# C# Program - Find all factors of a Number

Objective: Write a C# program to find all distinct factors (divisors) of a given natural number. The divisors of few numbers are given below:

``` Number: 10
Divisors: 1 2 5 10

Number: 15
Divisors: 1 3 5 15

Number: 100
Divisors: 1 2 4 5 10 20 25 50 100
```

### Method 1: Using iteration

One of the basic approach is to iterate from 1 to n and in each iteration check whether the number divides n. If it divides then print it.

```using System;

class MyProgram {
//method to print all divisors of a number
static void printDivisors(int n) {
Console.Write("Divisors of " + n + " are: ");
for(int i = 1; i <= n; i++) {
if(n%i == 0)
Console.Write(i + " ");
}
Console.WriteLine();
}

static void Main(string[] args) {
printDivisors(10);
printDivisors(50);
printDivisors(100);
}
}
```

The above code will give the following output:

```Divisors of 10 are: 1 2 5 10
Divisors of 50 are: 1 2 5 10 25 50
Divisors of 100 are: 1 2 4 5 10 20 25 50 100
```

### Method 2: Optimized Code

Instead of checking the divisibility of the given number from 1 to n, it is checked till square root of n. For a factor larger than square root of n, there must the a smaller factor which is already checked in the range of 1 to square root of n.

```using System;

class MyProgram {
//method to print all divisors of a number
static void printDivisors(int n) {
Console.Write("Divisors of " + n + " are: ");
//loop from 1 to Math.Sqrt(n)
for(int i = 1; i <= Math.Sqrt(n); i++) {
if(n%i == 0) {
if(n/i == i)
Console.Write(i + " ");
else
Console.Write(i + " " + n/i + " ");
}
}
Console.WriteLine();
}

static void Main(string[] args) {
printDivisors(10);
printDivisors(50);
printDivisors(100);
}
}
```

The above code will give the following output:

```Divisors of 10 are: 1 10 2 5
Divisors of 50 are: 1 50 2 25 5 10
Divisors of 100 are: 1 100 2 50 4 25 5 20 10
```

### Method 3: Optimized Code with sorted result

In the previous method, results are produced in a irregular fashion (printed in pairs - small number and large number). The result can be sorted by storing the larger number and print them later on. Consider the example below:

```using System;

class MyProgram {
//method to print all divisors of a number
static void printDivisors(int n) {
Console.Write("Divisors of " + n + " are: ");

//creating an array to store larger numbers
int[] arr = new int[n];
int j = 0;

//loop from 1 to Math.Sqrt(n)
for(int i = 1; i <= Math.Sqrt(n); i++) {
if(n%i == 0) {
if(n/i == i)
Console.Write(i + " ");
else {
Console.Write(i + " ");
//storing the large number of a pair
arr[j++] = n/i;
}
}
}

//printing stored large numbers of pairs
for(int i = j - 1; i >= 0; i--)
Console.Write(arr[i] + " ");
Console.WriteLine();
}

static void Main(string[] args) {
printDivisors(10);
printDivisors(50);
printDivisors(100);
}
}
```

The above code will give the following output:

```Divisors of 10 are: 1 2 5 10
Divisors of 50 are: 1 2 5 10 25 50
Divisors of 100 are: 1 2 4 5 10 20 25 50 100
```

### Method 4: Another Optimized Code

To produce the result in sorted order, we can iterate from 1 to square root of n and printing the number which divides n. After that we can iterate back (in reverse order) and printing the quotient of all numbers which divides n.

```using System;

class MyProgram {
//method to print all divisors of a number
static void printDivisors(int n) {
Console.Write("Divisors of " + n + " are: ");
//loop from 1 to Math.Sqrt(n)
int i;
for(i = 1; i <= Math.Sqrt(n); i++) {
if(n%i == 0)
Console.Write(i + " ");

//handing perfect squares
if(n/i == i) {
i--; break;
}
}

for(; i >= 1; i--) {
if(n%i == 0)
Console.Write(n/i + " ");
}
Console.WriteLine();
}

static void Main(string[] args) {
printDivisors(10);
printDivisors(50);
printDivisors(100);
}
}
```

The above code will give the following output:

```Divisors of 10 are: 1 2 5 10
Divisors of 50 are: 1 2 5 10 25 50
Divisors of 100 are: 1 2 4 5 10 20 25 50 100
```

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