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Question Number 106596 by mathdave last updated on 06/Aug/20

Answered by john santu last updated on 06/Aug/20

@JS@ (2^x −1)(x+3)(x−1)=0 → { ((x=−3; x=1 ; x = 0 } )) :}

$$\:\:\:\:\:@\mathrm{JS}@ \\ $$$$\left(\mathrm{2}^{\mathrm{x}} −\mathrm{1}\right)\left(\mathrm{x}+\mathrm{3}\right)\left(\mathrm{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\rightarrow\begin{cases}{\left.\mathrm{x}=−\mathrm{3};\:\mathrm{x}=\mathrm{1}\:;\:\mathrm{x}\:=\:\mathrm{0}\:\right\}\:}\end{cases} \\ $$$$ \\ $$

Answered by Rasheed.Sindhi last updated on 06/Aug/20

3.4^n +51 ^• 3 ∣ 3.4^n & 3∣ 51⇒3 ∣ (3.4^n +51) .....

$$\mathrm{3}.\mathrm{4}^{{n}} +\mathrm{51} \\ $$$$\:^{\bullet} \mathrm{3}\:\mid\:\mathrm{3}.\mathrm{4}^{{n}} \:\:\&\:\:\mathrm{3}\mid\:\mathrm{51}\Rightarrow\mathrm{3}\:\mid\:\left(\mathrm{3}.\mathrm{4}^{{n}} +\mathrm{51}\right) \\ $$$$..... \\ $$

Answered by JDamian last updated on 06/Aug/20

(1) ((3∙4^n +51)/3)=4^n +17 (4^n +17)mod 3 = (4^n mod 3)+(17 mod 3) = =(4 mod 3)^n + 2 = 1^n + 2 = 3 mod 3 = 0

$$\left(\mathrm{1}\right) \\ $$$$\frac{\mathrm{3}\centerdot\mathrm{4}^{{n}} +\mathrm{51}}{\mathrm{3}}=\mathrm{4}^{{n}} +\mathrm{17} \\ $$$$\left(\mathrm{4}^{{n}} +\mathrm{17}\right){mod}\:\mathrm{3}\:=\:\left(\mathrm{4}^{{n}} \:{mod}\:\mathrm{3}\right)+\left(\mathrm{17}\:{mod}\:\mathrm{3}\right)\:= \\ $$$$=\left(\mathrm{4}\:{mod}\:\mathrm{3}\right)^{{n}} \:+\:\mathrm{2}\:=\:\mathrm{1}^{{n}} +\:\mathrm{2}\:=\:\mathrm{3}\:{mod}\:\mathrm{3}\:=\:\mathrm{0} \\ $$

Answered by 1549442205PVT last updated on 06/Aug/20

We have: 4^n =(3+1)^n =3^n +C_n ^1 .3^(n−1) +C_n ^2 .3^(n−2) +... +...C^(n−1) .3+1=3m+1. ⇒4^n +17=3m+1+17=3m+18=3(m+6) ⇒3.4^n +51=3(4^n +17)=9(m+6)⋮9 Consequently,3.4^n +51 is divisible by both 3 and 9 (q.e.d)

$$\mathrm{We}\:\mathrm{have}: \\ $$$$\mathrm{4}^{\mathrm{n}} =\left(\mathrm{3}+\mathrm{1}\right)^{\mathrm{n}} =\mathrm{3}^{\mathrm{n}} +\mathrm{C}_{\mathrm{n}} ^{\mathrm{1}} .\mathrm{3}^{\mathrm{n}−\mathrm{1}} +\mathrm{C}_{\mathrm{n}} ^{\mathrm{2}} .\mathrm{3}^{\mathrm{n}−\mathrm{2}} +... \\ $$$$+...\mathrm{C}^{\mathrm{n}−\mathrm{1}} .\mathrm{3}+\mathrm{1}=\mathrm{3m}+\mathrm{1}. \\ $$$$\Rightarrow\mathrm{4}^{\mathrm{n}} +\mathrm{17}=\mathrm{3m}+\mathrm{1}+\mathrm{17}=\mathrm{3m}+\mathrm{18}=\mathrm{3}\left(\mathrm{m}+\mathrm{6}\right) \\ $$$$\Rightarrow\mathrm{3}.\mathrm{4}^{\mathrm{n}} +\mathrm{51}=\mathrm{3}\left(\mathrm{4}^{\mathrm{n}} +\mathrm{17}\right)=\mathrm{9}\left(\mathrm{m}+\mathrm{6}\right)\vdots\mathrm{9} \\ $$$$\mathrm{Consequently},\mathrm{3}.\mathrm{4}^{\mathrm{n}} +\mathrm{51}\:\mathrm{is}\:\:\mathrm{divisible}\:\mathrm{by} \\ $$$$\mathrm{both}\:\mathrm{3}\:\mathrm{and}\:\mathrm{9}\:\left(\mathrm{q}.\mathrm{e}.\mathrm{d}\right) \\ $$

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