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Even-Digit Palindromes Divisible by 11

Date: 12/08/97 at 14:14:27
From: Bob
Subject: Even-Digit Palindromes divisible by 11

When I play with a calculator or computer, I notice that every even-
digit palindromic number is divisible by 11:

6006/11 = 546

135531/11 = 12321

64899846/11 =  5899986

Could this be proved as a theorem?


Date: 12/08/97 at 16:51:35
From: Doctor Rob
Subject: Re: Even-Digit Palindromes divisible by 11

Yes, it can. Even-digit palindromes are of the following form. They
are a sum of terms, each consisting of a single digit times a power 
of 10 times a number which is one greater than an odd power of 10.


64899846 = 60000006 + 4000040 + 800800 + 99000
  = 6*10^0*(10^7+1) + 4*10^1*(10^5+1) + 8*10^2*(10^3+1) + 

The key is that 11 always divides evenly into 10^e+1 if e is odd.  
Prove this by proving that x+1 divides evenly into x^e+1 if e is odd, 
and then substitute x = 10. The quotient is 

  x^(e-1) - x^(e-2) + ... + x^2 - x + 1

This can also be proved using mathematical induction.

Since 11 divides evenly into 10^e+1, it divides evenly into every term
in the sum, and thus it divides evenly into the palindrome in 

-Doctor Rob,  The Math Forum
 Check out our web site!   

Date: 12/08/97 at 23:34:51
From: FTW1
Subject: Re: Even-Digit Palindromes divisible by 11

Thanks Dr. Rob!

Associated Topics:
High School Number Theory

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