# Number Theory Challenge: 4-digit palindromic numbers

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## 5 Replies - 7423 Views - Last Post: 21 April 2013 - 12:53 AM

### #1 jon.kiparsky

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# Number Theory Challenge: 4-digit palindromic numbers

Posted 18 April 2013 - 08:37 AM

Here's a fun problem to chew on.

I assert that all 4-digit palindromic numbers are multiples of 11. Can you prove or disprove this assertion?
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## Replies To: Number Theory Challenge: 4-digit palindromic numbers

### #2 sepp2k

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## Re: Number Theory Challenge: 4-digit palindromic numbers

Posted 18 April 2013 - 10:39 AM

Since the set of 4-digit palindromic numbers is finite (and not even very large), this could easily proven or disproven by exhaustive search, but here's a different approach:

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### #3 jon.kiparsky

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## Re: Number Theory Challenge: 4-digit palindromic numbers

Posted 18 April 2013 - 10:58 AM

Nice one, that's exactly the logic I used. But then I started working at it, and I believe it's further provable that this is true for any base b. That is, xFAAF/x11 = xebf exactly.

Oddly enough, this seems to be related to the fact that the digital sum of a multiple of (b-1) in base b always sums to b-1.

### #4 Nikitin

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## Re: Number Theory Challenge: 4-digit palindromic numbers

Posted 18 April 2013 - 12:43 PM

POPULAR

x*10^3 + y*10^2 + y*10 + x = 11 * (x*91 + y*10)

This post has been edited by Nikitin: 18 April 2013 - 12:45 PM

### #5 rodint

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## Re: Number Theory Challenge: 4-digit palindromic numbers

Posted 19 April 2013 - 09:15 PM

jon.kiparsky, on 18 April 2013 - 08:37 AM, said:

Here's a fun problem to chew on.

I assert that all 4-digit palindromic numbers are multiples of 11. Can you prove or disprove this assertion?

Let x = a*10^3 + b*10^2 + b*10 + a, where 0 < a,b < 10, a and b are integers.

x = a*1001 + b*110

Since 11|1001 and 11|110, 11|(a*1001 + b*110) for any integer a,b (and specifically for 0<a,b<10)

(This is from rule 7. http://primes.utm.ed...ge/divides.html )

Thus 11|x, which means x is a multiple of 11.

By construction of x, x is any 4-digit palindromic number, which means any 4-digit palindromic number is a multiple of 11.

### #6 Momerath

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## Re: Number Theory Challenge: 4-digit palindromic numbers

Posted 21 April 2013 - 12:53 AM

While your proof is good, your range is off. It should be 0 < a < 10 and 0 <= b < 10.