how to derive this formula?

Final formula = 9∗pow(10,n/2-1)

10^N -1 means 9999…N times.

We need A + rev(A) = 10^N -1.

So, if first digit of A is x, last digit will be 9-x, and same goes for second and second last and so on.

This means If N%2 != 0, ans is 0, as middle digit of odd length number will be the same, say x, but 2x != 9 in any case, so ans = 0.

First 9 is there because we cannot have leading zeroes, therefor 9 choices for first digit.

For remaining, we see that we can select any of the 10 digit for N/2 -1 places (rest will be determined by taking 9-corresponding digit from left side).

So, N/2-1 times 10 choices * 9 choicea for first digit gives 9*10^(N/2-1).

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Friend, whenever asking regarding a problem, do include problem name in question heading. :slight_smile:

great @taran_1407 thanks.

I think that since you help at least no laddu then also something like certificate or anything should be given to you contributors

No problem friend…

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