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Divisibility by 11 for all palindromes with an even number of digits 
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#1
May2510, 06:19 AM

P: 17

1. The problem statement, all variables and given/known data
Prove that every palindromic integer N in base 10 with an even number of digits is divisible by 11. Then prove that every palindromic integer in base k with an even number of digits is divisible by k+1. 2. Relevant equations palindromic means the number reads the same forwards and backwards 3. The attempt at a solution I tried a general representation of a palindromic integer with even digits of form D1*10^(2n1) + D2*10^(2n2) + ... +Dn * 10^n + Dn * 10^(n1) +...+ D2*10^1 +D1*10^0 which makes the number N equal to N= D1[10^(2n1) + 10^0] + D2 [10^(2n2) + 10^1] +...+Dn[10^n + 10^(n1)]. But I can't figure out how to prove divisibility by 11. Thanks for any help! 


#2
May2510, 07:01 AM

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P: 3,307

how about starting with a low form & seeing if you can generalise from there....
so start with: [tex] 10^3 a_0 + 10^2 a_1 + 10^1 a_1 + a_0 = (10^3+1)a_0 + (10^2+10)a_1 = (10^3+1)a_0 + 10 (11)a_1 [/tex] so for this case it remains to show a_0(10^3+1) is divisible by 11... 


#3
May2510, 07:48 AM

P: 17

Okay, thanks. I think I figured it out in base 10 by proving that you can factor out
(10^p +1) from each term, where p is some odd power, and then I proved by induction that 10^(2n1) +1 is divisible by 11. Any ideas on the general base n case? I can see that it works, just by using examples: some palindrome in base 16 with 6 digits is divisible by 17: 10*16^5 + 12*16^4 + 3*16^3 + 3*16^2+12*16^1 + 10*16^0 = 11285450 = 17 * 663850. but I have never done a proof like this for base n. Thanks, 


#4
May2510, 08:13 AM

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Divisibility by 11 for all palindromes with an even number of digits
Since the number has an even number of digits, let n be half the digits and prove by induction on n.
If n= 1, the number is of the form aa= a(11). Suppose any palindromic number with 2k digits is divisible by 11. let X be a palindromic number with 2k+ 2 digits. Let a be the leading digit (and so the ones digit). We can write X= a10^{2k1}+ M + a where M is now a palindromic number with k digits. X= a(10^{2k1}+ 1)+ M. M is divisible by 11 by the inductive hypothesis and 10^{2k1}+ 1 is divisible by 11 because, for m odd, x^{m}+ 1= (x+1)(x^{m1} x^{m2}+ ... + x 1. 


#5
May2510, 08:34 AM

P: 17

thank you. and for a base k palindromic number, can I just substitute k for 10?



#6
May2510, 08:43 AM

P: 17




#7
May2510, 01:14 PM

P: 312

Write down what the palindromic number means in base k. Express it modulo k+1. (Note that k is simply expressed modulo k+1.) You should finish up with 0 modulo k+1 if the number is to be divisible by k+1. (This includes 11 for base 10 as a special case of course.)
(If that isn't too clear just say so, and I'll expand it a bit.) 


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