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Question Number 172210 by Rasheed.Sindhi last updated on 24/Jun/22

$$\mathbf{Solve}\mathbf{in}\mathbb{N}:$$$$6^{\mathbf{x}}+6^{\mathbf{y}}=222$$

Commented by infinityaction last updated on 24/Jun/22

$$(1,3)and(3,1)$$

Commented by Rasheed.Sindhi last updated on 24/Jun/22

$$\mathrm{Correct}\mathrm{sir}!\mathrm{Process}\mathrm{please}!$$

Commented by infinityaction last updated on 22/Jul/22

$$x,y\in \mathbb{N}$$$$6^x+6^y=222$$$$6^4=1296⪈222$$$$soxandy⪇4$$$$andxandy⪈0$$$$0⪇x,y⪇4$$$$soxandyshouldbe1,2,3$$$$socombitionsofnumbers$$$$(x,y)=\left\{(1,3)(3,1)\right\}$$

Commented by Rasheed.Sindhi last updated on 24/Jun/22

$$\mathcal{T}han\mathcal{X}sir!$$

Answered by Rasheed.Sindhi last updated on 24/Jun/22

$$AnOtherWay$$$$6^x+6^y=222$$$$\Rightarrow 6^{x-1}+6^{y-1}=37$$$$37issumoftwopowersof6and$$$$oneiscertainlyodd\left(6^{0}isonlyodd\right)$$$$\therefore 7=36+1\mid 7=1+36$$$$▸6^{x-1}+6^{y-1}=36+1=6^2+6^0$$$$x-1=2\wedge y-1=0$$$$x=3\wedge y=1$$$$▸6^{x-1}+6^{y-1}=1+36=6^0+6^2$$$$x-1=0\wedge y-1=2$$$$x=1\wedge y=3$$$$(x,y)=(1,3),(3,1)$$

Commented by infinityaction last updated on 24/Jun/22

$$nicesir$$

Commented by Rasheed.Sindhi last updated on 24/Jun/22

🙏🙏🙏

Answered by kapoorshah last updated on 24/Jun/22

$$6^{x-y}.6^y+6^y=222$$$$6^y(6^{x-y}+1)=222$$$$6^y(6^{x-y}+1)=6.37$$$$$$$$6^y=6^1\Rightarrow y=1$$$$6^{x-1}+1=37$$$$6^{x-1}=6^2\Rightarrow x=3$$$$$$$$sothesolutions:x=3,y=1and$$$$x=1,y=3$$

Commented by Rasheed.Sindhi last updated on 24/Jun/22

$$\mathbb{T}\mathbf{han}\mathbb{k}\mathbf{s}\mathrm{sir}!$$

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