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From: greg@harvard.UUCP (Greg)
Newsgroups: net.math
Subject: Re: Nifty math problem
Message-ID: <779@harvard.UUCP>
Date: Sat, 15-Mar-86 17:37:53 EST
Article-I.D.: harvard.779
Posted: Sat Mar 15 17:37:53 1986
Date-Received: Mon, 17-Mar-86 04:17:52 EST
References: <2222@jhunix.UUCP>
Reply-To: greg@harvard.UUCP (Greg)
Distribution: net
Organization: Harvard
Lines: 22

In article <2222@jhunix.UUCP> ins_adsf@jhunix.UUCP (David S Fry) writes:
>
>	Here's a nifty yet simple number theory problem.  Prove that there
>is a unique palindromic prime number with an even number of digits.  Generalize
>to other bases.

The number 100....01, if it has an even number of digits, is divisible by
11.  Any palindrome with an even number of digits is the sum of such numbers.
Therefore all palindromes with evenly many digits are multiples of 11, so
the only such prime is 11 itself.

In base p-1, where p is prime, p is the only prime palindrome with evenly many
digits.  In other bases, all such numbers are composite.

Now for a more interesting problem in elementary number theory:

Let C(a,b) = a!/(b!*(a-b)!), and let p>3 be a prime number.
Show that:

C(p*a,p*b) = C(a,b) mod p^3
-- 
gregregreg