divisible by

Soroban

MHF Hall of Honor
May 2006
12,028
6,341
Lexington, MA (USA)
Hello, dapore!

I haven't solved it yet, but I know a few tricks . . .


Prove that:

. . \(\displaystyle \underbrace{x^{9999} + x^{8888} + x^{7777} + \hdots + x^{1111} + 1}_A\:\text{ is divisible by }\;\underbrace{x^9 + x^8 + x^7 + \hdots + x + 1}_B\)

Multiply \(\displaystyle A\) by \(\displaystyle \frac{x^{1111}-1}{x^{1111}-1}\!:\)

. . \(\displaystyle A \;=\;\frac{x^{1111}-1}{x^{1111}-1}\cdot\frac{x^{9999} + x^{8888} + x^{7777} + \hdots + x^{1111} + 1}{1} \) . \(\displaystyle =\;\frac{x^{11,110}-1}{x^{1111}-1} \)


Multiply \(\displaystyle B\) by \(\displaystyle \frac{x-1}{x-1}\!:\)

. . \(\displaystyle B \;=\; \frac{x-1}{x-1}\cdot\frac{x^9 + x^8 + x^7 + \hdots x + 1}{1} \;=\; \frac{x^{10}-1}{x-1}\)


Hence, we have:

. . \(\displaystyle \frac{A}{B} \;=\;\frac{x^{11,110} - 1}{x^{1111}-1}\cdot\frac{x-1}{x^{10}-1} \)



Does this inspire anyone? . . . Anyone? Anyone?

 
Aug 2007
3,171
860
USA
Only 10 terms? Try long division.

This will not only prove it but produce the result.

No, really. It will be fun. You've been saving that stack of scratch paper too long, anyway.
 
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Mar 2010
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\(\displaystyle \underbrace{x^{9999}+x^{8888}+x^{7777}+\hdots+x^{1111}+1}_{P(x)}\:\text{ is divisible by}\;\underbrace{x^9 + x^8 + x^7 + \hdots + x + 1}_{Q(x)}\)
We have \(\displaystyle P(x)= Q(x^{1111})\). Since \(\displaystyle Q(x) = 0\) for \(\displaystyle x_{k}=\cos\frac{2k\pi}{10}+i\sin\frac{2k\pi}{10}\), \(\displaystyle \qquad (k = 1, 2, 3, \dots, 9)\) and \(\displaystyle x_{k}^{1111}=\cos\frac{2222k\pi}{10}+i\sin\frac{2222k\pi}{10}=\cos(222k\pi+\frac{2k\pi}{10})+i\sin(222k\pi+\frac{2k\pi}{10})=x_{k}\), it follows that \(\displaystyle P(x^{k})=Q(x_{k}^{1111})=Q(x_{k})=0.\) Therefore, \(\displaystyle Q(x)|P(x)\). (Clapping)
 
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Apr 2010
384
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Canada
We have \(\displaystyle P(x)= Q(x^{1111})\). Since \(\displaystyle Q(x) = 0\) for \(\displaystyle x_{k}=\cos\frac{2k\pi}{10}+i\sin\frac{2k\pi}{10}\), \(\displaystyle \qquad (k = 1, 2, 3, \dots, 9)\) and \(\displaystyle x_{k}^{1111}=\cos\frac{2222k\pi}{10}+i\sin\frac{2222k\pi}{10}=\cos(222k\pi+\frac{2k\pi}{10})+i\sin(222k\pi+\frac{2k\pi}{10})=x_{k}\), it follows that \(\displaystyle P(x^{k})=Q(x_{k}^{1111})=Q(x_{k})=0.\) Therefore, \(\displaystyle Q(x)|P(x)\). (Clapping)
Okay I'm a little slow...but what?! We came up with \(\displaystyle x_k = e^{ i \frac{2k \pi }{10} }\) how?
 
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Mar 2010
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Okay I'm a little slow...but what?! We came up with \(\displaystyle x_k = e^{ i \frac{2k \pi }{10} }\) how?
We see that \(\displaystyle G(x) = \dfrac{x^{10}-1}{x-1}\) (by either using Soroban's trick or just recalling that \(\displaystyle a+ar+ar^2+ar^3+ \cdots+ar^{n}=\sum_{k=0}^{n}ar^k= a\left(\frac{1-r^{n+1}}{1-r}\right)\) and realising that \(\displaystyle G(x)\) is a case of that where \(\displaystyle a = 1\) and \(\displaystyle r = x\), and \(\displaystyle n = 9\)). We have \(\displaystyle G(x) = 0 \Rightarrow\dfrac{x^{10}-1}{x-1} = 0 \Rightarrow x^{10}-1 = 0 \Rightarrow x_{k} = e^{ i\frac{2k \pi }{10}}\), which are all roots of \(\displaystyle g(x)\), except of course \(\displaystyle x = 1\). Thus \(\displaystyle G(x) = 0[/Math] for \(\displaystyle x_k = e^{ i \frac{2k \pi }{10}}\)\)\(\displaystyle , \(\displaystyle (1\le{k}\le{9}).\)\)
 
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