C# Program - Reverse digits of a given Integer

In C#, the reverse of a given integer can be found out by using below mentioned methods.

Method 1: Using iteration

The method involves the following steps:

```Input: MyNum
Step 1: Initialize the RevNum = 0
Step 2: Iterate over MyNum while it is greater than zero.
Step 2a: Calculate remainder of MuNum / 10
Step 2b: Update RevNum by RevNum * 10 + remainder
Step 2c: Update MyNum by MyNum / 10
Step 3: Return RevNum
```

Example:

```Input: 564

RevNum: 0
Iteration 1:
Remainder: 564 % 10 = 4
RevNum: 0 * 10 + 4 = 4
MyNum: 564 / 10 = 56

Iteration 2:
Remainder: 56 % 10 = 6
RevNum: 4 * 10 + 6 = 46
MyNum: 56 / 10 = 5

Iteration 3:
Remainder: 5 % 10 = 5
RevNum: 46 * 10 + 5 = 465
MyNum: 5 / 10 = 0

return RevNum = 465
```

The below block of code shows the implementation of above concept:

```using System;

class MyProgram {

static int reverse(int MyNum) {
int RevNum = 0;
int remainder;

while(MyNum > 0){
remainder = MyNum % 10;
MyNum = MyNum / 10;
RevNum = RevNum * 10 + remainder;
}
return RevNum;
}

static void Main(string[] args) {
int x = 1285;
int y = 4567;
Console.WriteLine("Reverse of {0} is: {1}", x, reverse(x));
Console.WriteLine("Reverse of {0} is: {1}", y, reverse(y));
}
}
```

The above code will give the following output:

```Reverse of 1285 is: 5821
Reverse of 4567 is: 7654
```

Method 2: Using Recursion

The above result can also be achieved using recursive function:

```using System;

class MyProgram {
static int RevNum = 0;
static int BaseVal = 1;

static int reverse(int MyNum) {
if(MyNum > 0) {
reverse(MyNum/10);
RevNum += MyNum % 10 * BaseVal;
BaseVal *= 10;
}
return RevNum;
}

static void Main(string[] args) {
int x = 7902;
Console.WriteLine("Reverse of {0} is: {1}", x, reverse(x));
}
}
```

The above code will give the following output:

```Reverse of 7902 is: 2097
```

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