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

$$\mathrm{x},\mathrm{y},\mathrm{z}\in \mathbb{N}$$$$\mathrm{lcd}(\mathrm{x};\mathrm{y})=72$$$$\mathrm{lcd}(\mathrm{x};\mathrm{z})=600$$$$\mathrm{lcd}(\mathrm{y};\mathrm{z})=900$$$$\mathrm{Find}:1.(\mathrm{x};\mathrm{y};\mathrm{z})=?$$$$2.(\mathrm{x};\mathrm{y};\mathrm{z})=?$$$$3.(\mathrm{x};\mathrm{y};\mathrm{z})=?$$$$...$$$$\mathrm{a})15\mathrm{b})16\mathrm{c})24\mathrm{d})27\mathrm{e})64$$

Answered by A5T last updated on 26/Jan/25

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

Commented by hardmath last updated on 26/Jan/25

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

Commented by A5T last updated on 26/Jan/25

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

Commented by hardmath last updated on 26/Jan/25

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

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?

Commented by A5T last updated on 26/Jan/25

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

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?

Commented by A5T last updated on 26/Jan/25

$$\mathrm{lcm}\left(\mathrm{lowest}\mathrm{common}\mathrm{multiple}\right)$$$$\mathrm{For}\mathrm{x}={\mathrm{p}}_1^{{\mathrm{e}}_{11}}{\mathrm{p}}_2^{{\mathrm{e}}_{12}}{\mathrm{p}}_3^{{\mathrm{e}}_{13}}...{\mathrm{p}}_{\mathrm{i}}^{{\mathrm{e}}_{1\mathrm{i}}}\mathrm{and}\mathrm{y}={\mathrm{p}}_1^{{\mathrm{e}}_{21}}{\mathrm{p}}_2^{{\mathrm{e}}_{22}}{\mathrm{p}}_3^{{\mathrm{e}}_{23}}...{\mathrm{p}}_{\mathrm{i}}^{{\mathrm{e}}_{2\mathrm{i}}}$$$$\mathrm{when}{\mathrm{p}}_{\mathrm{i}}={\mathrm{p}}_{\mathrm{j}}\Rightarrow \mathrm{i}=\mathrm{j}\left(\mathrm{each}{\mathrm{p}}_{\mathrm{i}}\mathrm{is}\mathrm{distinct}\right)$$$$\mathrm{lcm}(\mathrm{x},\mathrm{y})={\mathrm{p}}_1^{\max ({\mathrm{e}}_{11},{\mathrm{e}}_{21})}{\mathrm{p}}_2^{\max ({\mathrm{e}}_{12},{\mathrm{e}}_{22})}{\mathrm{p}}_3^{\max ({\mathrm{e}}_{13},{\mathrm{e}}_{23})}...{\mathrm{p}}_{\mathrm{i}}^{\max ({\mathrm{e}}_{1\mathrm{i}},{\mathrm{e}}_{2\mathrm{i}})}$$$$$$$$\mathrm{I}\mathrm{applied}\mathrm{this}\mathrm{and}\mathrm{considered}\mathrm{the}\mathrm{necessary}\mathrm{cases}.$$$$\mathrm{Since}\mathrm{we}\mathrm{were}\mathrm{given}\mathrm{the}\mathrm{lcm}\mathrm{of}\mathrm{three}\mathrm{pairs}\mathrm{of}$$$$\mathrm{numbers},\mathrm{we}\mathrm{could}\mathrm{deduce},\mathrm{through}\mathrm{those},\mathrm{the}$$$$\mathrm{necessary}\mathrm{values}.$$$$\mathrm{Trying}\mathrm{to}\mathrm{explain}\mathrm{further}\mathrm{may}\mathrm{not}\mathrm{necessarily}$$$$\mathrm{make}\mathrm{it}\mathrm{more}\mathrm{clear}\mathrm{if}\mathrm{you}\mathrm{are}\mathrm{not}\mathrm{familiar}$$$$\mathrm{with}\mathrm{it}.$$

Commented by A5T last updated on 26/Jan/25

$$\mathrm{Since}\mathrm{the}\mathrm{lcm}\mathrm{of}\mathrm{the}\mathrm{pairs}\mathrm{of}\mathrm{numbers}\mathrm{contain}$$$$\mathrm{no}\mathrm{other}\mathrm{prime}\mathrm{factors}\mathrm{other}\mathrm{than}2,3\mathrm{or}5,\mathrm{the}$$$$\mathrm{numbers}\mathrm{are}\mathrm{of}\mathrm{the}\mathrm{form}{2}^{{\mathrm{e}}_{\mathrm{i}1}}{3}^{{\mathrm{e}}_{\mathrm{i}2}}{5}^{{\mathrm{e}}_{\mathrm{i}3}}$$$$\mathrm{x}=2^{{\mathrm{e}}_{11}}{3}^{{\mathrm{e}}_{12}}{5}^{{\mathrm{e}}_{13}};\mathrm{y}=2^{{\mathrm{e}}_{21}}{3}^{{\mathrm{e}}_{22}}{5}^{{\mathrm{e}}_{23}};\mathrm{z}=2^{{\mathrm{e}}_{31}}{3}^{{\mathrm{e}}_{32}}{5}^{{\mathrm{e}}_{33}}$$$$\mathrm{lcm}(\mathrm{x},\mathrm{y})=2^3\times 3^2\Rightarrow \max ({\mathrm{e}}_{21},{\mathrm{e}}_{11})=3$$$$\mathrm{But}\mathrm{from}\mathrm{lcm}(\mathrm{y},\mathrm{z})=2^2\times 3^2\times 5^2$$$$\max ({\mathrm{e}}_{21},{\mathrm{e}}_{31})=2\Rightarrow {\mathrm{e}}_{21}⩽2$$$$\Rightarrow \max ({\mathrm{e}}_{21},{\mathrm{e}}_{11})={\mathrm{e}}_{11}=3$$$$\Rightarrow \mathrm{x}=2^3{3}^{{\mathrm{e}}_{12}}{5}^{{\mathrm{e}}_{13}}$$$$\mathrm{Similarly},\mathrm{we}\mathrm{can}\mathrm{get}\mathrm{that}{\mathrm{e}}_{21}=2;{\mathrm{e}}_{33}=2.$$$$\mathrm{We}\mathrm{can}\mathrm{then}\mathrm{continue}\mathrm{to}\mathrm{find}\mathrm{possible}\mathrm{values}\mathrm{for}$$$$\mathrm{the}\mathrm{other}\mathrm{exponents}\left({\mathrm{e}}_{\mathrm{i}}\right).$$$$\mathrm{For}\mathrm{instance},\mathrm{we}\mathrm{get}\mathrm{that}{\mathrm{e}}_{13}={\mathrm{e}}_{23}=0\mathrm{because}$$$$\mathrm{lcm}(\mathrm{x},\mathrm{y})=2^3\times 3^2\times 5^0\Rightarrow \max ({\mathrm{e}}_{13},{\mathrm{e}}_{23})=0$$$$\Rightarrow {\mathrm{e}}_{13}={\mathrm{e}}_{23}=0$$

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