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Question-192463

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Question Number 192463 by Spillover last updated on 18/May/23
Answered by AST last updated on 18/May/23
420=2^2 ×3×5×7 264=2^3 ×3×11 Third number=2^a 3^b 5^c 7^d 11^e gcd(420,264,2^a 3^b 5^c 7^d 11^e )=2^(min(2,3,a)) 3^(min(1,1,b)) =2^2 ×3 ⇒min(2,3,a)=2⇒a≥2;min(1,1,b)=1⇒b≥1 Also,c,d,e≥0 lcm(420,264,2^a 3^b ...)=2^(max(2,3,a)) 3^(max(1,1,b)) ... =2^3 ×3^2 ×5×7×11 ⇒max(2,3,a)=3⇒a≤3 max(1,1,b)=2⇒b=2 max(1,0,c)=1⇒c≤1 (c=0,1) Also; d,e≤1 ⇒a∈{2,3},b∈{2};c,d,e∈{0,1} There are 2^4 =16 possible numbers.
$$\mathrm{420}=\mathrm{2}^{\mathrm{2}} ×\mathrm{3}×\mathrm{5}×\mathrm{7} \\ $$$$\mathrm{264}=\mathrm{2}^{\mathrm{3}} ×\mathrm{3}×\mathrm{11} \\ $$$${Third}\:{number}=\mathrm{2}^{{a}} \mathrm{3}^{{b}} \mathrm{5}^{{c}} \mathrm{7}^{{d}} \mathrm{11}^{{e}} \\ $$$${gcd}\left(\mathrm{420},\mathrm{264},\mathrm{2}^{{a}} \mathrm{3}^{{b}} \mathrm{5}^{{c}} \mathrm{7}^{{d}} \mathrm{11}^{{e}} \right)=\mathrm{2}^{{min}\left(\mathrm{2},\mathrm{3},{a}\right)} \mathrm{3}^{{min}\left(\mathrm{1},\mathrm{1},{b}\right)} =\mathrm{2}^{\mathrm{2}} ×\mathrm{3} \\ $$$$\Rightarrow{min}\left(\mathrm{2},\mathrm{3},{a}\right)=\mathrm{2}\Rightarrow{a}\geqslant\mathrm{2};{min}\left(\mathrm{1},\mathrm{1},{b}\right)=\mathrm{1}\Rightarrow{b}\geqslant\mathrm{1} \\ $$$${Also},{c},{d},{e}\geqslant\mathrm{0} \\ $$$${lcm}\left(\mathrm{420},\mathrm{264},\mathrm{2}^{{a}} \mathrm{3}^{{b}} …\right)=\mathrm{2}^{{max}\left(\mathrm{2},\mathrm{3},{a}\right)} \mathrm{3}^{{max}\left(\mathrm{1},\mathrm{1},{b}\right)} … \\ $$$$=\mathrm{2}^{\mathrm{3}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{5}×\mathrm{7}×\mathrm{11} \\ $$$$\Rightarrow{max}\left(\mathrm{2},\mathrm{3},{a}\right)=\mathrm{3}\Rightarrow{a}\leqslant\mathrm{3} \\ $$$${max}\left(\mathrm{1},\mathrm{1},{b}\right)=\mathrm{2}\Rightarrow{b}=\mathrm{2} \\ $$$${max}\left(\mathrm{1},\mathrm{0},{c}\right)=\mathrm{1}\Rightarrow{c}\leqslant\mathrm{1}\:\left({c}=\mathrm{0},\mathrm{1}\right) \\ $$$${Also};\:{d},{e}\leqslant\mathrm{1} \\ $$$$\Rightarrow{a}\in\left\{\mathrm{2},\mathrm{3}\right\},{b}\in\left\{\mathrm{2}\right\};{c},{d},{e}\in\left\{\mathrm{0},\mathrm{1}\right\} \\ $$$${There}\:{are}\:\mathrm{2}^{\mathrm{4}} =\mathrm{16}\:{possible}\:{numbers}. \\ $$
Commented by BaliramKumar last updated on 19/May/23
i think 8 numbers
$$\mathrm{i}\:\mathrm{think}\:\mathrm{8}\:\mathrm{numbers} \\ $$
Commented by Spillover last updated on 19/May/23
thank you.
$${thank}\:{you}. \\ $$
Answered by BaliramKumar last updated on 31/May/23
x = 2^2 ×3^2 [2^(0→1) , 5^(0→1) , 7^(0→1) , 11^(0→1) ] 420 = 2^2 ×3×5×7 264 = 2^2 ×3×2×11 27720 = 2^3 ×3^2 ×5×7×11 x = 2×2×2×2 = 16 value
$$ \\ $$$${x}\:=\:\mathrm{2}^{\mathrm{2}} ×\mathrm{3}^{\mathrm{2}} \left[\mathrm{2}^{\mathrm{0}\rightarrow\mathrm{1}} ,\:\mathrm{5}^{\mathrm{0}\rightarrow\mathrm{1}} ,\:\mathrm{7}^{\mathrm{0}\rightarrow\mathrm{1}} ,\:\mathrm{11}^{\mathrm{0}\rightarrow\mathrm{1}} \right] \\ $$$$\mathrm{420}\:=\:\mathrm{2}^{\mathrm{2}} ×\mathrm{3}×\mathrm{5}×\mathrm{7} \\ $$$$\mathrm{264}\:=\:\mathrm{2}^{\mathrm{2}} ×\mathrm{3}×\mathrm{2}×\mathrm{11} \\ $$$$\mathrm{27720}\:=\:\mathrm{2}^{\mathrm{3}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{5}×\mathrm{7}×\mathrm{11} \\ $$$$\mathrm{x}\:=\:\mathrm{2}×\mathrm{2}×\mathrm{2}×\mathrm{2}\:=\:\mathrm{16}\:\mathrm{value} \\ $$
Commented by AST last updated on 19/May/23
What about 72×1,72×5,... ?
