Question-113578

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Question Number 113578 by Algoritm last updated on 14/Sep/20

Commented by Algoritm last updated on 14/Sep/20

$$\mathrm{x}=\mathrm{1553}\:\:\:\mathrm{prove}\:\mathrm{that} \\ $$

Commented by Algoritm last updated on 14/Sep/20

$$\mathrm{step}\:\mathrm{by}\:\mathrm{step}\:\mathrm{solution}\:\mathrm{sir}\:\mathrm{please} \\ $$

Answered by 1549442205PVT last updated on 14/Sep/20

Solution:(2006!+((4012!)/(2006!)))=x(mod 4013) ⇔(2006!)^2 −2006!x+4012!=0(mod4013)(∗) ⇔4.(2006!)^2 −4.2006!x+4.4012!≡0 ⇔(2.2006!−x)^2 −x^2 +4.4012!≡0 ⇔(2.2006!−x)^2 =x^2 −4.4012!(mod4013)(1) For x=1553 we have x^2 =1553^2 =601.4013−4⇒1553^2 =−4(mod4013). On the other hands, Since 4013 is prime number,by Wilson′s theorem we have (4012!+1)⋮4013 It follows that 4012!=−1mod(4013)(2) ⇒4.4012!=−4(mod4013) From(1)(2)(3)we get (2.2006!−1553)^2 =1553^2 −4.4012!(mod4013) =−4−(−4)=0(mod4013) ⇒(2.2016!−1553)^2 =0(mod4013) ⇔2.2016!−1553=0(mod4013) The last equality is always true because of 2.2016!=2.(1.2.3....2016) ⋮1553 Thus,we proved that (1)is true for x=1553 or (∗)is true for x=1553. Therefore (2006!+((4012!)/(2006!)))=1553(mod 4013) (Q.E.D)

$$\mathrm{Solution}:\left(\mathrm{2006}!+\frac{\mathrm{4012}!}{\mathrm{2006}!}\right)=\mathrm{x}\left(\mathrm{mod}\:\mathrm{4013}\right) \\ $$$$\Leftrightarrow\left(\mathrm{2006}!\right)^{\mathrm{2}} −\mathrm{2006}!\mathrm{x}+\mathrm{4012}!=\mathrm{0}\left(\mathrm{mod4013}\right)\left(\ast\right) \\ $$$$\Leftrightarrow\mathrm{4}.\left(\mathrm{2006}!\right)^{\mathrm{2}} −\mathrm{4}.\mathrm{2006}!\mathrm{x}+\mathrm{4}.\mathrm{4012}!\equiv\mathrm{0} \\ $$$$\Leftrightarrow\left(\mathrm{2}.\mathrm{2006}!−\mathrm{x}\right)^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} +\mathrm{4}.\mathrm{4012}!\equiv\mathrm{0} \\ $$$$\Leftrightarrow\left(\mathrm{2}.\mathrm{2006}!−\mathrm{x}\right)^{\mathrm{2}} =\mathrm{x}^{\mathrm{2}} −\mathrm{4}.\mathrm{4012}!\left(\mathrm{mod4013}\right)\left(\mathrm{1}\right) \\ $$$$\mathrm{For}\:\mathrm{x}=\mathrm{1553}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{1553}^{\mathrm{2}} =\mathrm{601}.\mathrm{4013}−\mathrm{4}\Rightarrow\mathrm{1553}^{\mathrm{2}} =−\mathrm{4}\left(\mathrm{mod4013}\right). \\ $$$$\mathrm{On}\:\mathrm{the}\:\mathrm{other}\:\mathrm{hands}, \\ $$$$\mathrm{Since}\:\mathrm{4013}\:\mathrm{is}\:\mathrm{prime}\:\mathrm{number},\mathrm{by} \\ $$$$\mathrm{Wilson}'\mathrm{s}\:\mathrm{theorem}\:\mathrm{we}\:\mathrm{have}\:\left(\mathrm{4012}!+\mathrm{1}\right)\vdots\mathrm{4013} \\ $$$$\mathrm{It}\:\mathrm{follows}\:\mathrm{that}\:\mathrm{4012}!=−\mathrm{1mod}\left(\mathrm{4013}\right)\left(\mathrm{2}\right) \\ $$$$\Rightarrow\mathrm{4}.\mathrm{4012}!=−\mathrm{4}\left(\mathrm{mod4013}\right) \\ $$$$\mathrm{From}\left(\mathrm{1}\right)\left(\mathrm{2}\right)\left(\mathrm{3}\right)\mathrm{we}\:\mathrm{get} \\ $$$$\left(\mathrm{2}.\mathrm{2006}!−\mathrm{1553}\right)^{\mathrm{2}} =\mathrm{1553}^{\mathrm{2}} −\mathrm{4}.\mathrm{4012}!\left(\mathrm{mod4013}\right) \\ $$$$=−\mathrm{4}−\left(−\mathrm{4}\right)=\mathrm{0}\left(\mathrm{mod4013}\right) \\ $$$$\Rightarrow\left(\mathrm{2}.\mathrm{2016}!−\mathrm{1553}\right)^{\mathrm{2}} =\mathrm{0}\left(\mathrm{mod4013}\right) \\ $$$$\Leftrightarrow\mathrm{2}.\mathrm{2016}!−\mathrm{1553}=\mathrm{0}\left(\mathrm{mod4013}\right) \\ $$$$\mathrm{The}\:\mathrm{last}\:\mathrm{equality}\:\mathrm{is}\:\mathrm{always}\:\mathrm{true}\: \\ $$$$\mathrm{because}\:\mathrm{of}\:\mathrm{2}.\mathrm{2016}!=\mathrm{2}.\left(\mathrm{1}.\mathrm{2}.\mathrm{3}….\mathrm{2016}\right) \\ $$$$\vdots\mathrm{1553} \\ $$$$\mathrm{Thus},\mathrm{we}\:\mathrm{proved}\:\mathrm{that}\:\left(\mathrm{1}\right)\mathrm{is}\:\mathrm{true}\:\mathrm{for} \\ $$$$\mathrm{x}=\mathrm{1553}\:\mathrm{or}\:\left(\ast\right)\mathrm{is}\:\mathrm{true}\:\mathrm{for}\:\mathrm{x}=\mathrm{1553}. \\ $$$$\mathrm{Therefore}\:\left(\mathrm{2006}!+\frac{\mathrm{4012}!}{\mathrm{2006}!}\right)=\mathrm{1553}\left(\mathrm{mod}\:\mathrm{4013}\right) \\ $$$$\left(\boldsymbol{\mathrm{Q}}.\boldsymbol{\mathrm{E}}.\boldsymbol{\mathrm{D}}\right) \\ $$

