What are the divisors of 2561?

1, 13, 197, 2561

There is a total of 4 positive divisors.
The sum of these divisors is 2772.
The arithmetic mean is 693.

4 odd divisors

1, 13, 197, 2561

How to compute the divisors of 2561?

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 2561 by each of the numbers from 1 to 2561 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)

2561 / 1 = 2561 (the remainder is 0, so 1 is a divisor of 2561)
2561 / 2 = 1280.5 (the remainder is 1, so 2 is not a divisor of 2561)
2561 / 3 = 853.66666666667 (the remainder is 2, so 3 is not a divisor of 2561)
...
2561 / 2560 = 1.000390625 (the remainder is 1, so 2560 is not a divisor of 2561)
2561 / 2561 = 1 (the remainder is 0, so 2561 is a divisor of 2561)

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 2561 (i.e. 50.60632371552). 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 $\frac{N}{d} = D$ )

2561 / 1 = 2561 (the remainder is 0, so 1 and 2561 are divisors of 2561)
2561 / 2 = 1280.5 (the remainder is 1, so 2 is not a divisor of 2561)
2561 / 3 = 853.66666666667 (the remainder is 2, so 3 is not a divisor of 2561)
...
2561 / 49 = 52.265306122449 (the remainder is 13, so 49 is not a divisor of 2561)
2561 / 50 = 51.22 (the remainder is 11, so 50 is not a divisor of 2561)