What are the divisors of 1973?

1, 1973

There is a total of 2 positive divisors.
The sum of these divisors is 1974.
The arithmetic mean is 987.

2 odd divisors

1, 1973

How to compute the divisors of 1973?

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 1973 by each of the numbers from 1 to 1973 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)

1973 / 1 = 1973 (the remainder is 0, so 1 is a divisor of 1973)
1973 / 2 = 986.5 (the remainder is 1, so 2 is not a divisor of 1973)
1973 / 3 = 657.66666666667 (the remainder is 2, so 3 is not a divisor of 1973)
...
1973 / 1972 = 1.0005070993915 (the remainder is 1, so 1972 is not a divisor of 1973)
1973 / 1973 = 1 (the remainder is 0, so 1973 is a divisor of 1973)

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 1973 (i.e. 44.418464629026). 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 $\frac{N}{d} = D$ )

1973 / 1 = 1973 (the remainder is 0, so 1 and 1973 are divisors of 1973)
1973 / 2 = 986.5 (the remainder is 1, so 2 is not a divisor of 1973)
1973 / 3 = 657.66666666667 (the remainder is 2, so 3 is not a divisor of 1973)
...
1973 / 43 = 45.883720930233 (the remainder is 38, so 43 is not a divisor of 1973)
1973 / 44 = 44.840909090909 (the remainder is 37, so 44 is not a divisor of 1973)