**Problem :-**Write A C++ Program To Print A Reverse Order Of Any Number Using Loop

**Logic :-**Very Simple Logic Just Divide a number by 10 and then proceed for next statement

Sum=0

sum=sum*10+X

Like you can also solve the problem

while(n!=0)

{

rev=n%10;

System.out.print("Reversed Number = "+reverse);

n=n/10

}

but in this way number is printing actually not reversing .

If you Understood then try to solve given problem .

**Note :-**Always remember Do Not try to print the reverse Digit always try to modifiedLike you can also solve the problem

while(n!=0)

{

rev=n%10;

System.out.print("Reversed Number = "+reverse);

n=n/10

}

but in this way number is printing actually not reversing .

If you Understood then try to solve given problem .

Check this Geeksforgeeks Solution For " Reverse digit "

**Solution :-**
#include<iostream>

using namespace std;

int main()

{

//By-Ghanendra Yadav

int n,x,sum=0;

cout<<"Enter The Number To Be Reverse: \n";

cin>>n;

while(n>0)

{

x=n%10;

sum=sum*10+x;

n=n/10;

}

cout<<"\nThe Reverse Number is "<<sum;

}

using namespace std;

int main()

{

//By-Ghanendra Yadav

int n,x,sum=0;

cout<<"Enter The Number To Be Reverse: \n";

cin>>n;

while(n>0)

{

x=n%10;

sum=sum*10+x;

n=n/10;

}

cout<<"\nThe Reverse Number is "<<sum;

}

**Output:-**

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