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Question Number 216050 by hardmath last updated on 26/Jan/25

x,y,z ∈ N lcd(x;y)=72 lcd(x;z)=600 lcd(y;z)=900 Find: 1.(x;y;z)=? 2.(x;y;z)=? 3.(x;y;z)=? ... a)15 b)16 c)24 d)27 e)64

$$\mathrm{x},\mathrm{y},\mathrm{z}\:\in\:\mathbb{N} \\ $$$$\mathrm{lcd}\left(\mathrm{x};\mathrm{y}\right)=\mathrm{72} \\ $$$$\mathrm{lcd}\left(\mathrm{x};\mathrm{z}\right)=\mathrm{600} \\ $$$$\mathrm{lcd}\left(\mathrm{y};\mathrm{z}\right)=\mathrm{900} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{1}.\left(\mathrm{x};\mathrm{y};\mathrm{z}\right)=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}.\left(\mathrm{x};\mathrm{y};\mathrm{z}\right)=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}.\left(\mathrm{x};\mathrm{y};\mathrm{z}\right)=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:... \\ $$$$\left.\mathrm{a}\left.\right)\left.\mathrm{1}\left.\mathrm{5}\left.\:\:\:\:\:\mathrm{b}\right)\mathrm{16}\:\:\:\:\:\mathrm{c}\right)\mathrm{24}\:\:\:\:\:\mathrm{d}\right)\mathrm{27}\:\:\:\:\:\mathrm{e}\right)\mathrm{64} \\ $$

Answered by A5T last updated on 26/Jan/25

Do you mean gcd or lcm? gcd leads to a contradiction. lcm(x,y)=2^3 ×3^2 =72 lcm(y,z)=2^2 ×3^2 ×5^2 =900 lcm(z,x)=2^3 ×3×5^2 =600 ⇒x=2^3 k; y=3^2 l; z=5^2 m x=2^3 ×3^a ; y=3^2 ×2^b ; z=2^c ×3^d ×5^2 a∈{0,1}; b∈{0,1,2}; c∈{0,1,2}; d∈{0,1} There are five possible combinations (b,c): (0,2);(1,2);(2,0);(2,1);(2,2) a=0⇒d=1 and 5 possibilities for (b,c) a=1⇒ d∈{0,1} and 5 possibilities for (b,c) ⇒Total number of solutions=(1×5)+(2×5) =15

$$\mathrm{Do}\:\mathrm{you}\:\mathrm{mean}\:\mathrm{gcd}\:\mathrm{or}\:\mathrm{lcm}? \\ $$$$\mathrm{gcd}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{a}\:\mathrm{contradiction}. \\ $$$$\mathrm{lcm}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{2}^{\mathrm{3}} ×\mathrm{3}^{\mathrm{2}} =\mathrm{72} \\ $$$$\mathrm{lcm}\left(\mathrm{y},\mathrm{z}\right)=\mathrm{2}^{\mathrm{2}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{5}^{\mathrm{2}} =\mathrm{900} \\ $$$$\mathrm{lcm}\left(\mathrm{z},\mathrm{x}\right)=\mathrm{2}^{\mathrm{3}} ×\mathrm{3}×\mathrm{5}^{\mathrm{2}} =\mathrm{600} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{2}^{\mathrm{3}} \mathrm{k};\:\mathrm{y}=\mathrm{3}^{\mathrm{2}} \mathrm{l};\:\mathrm{z}=\mathrm{5}^{\mathrm{2}} \mathrm{m} \\ $$$$\mathrm{x}=\mathrm{2}^{\mathrm{3}} ×\mathrm{3}^{\mathrm{a}} ;\:\mathrm{y}=\mathrm{3}^{\mathrm{2}} ×\mathrm{2}^{\mathrm{b}} ;\:\mathrm{z}=\mathrm{2}^{\mathrm{c}} ×\mathrm{3}^{\mathrm{d}} ×\mathrm{5}^{\mathrm{2}} \\ $$$$\mathrm{a}\in\left\{\mathrm{0},\mathrm{1}\right\};\:\mathrm{b}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2}\right\};\:\mathrm{c}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2}\right\};\:\mathrm{d}\in\left\{\mathrm{0},\mathrm{1}\right\} \\ $$$$\mathrm{There}\:\mathrm{are}\:\mathrm{five}\:\mathrm{possible}\:\mathrm{combinations}\:\left(\mathrm{b},\mathrm{c}\right): \\ $$$$\left(\mathrm{0},\mathrm{2}\right);\left(\mathrm{1},\mathrm{2}\right);\left(\mathrm{2},\mathrm{0}\right);\left(\mathrm{2},\mathrm{1}\right);\left(\mathrm{2},\mathrm{2}\right) \\ $$$$\mathrm{a}=\mathrm{0}\Rightarrow\mathrm{d}=\mathrm{1}\:\mathrm{and}\:\mathrm{5}\:\mathrm{possibilities}\:\mathrm{for}\:\left(\mathrm{b},\mathrm{c}\right) \\ $$$$\mathrm{a}=\mathrm{1}\Rightarrow\:\mathrm{d}\in\left\{\mathrm{0},\mathrm{1}\right\}\:\mathrm{and}\:\mathrm{5}\:\mathrm{possibilities}\:\mathrm{for}\:\left(\mathrm{b},\mathrm{c}\right) \\ $$$$\Rightarrow\mathrm{Total}\:\mathrm{number}\:\mathrm{of}\:\mathrm{solutions}=\left(\mathrm{1}×\mathrm{5}\right)+\left(\mathrm{2}×\mathrm{5}\right) \\ $$$$=\mathrm{15} \\ $$

