What are the divisors of 1969?

1, 11, 179, 1969

There is a total of 4 positive divisors.
The sum of these divisors is 2160.
The arithmetic mean is 540.

4 odd divisors

1, 11, 179, 1969

How to compute the divisors of 1969?

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 1969 by each of the numbers from 1 to 1969 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)

1969 / 1 = 1969 (the remainder is 0, so 1 is a divisor of 1969)
1969 / 2 = 984.5 (the remainder is 1, so 2 is not a divisor of 1969)
1969 / 3 = 656.33333333333 (the remainder is 1, so 3 is not a divisor of 1969)
...
1969 / 1968 = 1.0005081300813 (the remainder is 1, so 1968 is not a divisor of 1969)
1969 / 1969 = 1 (the remainder is 0, so 1969 is a divisor of 1969)

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 1969 (i.e. 44.373415464668). 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 $\frac{N}{d} = D$ )

1969 / 1 = 1969 (the remainder is 0, so 1 and 1969 are divisors of 1969)
1969 / 2 = 984.5 (the remainder is 1, so 2 is not a divisor of 1969)
1969 / 3 = 656.33333333333 (the remainder is 1, so 3 is not a divisor of 1969)
...
1969 / 43 = 45.790697674419 (the remainder is 34, so 43 is not a divisor of 1969)
1969 / 44 = 44.75 (the remainder is 33, so 44 is not a divisor of 1969)