Commented by Algoritm last updated on 14/Sep/20

Mr you have to work without using the proven answer depending on the answer.

$$\mathrm{Mr}\:\mathrm{you}\:\mathrm{have}\:\mathrm{to}\:\mathrm{work}\:\mathrm{without}\:\mathrm{using}\:\mathrm{the}\:\mathrm{proven}\:\mathrm{answer}\:\mathrm{depending}\:\mathrm{on}\:\mathrm{the}\:\mathrm{answer}. \\ $$

Commented by Algoritm last updated on 14/Sep/20

$$\mathrm{thanks} \\ $$

Commented by 1549442205PVT last updated on 14/Sep/20

Thank,you are welcome.From above result we can prove that(2016!)^2 +1⋮4013 This is really an interesting Question

$$\mathrm{Thank},\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}.\mathrm{From}\:\mathrm{above}\:\mathrm{result} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{prove}\:\mathrm{that}\left(\mathrm{2016}!\right)^{\mathrm{2}} +\mathrm{1}\vdots\mathrm{4013} \\ $$$$\mathrm{This}\:\mathrm{is}\:\mathrm{really}\:\mathrm{an}\:\mathrm{interesting}\:\mathrm{Question} \\ $$

Commented by Algoritm last updated on 14/Sep/20

sir 2016!^2 +1 ⋮4013 ??? please explain in detale

$$\mathrm{sir}\:\:\mathrm{2016}!^{\mathrm{2}} +\mathrm{1}\:\vdots\mathrm{4013}\:\:\:???\:\mathrm{please}\:\mathrm{explain}\:\mathrm{in}\:\mathrm{detale} \\ $$

Commented by 1549442205PVT last updated on 15/Sep/20

2.2006!−1553≡0 ⇔(2.2006!)^2 ≡1553^2 ⇔4.2006!^2 ≡−4 ⇔2006!^2 ≡−1⇔(2006!)^2 +1=0(mod4013) ⇔[2006!^2 +1]⋮4013

$$\mathrm{2}.\mathrm{2006}!−\mathrm{1553}\equiv\mathrm{0} \\ $$$$\Leftrightarrow\left(\mathrm{2}.\mathrm{2006}!\right)^{\mathrm{2}} \equiv\mathrm{1553}^{\mathrm{2}} \Leftrightarrow\mathrm{4}.\mathrm{2006}!^{\mathrm{2}} \equiv−\mathrm{4} \\ $$$$\Leftrightarrow\mathrm{2006}!^{\mathrm{2}} \equiv−\mathrm{1}\Leftrightarrow\left(\mathrm{2006}!\right)^{\mathrm{2}} +\mathrm{1}=\mathrm{0}\left(\mathrm{mod4013}\right) \\ $$$$\Leftrightarrow\left[\mathrm{2006}!^{\mathrm{2}} +\mathrm{1}\right]\vdots\mathrm{4013} \\ $$

Commented by Algoritm last updated on 16/Sep/20

$$\mathrm{sir}\:\mathrm{please}\:\mathrm{help} \\ $$

Commented by Algoritm last updated on 16/Sep/20

$$\mathrm{2}.\left(\mathrm{2006}!\right)=\mathrm{x}\left(\mathrm{mod}\:\mathrm{4013}\right) \\ $$

Commented by 1549442205PVT last updated on 19/Sep/20

It had in the proof of above problem. It is followed from(1)(2)(3) x=1553

$$\mathrm{It}\:\mathrm{had}\:\mathrm{in}\:\mathrm{the}\:\:\mathrm{proof}\:\mathrm{of}\:\mathrm{above}\:\mathrm{problem}. \\ $$$$\mathrm{It}\:\mathrm{is}\:\mathrm{followed}\:\mathrm{from}\left(\mathrm{1}\right)\left(\mathrm{2}\right)\left(\mathrm{3}\right)\:\mathrm{x}=\mathrm{1553} \\ $$

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