Commented by hardmath last updated on 26/Jan/25

Sorry, dear professor - lcm

$$\mathrm{Sorry},\:\mathrm{dear}\:\mathrm{professor}\:-\:\mathrm{lcm} \\ $$

Commented by A5T last updated on 26/Jan/25

Is the answer above correct?

$$\mathrm{Is}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{above}\:\mathrm{correct}? \\ $$

Commented by hardmath last updated on 26/Jan/25

yes dear professor answer 15

$$\mathrm{yes}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{answer}\:\mathrm{15} \\ $$

Commented by hardmath last updated on 26/Jan/25

My dear professor, thank you very much for the nice solution, is there any other solution?

$$ \\ $$My dear professor, thank you very much for the nice solution, is there any other solution?

Commented by A5T last updated on 26/Jan/25

There could be other solutions.

$$\mathrm{There}\:\mathrm{could}\:\mathrm{be}\:\mathrm{other}\:\mathrm{solutions}. \\ $$

Commented by hardmath last updated on 26/Jan/25

It would be nice if there were other solutions, I don't fully understand this solution, is there a more succinct solution?

$$ \\ $$It would be nice if there were other solutions, I don't fully understand this solution, is there a more succinct solution?

Commented by A5T last updated on 26/Jan/25

lcm(lowest common multiple) For x=p_1 ^e_(11) p_2 ^e_(12) p_3 ^e_(13) ...p_i ^e_(1i) and y=p_1 ^e_(21) p_2 ^e_(22) p_3 ^e_(23) ...p_i ^e_(2i) when p_i =p_j ⇒i=j (each p_i is distinct) lcm(x,y)=p_1 ^(max(e_(11) ,e_(21) )) p_2 ^(max(e_(12) ,e_(22) )) p_3 ^(max(e_(13) ,e_(23) )) ...p_i ^(max(e_(1i) ,e_(2i) )) I applied this and considered the necessary cases. Since we were given the lcm of three pairs of numbers, we could deduce, through those, the necessary values. Trying to explain further may not necessarily make it more clear if you are not familiar with it.

