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        riddles >> putnam exam (pure math) >> Palindromic Prime with an even number of digits
        
(Message started by: mfirmata on Mar 15^th, 2005, 2:41pm)

Title: Palindromic Prime with an even number of digits
Post by mfirmata on Mar 15^th, 2005, 2:41pm

In a Clifford Pickover tear-a-day he asks, "What is the only palindromic prime number with an even number of digits?"

His answer is 11. 11 is certainly an example of a palindromic prime with an even number of digits, but how can he be sure it's the only one? What if there exists a prime number with, say, 100 billion and 2 digits in it that is also palindromic. Is there a proof for what he states?

At best, I'm a recreational mathematician, so this type of thing is out of league (not understanding a proof, just attempting one.)

I tried to e-mail Clifford Pickover (I've found other problems he posed whose answers were either wrong or else the problem was poorly stated) and it bounced back. I guess the story could be that he simply didn't apend, "under 1,000" to his question or something careless like that.

I apologize if this is in the wrong forum. Thanks

Title: Re: Palindromic Prime with an even number of digit
Post by Grimbal on Mar 15^th, 2005, 2:52pm

I think it is true.

11 is a multiple of 11,
1001 is a multiple of 11,
100001 is a multiple of 11,
etc.
So, for instance, 314159951413 is a multiple of 11

Title: Re: Palindromic Prime with an even number of digit
Post by Icarus on Mar 15^th, 2005, 5:35pm

Grimbal is correct. (xⁿ + 1) = (x + 1)(x^n-1 - x^n-2 + ... - x + 1) when n is odd. In particular, when x =10 we get that 11 divides 10ⁿ + 1 for all odd n.

Since even digit palindromic numbers are all of the form [sum] a_i(10^2i-1 + 1), where the a_i are the digits of the palindrome, they are all divisible by 11.

So the only one that can be prime is 11 itself.

Title: Re: Palindromic Prime with an even number of digit
Post by markr on Mar 15^th, 2005, 9:58pm

A test for divisibility by 11 is to add the digits of a number (alternating the sign of each digit as you go). If the result is a multiple of 11, then so is the original number. In a palindrome of even length, the sum would be zero.