Question Number 181837 by mr W last updated on 01/Dec/22
$${find}\:{integers}\:{a}>{b}>{c}>\mathrm{0}\:{such}\:{that} \\ $$$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{2}}{{b}}+\frac{\mathrm{3}}{{c}}=\mathrm{1} \\ $$
Commented by SEKRET last updated on 01/Dec/22
$$\left(\mathrm{2};\mathrm{5};\mathrm{30}\right)\:\left(\mathrm{2};\mathrm{6};\mathrm{18}\right)\:\left(\mathrm{2};\mathrm{7};\mathrm{14}\right)\:\left(\mathrm{2};\mathrm{8};\mathrm{12}\right) \\ $$$$\left(\mathrm{2};\mathrm{10};\mathrm{10}\right)\:\left(\mathrm{2};\mathrm{12};\mathrm{9}\right)\left(\mathrm{2};\mathrm{16};\mathrm{8}\right)\left(\mathrm{2};\mathrm{28};\mathrm{7}\right) \\ $$$$\left(\mathrm{3};\mathrm{4};\mathrm{18}\right)\left(\mathrm{3};\mathrm{6};\mathrm{9}\right)\left(\mathrm{3};\mathrm{12};\mathrm{6}\right)\left(\mathrm{3};\mathrm{30};\mathrm{5}\right) \\ $$$$\left(\mathrm{4};\mathrm{3};\mathrm{36}\right)\left(\mathrm{4};\mathrm{4};\mathrm{12}\right)\left(\mathrm{4};\mathrm{8};\mathrm{6}\right)\left(\mathrm{5};\mathrm{4};\mathrm{10}\right) \\ $$$$\left(\mathrm{5};\mathrm{10};\mathrm{5}\right)\left(\mathrm{5};\mathrm{40};\mathrm{4}\right)\left(\mathrm{15};\mathrm{6};\mathrm{5}\right)\left(\mathrm{20};\mathrm{10};\mathrm{4}\right) \\ $$$$\left(\mathrm{30};\mathrm{3};\mathrm{10}\right)\left(\mathrm{36};\mathrm{9};\mathrm{4}\right)\left(\mathrm{6};\mathrm{3};\mathrm{18}\right)\left(\mathrm{6};\mathrm{4};\mathrm{9}\right) \\ $$$$\left(\mathrm{6};\mathrm{6};\mathrm{6}\right)\left(\mathrm{6};\mathrm{24};\mathrm{4}\right)\left(\mathrm{8};\mathrm{4};\mathrm{8}\right)\left(\mathrm{8};\mathrm{16};\mathrm{4}\right)\left(\mathrm{10};\mathrm{5};\mathrm{6}\right) \\ $$$$\left(\mathrm{12};\mathrm{3};\mathrm{2}\right)\left(\mathrm{12};\mathrm{12};\mathrm{4}\right)\left(\mathrm{14};\mathrm{4};\mathrm{7}\right)….. \\ $$$$ \\ $$
Commented by mr W last updated on 01/Dec/22
$${only}\:{solutions}\:{with}\:{a}>{b}>{c}>\mathrm{0}\:{are}\: \\ $$$${requested}. \\ $$
Answered by prakash jain last updated on 01/Dec/22
$${a}=\mathrm{2} \\ $$$${b}=\mathrm{8} \\ $$$${c}=\mathrm{12} \\ $$
Commented by mr W last updated on 01/Dec/22
$${thanks}\:{for}\:{trying}\:{sir}! \\ $$$${but}\:{only}\:{solutions}\:{with}\:{a}>{b}>{c}>\mathrm{0}\:{are}\: \\ $$$${requested}. \\ $$
Answered by MJS_new last updated on 01/Dec/22
$${a}>{b}>{c}>\mathrm{0}\:\Rightarrow\:{c}>\mathrm{3} \\ $$$${a}=\frac{{bc}}{{bc}−\mathrm{3}{b}−\mathrm{2}{c}} \\ $$$${a}>{b}\:\Rightarrow\:\frac{{c}}{{bc}−\mathrm{3}{b}−\mathrm{2}{c}}>\mathrm{1}\wedge{c}>\mathrm{3} \\ $$$$\Leftrightarrow \\ $$$${c}<{b}<\frac{\mathrm{3}{c}}{{c}−\mathrm{3}}\wedge{c}>\mathrm{3}\:\Rightarrow\:{c}=\mathrm{4}\wedge\mathrm{4}<{b}<\mathrm{12}\:\vee\:{c}=\mathrm{5}\wedge\mathrm{5}<{b}<\mathrm{7}.\mathrm{5} \\ $$$$\mathrm{not}\:\mathrm{many}\:\mathrm{possibilities}\:\mathrm{to}\:\mathrm{try}\:\mathrm{out} \\ $$$$\Rightarrow \\ $$$${a}=\mathrm{36}\wedge{b}=\mathrm{9}\wedge{c}=\mathrm{4} \\ $$$${a}=\mathrm{20}\wedge{b}=\mathrm{10}\wedge{c}=\mathrm{4} \\ $$$${a}=\mathrm{15}\wedge{b}=\mathrm{6}\wedge{c}=\mathrm{5} \\ $$$$\mathrm{are}\:\mathrm{the}\:\mathrm{only}\:\mathrm{solutions} \\ $$
