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Find-out-x-such-that-lcm-50-80-x-2800-lcm-56-84-x-840-




Question Number 108264 by Rasheed.Sindhi last updated on 15/Aug/20
Find out x, such that:  lcm(50,80,x)=2800 ∧ lcm(56,84,x)=840
$${Find}\:{out}\:{x},\:{such}\:{that}: \\ $$$$\mathrm{lcm}\left(\mathrm{50},\mathrm{80},{x}\right)=\mathrm{2800}\:\wedge\:\mathrm{lcm}\left(\mathrm{56},\mathrm{84},{x}\right)=\mathrm{840} \\ $$$$\:\:\:\:\:\:\:\:\:\:\: \\ $$
Answered by mr W last updated on 15/Aug/20
50=2×5^2   80=2^4 ×5  2800=2^4 ×5^2 ×7  from lcm(50,80,x)=2800  ⇒x=2^(0..4) ×5^(0..2) ×7^1      ...(i)    56=2^3 ×7  84=2^2 ×3×7  840=2^3 ×3×5×7  from lcm(56,84,x)=840  ⇒x=2^(0..3) ×3^(0..1) ×5^1 ×7^(0..1)    ...(ii)    from (i) and (ii):  ⇒x=2^(0..3) ×3^0 ×5^1 ×7^1 =35×2^(0..3)    =(35, 70, 140, 280)
$$\mathrm{50}=\mathrm{2}×\mathrm{5}^{\mathrm{2}} \\ $$$$\mathrm{80}=\mathrm{2}^{\mathrm{4}} ×\mathrm{5} \\ $$$$\mathrm{2800}=\mathrm{2}^{\mathrm{4}} ×\mathrm{5}^{\mathrm{2}} ×\mathrm{7} \\ $$$${from}\:{lcm}\left(\mathrm{50},\mathrm{80},{x}\right)=\mathrm{2800} \\ $$$$\Rightarrow{x}=\mathrm{2}^{\mathrm{0}..\mathrm{4}} ×\mathrm{5}^{\mathrm{0}..\mathrm{2}} ×\mathrm{7}^{\mathrm{1}} \:\:\:\:\:…\left({i}\right) \\ $$$$ \\ $$$$\mathrm{56}=\mathrm{2}^{\mathrm{3}} ×\mathrm{7} \\ $$$$\mathrm{84}=\mathrm{2}^{\mathrm{2}} ×\mathrm{3}×\mathrm{7} \\ $$$$\mathrm{840}=\mathrm{2}^{\mathrm{3}} ×\mathrm{3}×\mathrm{5}×\mathrm{7} \\ $$$${from}\:{lcm}\left(\mathrm{56},\mathrm{84},{x}\right)=\mathrm{840} \\ $$$$\Rightarrow{x}=\mathrm{2}^{\mathrm{0}..\mathrm{3}} ×\mathrm{3}^{\mathrm{0}..\mathrm{1}} ×\mathrm{5}^{\mathrm{1}} ×\mathrm{7}^{\mathrm{0}..\mathrm{1}} \:\:\:…\left({ii}\right) \\ $$$$ \\ $$$${from}\:\left({i}\right)\:{and}\:\left({ii}\right): \\ $$$$\Rightarrow{x}=\mathrm{2}^{\mathrm{0}..\mathrm{3}} ×\mathrm{3}^{\mathrm{0}} ×\mathrm{5}^{\mathrm{1}} ×\mathrm{7}^{\mathrm{1}} =\mathrm{35}×\mathrm{2}^{\mathrm{0}..\mathrm{3}} \: \\ $$$$=\left(\mathrm{35},\:\mathrm{70},\:\mathrm{140},\:\mathrm{280}\right) \\ $$
Commented by Rasheed.Sindhi last updated on 16/Aug/20
Thanks mr W sir!
$$\mathcal{T}{hanks}\:{mr}\:{W}\:{sir}! \\ $$
Answered by floor(10²Eta[1]) last updated on 16/Aug/20
lcm(a,b,c)=lcm(lcm(a,b),c)  (1):lcm(50,80,x)=lcm(lcm(50,80),x)  =lcm(400,x)=2800  (2):lcm(56,84,x)=lcm(lcm(56,84),x)  =lcm(168,x)=840  (3):400=2^4 .5^2   2800=2^4 .5^2 .7  ⇒x=2^({0,1,2,3,4}) .5^({0,1,2}) .7  (4):168=2^3 .3.7  840=2^3 .3.5.7  ⇒x=2^({0,1,2,3}) .3^({0,1}) .5.7^({0,1})   ⇒x=2^({0,1,2,3}) .5.7  x={35,70,140,280}
$$\mathrm{lcm}\left(\mathrm{a},\mathrm{b},\mathrm{c}\right)=\mathrm{lcm}\left(\mathrm{lcm}\left(\mathrm{a},\mathrm{b}\right),\mathrm{c}\right) \\ $$$$\left(\mathrm{1}\right):\mathrm{lcm}\left(\mathrm{50},\mathrm{80},\mathrm{x}\right)=\mathrm{lcm}\left(\mathrm{lcm}\left(\mathrm{50},\mathrm{80}\right),\mathrm{x}\right) \\ $$$$=\mathrm{lcm}\left(\mathrm{400},\mathrm{x}\right)=\mathrm{2800} \\ $$$$\left(\mathrm{2}\right):\mathrm{lcm}\left(\mathrm{56},\mathrm{84},\mathrm{x}\right)=\mathrm{lcm}\left(\mathrm{lcm}\left(\mathrm{56},\mathrm{84}\right),\mathrm{x}\right) \\ $$$$=\mathrm{lcm}\left(\mathrm{168},\mathrm{x}\right)=\mathrm{840} \\ $$$$\left(\mathrm{3}\right):\mathrm{400}=\mathrm{2}^{\mathrm{4}} .\mathrm{5}^{\mathrm{2}} \\ $$$$\mathrm{2800}=\mathrm{2}^{\mathrm{4}} .\mathrm{5}^{\mathrm{2}} .\mathrm{7} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{2}^{\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}\right\}} .\mathrm{5}^{\left\{\mathrm{0},\mathrm{1},\mathrm{2}\right\}} .\mathrm{7} \\ $$$$\left(\mathrm{4}\right):\mathrm{168}=\mathrm{2}^{\mathrm{3}} .\mathrm{3}.\mathrm{7} \\ $$$$\mathrm{840}=\mathrm{2}^{\mathrm{3}} .\mathrm{3}.\mathrm{5}.\mathrm{7} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{2}^{\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3}\right\}} .\mathrm{3}^{\left\{\mathrm{0},\mathrm{1}\right\}} .\mathrm{5}.\mathrm{7}^{\left\{\mathrm{0},\mathrm{1}\right\}} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{2}^{\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3}\right\}} .\mathrm{5}.\mathrm{7} \\ $$$$\mathrm{x}=\left\{\mathrm{35},\mathrm{70},\mathrm{140},\mathrm{280}\right\} \\ $$
Commented by Rasheed.Sindhi last updated on 16/Aug/20
Thanks floor sir!
$$\mathcal{T}{hanks}\:{floor}\:{sir}! \\ $$

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