Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 114290 by mathdave last updated on 18/Sep/20

Answered by 1549442205PVT last updated on 18/Sep/20

From the hypothesis abc^(−) ,bca^(−) ,cab^(−)   form a G.P we infer abc^(−) ×cab^(−) =(bca^(−) )^2   ⇔((100a+10b+c)(100c+10a+b)=  (100b+10c+a)^2   ⇔10000ac+1000bc+100c^2 +1000a^2   +100ab+10ac+100ab+10b^2 +bc  =10000b^2 +100c^2 +a^2 +2000bc+200ab+20ac  ⇔9990ac−999bc−9990b^2 +999a^2 =0  ⇔a^2 +10ac=10b^2 +bc  ⇔(a+5c)^2 =25c^2 +10b^2 +bc  ⇔100(a+5c)^2 =2500c^2 +100bc+1000b^2   ⇔(10a+50c)^2 =(50c+b)^2 +999b^2   ⇔(10a+100c+b)(10a−b)=999b^2 (2)  =37.27.b^2 ⇒LHS⋮37  i)If b=9 then it is easy to see that   10a−b isn′t divisible 37.Hence  10a+100c+9⋮37.Since 10a+100c+9>111  and(10a+100c+9)=37k⇒37k−9⋮10  ⇒37k=999⇒10a−9=81⇒a=9  +)If 10.9+100c+9=999⇒c=9  We get abc^(−) =999  ii)If b=8 then 10a−8 isn′t divisible 37  10a+100c+8=37k⇒37k−8⋮10  this is impossible since 37k∈{259,629,999}  By similar argument for the cases  b=1,2,3,4,5,6,7 we see that ∄a,c  satisfying(2)  Thus,there is only one three−digits    number abc^(−) =999 satisfying the  condition of our problem

