What are the divisors of 1971?

1, 3, 9, 27, 73, 219, 657, 1971

There is a total of 8 positive divisors.
The sum of these divisors is 2960.
The arithmetic mean is 370.

8 odd divisors

1, 3, 9, 27, 73, 219, 657, 1971

How to compute the divisors of 1971?

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 1971 by each of the numbers from 1 to 1971 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)

1971 / 1 = 1971 (the remainder is 0, so 1 is a divisor of 1971)
1971 / 2 = 985.5 (the remainder is 1, so 2 is not a divisor of 1971)
1971 / 3 = 657 (the remainder is 0, so 3 is a divisor of 1971)
...
1971 / 1970 = 1.0005076142132 (the remainder is 1, so 1970 is not a divisor of 1971)
1971 / 1971 = 1 (the remainder is 0, so 1971 is a divisor of 1971)

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 1971 (i.e. 44.395945760846). 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 $\frac{N}{d} = D$ )

1971 / 1 = 1971 (the remainder is 0, so 1 and 1971 are divisors of 1971)
1971 / 2 = 985.5 (the remainder is 1, so 2 is not a divisor of 1971)
1971 / 3 = 657 (the remainder is 0, so 3 and 657 are divisors of 1971)
...
1971 / 43 = 45.837209302326 (the remainder is 36, so 43 is not a divisor of 1971)
1971 / 44 = 44.795454545455 (the remainder is 35, so 44 is not a divisor of 1971)