$${What}\:{about}\:\mathrm{72}×\mathrm{1},\mathrm{72}×\mathrm{5},…\:? \\ $$
Commented by BaliramKumar last updated on 19/May/23
Yes Thanks
$$\mathrm{Yes}\:\:\:\mathrm{Thanks} \\ $$
Commented by AST last updated on 19/May/23
⇒16 numbers
$$\Rightarrow\mathrm{16}\:{numbers} \\ $$
Commented by Spillover last updated on 19/May/23
thank you
$${thank}\:{you} \\ $$
Answered by Spillover last updated on 01/Feb/24
LCM=27720=2^3 ×3^2 ×5×11 GCF=12=2^2 ×3 1^(st) no=420=2^2 ×3×5×7 2^(nd) no=264=2^3 ×3×11 3^(rd ) no may have 5,7 or 11 as it prime factors but 2 and 3 must be contained Possible value of the third number 36=2^2 ×3^2 72=2^3 ×3^2 180=2^2 ×3^2 ×5 360=2^3 ×3^2 ×5 252=2^2 ×3^2 ×7 504=2^3 ×3^2 ×7 396=2^2 ×3^2 ×11 792=2^3 ×3^2 ×5×7 1260=2^2 ×3^2 ×5×7 1980=2^2 ×3^2 ×5×11 2772=2^2 ×3^2 ×7×11 2520=2^3 ×3^2 ×5×7 3960=2^3 ×3^2 ×5×11 5544=2^3 ×3^2 ×7×11 13860=2^2 ×3^2 ×5×7×11 19/5/2023
$${LCM}=\mathrm{27720}=\mathrm{2}^{\mathrm{3}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{5}×\mathrm{11}\:\:\:\: \\ $$$${GCF}=\mathrm{12}=\mathrm{2}^{\mathrm{2}} ×\mathrm{3} \\ $$$$\mathrm{1}^{{st}} \:{no}=\mathrm{420}=\mathrm{2}^{\mathrm{2}} ×\mathrm{3}×\mathrm{5}×\mathrm{7} \\ $$$$\mathrm{2}^{{nd}} \:{no}=\mathrm{264}=\mathrm{2}^{\mathrm{3}} ×\mathrm{3}×\mathrm{11} \\ $$$$\mathrm{3}^{{rd}\:} {no}\:{may}\:{have}\:\mathrm{5},\mathrm{7}\:{or}\:\mathrm{11}\:{as}\:{it}\:{prime}\: \\ $$$${factors}\:{but}\:\mathrm{2}\:{and}\:\mathrm{3}\:{must}\:{be}\:{contained} \\ $$$$\underline{{Possible}\:{value}\:{of}\:{the}\:{third}\:{number}} \\ $$$$\mathrm{36}=\mathrm{2}^{\mathrm{2}} ×\mathrm{3}^{\mathrm{2}} \\ $$$$\mathrm{72}=\mathrm{2}^{\mathrm{3}} ×\mathrm{3}^{\mathrm{2}} \\ $$$$\mathrm{180}=\mathrm{2}^{\mathrm{2}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{5} \\ $$$$\mathrm{360}=\mathrm{2}^{\mathrm{3}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{5} \\ $$$$\mathrm{252}=\mathrm{2}^{\mathrm{2}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{7} \\ $$$$\mathrm{504}=\mathrm{2}^{\mathrm{3}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{7} \\ $$$$\mathrm{396}=\mathrm{2}^{\mathrm{2}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{11} \\ $$$$\mathrm{792}=\mathrm{2}^{\mathrm{3}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{5}×\mathrm{7} \\ $$$$\mathrm{1260}=\mathrm{2}^{\mathrm{2}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{5}×\mathrm{7} \\ $$$$\mathrm{1980}=\mathrm{2}^{\mathrm{2}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{5}×\mathrm{11} \\ $$$$\mathrm{2772}=\mathrm{2}^{\mathrm{2}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{7}×\mathrm{11} \\ $$$$\mathrm{2520}=\mathrm{2}^{\mathrm{3}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{5}×\mathrm{7} \\ $$$$\mathrm{3960}=\mathrm{2}^{\mathrm{3}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{5}×\mathrm{11} \\ $$$$\mathrm{5544}=\mathrm{2}^{\mathrm{3}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{7}×\mathrm{11} \\ $$$$\mathrm{13860}=\mathrm{2}^{\mathrm{2}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{5}×\mathrm{7}×\mathrm{11} \\ $$$$\mathrm{19}/\mathrm{5}/\mathrm{2023} \\ $$

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