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if-x-and-y-are-positive-integers-with-2010-2011-lt-x-y-lt-2011-2012-then-compute-the-minimum-value-for-x-y-and-the-values-of-x-and-y-which-achieves-this-minimum-




Question Number 158444 by HongKing last updated on 04/Nov/21
if  x  and  y  are positive integers with  ((2010)/(2011)) < (x/y) < ((2011)/(2012))  then compute the  minimum value for  x+y  and the  values of  x  and  y  which achieves  this minimum
$$\mathrm{if}\:\:\boldsymbol{\mathrm{x}}\:\:\mathrm{and}\:\:\boldsymbol{\mathrm{y}}\:\:\mathrm{are}\:\mathrm{positive}\:\mathrm{integers}\:\mathrm{with} \\ $$$$\frac{\mathrm{2010}}{\mathrm{2011}}\:<\:\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{y}}}\:<\:\frac{\mathrm{2011}}{\mathrm{2012}}\:\:\mathrm{then}\:\mathrm{compute}\:\mathrm{the} \\ $$$$\mathrm{minimum}\:\mathrm{value}\:\mathrm{for}\:\:\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}\:\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{values}\:\mathrm{of}\:\:\boldsymbol{\mathrm{x}}\:\:\mathrm{and}\:\:\boldsymbol{\mathrm{y}}\:\:\mathrm{which}\:\mathrm{achieves} \\ $$$$\mathrm{this}\:\mathrm{minimum} \\ $$
Commented by mr W last updated on 04/Nov/21
(x/y)=((4021)/(4023))
$$\frac{{x}}{{y}}=\frac{\mathrm{4021}}{\mathrm{4023}} \\ $$
Commented by HongKing last updated on 04/Nov/21
how my dear Ser solution please
$$\mathrm{how}\:\mathrm{my}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er}\:\mathrm{solution}\:\mathrm{please} \\ $$
Commented by Rasheed.Sindhi last updated on 04/Nov/21
Sir mr W , when you have some time  pl see my answer to Q#158124.  Actually my answer doesn′t match  the answer of the questioner...  Thanks in advance sir!
$${Sir}\:{mr}\:{W}\:,\:{when}\:{you}\:{have}\:{some}\:{time} \\ $$$${pl}\:{see}\:{my}\:{answer}\:{to}\:{Q}#\mathrm{158124}. \\ $$$${Actually}\:{my}\:{answer}\:{doesn}'{t}\:{match} \\ $$$${the}\:{answer}\:{of}\:{the}\:{questioner}… \\ $$$$\mathcal{T}{hanks}\:{in}\:{advance}\:{sir}! \\ $$
Commented by mr W last updated on 04/Nov/21
i have checked. your answer is correct.
$${i}\:{have}\:{checked}.\:{your}\:{answer}\:{is}\:{correct}. \\ $$
Commented by Rasheed.Sindhi last updated on 05/Nov/21
THANKS A LOT   SIR!
$$\mathcal{THANKS}\:\mathcal{A}\:\mathcal{LOT}\:\:\:\mathcal{SIR}! \\ $$
Answered by mr W last updated on 04/Nov/21
x must fulfill  ⌈((2011x)/(2010))⌉−⌊((2012x)/(2011))⌋≥2  we get x_(min) =4021  y_(min) =⌊((2012×4021)/(2011))⌋+1=4023  (x+y)_(min) =4021+4023=8044    (x/y)=((4021)/(4023)), ((6031)/(6034)), ((6032)/(6035)), ((8041)/(8045)), ...
