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Question Number 181837 by mr W last updated on 01/Dec/22

find integers a>b>c>0 such that (1/a)+(2/b)+(3/c)=1

$${find}\:{integers}\:{a}>{b}>{c}>\mathrm{0}\:{such}\:{that} \\ $$ $$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{2}}{{b}}+\frac{\mathrm{3}}{{c}}=\mathrm{1} \\ $$

Commented bySEKRET last updated on 01/Dec/22

(2;5;30) (2;6;18) (2;7;14) (2;8;12) (2;10;10) (2;12;9)(2;16;8)(2;28;7) (3;4;18)(3;6;9)(3;12;6)(3;30;5) (4;3;36)(4;4;12)(4;8;6)(5;4;10) (5;10;5)(5;40;4)(15;6;5)(20;10;4) (30;3;10)(36;9;4)(6;3;18)(6;4;9) (6;6;6)(6;24;4)(8;4;8)(8;16;4)(10;5;6) (12;3;2)(12;12;4)(14;4;7).....

$$\left(\mathrm{2};\mathrm{5};\mathrm{30}\right)\:\left(\mathrm{2};\mathrm{6};\mathrm{18}\right)\:\left(\mathrm{2};\mathrm{7};\mathrm{14}\right)\:\left(\mathrm{2};\mathrm{8};\mathrm{12}\right) \\ $$ $$\left(\mathrm{2};\mathrm{10};\mathrm{10}\right)\:\left(\mathrm{2};\mathrm{12};\mathrm{9}\right)\left(\mathrm{2};\mathrm{16};\mathrm{8}\right)\left(\mathrm{2};\mathrm{28};\mathrm{7}\right) \\ $$ $$\left(\mathrm{3};\mathrm{4};\mathrm{18}\right)\left(\mathrm{3};\mathrm{6};\mathrm{9}\right)\left(\mathrm{3};\mathrm{12};\mathrm{6}\right)\left(\mathrm{3};\mathrm{30};\mathrm{5}\right) \\ $$ $$\left(\mathrm{4};\mathrm{3};\mathrm{36}\right)\left(\mathrm{4};\mathrm{4};\mathrm{12}\right)\left(\mathrm{4};\mathrm{8};\mathrm{6}\right)\left(\mathrm{5};\mathrm{4};\mathrm{10}\right) \\ $$ $$\left(\mathrm{5};\mathrm{10};\mathrm{5}\right)\left(\mathrm{5};\mathrm{40};\mathrm{4}\right)\left(\mathrm{15};\mathrm{6};\mathrm{5}\right)\left(\mathrm{20};\mathrm{10};\mathrm{4}\right) \\ $$ $$\left(\mathrm{30};\mathrm{3};\mathrm{10}\right)\left(\mathrm{36};\mathrm{9};\mathrm{4}\right)\left(\mathrm{6};\mathrm{3};\mathrm{18}\right)\left(\mathrm{6};\mathrm{4};\mathrm{9}\right) \\ $$ $$\left(\mathrm{6};\mathrm{6};\mathrm{6}\right)\left(\mathrm{6};\mathrm{24};\mathrm{4}\right)\left(\mathrm{8};\mathrm{4};\mathrm{8}\right)\left(\mathrm{8};\mathrm{16};\mathrm{4}\right)\left(\mathrm{10};\mathrm{5};\mathrm{6}\right) \\ $$ $$\left(\mathrm{12};\mathrm{3};\mathrm{2}\right)\left(\mathrm{12};\mathrm{12};\mathrm{4}\right)\left(\mathrm{14};\mathrm{4};\mathrm{7}\right)..... \\ $$ $$ \\ $$

Commented bymr W last updated on 01/Dec/22

only solutions with a>b>c>0 are requested.

$${only}\:{solutions}\:{with}\:{a}>{b}>{c}>\mathrm{0}\:{are}\: \\ $$ $${requested}. \\ $$

Answered by prakash jain last updated on 01/Dec/22

a=2 b=8 c=12

$${a}=\mathrm{2} \\ $$ $${b}=\mathrm{8} \\ $$ $${c}=\mathrm{12} \\ $$

Commented bymr W last updated on 01/Dec/22

thanks for trying sir! but only solutions with a>b>c>0 are requested.