$$\mathrm{From}\:\mathrm{the}\:\mathrm{hypothesis}\:\overline {\mathrm{abc}},\overline {\mathrm{bca}},\overline {\mathrm{cab}} \\ $$$$\mathrm{form}\:\mathrm{a}\:\mathrm{G}.\mathrm{P}\:\mathrm{we}\:\mathrm{infer}\:\overline {\mathrm{abc}}×\overline {\mathrm{cab}}=\left(\overline {\mathrm{bca}}\right)^{\mathrm{2}} \\ $$$$\Leftrightarrow\left(\left(\mathrm{100a}+\mathrm{10b}+\mathrm{c}\right)\left(\mathrm{100c}+\mathrm{10a}+\mathrm{b}\right)=\right. \\ $$$$\left(\mathrm{100b}+\mathrm{10c}+\mathrm{a}\right)^{\mathrm{2}} \\ $$$$\Leftrightarrow\mathrm{10000ac}+\mathrm{1000bc}+\mathrm{100c}^{\mathrm{2}} +\mathrm{1000a}^{\mathrm{2}} \\ $$$$+\mathrm{100ab}+\mathrm{10ac}+\mathrm{100ab}+\mathrm{10b}^{\mathrm{2}} +\mathrm{bc} \\ $$$$=\mathrm{10000b}^{\mathrm{2}} +\mathrm{100c}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} +\mathrm{2000bc}+\mathrm{200ab}+\mathrm{20ac} \\ $$$$\Leftrightarrow\mathrm{9990ac}−\mathrm{999bc}−\mathrm{9990b}^{\mathrm{2}} +\mathrm{999a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{a}^{\mathrm{2}} +\mathrm{10ac}=\mathrm{10b}^{\mathrm{2}} +\mathrm{bc} \\ $$$$\Leftrightarrow\left(\mathrm{a}+\mathrm{5c}\right)^{\mathrm{2}} =\mathrm{25c}^{\mathrm{2}} +\mathrm{10b}^{\mathrm{2}} +\mathrm{bc} \\ $$$$\Leftrightarrow\mathrm{100}\left(\mathrm{a}+\mathrm{5c}\right)^{\mathrm{2}} =\mathrm{2500c}^{\mathrm{2}} +\mathrm{100bc}+\mathrm{1000b}^{\mathrm{2}} \\ $$$$\Leftrightarrow\left(\mathrm{10a}+\mathrm{50c}\right)^{\mathrm{2}} =\left(\mathrm{50c}+\mathrm{b}\right)^{\mathrm{2}} +\mathrm{999b}^{\mathrm{2}} \\ $$$$\Leftrightarrow\left(\mathrm{10a}+\mathrm{100c}+\mathrm{b}\right)\left(\mathrm{10a}−\mathrm{b}\right)=\mathrm{999b}^{\mathrm{2}} \left(\mathrm{2}\right) \\ $$$$=\mathrm{37}.\mathrm{27}.\mathrm{b}^{\mathrm{2}} \Rightarrow\mathrm{LHS}\vdots\mathrm{37} \\ $$$$\left.\mathrm{i}\right)\mathrm{If}\:\mathrm{b}=\mathrm{9}\:\mathrm{then}\:\mathrm{it}\:\mathrm{is}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{see}\:\mathrm{that}\: \\ $$$$\mathrm{10a}−\mathrm{b}\:\mathrm{isn}'\mathrm{t}\:\mathrm{divisible}\:\mathrm{37}.\mathrm{Hence} \\ $$$$\mathrm{10a}+\mathrm{100c}+\mathrm{9}\vdots\mathrm{37}.\mathrm{Since}\:\mathrm{10a}+\mathrm{100c}+\mathrm{9}>\mathrm{111} \\ $$$$\mathrm{and}\left(\mathrm{10a}+\mathrm{100c}+\mathrm{9}\right)=\mathrm{37k}\Rightarrow\mathrm{37k}−\mathrm{9}\vdots\mathrm{10} \\ $$$$\Rightarrow\mathrm{37k}=\mathrm{999}\Rightarrow\mathrm{10a}−\mathrm{9}=\mathrm{81}\Rightarrow\mathrm{a}=\mathrm{9} \\ $$$$\left.+\right)\mathrm{If}\:\mathrm{10}.\mathrm{9}+\mathrm{100c}+\mathrm{9}=\mathrm{999}\Rightarrow\mathrm{c}=\mathrm{9} \\ $$$$\mathrm{We}\:\mathrm{get}\:\overline {\mathrm{abc}}=\mathrm{999} \\ $$$$\left.\mathrm{ii}\right)\mathrm{If}\:\mathrm{b}=\mathrm{8}\:\mathrm{then}\:\mathrm{10a}−\mathrm{8}\:\mathrm{isn}'\mathrm{t}\:\mathrm{divisible}\:\mathrm{37} \\ $$$$\mathrm{10a}+\mathrm{100c}+\mathrm{8}=\mathrm{37k}\Rightarrow\mathrm{37k}−\mathrm{8}\vdots\mathrm{10} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{impossible}\:\mathrm{since}\:\mathrm{37k}\in\left\{\mathrm{259},\mathrm{629},\mathrm{999}\right\} \\ $$$$\mathrm{By}\:\mathrm{similar}\:\mathrm{argument}\:\mathrm{for}\:\mathrm{the}\:\mathrm{cases} \\ $$$$\mathrm{b}=\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7}\:\mathrm{we}\:\mathrm{see}\:\mathrm{that}\:\nexists\mathrm{a},\mathrm{c} \\ $$$$\mathrm{satisfying}\left(\mathrm{2}\right) \\ $$$$\mathrm{Thus},\mathrm{there}\:\mathrm{is}\:\mathrm{only}\:\mathrm{one}\:\mathrm{three}−\mathrm{digits}\:\: \\ $$$$\mathrm{number}\:\overline {\mathrm{abc}}=\mathrm{999}\:\mathrm{satisfying}\:\mathrm{the} \\ $$$$\mathrm{condition}\:\mathrm{of}\:\mathrm{our}\:\mathrm{problem} \\ $$

Commented by Rasheed.Sindhi last updated on 18/Sep/20

a,ar,ar^2 ...is GP  only when r≠1  ...

$${a},{ar},{ar}^{\mathrm{2}} ...{is}\:{GP}\:\:{only}\:{when}\:{r}\neq\mathrm{1} \\ $$$$... \\ $$

Commented by mr W last updated on 18/Sep/20

111  222  333  ....  999  are obvious solutions.

$$\mathrm{111} \\ $$$$\mathrm{222} \\ $$$$\mathrm{333} \\ $$$$.... \\ $$$$\mathrm{999} \\ $$$${are}\:{obvious}\:{solutions}. \\ $$

Commented by 1549442205PVT last updated on 18/Sep/20

Thank Sir.I mistaked

$$\mathrm{Thank}\:\mathrm{Sir}.\mathrm{I}\:\mathrm{mistaked} \\ $$

Commented by 1549442205PVT last updated on 20/Sep/20

We assume that this is a special  geometry progression with q=1

$$\mathrm{We}\:\mathrm{assume}\:\mathrm{that}\:\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{special} \\ $$$$\mathrm{geometry}\:\mathrm{progression}\:\mathrm{with}\:\mathrm{q}=\mathrm{1} \\ $$

Commented by Rasheed.Sindhi last updated on 20/Sep/20

Ok sir!

$${Ok}\:{sir}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com