# Number of $n$-digit palindromes

How can one count the number of all $$n$$-digit palindromes? Is there any recurrence for that?

I'm not sure if my reasoning is right, but I thought that:

For $$n=1$$, we have $$10$$ such numbers (including $$0$$).
For $$n=2$$, we obviously have $$9$$ possibilities.
For $$n=3$$, we can choose 'extreme digits' in $$9$$ ways. Then there are $$10$$ possibilities for digits in the middle.
For n=4, again we choose extreme digits in $$9$$ ways and middle digits in $$10$$ ways
and, so on.

It seems that for even lengths of numbers we have $$9 \cdot 10^{\frac{n}{2}-1}$$ palindromes and for odd lengths $$9 \cdot 10^{n-2}$$. But this is certainly not even close to a proper solution of this problem.

How do I proceed?

• Hint: How many $7$-digit or $8$-digit palindromes start with $1000$? How many start with $1001$? Jan 26, 2013 at 20:52

Details depend on whether for example $0110$ counts as a $4$-digit palindrome. We will suppose it doesn't. This makes things a little harder.
If $n$ is even, say $n=2m$, the first digit can be any of $9$, then the next $m-1$ can be any of $10$, and then the rest are determined. So there are $9\cdot 10^{m-1}$ palindromes with $2m$ digits.
If $n$ is odd, say $n=2m+1$, then the same sort of reasoning yields the answer $9\cdot 10^{m}$.
An n-digit number abc...xyz can be mapped to the 2n-digit palindrome abc...xyzzyx...bca, and to the (2n-1)-digit palindrome abc...xyzyx...bca. So the number of 2n-digit palindromes and (2n-1)-digit palindromes is simply the number of n-digit numbers: $9 \times 10^{n-1}$.