Commented by mr W last updated on 02/Dec/22
$${thanks}\:{alot}\:{sir}! \\ $$
Answered by mr W last updated on 02/Dec/22
$${a}>{b}>{c}>\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}}<\frac{\mathrm{1}}{{b}}<\frac{\mathrm{1}}{{c}} \\ $$$$\mathrm{1}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{2}}{{b}}+\frac{\mathrm{3}}{{c}}<\frac{\mathrm{1}}{{c}}+\frac{\mathrm{2}}{{c}}+\frac{\mathrm{3}}{{c}}=\frac{\mathrm{6}}{{c}} \\ $$$$\Rightarrow{c}<\mathrm{6} \\ $$$$\frac{\mathrm{3}}{{c}}=\mathrm{1}−\frac{\mathrm{1}}{{a}}−\frac{\mathrm{2}}{{b}}<\mathrm{1} \\ $$$$\Rightarrow{c}>\mathrm{3} \\ $$$$\Rightarrow{c}\:{may}\:{only}\:{be}\:\mathrm{4}\:{or}\:\mathrm{5}. \\ $$$$\underline{\boldsymbol{{with}}\:\boldsymbol{{c}}=\mathrm{4}:} \\ $$$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{2}}{{b}}=\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{2}}{{b}}<\frac{\mathrm{1}}{{b}}+\frac{\mathrm{2}}{{b}}=\frac{\mathrm{3}}{{b}} \\ $$$$\Rightarrow{b}<\mathrm{12} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{2}}{{b}}>\frac{\mathrm{2}}{{b}} \\ $$$$\Rightarrow{b}>\mathrm{8} \\ $$$$\Rightarrow{b}\:{may}\:{only}\:{be}\:\mathrm{9},\:\mathrm{10}\:{or}\:\mathrm{11}. \\ $$$${b}=\mathrm{9}:\:\frac{\mathrm{1}}{{a}}=\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{2}}{\mathrm{9}}=\frac{\mathrm{1}}{\mathrm{36}}\:\Rightarrow{a}=\mathrm{36}\:\checkmark \\ $$$${b}=\mathrm{10}:\:\frac{\mathrm{1}}{{a}}=\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{2}}{\mathrm{10}}=\frac{\mathrm{1}}{\mathrm{20}}\:\Rightarrow{a}=\mathrm{20}\:\checkmark \\ $$$${b}=\mathrm{11}:\:\frac{\mathrm{1}}{{a}}=\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{2}}{\mathrm{11}}=\frac{\mathrm{7}}{\mathrm{44}}\:\Rightarrow{a}=\frac{\mathrm{44}}{\mathrm{7}}\:× \\ $$$$\underline{\boldsymbol{{with}}\:\boldsymbol{{c}}=\mathrm{5}:} \\ $$$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{2}}{{b}}=\mathrm{1}−\frac{\mathrm{3}}{\mathrm{5}}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\frac{\mathrm{2}}{\mathrm{5}}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{2}}{{b}}<\frac{\mathrm{1}}{{b}}+\frac{\mathrm{2}}{{b}}=\frac{\mathrm{3}}{{b}} \\ $$$$\Rightarrow{b}<\frac{\mathrm{15}}{\mathrm{2}}=\mathrm{7}.\mathrm{5} \\ $$$$\frac{\mathrm{2}}{\mathrm{5}}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{2}}{{b}}>\frac{\mathrm{2}}{{b}} \\ $$$$\Rightarrow{b}>\mathrm{5} \\ $$$$\Rightarrow{b}\:{may}\:{only}\:{be}\:\mathrm{6}\:{or}\:\mathrm{7}. \\ $$$${b}=\mathrm{6}:\:\frac{\mathrm{1}}{{a}}=\frac{\mathrm{2}}{\mathrm{5}}−\frac{\mathrm{2}}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{15}}\:\Rightarrow{a}=\mathrm{15}\:\checkmark \\ $$$${b}=\mathrm{7}:\:\frac{\mathrm{1}}{{a}}=\frac{\mathrm{2}}{\mathrm{5}}−\frac{\mathrm{2}}{\mathrm{7}}=\frac{\mathrm{4}}{\mathrm{35}}\:\Rightarrow{a}=\frac{\mathrm{35}}{\mathrm{4}}\:× \\ $$$${summary}: \\ $$$$\left({a},{b},{c}\right)=\left(\mathrm{36},\mathrm{9},\mathrm{4}\right),\:\left(\mathrm{20},\mathrm{10},\mathrm{4}\right),\:\left(\mathrm{15},\mathrm{6},\mathrm{5}\right) \\ $$