What are the divisors of 1701?

1, 3, 7, 9, 21, 27, 63, 81, 189, 243, 567, 1701

There is a total of 12 positive divisors.
The sum of these divisors is 2912.
The arithmetic mean is 242.66666666667.

12 odd divisors

1, 3, 7, 9, 21, 27, 63, 81, 189, 243, 567, 1701

How to compute the divisors of 1701?

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 1701 by each of the numbers from 1 to 1701 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)

1701 / 1 = 1701 (the remainder is 0, so 1 is a divisor of 1701)
1701 / 2 = 850.5 (the remainder is 1, so 2 is not a divisor of 1701)
1701 / 3 = 567 (the remainder is 0, so 3 is a divisor of 1701)
...
1701 / 1700 = 1.0005882352941 (the remainder is 1, so 1700 is not a divisor of 1701)
1701 / 1701 = 1 (the remainder is 0, so 1701 is a divisor of 1701)

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 1701 (i.e. 41.243181254603). 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 $\frac{N}{d} = D$ )

1701 / 1 = 1701 (the remainder is 0, so 1 and 1701 are divisors of 1701)
1701 / 2 = 850.5 (the remainder is 1, so 2 is not a divisor of 1701)
1701 / 3 = 567 (the remainder is 0, so 3 and 567 are divisors of 1701)
...
1701 / 40 = 42.525 (the remainder is 21, so 40 is not a divisor of 1701)
1701 / 41 = 41.487804878049 (the remainder is 20, so 41 is not a divisor of 1701)