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Difference between revisions of "User talk:Bobthesmartypants/Sandbox"

(→‎Bobthesmartypants's Sandbox)
(→‎Bobthesmartypants's Sandbox)
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==Bobthesmartypants's Sandbox==
 
==Bobthesmartypants's Sandbox==
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==Solution 1==
 
<asy>
 
<asy>
 
path Q;
 
path Q;
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Therefore, <math>\angle PBA=\angle PDA</math>. <math>\Box</math>
 
Therefore, <math>\angle PBA=\angle PDA</math>. <math>\Box</math>
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 +
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==Solution 2==
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Note that <math>\dfrac{1}{n}</math> is rational and <math>n</math> is not divisible by <math>2</math> nor <math>5</math> because <math>n>11</math>.
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This means the decimal representation of <math>\dfrac{1}{n}</math> is a repeating decimal.
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Let us set <math>a_1a_2\cdots a_x</math> as the block that repeats in the repeating decimal: <math>\dfrac{1}{n}=0.\overline{a_1a_2\cdots a_x}</math>.
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(The block <math>a_1a_2\cdots a_x</math> is written without the usual overline used to signify one number so as not to confuse it with the notation for repeating decimal)
 +
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The fractional representation of this repeating decimal would be <math>\dfrac{1}{n}=\dfrac{a_1a_2\cdots a_x}{10^x-1}</math>.
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Taking the reciprocal of both sides you get <math>n=\dfrac{10^x-1}{a_1a_2\cdots a_x}</math>.
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 +
Multiplying both sides by <math>a_1a_2\cdots a_n</math> gives <math>n(a_1a_2\cdots a_x)=10^x-1</math>.
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Since <math>10^x-1=9\times \underbrace{111\cdots 111}_{x\text{ times}}</math> we divide <math>9</math> on both sides of the equation to get <math>\dfrac{n(a_1a_2\cdots a_x)}{9}=\underbrace{111\cdots 111}_{x\text{ times}}</math>.
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 +
Because <math>n</math> is not divisible by <math>3</math> (therefore <math>9</math>) since <math>n>11</math> and <math>n</math> is prime, it follows that <math>n|\underbrace{111\cdots 111}_{x\text{ times}}</math>. <math>\Box</math>

Revision as of 22:20, 8 October 2013

Bobthesmartypants's Sandbox

Solution 1

[asy] path Q; Q=(0,0)--(1,2)--(5,2)--(4,0)--cycle; draw(Q); draw((0,0)--(1.5,1)); label("D",(0,0),S); draw((1,2)--(1.5,1)); label("A",(1,2),N); draw((5,2)--(1.5,1)); label("B",(5,2),N); draw((4,0)--(1.5,1)); label("C",(4,0),S); draw((2,0)--(1.5,1),linetype("8 8")); label("E",(2,0),S); draw((2/3,4/3)--(1.5,1),linetype("8 8")); label("F",(2/3,4/3),W); label("P",(1.5,1),NNE); [/asy]

First, continue $\overline{AP}$ to hit $\overline{CD}$ at $E$. Also continue $\overline{CP}$ to hit $\overline{AD}$ at $F$.

We have that $\angle PAB=\angle PCB$. Because $\overline{AB}\parallel\overline{CD}$, we have $\angle PAB=\angle PED$.

Similarly, because $\overline{AD}\parallel\overline{BC}$, we have $\angle PCB=\angle PFD$.

Therefore, $\angle PAB=\angle PED=\angle PCB=\angle PFD$.

We also have that $\angle ADC=\angle ABC$ because $ABCD$ is a parallelogram, and $\angle APC=\angle FPE$.

Therefore, $ABCP\sim FDEP$. This means that $\dfrac{FD}{AB}=\dfrac{FP}{AP}=\dfrac{DP}{BP}$, so $\Delta ABP\sim\Delta FDP$.

Therefore, $\angle PBA=\angle PDA$. $\Box$


Solution 2

Note that $\dfrac{1}{n}$ is rational and $n$ is not divisible by $2$ nor $5$ because $n>11$.

This means the decimal representation of $\dfrac{1}{n}$ is a repeating decimal.

Let us set $a_1a_2\cdots a_x$ as the block that repeats in the repeating decimal: $\dfrac{1}{n}=0.\overline{a_1a_2\cdots a_x}$.

(The block $a_1a_2\cdots a_x$ is written without the usual overline used to signify one number so as not to confuse it with the notation for repeating decimal)

The fractional representation of this repeating decimal would be $\dfrac{1}{n}=\dfrac{a_1a_2\cdots a_x}{10^x-1}$.

Taking the reciprocal of both sides you get $n=\dfrac{10^x-1}{a_1a_2\cdots a_x}$.

Multiplying both sides by $a_1a_2\cdots a_n$ gives $n(a_1a_2\cdots a_x)=10^x-1$.

Since $10^x-1=9\times \underbrace{111\cdots 111}_{x\text{ times}}$ we divide $9$ on both sides of the equation to get $\dfrac{n(a_1a_2\cdots a_x)}{9}=\underbrace{111\cdots 111}_{x\text{ times}}$.

Because $n$ is not divisible by $3$ (therefore $9$) since $n>11$ and $n$ is prime, it follows that $n|\underbrace{111\cdots 111}_{x\text{ times}}$. $\Box$

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