Question Number 56301 by naka3546 last updated on 13/Mar/19 | ||
$${Minimum}\:\:{value}\:\:{of}\:\:{b}\:\:{that}\:\:{satisfy}\:\:{the} \\ $$ $${following}\:\:{inequality}\:\: \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{53}}{\mathrm{201}}\:\:<\:\:\frac{{a}}{{b}}\:\:<\:\:\frac{\mathrm{4}}{\mathrm{15}}\:\:\:\:\:{is}\:\:\:... \\ $$ | ||
Commented bynaka3546 last updated on 13/Mar/19 | ||
$${a},\:{b}\:\:{are}\:\:{positive}\:{integers}\:. \\ $$ | ||
Answered by tanmay.chaudhury50@gmail.com last updated on 13/Mar/19 | ||
$$\mathrm{201}=\mathrm{3}×\mathrm{67}\:\:\:\:\mathrm{15}=\mathrm{3}×\mathrm{5} \\ $$ $$\frac{\mathrm{53}}{\mathrm{201}}=\frac{\mathrm{265}}{\mathrm{1005}}\:\:{and}\:\:\frac{\mathrm{4}}{\mathrm{15}}=\frac{\mathrm{4}×\mathrm{67}}{\mathrm{15}×\mathrm{67}}=\frac{\mathrm{268}}{\mathrm{1005}} \\ $$ $$\frac{{a}}{{b}}=\frac{\mathrm{266}}{\mathrm{1005}}\:\leftarrow{first}\:{answer} \\ $$ $$\frac{{a}}{{b}}=\frac{\mathrm{267}}{\mathrm{1005}}\leftarrow{second}\:{answer} \\ $$ $$\boldsymbol{{so}}\:\boldsymbol{{b}}=\mathrm{1005} \\ $$ | ||
Commented bymr W last updated on 14/Mar/19 | ||
$${with}\:{mininmum}\:{b}:\:\frac{\mathrm{9}}{\mathrm{34}} \\ $$ $${the}\:{next}\:{are}\:\frac{\mathrm{13}}{\mathrm{49}},\:\frac{\mathrm{14}}{\mathrm{53}},\:\frac{\mathrm{17}}{\mathrm{64}},\frac{\mathrm{19}}{\mathrm{72}}.... \\ $$ | ||
Commented bynaka3546 last updated on 14/Mar/19 | ||
$${how}\:\:{to}\:\:{get}\:\:{it},\:{sir}. \\ $$ | ||
Commented byMJS last updated on 14/Mar/19 | ||
$$\frac{\mathrm{53}}{\mathrm{201}}<\frac{{a}}{{b}}<\frac{\mathrm{4}}{\mathrm{15}} \\ $$ $$\frac{\mathrm{53}}{\mathrm{201}}{b}<{a}<\frac{\mathrm{4}}{\mathrm{15}}{b} \\ $$ $${b}\neq\mathrm{15}{n}\:\wedge\:{b}\neq\mathrm{201}{n} \\ $$ $$\lceil\frac{\mathrm{53}}{\mathrm{201}}{b}\rceil>\lfloor\frac{\mathrm{4}}{\mathrm{15}}{b}\rfloor\:\mathrm{xor}\:\lceil\frac{\mathrm{53}}{\mathrm{201}}{b}\rceil=\lfloor\frac{\mathrm{4}}{\mathrm{15}}{b}\rfloor={a} \\ $$ $$\mathrm{now}\:\mathrm{we}\:\mathrm{must}\:\mathrm{try} \\ $$ | ||
Commented byKunal12588 last updated on 13/Mar/19 | ||
$${sir}\:\frac{\mathrm{53}}{\mathrm{201}}<\frac{\mathrm{33}}{\mathrm{125}}<\frac{\mathrm{4}}{\mathrm{15}} \\ $$ $${don}'{t}\:{know}\:{there}\:{can}\:{be}\:{smaller}\:{answers}. \\ $$ | ||
Commented bynaka3546 last updated on 14/Mar/19 | ||
$$\frac{\mathrm{9}}{\mathrm{34}}\:\:{may}\:\:{be},\:\:{I}\:\:{don}'{t}\:\:{know}\:{about}\:\:{it}. \\ $$ | ||
