What are the divisors of 1984?

1, 2, 4, 8, 16, 31, 32, 62, 64, 124, 248, 496, 992, 1984

There is a total of 14 positive divisors.
The sum of these divisors is 4064.
The arithmetic mean is 290.28571428571.

12 even divisors

2, 4, 8, 16, 32, 62, 64, 124, 248, 496, 992, 1984

2 odd divisors

1, 31

How to compute the divisors of 1984?

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 1984 by each of the numbers from 1 to 1984 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)

1984 / 1 = 1984 (the remainder is 0, so 1 is a divisor of 1984)
1984 / 2 = 992 (the remainder is 0, so 2 is a divisor of 1984)
1984 / 3 = 661.33333333333 (the remainder is 1, so 3 is not a divisor of 1984)
...
1984 / 1983 = 1.0005042864347 (the remainder is 1, so 1983 is not a divisor of 1984)
1984 / 1984 = 1 (the remainder is 0, so 1984 is a divisor of 1984)

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 1984 (i.e. 44.54211490264). 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 $\frac{N}{d} = D$ )

1984 / 1 = 1984 (the remainder is 0, so 1 and 1984 are divisors of 1984)
1984 / 2 = 992 (the remainder is 0, so 2 and 992 are divisors of 1984)
1984 / 3 = 661.33333333333 (the remainder is 1, so 3 is not a divisor of 1984)
...
1984 / 43 = 46.139534883721 (the remainder is 6, so 43 is not a divisor of 1984)
1984 / 44 = 45.090909090909 (the remainder is 4, so 44 is not a divisor of 1984)