What are the divisors of 1203?

1, 3, 401, 1203

There is a total of 4 positive divisors.
The sum of these divisors is 1608.
The arithmetic mean is 402.

4 odd divisors

1, 3, 401, 1203

How to compute the divisors of 1203?

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 1203 by each of the numbers from 1 to 1203 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)

1203 / 1 = 1203 (the remainder is 0, so 1 is a divisor of 1203)
1203 / 2 = 601.5 (the remainder is 1, so 2 is not a divisor of 1203)
1203 / 3 = 401 (the remainder is 0, so 3 is a divisor of 1203)
...
1203 / 1202 = 1.0008319467554 (the remainder is 1, so 1202 is not a divisor of 1203)
1203 / 1203 = 1 (the remainder is 0, so 1203 is a divisor of 1203)

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 1203 (i.e. 34.684290392049). 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 $\frac{N}{d} = D$ )

1203 / 1 = 1203 (the remainder is 0, so 1 and 1203 are divisors of 1203)
1203 / 2 = 601.5 (the remainder is 1, so 2 is not a divisor of 1203)
1203 / 3 = 401 (the remainder is 0, so 3 and 401 are divisors of 1203)
...
1203 / 33 = 36.454545454545 (the remainder is 15, so 33 is not a divisor of 1203)
1203 / 34 = 35.382352941176 (the remainder is 13, so 34 is not a divisor of 1203)