# Dart: Reverse an integer number

In this challenge, we are going to reverse an integer number. For example, if you got the digits **321**, reversing
them would give you **123**

This solution will use a while loop, modulo, and integer division.

First, we need to obtain the last digit of the given number, which is always the remainder when dividing it by **10**.
For example, to get the **1** from the number **321**:

When we divide by 10, the remainder is always the last digit

Now that we know how to get the last digit of the given number, we need to store that digit in a variable and remove it from the given number.

`int number = 321;`

int lastDigit = number % 10;

// TODO: remove the digit from the given number

number = ???????????;

Removing the last digit from the given number is straightforward. We divide by **10**, and the quotient will
not contain the last digit:

The quotient or final answer does not include the last digit

So our code would look like this:

`int number = 321;`

int lastDigit = number % 10;

number ~/= 10;

Now, we need to repeat this code as long as the number is not equal to **0**, so we'll add a loop:

`int number = 321;`

// loop while the number is not 0

while (number != 0) {

int lastDigit = number % 10;

number ~/= 10;

}

Finally, we must create a variable `int reversed`

to add the last digit. Since we need to add the digit to the right
side of the previous digit each iteration, and we can't use concatenation, we'll multiply the reversed number
by **10** to add the digit. The final code will look like this:

I have noticed that some people say this code does not work if the first digit of the given number is **0**. For
example, the number **0123** should be reversed to **3210**. Well, the answer is that when storing **0123** in an
integer variable, it automatically becomes **123**, as the leading zero has no value. Therefore, when reversed,
it becomes **321**.