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Math Help

fullerene

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well... this is borderline science, but doesn't quite fit. I've got this ridiculous math problem due tomorrow, and I have absolutely no idea what to do with it. I figured hell... what better place than somewhere filled with a bunch of NTs. So here it is.

We're supposed to prove that every palindromic number (same value read forwards and backwards), in base 10, with an even number of digits is divisible by 11. Then we have to prove that every integer whose base k representation is palindromic with an even length is divisible by k + 1.

...anyone good at math have any idea what to do about something like that? I don't know where to start... but we're in the chapter on modular arithmetic, if that helps.
 

NoID10ts

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I'm great at math, but you'll have to post all of that in English before I can help you.

:D

Aren't you glad I stopped and took the time out of my busy schedule to reply to you?
 

Decaf

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What level math is this for?
 

fullerene

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It's actually a freshman level class... it just goes by the name "concepts of mathematics."

Like discrete... only more in-depth. It's like a catch-all for applying all the math you learn before calculus.
 

FusionKnight

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I suppose you could start by factoring all the palindromes, and see what relationships you find...
 

fullerene

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well... I'm willing to bet you'd find 11*(stuff) in all of them, lol.

it doesn't matter, though. It was due today and I just left that one out... luckily it was only a homework problem. I just threw it on here as a last resort. I know there are a few science grad students around, so I was hoping that someone had seen something like this before.
 
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Oh crap... I just did the whole first part before I realized you didn't need it anymore... Well... Here it is anyways:

Two digit palindromes we know, because all two digit palindromes are numbers in the 11 times table (multiplied by 1-9).

Four digit:
1001 is the first four digit palindrome.
Every other four digit palindrome meets this rule:
x(1001) + y(110), where x is any integer from 1 to 9 and y is any integer from 0 to 9.
1001 is divisible by 11, and 110 is divisible by 11, so all four digit palindromes are divisible by 11.

The rules for all other even-number-of-digits palindromes will be similar:
x(100001) + y(11110), x(10000001) + y(1111110), etc.
So if n=number of digits (assuming it's an even number, 4 or greater), the common rule is:
x(10^(n-1) + 1) + y(110((10^0)+(10^2)+(10^4)+...(10^n-4))

10^(n-1)+1 will always be divisible by 11:
10^(n-1)+1= 11 x 10^(n-2)

And, of course, (110((10^0)+(10^2)+(10^4)+...(10^n-4)) will always be divisible by 11 because 110 is divisible by 11...

I'm not sure if there's a simpler way to do it, but there's part 1... And to think I despise math. I think I'd like it more if I got to do fun problems like this instead of boring, nitpick-y AP Statistics...
 

fullerene

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lol thanks anyway.

...eh what the hell, it makes interesting discussion at this point.

Doesn't that neglect a... er... whole lot of numbers, though? Maybe I just missed it, but what about 123321, or any number with more than two digits? And wouldn't numbers of the form "y(110((10^0)+(10^2)+(10^4)+...(10^n-4))" come in blocks of two digits? 110*10^0 gives you 110, and 110*10^2 gives you the 11000, but even if you change the "y" to a different integer for every term (so that you could get more than two digits in there), how do you justify that 20020 is divisible by 11? I don't think it fits the form you outlined...
 
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Damn it... My brain wiped out on me and I didn't realize that that rule only applies to numbers where the MIDDLE bit of the palindrome is all the same number (12222221, 3999999999999993, etc.). Oh well... If you ever need to prove that all even-digit palindromes where all but the first and last digits are the same are divisible by 11, you're set. :)

Oh well. It did seem a bit too easy...
 

fullerene

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:D. Thanks anyway... I really do appreciate the attempt.
 
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