$$\mathrm{lcm}\left(\mathrm{lowest}\:\mathrm{common}\:\mathrm{multiple}\right) \\ $$$$\mathrm{For}\:\mathrm{x}=\mathrm{p}_{\mathrm{1}} ^{\mathrm{e}_{\mathrm{11}} } \mathrm{p}_{\mathrm{2}} ^{\mathrm{e}_{\mathrm{12}} } \mathrm{p}_{\mathrm{3}} ^{\mathrm{e}_{\mathrm{13}} } ...\mathrm{p}_{\mathrm{i}} ^{\mathrm{e}_{\mathrm{1i}} } \:\mathrm{and}\:\mathrm{y}=\mathrm{p}_{\mathrm{1}} ^{\mathrm{e}_{\mathrm{21}} } \mathrm{p}_{\mathrm{2}} ^{\mathrm{e}_{\mathrm{22}} } \mathrm{p}_{\mathrm{3}} ^{\mathrm{e}_{\mathrm{23}} } ...\mathrm{p}_{\mathrm{i}} ^{\mathrm{e}_{\mathrm{2i}} } \\ $$$$\mathrm{when}\:\mathrm{p}_{\mathrm{i}} =\mathrm{p}_{\mathrm{j}} \Rightarrow\mathrm{i}=\mathrm{j}\:\left(\mathrm{each}\:\mathrm{p}_{\mathrm{i}} \:\mathrm{is}\:\mathrm{distinct}\right) \\ $$$$\mathrm{lcm}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{p}_{\mathrm{1}} ^{\mathrm{max}\left(\mathrm{e}_{\mathrm{11}} ,\mathrm{e}_{\mathrm{21}} \right)} \mathrm{p}_{\mathrm{2}} ^{\mathrm{max}\left(\mathrm{e}_{\mathrm{12}} ,\mathrm{e}_{\mathrm{22}} \right)} \mathrm{p}_{\mathrm{3}} ^{\mathrm{max}\left(\mathrm{e}_{\mathrm{13}} ,\mathrm{e}_{\mathrm{23}} \right)} ...\mathrm{p}_{\mathrm{i}} ^{\mathrm{max}\left(\mathrm{e}_{\mathrm{1i}} ,\mathrm{e}_{\mathrm{2i}} \right)} \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{applied}\:\mathrm{this}\:\mathrm{and}\:\mathrm{considered}\:\mathrm{the}\:\mathrm{necessary}\:\mathrm{cases}. \\ $$$$\mathrm{Since}\:\mathrm{we}\:\mathrm{were}\:\mathrm{given}\:\mathrm{the}\:\mathrm{lcm}\:\mathrm{of}\:\mathrm{three}\:\mathrm{pairs}\:\mathrm{of} \\ $$$$\mathrm{numbers},\:\mathrm{we}\:\mathrm{could}\:\mathrm{deduce},\:\mathrm{through}\:\mathrm{those},\:\mathrm{the} \\ $$$$\mathrm{necessary}\:\mathrm{values}.\: \\ $$$$\mathrm{Trying}\:\mathrm{to}\:\mathrm{explain}\:\mathrm{further}\:\mathrm{may}\:\mathrm{not}\:\mathrm{necessarily} \\ $$$$\mathrm{make}\:\mathrm{it}\:\mathrm{more}\:\mathrm{clear}\:\mathrm{if}\:\mathrm{you}\:\mathrm{are}\:\mathrm{not}\:\mathrm{familiar} \\ $$$$\mathrm{with}\:\mathrm{it}. \\ $$

Commented by A5T last updated on 26/Jan/25

Since the lcm of the pairs of numbers contain no other prime factors other than 2, 3 or 5, the numbers are of the form 2^e_(i1) 3^e_(i2) 5^e_(i3) x=2^e_(11) 3^e_(12) 5^e_(13) ; y=2^e_(21) 3^e_(22) 5^e_(23) ; z=2^e_(31) 3^e_(32) 5^e_(33) lcm(x,y)=2^3 ×3^2 ⇒max(e_(21) , e_(11) )=3 But from lcm(y,z)=2^2 ×3^2 ×5^2 max(e_(21) , e_(31) )=2⇒e_(21) ≤2 ⇒max(e_(21) ,e_(11) )=e_(11) =3 ⇒x=2^3 3^e_(12) 5^e_(13) Similarly, we can get that e_(21) =2; e_(33) =2 . We can then continue to find possible values for the other exponents(e_i ) . For instance, we get that e_(13) =e_(23) =0 because lcm(x,y)=2^3 ×3^2 ×5^0 ⇒max(e_(13) ,e_(23) )=0 ⇒e_(13) =e_(23) =0

