What are the divisors of 650?

1, 2, 5, 10, 13, 25, 26, 50, 65, 130, 325, 650

There is a total of 12 positive divisors.
The sum of these divisors is 1302.
The arithmetic mean is 108.5.

6 even divisors

2, 10, 26, 50, 130, 650

6 odd divisors

1, 5, 13, 25, 65, 325

How to compute the divisors of 650?

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 650 by each of the numbers from 1 to 650 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)

650 / 1 = 650 (the remainder is 0, so 1 is a divisor of 650)
650 / 2 = 325 (the remainder is 0, so 2 is a divisor of 650)
650 / 3 = 216.66666666667 (the remainder is 2, so 3 is not a divisor of 650)
...
650 / 649 = 1.0015408320493 (the remainder is 1, so 649 is not a divisor of 650)
650 / 650 = 1 (the remainder is 0, so 650 is a divisor of 650)

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 650 (i.e. 25.495097567964). 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 $\frac{N}{d} = D$ )

650 / 1 = 650 (the remainder is 0, so 1 and 650 are divisors of 650)
650 / 2 = 325 (the remainder is 0, so 2 and 325 are divisors of 650)
650 / 3 = 216.66666666667 (the remainder is 2, so 3 is not a divisor of 650)
...
650 / 24 = 27.083333333333 (the remainder is 2, so 24 is not a divisor of 650)
650 / 25 = 26 (the remainder is 0, so 25 and 26 are divisors of 650)