What are the divisors of 1764?

1, 2, 3, 4, 6, 7, 9, 12, 14, 18, 21, 28, 36, 42, 49, 63, 84, 98, 126, 147, 196, 252, 294, 441, 588, 882, 1764

There is a total of 27 positive divisors.
The sum of these divisors is 5187.
The arithmetic mean is 192.11111111111.

18 even divisors

2, 4, 6, 12, 14, 18, 28, 36, 42, 84, 98, 126, 196, 252, 294, 588, 882, 1764

9 odd divisors

1, 3, 7, 9, 21, 49, 63, 147, 441

How to compute the divisors of 1764?

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 1764 by each of the numbers from 1 to 1764 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)

1764 / 1 = 1764 (the remainder is 0, so 1 is a divisor of 1764)
1764 / 2 = 882 (the remainder is 0, so 2 is a divisor of 1764)
1764 / 3 = 588 (the remainder is 0, so 3 is a divisor of 1764)
...
1764 / 1763 = 1.0005672149745 (the remainder is 1, so 1763 is not a divisor of 1764)
1764 / 1764 = 1 (the remainder is 0, so 1764 is a divisor of 1764)

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 1764 (i.e. 42). 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 $\frac{N}{d} = D$ )

1764 / 1 = 1764 (the remainder is 0, so 1 and 1764 are divisors of 1764)
1764 / 2 = 882 (the remainder is 0, so 2 and 882 are divisors of 1764)
1764 / 3 = 588 (the remainder is 0, so 3 and 588 are divisors of 1764)
...
1764 / 41 = 43.024390243902 (the remainder is 1, so 41 is not a divisor of 1764)
1764 / 42 = 42 (the remainder is 0, so 42 and 42 are divisors of 1764)