$${thanks}\:{for}\:{trying}\:{sir}! \\ $$ $${but}\:{only}\:{solutions}\:{with}\:{a}>{b}>{c}>\mathrm{0}\:{are}\: \\ $$ $${requested}. \\ $$

Answered by MJS_new last updated on 01/Dec/22

a>b>c>0 ⇒ c>3 a=((bc)/(bc−3b−2c)) a>b ⇒ (c/(bc−3b−2c))>1∧c>3 ⇔ c<b<((3c)/(c−3))∧c>3 ⇒ c=4∧4<b<12 ∨ c=5∧5<b<7.5 not many possibilities to try out ⇒ a=36∧b=9∧c=4 a=20∧b=10∧c=4 a=15∧b=6∧c=5 are the only solutions

$${a}>{b}>{c}>\mathrm{0}\:\Rightarrow\:{c}>\mathrm{3} \\ $$ $${a}=\frac{{bc}}{{bc}−\mathrm{3}{b}−\mathrm{2}{c}} \\ $$ $${a}>{b}\:\Rightarrow\:\frac{{c}}{{bc}−\mathrm{3}{b}−\mathrm{2}{c}}>\mathrm{1}\wedge{c}>\mathrm{3} \\ $$ $$\Leftrightarrow \\ $$ $${c}<{b}<\frac{\mathrm{3}{c}}{{c}−\mathrm{3}}\wedge{c}>\mathrm{3}\:\Rightarrow\:{c}=\mathrm{4}\wedge\mathrm{4}<{b}<\mathrm{12}\:\vee\:{c}=\mathrm{5}\wedge\mathrm{5}<{b}<\mathrm{7}.\mathrm{5} \\ $$ $$\mathrm{not}\:\mathrm{many}\:\mathrm{possibilities}\:\mathrm{to}\:\mathrm{try}\:\mathrm{out} \\ $$ $$\Rightarrow \\ $$ $${a}=\mathrm{36}\wedge{b}=\mathrm{9}\wedge{c}=\mathrm{4} \\ $$ $${a}=\mathrm{20}\wedge{b}=\mathrm{10}\wedge{c}=\mathrm{4} \\ $$ $${a}=\mathrm{15}\wedge{b}=\mathrm{6}\wedge{c}=\mathrm{5} \\ $$ $$\mathrm{are}\:\mathrm{the}\:\mathrm{only}\:\mathrm{solutions} \\ $$

Commented bymr W last updated on 02/Dec/22

thanks alot sir!

$${thanks}\:{alot}\:{sir}! \\ $$

Answered by mr W last updated on 02/Dec/22

a>b>c>0 ⇒(1/a)<(1/b)<(1/c) 1=(1/a)+(2/b)+(3/c)<(1/c)+(2/c)+(3/c)=(6/c) ⇒c<6 (3/c)=1−(1/a)−(2/b)<1 ⇒c>3 ⇒c may only be 4 or 5. with c=4: (1/a)+(2/b)=1−(3/4)=(1/4) (1/4)=(1/a)+(2/b)<(1/b)+(2/b)=(3/b) ⇒b<12 (1/4)=(1/a)+(2/b)>(2/b) ⇒b>8 ⇒b may only be 9, 10 or 11. b=9: (1/a)=(1/4)−(2/9)=(1/(36)) ⇒a=36 ✓ b=10: (1/a)=(1/4)−(2/(10))=(1/(20)) ⇒a=20 ✓ b=11: (1/a)=(1/4)−(2/(11))=(7/(44)) ⇒a=((44)/7) × with c=5: (1/a)+(2/b)=1−(3/5)=(2/5) (2/5)=(1/a)+(2/b)<(1/b)+(2/b)=(3/b) ⇒b<((15)/2)=7.5 (2/5)=(1/a)+(2/b)>(2/b) ⇒b>5 ⇒b may only be 6 or 7. b=6: (1/a)=(2/5)−(2/6)=(1/(15)) ⇒a=15 ✓ b=7: (1/a)=(2/5)−(2/7)=(4/(35)) ⇒a=((35)/4) × summary: (a,b,c)=(36,9,4), (20,10,4), (15,6,5)

