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Question Number 33089 by mondodotto@gmail.com last updated on 10/Apr/18

$$\mathbf{T}\mathrm{he}\mathbf{LCM}\mathbf{and}\mathbf{GCF}\mathbf{of}\mathbf{three}\mathbf{numbers}\mathbf{is}$$$$360\mathbf{and}6\mathbf{respectively}.\mathbf{if}\mathbf{the}$$$$\mathbf{two}\mathbf{numbers}\mathbf{are}18\mathbf{and}60.$$$$\mathbf{find}\mathbf{the}\mathbf{third}\mathbf{number}.$$

Commented by Rasheed.Sindhi last updated on 10/Apr/18

$$24,72,120,360$$

Commented by mondodotto@gmail.com last updated on 10/Apr/18

$$\mathbf{please}\mathbf{show}\mathbf{workings}$$

Commented by mondodotto@gmail.com last updated on 10/Apr/18

$$\mathbf{please}\mathbf{help}$$

Commented by mondodotto@gmail.com last updated on 10/Apr/18

$$\mathrm{solution}\mathrm{please}$$

Commented by mondodotto@gmail.com last updated on 10/Apr/18

$$\mathrm{also}\mathrm{answers}\mathrm{are}\mathrm{not}\mathrm{true}\mathrm{because}\mathrm{i}\mathrm{have}\mathrm{tried}\mathrm{to}\mathrm{prove}\mathrm{i}\mathrm{get}\mathrm{LCM}=720\mathrm{instead}\mathrm{of}360\mathrm{but}\mathrm{GCF}\mathrm{is}\mathrm{correct}$$

Answered by MJS last updated on 10/Apr/18

$$18=2\times 3^2$$$$60=2^2\times 3\times 5$$$$6=2\times 3\Rightarrow x=2^l\times 3^m;l,m>0$$$$360=2^3\times 3^2\times 5\Rightarrow x=2^3\times 3^m\times 5^n;0⩽m⩽2;0⩽n⩽1$$$$x_1=2^3\times 3=24$$$$x_2=2^3\times 3^2=72$$$$x_3=2^3\times 3\times 5=120$$$$x_4=2^3\times 3^2\times 5=360$$

Commented by mondodotto@gmail.com last updated on 10/Apr/18

$$120\mathrm{and}360\mathrm{proved}\mathrm{correct}$$$$\mathrm{but}24\mathrm{and}72\mathrm{are}\mathrm{not}$$

Commented by MJS last updated on 10/Apr/18

$$\mathrm{then}\mathrm{your}\mathrm{proof}\mathrm{is}\mathrm{wrong}$$$$\mathrm{the}gcd(18;60;x);x\in \{24;72;120;360\}$$$$\mathrm{is}6\mathrm{because}\mathrm{the}gcd(18;60)=6$$$$\mathrm{and}\mathrm{it}{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{get}\mathrm{higher}\mathrm{adding}$$$$\mathrm{a}{3}^{\mathrm{rd}}\mathrm{number}$$$$$$$$lcm(18;24;60)=$$$$=lcm(2\times 3^2;2^3\times 3;2^2\times 3\times 5)=360$$$$lcm(18;60;72)=$$$$=lcm(2\times 3^2;2^2\times 3\times 5;2^3\times 3^2)=360$$$$lcm(18;60;120)=$$$$=lcm(2\times 3^2;2^2\times 3\times 5;2^3\times 3\times 5)=360$$$$lcm(18;60;360)=$$$$=lcm(2\times 3^2;2^2\times 3\times 5;2^3\times 3^2\times 5)=360$$

Commented by mondodotto@gmail.com last updated on 10/Apr/18

$$\mathrm{thanx}\mathrm{now}\mathrm{i}\mathrm{understand}$$

Commented by MJS last updated on 10/Apr/18

$${\mathrm{you}}^{\prime }\mathrm{re}\mathrm{welcome}$$

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