Answered by mr W last updated on 15/Mar/19 | ||
$${this}\:{is}\:{my}\:{method}: \\ $$ $$\frac{\mathrm{53}}{\mathrm{201}}\:\:<\:\:\frac{{a}}{{b}}\:\:<\:\:\frac{\mathrm{4}}{\mathrm{15}} \\ $$ $$\Rightarrow\:\frac{\mathrm{201}}{\mathrm{53}}>\frac{{b}}{{a}}>\frac{\mathrm{15}}{\mathrm{4}} \\ $$ $$\Rightarrow\:\frac{\mathrm{15}{a}}{\mathrm{4}}<{b}<\frac{\mathrm{201}{a}}{\mathrm{53}} \\ $$ $$\Rightarrow\lfloor\frac{\mathrm{15}{a}}{\mathrm{4}}\rfloor+\mathrm{1}\leqslant{b}\leqslant\lceil\frac{\mathrm{201}{a}}{\mathrm{53}}\rceil−\mathrm{1} \\ $$ $$ \\ $$ $${for}\:{a}\:{given}\:{value}\:{of}\:{a}\:{we}\:{calculate} \\ $$ $$\lfloor\frac{\mathrm{15}{a}}{\mathrm{4}}\rfloor+\mathrm{1}\:{and}\:\lceil\frac{\mathrm{201}{a}}{\mathrm{53}}\rceil−\mathrm{1}. \\ $$ $${if}\:\lfloor\frac{\mathrm{15}{a}}{\mathrm{4}}\rfloor+\mathrm{1}\leqslant\lceil\frac{\mathrm{201}{a}}{\mathrm{53}}\rceil−\mathrm{1},\:{then}\:{b}\:{exists}, \\ $$ $${and}\:{b}=\lfloor\frac{\mathrm{15}{a}}{\mathrm{4}}\rfloor+\mathrm{1}...\lceil\frac{\mathrm{201}{a}}{\mathrm{53}}\rceil−\mathrm{1}. \\ $$ $$ \\ $$ $${if}\:{we}\:{start}\:{with}\:{a}=\mathrm{1},\mathrm{2},\mathrm{3},...\:{we}'{ll}\:{get}\:{all} \\ $$ $${possible}\:\left({infinite}\right)\:{solutions}: \\ $$ $${a}=\mathrm{9}\Rightarrow{b}=\mathrm{34} \\ $$ $${a}=\mathrm{13}\Rightarrow{b}=\mathrm{49} \\ $$ $${a}=\mathrm{14}\Rightarrow{b}=\mathrm{53} \\ $$ $${a}=\mathrm{17}\Rightarrow{b}=\mathrm{64} \\ $$ $${a}=\mathrm{19}\Rightarrow{b}=\mathrm{72} \\ $$ $${a}=\mathrm{21}\Rightarrow{b}=\mathrm{79} \\ $$ $${a}=\mathrm{22}\Rightarrow{b}=\mathrm{83} \\ $$ $${a}=\mathrm{23}\Rightarrow{b}=\mathrm{87} \\ $$ $${a}=\mathrm{24}\Rightarrow{b}=\mathrm{91} \\ $$ $${a}=\mathrm{25}\Rightarrow{b}=\mathrm{94} \\ $$ $${a}=\mathrm{29}\Rightarrow{b}=\mathrm{109} \\ $$ $${a}=\mathrm{30}\Rightarrow{b}=\mathrm{113} \\ $$ $${a}=\mathrm{31}\Rightarrow{b}=\mathrm{117} \\ $$ $${a}=\mathrm{32}\Rightarrow{b}=\mathrm{121} \\ $$ $${a}=\mathrm{33}\Rightarrow{b}=\mathrm{124},\:\mathrm{125} \\ $$ $${a}=\mathrm{35}\Rightarrow{b}=\mathrm{132} \\ $$ $$...... \\ $$ $${a}=\mathrm{117}\Rightarrow{b}=\mathrm{439},\:\mathrm{440},\:\mathrm{442},\:\mathrm{443} \\ $$ $$...... \\ $$ | ||
Commented bymr W last updated on 15/Mar/19 | ||
$${is}\:{there}\:{an}\:{easier}\:{way}\:{than}\:{this}? \\ $$ | ||
Commented byMJS last updated on 16/Mar/19 | ||
$$\mathrm{no}. \\ $$ | ||