Find the reverse of given number

        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