$${x}\:{must}\:{fulfill} \\ $$$$\lceil\frac{\mathrm{2011}{x}}{\mathrm{2010}}\rceil−\lfloor\frac{\mathrm{2012}{x}}{\mathrm{2011}}\rfloor\geqslant\mathrm{2} \\ $$$${we}\:{get}\:{x}_{{min}} =\mathrm{4021} \\ $$$${y}_{{min}} =\lfloor\frac{\mathrm{2012}×\mathrm{4021}}{\mathrm{2011}}\rfloor+\mathrm{1}=\mathrm{4023} \\ $$$$\left({x}+{y}\right)_{{min}} =\mathrm{4021}+\mathrm{4023}=\mathrm{8044} \\ $$$$ \\ $$$$\frac{{x}}{{y}}=\frac{\mathrm{4021}}{\mathrm{4023}},\:\frac{\mathrm{6031}}{\mathrm{6034}},\:\frac{\mathrm{6032}}{\mathrm{6035}},\:\frac{\mathrm{8041}}{\mathrm{8045}},\:… \\ $$
Commented by Rasheed.Sindhi last updated on 06/Nov/21
((2010)/(2011)) < (x/y) < ((2011)/(2012))  Sir in this case :  x_(min) =2010+2011=4021  y_(min) =2011+2012=4023  Is it a coincidence   or generally if (a/b)<(x/y)<(c/d)  then     x_(min) =a+b & y_(min) =c+d  and (x+y)_(min) =a+b+c+d  ?
$$\frac{\mathrm{2010}}{\mathrm{2011}}\:<\:\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{y}}}\:<\:\frac{\mathrm{2011}}{\mathrm{2012}} \\ $$$${Sir}\:{in}\:{this}\:{case}\:: \\ $$$${x}_{{min}} =\mathrm{2010}+\mathrm{2011}=\mathrm{4021} \\ $$$${y}_{{min}} =\mathrm{2011}+\mathrm{2012}=\mathrm{4023} \\ $$$${Is}\:{it}\:{a}\:{coincidence}\: \\ $$$${or}\:{generally}\:{if}\:\frac{{a}}{{b}}<\frac{{x}}{{y}}<\frac{{c}}{{d}} \\ $$$${then}\:\:\:\:\:{x}_{{min}} ={a}+{b}\:\&\:{y}_{{min}} ={c}+{d} \\ $$$${and}\:\left({x}+{y}\right)_{{min}} ={a}+{b}+{c}+{d}\:\:? \\ $$
Commented by mr W last updated on 06/Nov/21
a very nice thought sir!  but i think here it′s just a coincidence.    with x=a+c and y=b+d  it fulfills indeed  (a/b)<(x/y)<(c/d)  but x=a+c mustn′t be x_(min)  and  y=b+d mustn′t be y_(min) . for example  when gcd(a+c,b+d)≠1.
$${a}\:{very}\:{nice}\:{thought}\:{sir}! \\ $$$${but}\:{i}\:{think}\:{here}\:{it}'{s}\:{just}\:{a}\:{coincidence}. \\ $$$$ \\ $$$${with}\:{x}={a}+{c}\:{and}\:{y}={b}+{d} \\ $$$${it}\:{fulfills}\:{indeed} \\ $$$$\frac{{a}}{{b}}<\frac{{x}}{{y}}<\frac{{c}}{{d}} \\ $$$${but}\:{x}={a}+{c}\:{mustn}'{t}\:{be}\:{x}_{{min}} \:{and} \\ $$$${y}={b}+{d}\:{mustn}'{t}\:{be}\:{y}_{{min}} .\:{for}\:{example} \\ $$$${when}\:{gcd}\left({a}+{c},{b}+{d}\right)\neq\mathrm{1}. \\ $$
Commented by mr W last updated on 06/Nov/21
an example  ((13)/(75))<(x/y)<(9/(50))  x_(min) =3, y_(min) =17  ((13)/(75))<(3/(17))<(9/(50))
$${an}\:{example} \\ $$$$\frac{\mathrm{13}}{\mathrm{75}}<\frac{{x}}{{y}}<\frac{\mathrm{9}}{\mathrm{50}} \\ $$$${x}_{{min}} =\mathrm{3},\:{y}_{{min}} =\mathrm{17} \\ $$$$\frac{\mathrm{13}}{\mathrm{75}}<\frac{\mathrm{3}}{\mathrm{17}}<\frac{\mathrm{9}}{\mathrm{50}} \\ $$
Commented by Rasheed.Sindhi last updated on 06/Nov/21
ㄒ卄卂几Ҝ丂 爪尺 山 丂丨尺!

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