A prime number which is also a palindrome. I don't think there's any mathematical significance; it's just a curiosity.

In base 10, the first twenty palindromic primes are:

2, 3, 5, 7, 11, 101, 131, 151, 181, 191, 313, 353, 373, 383, 727, 757, 787, 797, 919, 929
A slightly larger one: 167474761

(Hey, I'm not a geek, I just pretend!)

Update Nov 5 2000:

A nice pyramid of palindromic primes;

```                    2
30203
133020331
1713302033171
12171330203317121
151217133020331712151
1815121713302033171215181
16181512171330203317121518161
331618151217133020331712151816133
9333161815121713302033171215181613339
11933316181512171330203317121518161333911
```

And as a final snack: 10^35352 + 2049402*10^17673 + 1 is a palindromic prime with a 35353 digits - which again is a palindromic prime! Whee!!

Prove that all palindromic prime numbers greater than 11 have an odd number of digits.

Rough proof outline (need pure mathematicians to make it rigorous):
• 1001 = 91 * 11; 100,001 = 9,091 * 11; 10,000,001 = 909,091 * 11 (this can be proved with mathematical induction)
• All palindromic numbers with an even number of digits can be trivially represented by sums of terms of the form 1000..0001 * 10^x. For example, 13377331 = 1 * 10000001 + 3 * 100001 * 10 + 3 * 1001 * 100 + 7 * 11 * 1000.
• All these terms are multiples of 11; the sum also must be a multiple of 11 and is therefore not prime. The contrapositive of this is that any palindrome that is not a multiple of 11 must have an odd number of digits; QED.

To put it another way, ariels writes: "what you prove there stems from the following test for divisibility by 11: add the digits, alternating `+' and `-' signs; the number is prime iff this `+-sum' of the digits is too."

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