$${a}>{b}>{c}>\mathrm{0} \\ $$ $$\Rightarrow\frac{\mathrm{1}}{{a}}<\frac{\mathrm{1}}{{b}}<\frac{\mathrm{1}}{{c}} \\ $$ $$\mathrm{1}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{2}}{{b}}+\frac{\mathrm{3}}{{c}}<\frac{\mathrm{1}}{{c}}+\frac{\mathrm{2}}{{c}}+\frac{\mathrm{3}}{{c}}=\frac{\mathrm{6}}{{c}} \\ $$ $$\Rightarrow{c}<\mathrm{6} \\ $$ $$\frac{\mathrm{3}}{{c}}=\mathrm{1}−\frac{\mathrm{1}}{{a}}−\frac{\mathrm{2}}{{b}}<\mathrm{1} \\ $$ $$\Rightarrow{c}>\mathrm{3} \\ $$ $$\Rightarrow{c}\:{may}\:{only}\:{be}\:\mathrm{4}\:{or}\:\mathrm{5}. \\ $$ $$\underline{\boldsymbol{{with}}\:\boldsymbol{{c}}=\mathrm{4}:} \\ $$ $$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{2}}{{b}}=\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$ $$\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{2}}{{b}}<\frac{\mathrm{1}}{{b}}+\frac{\mathrm{2}}{{b}}=\frac{\mathrm{3}}{{b}} \\ $$ $$\Rightarrow{b}<\mathrm{12} \\ $$ $$\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{2}}{{b}}>\frac{\mathrm{2}}{{b}} \\ $$ $$\Rightarrow{b}>\mathrm{8} \\ $$ $$\Rightarrow{b}\:{may}\:{only}\:{be}\:\mathrm{9},\:\mathrm{10}\:{or}\:\mathrm{11}. \\ $$ $${b}=\mathrm{9}:\:\frac{\mathrm{1}}{{a}}=\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{2}}{\mathrm{9}}=\frac{\mathrm{1}}{\mathrm{36}}\:\Rightarrow{a}=\mathrm{36}\:\checkmark \\ $$ $${b}=\mathrm{10}:\:\frac{\mathrm{1}}{{a}}=\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{2}}{\mathrm{10}}=\frac{\mathrm{1}}{\mathrm{20}}\:\Rightarrow{a}=\mathrm{20}\:\checkmark \\ $$ $${b}=\mathrm{11}:\:\frac{\mathrm{1}}{{a}}=\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{2}}{\mathrm{11}}=\frac{\mathrm{7}}{\mathrm{44}}\:\Rightarrow{a}=\frac{\mathrm{44}}{\mathrm{7}}\:× \\ $$ $$\underline{\boldsymbol{{with}}\:\boldsymbol{{c}}=\mathrm{5}:} \\ $$ $$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{2}}{{b}}=\mathrm{1}−\frac{\mathrm{3}}{\mathrm{5}}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$ $$\frac{\mathrm{2}}{\mathrm{5}}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{2}}{{b}}<\frac{\mathrm{1}}{{b}}+\frac{\mathrm{2}}{{b}}=\frac{\mathrm{3}}{{b}} \\ $$ $$\Rightarrow{b}<\frac{\mathrm{15}}{\mathrm{2}}=\mathrm{7}.\mathrm{5} \\ $$ $$\frac{\mathrm{2}}{\mathrm{5}}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{2}}{{b}}>\frac{\mathrm{2}}{{b}} \\ $$ $$\Rightarrow{b}>\mathrm{5} \\ $$ $$\Rightarrow{b}\:{may}\:{only}\:{be}\:\mathrm{6}\:{or}\:\mathrm{7}. \\ $$ $${b}=\mathrm{6}:\:\frac{\mathrm{1}}{{a}}=\frac{\mathrm{2}}{\mathrm{5}}−\frac{\mathrm{2}}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{15}}\:\Rightarrow{a}=\mathrm{15}\:\checkmark \\ $$ $${b}=\mathrm{7}:\:\frac{\mathrm{1}}{{a}}=\frac{\mathrm{2}}{\mathrm{5}}−\frac{\mathrm{2}}{\mathrm{7}}=\frac{\mathrm{4}}{\mathrm{35}}\:\Rightarrow{a}=\frac{\mathrm{35}}{\mathrm{4}}\:× \\ $$ $${summary}: \\ $$ $$\left({a},{b},{c}\right)=\left(\mathrm{36},\mathrm{9},\mathrm{4}\right),\:\left(\mathrm{20},\mathrm{10},\mathrm{4}\right),\:\left(\mathrm{15},\mathrm{6},\mathrm{5}\right) \\ $$

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