For any given number we would need to return the number in reverse, with the starting place to the end place. For example, for an input of 1423 we would need to return 3241 which is its exact reverse. To achieve this, we would use the same principle we use for splitting the number; the Divisibility by 10 rule.

```
public static int GetReversedNumber(int n)
{
int r, sum = 0, input = n;
if (n < 10) return n;
while (n != 0)
{
r = n % 10;
sum = sum * 10 + r;
n = n / 10;
}
return sum;
}
```

**How it works:**

The logic starts in the same way as splitting the numbers, but for every time a remainder is obtained, it is added up on to a sum and before adding up the remainder the sum is multiplied with a 10 so that it grows in one place in the number.

```
For an input 1423:
sum = 0*10 + 3 = 3
sum = 3*10 + 2 = 32
sum = 32*10 + 4 = 324
sum = 324*10 + 1 = 3241
```

Loops • Added one month ago