CareerCup

Perhaps as 999999 overflows to 000000 , is the answer 1 ?

11 always irrespective of the number of digits
00 - 11
090 - 101
0990 - 1001
09990 - 10001
etc.,

Depends on the number of digits in the odometer, the answer is different.
1 digit -> 1, from n->n+1
2 digits -> 11, from nn->(n+1)(n+1)
3 digits -> 10, from xnx -> x(n+1)x
4+ digits -> 11, from n99n->(n+1)00(n+1)

@Jay has got the right answer

000 - 010 = 10 miles

are not we suppose to write the code to solve this problem? Input is given as a past number and based on that next number should be determined? If it's just for this one tricky question, the question loses it's charm.

The way I saw it,

if the current reading is 12321
then we start at the middle digit, in this case 3
and add 1 to it
then the new number becomes 12421 which is the next smallest palindromic number.

If the number of digits is even like 123321
then we take two middle digits in this case 3 and 3
and add 1 to both which becomes 124421.
In case of 9's we process it a little differently

eg 1239321
here middle digit is 9
so instead of adding 1 to it, we put 0 in that position
and look at next and previous digits of 9 in that number.
In this case they are 3 and 3.
so we can add 1s to both of them, so we put 1s in those places
and then append the 0s to complete the number.
Here final number will be 10100 which gives us palindromic number 1249421
Number 99999... is a special case. Just adding 2 to this number will give us the answer.
Here's the C code for it.
Please let me know if I missed something.
(The code assumes that we get he reading in the form of an array of integers)

double palindRead( const int * reading,
	                        const int n ) //size of the array
{
	//if the odometer reading is just 1 digit
        if(n < 2) {
                if(reading[0] == 9) {
                        return 2;
                } else {
                        return 1;
                }
        }

        int index1, index2;
        double result;

        index1 = n/2;
        if(n%2 != 0) {
                index2 = index1;
                result = pow(10,(n-index2-1));
        } else {
                index2 = index1 - 1;
                result = pow(10, (n-index2-1)) + pow(10, (n-index1-1));
        }
        printf("%f\n", result);
        printf("%d\n", index1);
        printf("%d\n", index2);
        while(index1 < n) {
                if(reading[index1]+1 < 10
                   && reading[index2]+1 < 10) {
                        break;
                } else if(index1 == n-1
                          && reading[index1] == 9) {
                        return 2;
                } else {
                        index2--;
                        index1++;
                        result = pow(10, (n-index2-1)) + pow(10, (n-index1-1));
                }
        }

        return result;

}