Can Palindromes in Base 7 be Divisible by 8?

[1+1=1]

suppose x=(bn,bn-1...b2,b1,b0)base b . show that x is divisible by (b-1)is divisible by the summation of bi, i=0 up to n. (i wish i could find out how to write it in the notation that you guys use on here)

also, i need to show that every bsae 7 palindrome interger w/ an even number of digits is divisible by 8.

what i know so far...

could i show both of these by induction? also, i know that a palindrome is a number that reads forwards and backwards is the same. i.e. 7117=7117 a palindrome. the first one i honestly do not know where to get started, but could induction as i suggest show it? can anyone offer advise?

 

[AKG]

suppose x=(bn,bn-1...b2,b1,b0)base b . show that x is divisible by (b-1)is divisible by the summation of bi, i=0 up to n.

You're not even trying to make sense here. It'd be nice if you used proper spelling and grammar, and didn't say stuff like "= to". Anyways, what do you mean by, "x is divisible by (b-1) is divisible by the summation..."? Do you mean x is divisible by (b-1), and (b-1) is divisible by the summation, or what? And what do you mean by "base b"?

(i wish i could find out how to write it in the notation that you guys use on here)

You can. Click on the following lines; each one gives you the code used to generate the text:

$$x = (b_n,\ b_{n-1},\ \dots,\ b_1,\ b_0)$$

$$\sum _{i = 0} ^n b_i$$

Also, search this forum for "LaTeX", and there's a thread all about how to use it. Also, you can check out this pdf file.

also, i need to show that every bsae 7 palindrome interger w/ an even number of digits is divisible by 8.

A base 7 palindrome with an even number of digits, 2N, can be written as:

$$a_1a_2\dots a_Na_N\dots a_2a_1 = \sum _{i=1} ^{N} a_i(7^{i-1} + 7^{2N-i})$$

I don't know what good induction will do you (or the above, just typing what comes to mind). Have you tried to work on this problem? Have you looked for useful theorems in your book, does it say anything about palindromes?

Consider the palindromes:

xx, x00x, x0000x, x000000x, etc. For these to work, it must be that:

$$\forall x,n \in \mathbb{W},\ 0 \leq x < 8,\ x + x(7^{2n-1}) \equiv 0 (\mbox{mod }8)$$

This you can prove easily by induction, so I suppose induction is worthwhile after all. Now, for any palindrome:

xyz...abccba...zyx = x000...000x + 0y000...000y0 + ... + 000...cc...000
= x000...000x + 10(y000...000y) + 100(z000...000z) + ... + 1000...00(cc).

Now, all those things, x000..000x, y000...000y,... cc are all congruent to 0 (mod 8) by the inductive proof given above. Add 100 numbers together that are congruent to 0 (mod 8), e.g. 100 * (z000...000z), and you get a number congruent to 0 (mod 8). Then add that to other numbers, like x000...000x and 1000...000(cc) which are also congruent to 0 (mod 8), and you still get a number congruent to 0 (mod 8), Q.E.D.

 

[1+1=1]

You're not even trying to make sense here.

i would make sense if i could write in latex, but i don't know how. i never said i wasthe smartest person, that is why i am asking for help. thank you very much for helping me on these problems. i know i ask a lot of questions but i don't have a lot to offer for the good of this site, meaning posting replies to questions asked by others. again, forgive me about the notation on the letters and all, i need to get latex and find out how to use it. I'm sorry I'm not advanced...

 

[1+1=1]

$$\{sqrt4}(\in)$$

 

[AKG]

i would make sense if i could write in latex, but i don't know how. i never said i wasthe smartest person, that is why i am asking for help. thank you very much for helping me on these problems. i know i ask a lot of questions but i don't have a lot to offer for the good of this site, meaning posting replies to questions asked by others.

That's not a problem, but if you're going to ask questions, don't make it hard on the reader. LaTeX is one thing, but you should consider improving your basic English. Especially for people who could help you but know English as a second language, some of your posts might seem very confusing. Also, (a,b,c) isn't standard notation for gcd(a,b,c), so I'd advise that you refrain from using it, and actually write out gcd(a,b,c). Make it as easy as possible for others to help you. Also, I don't really know any theorems or have any formal education in number theory, but a little effort is all you need. You were on the right track in terms of using induction for that last question, you probably could have just taken some time and solved it, or solved more of it, yourself. Don't feel hesitant to ask questions, but at the same time, you won't learn anything without trying.

i need to get latex and find out how to use it.

There is no "getting" LaTeX. It's something you can use on these forums, just like the [ b ] [/ b ] tags, e.g. bold. For example, if you want 6 to the exponent 2, you can write:

[ tex ]6^2[/ tex ]. Of course, get rid of those spaces inside the [], so you have:

$$6^2$$

You can click on the above to see what it should look like without the spaces. Also, I explicitly gave you 3 ways to learn about LaTeX, so I don't see what the problem is. Check out that pdf file I showed you, go around the site and click on things written in latex to see how they were written, and search for "LaTeX", there's a whole thread on various things you can do with LaTeX. Another thing I find pretty helpful, once you've got the hang of it a bit, is go to Mathworld.com, and search for something, and on the search result page, each search result gives you the first few lines or so of the article. If there is some LaTeX in those lines, you can see what code they used, and copy it.

 

[1+1=1]

i do have the pdf file now and am experimenting with the notation on latex. i thought it was an actual program on the computer that needed to be installed.

$$\bigcup$$ hehe just trying it!

 

[AKG]

$$\{sqrt4}(\in)$$

You're going to want to put brace brackets around the radicand, like so:

$$\sqrt{4}(\in)$$

Click on the above and compare to what you wrote to see how to do it.