 2557

How to compute the divisors of 2557?

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 2557 by each of the numbers from 1 to 2557 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)

2557 / 1 = 2557 (the remainder is 0, so 1 is a divisor of 2557)
2557 / 2 = 1278.5 (the remainder is 1, so 2 is not a divisor of 2557)
2557 / 3 = 852.33333333333 (the remainder is 1, so 3 is not a divisor of 2557)
...
2557 / 2556 = 1.0003912363067 (the remainder is 1, so 2556 is not a divisor of 2557)
2557 / 2557 = 1 (the remainder is 0, so 2557 is a divisor of 2557)

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 2557 (i.e. 50.566787519082). 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 $\frac{N}{d} = D$ )

2557 / 1 = 2557 (the remainder is 0, so 1 and 2557 are divisors of 2557)
2557 / 2 = 1278.5 (the remainder is 1, so 2 is not a divisor of 2557)
2557 / 3 = 852.33333333333 (the remainder is 1, so 3 is not a divisor of 2557)
...
2557 / 49 = 52.183673469388 (the remainder is 9, so 49 is not a divisor of 2557)
2557 / 50 = 51.14 (the remainder is 7, so 50 is not a divisor of 2557)