Divisibility by 11 for all palindromes with an even number of digits

In summary: The palindromic number in base k means that it reads the same forwards and backwards. For example, the number 1234 would be a palindromic number in base 10 because it reads the same forwards and backwards (1, 2, 3, 4, 3, 2, 1). The number 1234 in base 16 would be 12, 34, 4, since it would read the same forwards and backwards (1, 2, 3, 4, 5, 6, 7, 8, 9).
  • #1
ripcity4545
16
0

Homework Statement



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.


Homework Equations



palindromic means the number reads the same forwards and backwards


The Attempt at a Solution



I tried a general representation of a palindromic integer with even digits of form

D1*10^(2n-1) + D2*10^(2n-2) + ... +Dn * 10^n + Dn * 10^(n-1) +...+ D2*10^1 +D1*10^0

which makes the number N equal to

N= D1[10^(2n-1) + 10^0] + D2 [10^(2n-2) + 10^1] +...+Dn[10^n + 10^(n-1)].

But I can't figure out how to prove divisibility by 11. Thanks for any help!
 
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  • #2
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
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^(2n-1) +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
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= a102k-1+ M + a where M is now a palindromic number with k digits.
X= a(102k-1+ 1)+ M. M is divisible by 11 by the inductive hypothesis and 102k-1+ 1 is divisible by 11 because, for m odd, xm+ 1= (x+1)(xm-1- xm-2+ ... + x- 1.
 
  • #5
thank you. and for a base k palindromic number, can I just substitute k for 10?
 
  • #6
ripcity4545 said:
thank you. and for a base k palindromic number, can I just substitute k for 10?

nope, that doesn't work. I am definitely stuck on proving that a base k palindromic number with even digits is divisible by k+1.
 
  • #7
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.)
 
Last edited:

1. How do you determine if a palindrome with an even number of digits is divisible by 11?

To determine if a palindrome with an even number of digits is divisible by 11, you can use the divisibility rule for 11 which states that if the sum of the odd-positioned digits is equal to the sum of the even-positioned digits, then the number is divisible by 11.

2. Can you provide an example of a palindrome with an even number of digits that is divisible by 11?

Yes, an example would be the number 913391, which has an even number of digits (6) and follows the divisibility rule for 11 (9+3+1=3+9+1).

3. Are there any exceptions to the divisibility rule for 11 when it comes to palindromes with an even number of digits?

No, the divisibility rule for 11 applies to all palindromes with an even number of digits. If the sum of the odd-positioned digits is not equal to the sum of the even-positioned digits, then the number is not divisible by 11.

4. How does this rule apply to larger palindromes with an even number of digits?

The rule for divisibility by 11 for palindromes with an even number of digits can be applied to larger numbers by breaking them down into smaller chunks of two digits and checking if each chunk follows the rule. If all the chunks follow the rule, then the entire number is divisible by 11.

5. Is there a connection between the divisibility by 11 and the fact that palindromes with an even number of digits have a 0 in the middle?

Yes, there is a connection. The divisibility rule for 11 works because the sum of the odd-positioned digits and the even-positioned digits are always equal when there is a 0 in the middle. If there was no 0 in the middle, the rule would not apply and the number may not be divisible by 11.

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