What are the divisors of 1981?

1, 7, 283, 1981

There is a total of 4 positive divisors.
The sum of these divisors is 2272.
The arithmetic mean is 568.

4 odd divisors

1, 7, 283, 1981

How to compute the divisors of 1981?

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 1981 by each of the numbers from 1 to 1981 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)

1981 / 1 = 1981 (the remainder is 0, so 1 is a divisor of 1981)
1981 / 2 = 990.5 (the remainder is 1, so 2 is not a divisor of 1981)
1981 / 3 = 660.33333333333 (the remainder is 1, so 3 is not a divisor of 1981)
...
1981 / 1980 = 1.0005050505051 (the remainder is 1, so 1980 is not a divisor of 1981)
1981 / 1981 = 1 (the remainder is 0, so 1981 is a divisor of 1981)

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 1981 (i.e. 44.508426168536). 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 $\frac{N}{d} = D$ )

1981 / 1 = 1981 (the remainder is 0, so 1 and 1981 are divisors of 1981)
1981 / 2 = 990.5 (the remainder is 1, so 2 is not a divisor of 1981)
1981 / 3 = 660.33333333333 (the remainder is 1, so 3 is not a divisor of 1981)
...
1981 / 43 = 46.06976744186 (the remainder is 3, so 43 is not a divisor of 1981)
1981 / 44 = 45.022727272727 (the remainder is 1, so 44 is not a divisor of 1981)