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Question Number 21236 by Joel577 last updated on 17/Sep/17

If (1/a) + (1/(2a)) + (1/(3a)) = (1/(b^2 − 2b)) a and b are positive integers Find minimum value of a + b

$$\mathrm{If}\:\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{\mathrm{2}{a}}\:+\:\frac{\mathrm{1}}{\mathrm{3}{a}}\:=\:\frac{\mathrm{1}}{{b}^{\mathrm{2}} \:−\:\mathrm{2}{b}} \\ $$$${a}\:\mathrm{and}\:{b}\:\mathrm{are}\:\mathrm{positive}\:\mathrm{integers} \\ $$$$\mathrm{Find}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:{a}\:+\:{b} \\ $$

Answered by mrW1 last updated on 17/Sep/17

(1/a)(1+(1/2)+(1/3))=(1/(b(b−2))) ((11)/(6a))=(1/(b(b−2))) a=((11b(b−2))/6)=11k b(b−2)=6k b^2 −2b−6k=0 b=((2+(√(4+4×6k)))/2)=1+(√(1+6k)) ⇒ k=2i(3i±1) with i=1,2,3...... ⇒b=6i+1±1 ⇒a=22i(3i±1) k=4, 8, 20, 28, 48,60... b=6, 8, 12, 14, 18, 20... a=44, 88, 132, 154, 198, 220... k_(min) =4 b_(min) =1+5=6 a_(min) =11×4=44 ⇒(a+b)_(min) =50

$$\frac{\mathrm{1}}{\mathrm{a}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{1}}{\mathrm{b}\left(\mathrm{b}−\mathrm{2}\right)} \\ $$$$\frac{\mathrm{11}}{\mathrm{6a}}=\frac{\mathrm{1}}{\mathrm{b}\left(\mathrm{b}−\mathrm{2}\right)} \\ $$$$\mathrm{a}=\frac{\mathrm{11b}\left(\mathrm{b}−\mathrm{2}\right)}{\mathrm{6}}=\mathrm{11k} \\ $$$$\mathrm{b}\left(\mathrm{b}−\mathrm{2}\right)=\mathrm{6k} \\ $$$$\mathrm{b}^{\mathrm{2}} −\mathrm{2b}−\mathrm{6k}=\mathrm{0} \\ $$$$\mathrm{b}=\frac{\mathrm{2}+\sqrt{\mathrm{4}+\mathrm{4}×\mathrm{6k}}}{\mathrm{2}}=\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{6k}} \\ $$$$\Rightarrow\:\mathrm{k}=\mathrm{2i}\left(\mathrm{3i}\pm\mathrm{1}\right)\:\mathrm{with}\:\mathrm{i}=\mathrm{1},\mathrm{2},\mathrm{3}...... \\ $$$$\Rightarrow\mathrm{b}=\mathrm{6i}+\mathrm{1}\pm\mathrm{1} \\ $$$$\Rightarrow\mathrm{a}=\mathrm{22i}\left(\mathrm{3i}\pm\mathrm{1}\right) \\ $$$$ \\ $$$$\mathrm{k}=\mathrm{4},\:\mathrm{8},\:\mathrm{20},\:\mathrm{28},\:\mathrm{48},\mathrm{60}... \\ $$$$\mathrm{b}=\mathrm{6},\:\mathrm{8},\:\mathrm{12},\:\mathrm{14},\:\mathrm{18},\:\mathrm{20}... \\ $$$$\mathrm{a}=\mathrm{44},\:\mathrm{88},\:\mathrm{132},\:\mathrm{154},\:\mathrm{198},\:\mathrm{220}... \\ $$$$ \\ $$$$\mathrm{k}_{\mathrm{min}} =\mathrm{4} \\ $$$$\mathrm{b}_{\mathrm{min}} =\mathrm{1}+\mathrm{5}=\mathrm{6} \\ $$$$\mathrm{a}_{\mathrm{min}} =\mathrm{11}×\mathrm{4}=\mathrm{44} \\ $$$$\Rightarrow\left(\mathrm{a}+\mathrm{b}\right)_{\mathrm{min}} =\mathrm{50} \\ $$

Commented by $@ty@m last updated on 17/Sep/17

Will you pl. explain blue part?

$${Will}\:{you}\:{pl}.\:{explain}\:{blue}\:{part}? \\ $$

Commented by mrW1 last updated on 17/Sep/17

b=1+(√(1+6k)) 1+6k=j^2 k=((j^2 −1)/6)=(((j−1)(j+1))/(2×3)) j must be odd, i.e. j=2m+1 ⇒k=((2m(2m+2))/(2×3))=((2m(m+1))/3) m must be 3i or 3i−1 ⇒k=2i(3i+1) or 2i(3i−1) ⇒k=2i(3i±1) b=1+(√(1+6k))=1+(√(1+6×2i(3i±1))) =1+(√((6i±1)^2 )) =1+6i±1 a=11k=22i(3i±1)

$$\mathrm{b}=\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{6k}} \\ $$$$\mathrm{1}+\mathrm{6k}=\mathrm{j}^{\mathrm{2}} \\ $$$$\mathrm{k}=\frac{\mathrm{j}^{\mathrm{2}} −\mathrm{1}}{\mathrm{6}}=\frac{\left(\mathrm{j}−\mathrm{1}\right)\left(\mathrm{j}+\mathrm{1}\right)}{\mathrm{2}×\mathrm{3}} \\ $$$$\mathrm{j}\:\mathrm{must}\:\mathrm{be}\:\mathrm{odd},\:\mathrm{i}.\mathrm{e}.\:\mathrm{j}=\mathrm{2m}+\mathrm{1} \\ $$$$\Rightarrow\mathrm{k}=\frac{\mathrm{2m}\left(\mathrm{2m}+\mathrm{2}\right)}{\mathrm{2}×\mathrm{3}}=\frac{\mathrm{2m}\left(\mathrm{m}+\mathrm{1}\right)}{\mathrm{3}} \\ $$$$\mathrm{m}\:\mathrm{must}\:\mathrm{be}\:\mathrm{3i}\:\mathrm{or}\:\mathrm{3i}−\mathrm{1} \\ $$$$\Rightarrow\mathrm{k}=\mathrm{2i}\left(\mathrm{3i}+\mathrm{1}\right)\:\mathrm{or}\:\mathrm{2i}\left(\mathrm{3i}−\mathrm{1}\right) \\ $$$$\Rightarrow\mathrm{k}=\mathrm{2i}\left(\mathrm{3i}\pm\mathrm{1}\right) \\ $$$$\mathrm{b}=\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{6k}}=\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{6}×\mathrm{2i}\left(\mathrm{3i}\pm\mathrm{1}\right)} \\ $$$$=\mathrm{1}+\sqrt{\left(\mathrm{6i}\pm\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\mathrm{1}+\mathrm{6i}\pm\mathrm{1} \\ $$$$\mathrm{a}=\mathrm{11k}=\mathrm{22i}\left(\mathrm{3i}\pm\mathrm{1}\right) \\ $$

Commented by Joel577 last updated on 18/Sep/17

understood. thank you very much

$${understood}.\:{thank}\:{you}\:{very}\:{much} \\ $$

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