What are the divisors of 1206?

1, 2, 3, 6, 9, 18, 67, 134, 201, 402, 603, 1206

There is a total of 12 positive divisors.
The sum of these divisors is 2652.
The arithmetic mean is 221.

6 even divisors

2, 6, 18, 134, 402, 1206

6 odd divisors

1, 3, 9, 67, 201, 603

How to compute the divisors of 1206?

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 1206 by each of the numbers from 1 to 1206 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)

1206 / 1 = 1206 (the remainder is 0, so 1 is a divisor of 1206)
1206 / 2 = 603 (the remainder is 0, so 2 is a divisor of 1206)
1206 / 3 = 402 (the remainder is 0, so 3 is a divisor of 1206)
...
1206 / 1205 = 1.0008298755187 (the remainder is 1, so 1205 is not a divisor of 1206)
1206 / 1206 = 1 (the remainder is 0, so 1206 is a divisor of 1206)

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 1206 (i.e. 34.727510708371). 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 $\frac{N}{d} = D$ )

1206 / 1 = 1206 (the remainder is 0, so 1 and 1206 are divisors of 1206)
1206 / 2 = 603 (the remainder is 0, so 2 and 603 are divisors of 1206)
1206 / 3 = 402 (the remainder is 0, so 3 and 402 are divisors of 1206)
...
1206 / 33 = 36.545454545455 (the remainder is 18, so 33 is not a divisor of 1206)
1206 / 34 = 35.470588235294 (the remainder is 16, so 34 is not a divisor of 1206)