What are the divisors of 1977?

1, 3, 659, 1977

There is a total of 4 positive divisors.
The sum of these divisors is 2640.
The arithmetic mean is 660.

4 odd divisors

1, 3, 659, 1977

How to compute the divisors of 1977?

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 1977 by each of the numbers from 1 to 1977 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)

1977 / 1 = 1977 (the remainder is 0, so 1 is a divisor of 1977)
1977 / 2 = 988.5 (the remainder is 1, so 2 is not a divisor of 1977)
1977 / 3 = 659 (the remainder is 0, so 3 is a divisor of 1977)
...
1977 / 1976 = 1.0005060728745 (the remainder is 1, so 1976 is not a divisor of 1977)
1977 / 1977 = 1 (the remainder is 0, so 1977 is a divisor of 1977)

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 1977 (i.e. 44.463468150831). 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 $\frac{N}{d} = D$ )

1977 / 1 = 1977 (the remainder is 0, so 1 and 1977 are divisors of 1977)
1977 / 2 = 988.5 (the remainder is 1, so 2 is not a divisor of 1977)
1977 / 3 = 659 (the remainder is 0, so 3 and 659 are divisors of 1977)
...
1977 / 43 = 45.976744186047 (the remainder is 42, so 43 is not a divisor of 1977)
1977 / 44 = 44.931818181818 (the remainder is 41, so 44 is not a divisor of 1977)