$$\mathrm{Since}\:\mathrm{the}\:\mathrm{lcm}\:\mathrm{of}\:\mathrm{the}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{numbers}\:\mathrm{contain} \\ $$$$\mathrm{no}\:\mathrm{other}\:\mathrm{prime}\:\mathrm{factors}\:\mathrm{other}\:\mathrm{than}\:\mathrm{2},\:\mathrm{3}\:\mathrm{or}\:\mathrm{5},\:\mathrm{the} \\ $$$$\mathrm{numbers}\:\mathrm{are}\:\mathrm{of}\:\mathrm{the}\:\mathrm{form}\:\mathrm{2}^{\mathrm{e}_{\mathrm{i1}} } \mathrm{3}^{\mathrm{e}_{\mathrm{i2}} } \mathrm{5}^{\mathrm{e}_{\mathrm{i3}} } \\ $$$$\mathrm{x}=\mathrm{2}^{\mathrm{e}_{\mathrm{11}} } \mathrm{3}^{\mathrm{e}_{\mathrm{12}} } \mathrm{5}^{\mathrm{e}_{\mathrm{13}} } ;\:\mathrm{y}=\mathrm{2}^{\mathrm{e}_{\mathrm{21}} } \mathrm{3}^{\mathrm{e}_{\mathrm{22}} } \mathrm{5}^{\mathrm{e}_{\mathrm{23}} } ;\:\mathrm{z}=\mathrm{2}^{\mathrm{e}_{\mathrm{31}} } \mathrm{3}^{\mathrm{e}_{\mathrm{32}} } \mathrm{5}^{\mathrm{e}_{\mathrm{33}} } \\ $$$$\mathrm{lcm}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{2}^{\mathrm{3}} ×\mathrm{3}^{\mathrm{2}} \Rightarrow\mathrm{max}\left(\mathrm{e}_{\mathrm{21}} ,\:\mathrm{e}_{\mathrm{11}} \right)=\mathrm{3} \\ $$$$\mathrm{But}\:\mathrm{from}\:\mathrm{lcm}\left(\mathrm{y},\mathrm{z}\right)=\mathrm{2}^{\mathrm{2}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{5}^{\mathrm{2}} \\ $$$$\mathrm{max}\left(\mathrm{e}_{\mathrm{21}} ,\:\mathrm{e}_{\mathrm{31}} \right)=\mathrm{2}\Rightarrow\mathrm{e}_{\mathrm{21}} \leqslant\mathrm{2}\: \\ $$$$\Rightarrow\mathrm{max}\left(\mathrm{e}_{\mathrm{21}} ,\mathrm{e}_{\mathrm{11}} \right)=\mathrm{e}_{\mathrm{11}} =\mathrm{3} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{2}^{\mathrm{3}} \mathrm{3}^{\mathrm{e}_{\mathrm{12}} } \mathrm{5}^{\mathrm{e}_{\mathrm{13}} } \\ $$$$\mathrm{Similarly},\:\mathrm{we}\:\mathrm{can}\:\mathrm{get}\:\mathrm{that}\:\mathrm{e}_{\mathrm{21}} =\mathrm{2};\:\mathrm{e}_{\mathrm{33}} =\mathrm{2}\:. \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{then}\:\mathrm{continue}\:\mathrm{to}\:\mathrm{find}\:\mathrm{possible}\:\mathrm{values}\:\mathrm{for} \\ $$$$\mathrm{the}\:\mathrm{other}\:\mathrm{exponents}\left(\mathrm{e}_{\mathrm{i}} \right)\:. \\ $$$$\mathrm{For}\:\mathrm{instance},\:\mathrm{we}\:\mathrm{get}\:\mathrm{that}\:\mathrm{e}_{\mathrm{13}} =\mathrm{e}_{\mathrm{23}} =\mathrm{0}\:\mathrm{because} \\ $$$$\mathrm{lcm}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{2}^{\mathrm{3}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{5}^{\mathrm{0}} \Rightarrow\mathrm{max}\left(\mathrm{e}_{\mathrm{13}} ,\mathrm{e}_{\mathrm{23}} \right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{e}_{\mathrm{13}} =\mathrm{e}_{\mathrm{23}} =\mathrm{0} \\ $$

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