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

Solve in N: 6^x +6^y =222

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

Commented by infinityaction last updated on 24/Jun/22

(1,3) and (3,1)

$$\left(\mathrm{1},\mathrm{3}\right)\:{and}\:\left(\mathrm{3},\mathrm{1}\right) \\ $$

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

Correct sir! Process please!

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

Commented by infinityaction last updated on 22/Jul/22

x,y∈N 6^x +6^(y ) = 222 6^4 = 1296 ⪈ 222 so x and y ⪇ 4 and x and y ⪈0 0⪇x,y⪇4 so x and y should be 1,2,3 so combitions of numbers (x,y) = {(1,3) (3,1)}

$${x},{y}\in\mathbb{N} \\ $$$$\:\mathrm{6}^{{x}} +\mathrm{6}^{{y}\:} \:=\:\mathrm{222} \\ $$$$\mathrm{6}^{\mathrm{4}} \:=\:\mathrm{1296}\:\gneq\:\mathrm{222} \\ $$$${so}\:{x}\:{and}\:{y}\:\lneq\:\mathrm{4} \\ $$$$\:\:{and}\:{x}\:{and}\:{y}\:\gneq\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{0}\lneq{x},{y}\lneq\mathrm{4} \\ $$$${so}\:{x}\:{and}\:{y}\:{should}\:{be}\:\mathrm{1},\mathrm{2},\mathrm{3} \\ $$$$\:{so}\:\:{combitions}\:{of}\:{numbers} \\ $$$$\:\left({x},{y}\right)\:=\:\left\{\left(\mathrm{1},\mathrm{3}\right)\:\left(\mathrm{3},\mathrm{1}\right)\right\} \\ $$

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

ThanX sir!

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

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

AnOther Way 6^x +6^y =222 ⇒6^(x−1) +6^(y−1) =37 37 is sum of two powers of 6 and one is certainly odd(6^0 is only odd) ∴ 7=36+1 ∣ 7=1+36 ▶6^(x−1) +6^(y−1) =36+1=6^2 +6^0 x−1=2 ∧ y−1=0 x=3 ∧ y=1 ▶ 6^(x−1) +6^(y−1) =1+36=6^0 +6^2 x−1=0 ∧ y−1=2 x=1 ∧ y=3 (x,y)=(1,3),(3,1)

$${AnOther}\:{Way} \\ $$$$\mathrm{6}^{{x}} +\mathrm{6}^{{y}} =\mathrm{222} \\ $$$$\Rightarrow\mathrm{6}^{{x}−\mathrm{1}} +\mathrm{6}^{{y}−\mathrm{1}} =\mathrm{37} \\ $$$$\mathrm{37}\:{is}\:{sum}\:{of}\:{two}\:{powers}\:{of}\:\mathrm{6}\:{and} \\ $$$${one}\:{is}\:{certainly}\:{odd}\left(\mathrm{6}^{\mathrm{0}} \:{is}\:{only}\:{odd}\right) \\ $$$$\therefore\:\mathrm{7}=\mathrm{36}+\mathrm{1}\:\mid\:\mathrm{7}=\mathrm{1}+\mathrm{36} \\ $$$$\blacktriangleright\mathrm{6}^{{x}−\mathrm{1}} +\mathrm{6}^{{y}−\mathrm{1}} =\mathrm{36}+\mathrm{1}=\mathrm{6}^{\mathrm{2}} +\mathrm{6}^{\mathrm{0}} \\ $$$$\:\:\:\:\:{x}−\mathrm{1}=\mathrm{2}\:\wedge\:{y}−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:{x}=\mathrm{3}\:\wedge\:{y}=\mathrm{1} \\ $$$$\blacktriangleright\:\mathrm{6}^{{x}−\mathrm{1}} +\mathrm{6}^{{y}−\mathrm{1}} =\mathrm{1}+\mathrm{36}=\mathrm{6}^{\mathrm{0}} +\mathrm{6}^{\mathrm{2}} \\ $$$$\:\:\:\:\:{x}−\mathrm{1}=\mathrm{0}\:\wedge\:{y}−\mathrm{1}=\mathrm{2} \\ $$$$\:\:\:\:\:{x}=\mathrm{1}\:\wedge\:{y}=\mathrm{3} \\ $$$$\left({x},{y}\right)=\left(\mathrm{1},\mathrm{3}\right),\left(\mathrm{3},\mathrm{1}\right) \\ $$

Commented by infinityaction last updated on 24/Jun/22

nice sir

$${nice}\:{sir} \\ $$

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 ⇒ y = 1 6^(x − 1) +1 = 37 6^(x − 1) = 6^2 ⇒ x = 3 so the solutions : x = 3, y = 1 and x = 1, y = 3

$$\mathrm{6}^{{x}\:−\:{y}} .\mathrm{6}^{{y}} \:+\:\mathrm{6}^{{y}} \:=\:\mathrm{222} \\ $$$$\mathrm{6}^{{y}} \:\left(\mathrm{6}^{{x}\:−\:{y}} \:+\:\mathrm{1}\right)\:=\:\mathrm{222} \\ $$$$\mathrm{6}^{{y}} \:\left(\mathrm{6}^{{x}\:−\:{y}} \:+\:\mathrm{1}\right)\:=\:\mathrm{6}\:.\:\mathrm{37} \\ $$$$ \\ $$$$\mathrm{6}^{{y}} \:=\:\mathrm{6}^{\mathrm{1}} \:\Rightarrow\:{y}\:=\:\mathrm{1} \\ $$$$\mathrm{6}^{{x}\:−\:\mathrm{1}} \:+\mathrm{1}\:=\:\mathrm{37} \\ $$$$\mathrm{6}^{{x}\:−\:\mathrm{1}} \:=\:\mathrm{6}^{\mathrm{2}} \:\Rightarrow\:{x}\:=\:\mathrm{3} \\ $$$$ \\ $$$${so}\:{the}\:{solutions}\::\:{x}\:=\:\mathrm{3},\:{y}\:=\:\mathrm{1}\:{and}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\:=\:\mathrm{1},\:{y}\:=\:\mathrm{3} \\ $$

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

Thanks sir!

$$\mathbb{T}\boldsymbol{\mathrm{han}}\Bbbk\boldsymbol{\mathrm{s}}\:\mathrm{sir}! \\ $$

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