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Question Number 22407 by mondodotto@gmail.com last updated on 17/Oct/17

Answered by $@ty@m last updated on 18/Oct/17

∵ GCF of the three numbers is 12 ∴ third number∈{12, 24,36,48,...} Out of which only 24 and 72 satisfies the condition LCM(12,36, third no.)=72

$$\because\:{GCF}\:{of}\:{the}\:{three}\:{numbers}\: \\ $$$${is}\:\mathrm{12} \\ $$$$\therefore\:{third}\:{number}\in\left\{\mathrm{12},\:\mathrm{24},\mathrm{36},\mathrm{48},...\right\} \\ $$$${Out}\:{of}\:{which}\:{only}\:\mathrm{24}\:{and}\:\mathrm{72}\:{satisfies} \\ $$$${the}\:{condition}\:{LCM}\left(\mathrm{12},\mathrm{36},\:{third}\:{no}.\right)=\mathrm{72} \\ $$

Commented by NECx last updated on 17/Oct/17

workings please

$${workings}\:{please} \\ $$$$ \\ $$

Answered by Rasheed.Sindhi last updated on 17/Oct/17

72=2^3 .3^(2 ) ,12=2^2 .3 ,36=2^2 .3^2 L^�^� et the third number is 2^m .3^n lcm(2^2 .3, 2^2 .3^2 , 2^m .3^n ) =2^(max(2,2,m)) ×3^(max(1,2,n)) =2^3 .3^2 max(2,2,m)=3⇒m=3....(i) max(1,2,n)=2⇒n≤2......(ii) gcd(2^2 .3, 2^2 .3^2 , 2^m .3^n ) =2^(min(2,2,m)) ×3^(min(1,2,n)) =2^2 ×3 min(2,2,m)=2⇒m≥2...(iii) min(1,2,n)=1⇒n≥1....(iv) (i) & (iii): m=3 ∧ m≥2⇒m=3 (ii) & (iv): n≤2 ∧ n≥1⇒n=1,2 The third number: 2^m .3^n =2^3 .3 or 2^3 .3^2 2^m .3^n =24 or 72

$$\mathrm{72}=\mathrm{2}^{\mathrm{3}} .\mathrm{3}^{\mathrm{2}\:} ,\mathrm{12}=\mathrm{2}^{\mathrm{2}} .\mathrm{3}\:,\mathrm{36}=\mathrm{2}^{\mathrm{2}} .\mathrm{3}^{\mathrm{2}} \\ $$$$\bar {\mathrm{L}et}\:\mathrm{the}\:\mathrm{third}\:\mathrm{number}\:\mathrm{is}\:\mathrm{2}^{\mathrm{m}} .\mathrm{3}^{\mathrm{n}} \\ $$$$\mathrm{lcm}\left(\mathrm{2}^{\mathrm{2}} .\mathrm{3},\:\mathrm{2}^{\mathrm{2}} .\mathrm{3}^{\mathrm{2}} ,\:\mathrm{2}^{\mathrm{m}} .\mathrm{3}^{\mathrm{n}} \right) \\ $$$$\:\:\:=\mathrm{2}^{\mathrm{max}\left(\mathrm{2},\mathrm{2},\mathrm{m}\right)} ×\mathrm{3}^{\mathrm{max}\left(\mathrm{1},\mathrm{2},\mathrm{n}\right)} =\mathrm{2}^{\mathrm{3}} .\mathrm{3}^{\mathrm{2}} \\ $$$$\:\:\mathrm{max}\left(\mathrm{2},\mathrm{2},\mathrm{m}\right)=\mathrm{3}\Rightarrow\mathrm{m}=\mathrm{3}....\left(\mathrm{i}\right) \\ $$$$\:\:\mathrm{max}\left(\mathrm{1},\mathrm{2},\mathrm{n}\right)=\mathrm{2}\Rightarrow\mathrm{n}\leqslant\mathrm{2}......\left(\mathrm{ii}\right) \\ $$$$\mathrm{gcd}\left(\mathrm{2}^{\mathrm{2}} .\mathrm{3},\:\mathrm{2}^{\mathrm{2}} .\mathrm{3}^{\mathrm{2}} ,\:\mathrm{2}^{\mathrm{m}} .\mathrm{3}^{\mathrm{n}} \right) \\ $$$$\:\:=\mathrm{2}^{\mathrm{min}\left(\mathrm{2},\mathrm{2},\mathrm{m}\right)} ×\mathrm{3}^{\mathrm{min}\left(\mathrm{1},\mathrm{2},\mathrm{n}\right)} =\mathrm{2}^{\mathrm{2}} ×\mathrm{3} \\ $$$$\:\:\:\mathrm{min}\left(\mathrm{2},\mathrm{2},\mathrm{m}\right)=\mathrm{2}\Rightarrow\mathrm{m}\geqslant\mathrm{2}...\left(\mathrm{iii}\right) \\ $$$$\:\:\:\:\mathrm{min}\left(\mathrm{1},\mathrm{2},\mathrm{n}\right)=\mathrm{1}\Rightarrow\mathrm{n}\geqslant\mathrm{1}....\left(\mathrm{iv}\right) \\ $$$$\left(\mathrm{i}\right)\:\&\:\left(\mathrm{iii}\right): \\ $$$$\mathrm{m}=\mathrm{3}\:\wedge\:\mathrm{m}\geqslant\mathrm{2}\Rightarrow\mathrm{m}=\mathrm{3} \\ $$$$\left(\mathrm{ii}\right)\:\&\:\left(\mathrm{iv}\right): \\ $$$$\mathrm{n}\leqslant\mathrm{2}\:\wedge\:\mathrm{n}\geqslant\mathrm{1}\Rightarrow\mathrm{n}=\mathrm{1},\mathrm{2} \\ $$$$\mathrm{The}\:\mathrm{third}\:\mathrm{number}: \\ $$$$\mathrm{2}^{\mathrm{m}} .\mathrm{3}^{\mathrm{n}} =\mathrm{2}^{\mathrm{3}} .\mathrm{3}\:\:\mathrm{or}\:\:\mathrm{2}^{\mathrm{3}} .\mathrm{3}^{\mathrm{2}} \\ $$$$\mathrm{2}^{\mathrm{m}} .\mathrm{3}^{\mathrm{n}} =\mathrm{24}\:\:\mathrm{or}\:\:\mathrm{72} \\ $$$$ \\ $$

Commented by $@ty@m last updated on 18/Oct/17

Nice.. In a systematic way...

$${Nice}.. \\ $$$${In}\:{a}\:{systematic}\:{way}... \\ $$

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