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Question Number 184431 by SulaymonNorboyev last updated on 06/Jan/23

If { ((a+(1/b)=(7/3))),((b+(1/c)=4)),((c+(1/a)=1)) :} find 2022∙abc

$${If}\:\:\:\begin{cases}{\mathrm{a}+\frac{\mathrm{1}}{\mathrm{b}}=\frac{\mathrm{7}}{\mathrm{3}}}\\{\mathrm{b}+\frac{\mathrm{1}}{\mathrm{c}}=\mathrm{4}}\\{\mathrm{c}+\frac{\mathrm{1}}{\mathrm{a}}=\mathrm{1}}\end{cases}\:\:\:\mathrm{find}\:\:\mathrm{2022}\centerdot\mathrm{abc} \\ $$

Answered by mr W last updated on 06/Jan/23

c=1−(1/a)=((a−1)/a) b=4−(a/(a−1))=((3a−4)/(a−1)) ⇒abc=a×((3a−4)/(a−1))×((a−1)/a)=3a−4 a+((a−1)/(3a−4))=(7/3) 9a^2 −30a+25=0 (3a−5)^2 =0 ⇒a=(5/3) abc=3a−4=5−4=1 2023abc=2023 ✓

$${c}=\mathrm{1}−\frac{\mathrm{1}}{{a}}=\frac{{a}−\mathrm{1}}{{a}} \\ $$$${b}=\mathrm{4}−\frac{{a}}{{a}−\mathrm{1}}=\frac{\mathrm{3}{a}−\mathrm{4}}{{a}−\mathrm{1}} \\ $$$$\Rightarrow{abc}={a}×\frac{\mathrm{3}{a}−\mathrm{4}}{{a}−\mathrm{1}}×\frac{{a}−\mathrm{1}}{{a}}=\mathrm{3}{a}−\mathrm{4} \\ $$$$ \\ $$$${a}+\frac{{a}−\mathrm{1}}{\mathrm{3}{a}−\mathrm{4}}=\frac{\mathrm{7}}{\mathrm{3}} \\ $$$$\mathrm{9}{a}^{\mathrm{2}} −\mathrm{30}{a}+\mathrm{25}=\mathrm{0} \\ $$$$\left(\mathrm{3}{a}−\mathrm{5}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{a}=\frac{\mathrm{5}}{\mathrm{3}} \\ $$$${abc}=\mathrm{3}{a}−\mathrm{4}=\mathrm{5}−\mathrm{4}=\mathrm{1} \\ $$$$\mathrm{2023}{abc}=\mathrm{2023}\:\checkmark \\ $$

Answered by ajfour last updated on 06/Jan/23

(a+(1/b))(b+(1/c))(c+(1/a))=pqr abc+b+a+(1/c)+c+(1/a)+(1/b)+(1/(abc)) =pqr abc+(1/(abc))+(p+q+r)=pqr (abc)((1/(abc)))=1 ⇒ abc, (1/(abc))=((pqr−(p+q+r))/2) ±(√({((pqr−(p+q+r))/2)}^2 −1)) =((14)/3)−((11)/3)±(√(1−1)) =1

$$\left({a}+\frac{\mathrm{1}}{{b}}\right)\left({b}+\frac{\mathrm{1}}{{c}}\right)\left({c}+\frac{\mathrm{1}}{{a}}\right)={pqr} \\ $$$${abc}+{b}+{a}+\frac{\mathrm{1}}{{c}}+{c}+\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{abc}} \\ $$$$\:\:\:={pqr} \\ $$$${abc}+\frac{\mathrm{1}}{{abc}}+\left({p}+{q}+{r}\right)={pqr} \\ $$$$\left({abc}\right)\left(\frac{\mathrm{1}}{{abc}}\right)=\mathrm{1} \\ $$$$\Rightarrow\:{abc},\:\frac{\mathrm{1}}{{abc}}=\frac{{pqr}−\left({p}+{q}+{r}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\pm\sqrt{\left\{\frac{{pqr}−\left({p}+{q}+{r}\right)}{\mathrm{2}}\right\}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\:\:\:\:=\frac{\mathrm{14}}{\mathrm{3}}−\frac{\mathrm{11}}{\mathrm{3}}\pm\sqrt{\mathrm{1}−\mathrm{1}}\:\:=\mathrm{1} \\ $$

Commented by mr W last updated on 06/Jan/23

very smart approach!

$${very}\:{smart}\:{approach}! \\ $$

Commented by manxsol last updated on 07/Jan/23

(abc)((1/(abc)))=1 to the toolbox

$$ \\ $$$$\left({abc}\right)\left(\frac{\mathrm{1}}{{abc}}\right)=\mathrm{1}\:\:\:{to}\:{the}\:{toolbox} \\ $$$$ \\ $$

Commented by manxsol last updated on 07/Jan/23

a surreal answer

$${a}\:{surreal}\:{answer} \\ $$

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