# 11 is Only Palindromic Prime with Even Number of Digits

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## Theorem

$11$ is the only palindromic prime with an even number of digits when expressed in decimal notation.

## Proof

Let $P$ be a palindromic number with $2 n$ digits:

- $P = \sqbrk {a_{2 n - 1} a_{2 n - 2} \ldots a_2 a_1 a_0}_{10}$

Thus:

- $P = \displaystyle \sum_{j \mathop = 0}^{n - 1} a_j + 10^{2 n - 1 - j}$

Consider the summation:

- $S = \displaystyle \sum_{k \mathop = 0}^{2 n - 1} \paren {-1}^k a_k$

As $a_k = a_{2 n - 1 - k}$ we have:

\(\displaystyle S\) | \(=\) | \(\displaystyle \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k a_k + \paren {-1}^{2 n - 1 - k} a_k\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{k \mathop = 0}^{n - 1} a_k \paren {\paren {-1}^k + \paren {-1}^{2 n - 1 - k} }\) |

As $2 n - 1$ is odd, it follows that $k$ and $2 n - 1 - k$ are of opposite parity.

Thus:

- $\paren {-1}^k = -\paren {-1}^{2 n - 1 - k}$

and it follows that:

- $S = 0$

It follows by Divisibility by 11 that $P$ is divisible by $11$.

Thus, except for $11$ itself, a palindromic number with an even number of digits cannot be prime.

The result follows.

$\blacksquare$

## Sources

- 1986: David Wells:
*Curious and Interesting Numbers*... (previous) ... (next): $11$ - 1997: David Wells:
*Curious and Interesting Numbers*(2nd ed.) ... (previous) ... (next): $11$