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?

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

2 Answers 2


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 $9\cdot 10^{m}$.

  • $\begingroup$ Thanks. I see I was wrong about odd ns. $\endgroup$
    – Hagrid
    Jan 26, 2013 at 21:05
  • $\begingroup$ @Hagrid: In the odd case, the "middle" person can be anything. $\endgroup$ Jan 26, 2013 at 22:07

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: $9 \times 10^{n-1}$.


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