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Minimum-value-of-b-that-satisfy-the-following-inequality-53-201-lt-a-b-lt-4-15-is-

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Question Number 56301 by naka3546 last updated on 13/Mar/19
Minimum value of b that satisfy the following inequality ((53)/(201)) < (a/b) < (4/(15)) is ...
$${Minimum}\:\:{value}\:\:{of}\:\:{b}\:\:{that}\:\:{satisfy}\:\:{the} \\ $$$${following}\:\:{inequality}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{53}}{\mathrm{201}}\:\:<\:\:\frac{{a}}{{b}}\:\:<\:\:\frac{\mathrm{4}}{\mathrm{15}}\:\:\:\:\:{is}\:\:\:… \\ $$
Commented by naka3546 last updated on 13/Mar/19
a, b are positive integers .
$${a},\:{b}\:\:{are}\:\:{positive}\:{integers}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 13/Mar/19
201=3×67 15=3×5 ((53)/(201))=((265)/(1005)) and (4/(15))=((4×67)/(15×67))=((268)/(1005)) (a/b)=((266)/(1005)) ←first answer (a/b)=((267)/(1005))←second answer so b=1005
$$\mathrm{201}=\mathrm{3}×\mathrm{67}\:\:\:\:\mathrm{15}=\mathrm{3}×\mathrm{5} \\ $$$$\frac{\mathrm{53}}{\mathrm{201}}=\frac{\mathrm{265}}{\mathrm{1005}}\:\:{and}\:\:\frac{\mathrm{4}}{\mathrm{15}}=\frac{\mathrm{4}×\mathrm{67}}{\mathrm{15}×\mathrm{67}}=\frac{\mathrm{268}}{\mathrm{1005}} \\ $$$$\frac{{a}}{{b}}=\frac{\mathrm{266}}{\mathrm{1005}}\:\leftarrow{first}\:{answer} \\ $$$$\frac{{a}}{{b}}=\frac{\mathrm{267}}{\mathrm{1005}}\leftarrow{second}\:{answer} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{b}}=\mathrm{1005} \\ $$
Commented by mr W last updated on 14/Mar/19
with mininmum b: (9/(34)) the next are ((13)/(49)), ((14)/(53)), ((17)/(64)),((19)/(72))....
$${with}\:{mininmum}\:{b}:\:\frac{\mathrm{9}}{\mathrm{34}} \\ $$$${the}\:{next}\:{are}\:\frac{\mathrm{13}}{\mathrm{49}},\:\frac{\mathrm{14}}{\mathrm{53}},\:\frac{\mathrm{17}}{\mathrm{64}},\frac{\mathrm{19}}{\mathrm{72}}…. \\ $$
Commented by naka3546 last updated on 14/Mar/19
how to get it, sir.
$${how}\:\:{to}\:\:{get}\:\:{it},\:{sir}. \\ $$
Commented by MJS last updated on 14/Mar/19
((53)/(201))<(a/b)<(4/(15)) ((53)/(201))b<a<(4/(15))b b≠15n ∧ b≠201n ⌈((53)/(201))b⌉>⌊(4/(15))b⌋ xor ⌈((53)/(201))b⌉=⌊(4/(15))b⌋=a now we must try
$$\frac{\mathrm{53}}{\mathrm{201}}<\frac{{a}}{{b}}<\frac{\mathrm{4}}{\mathrm{15}} \\ $$$$\frac{\mathrm{53}}{\mathrm{201}}{b}<{a}<\frac{\mathrm{4}}{\mathrm{15}}{b} \\ $$$${b}\neq\mathrm{15}{n}\:\wedge\:{b}\neq\mathrm{201}{n} \\ $$$$\lceil\frac{\mathrm{53}}{\mathrm{201}}{b}\rceil>\lfloor\frac{\mathrm{4}}{\mathrm{15}}{b}\rfloor\:\mathrm{xor}\:\lceil\frac{\mathrm{53}}{\mathrm{201}}{b}\rceil=\lfloor\frac{\mathrm{4}}{\mathrm{15}}{b}\rfloor={a} \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{must}\:\mathrm{try} \\ $$
Commented by Kunal12588 last updated on 13/Mar/19
sir ((53)/(201))<((33)/(125))<(4/(15)) don′t know there can be smaller answers.
$${sir}\:\frac{\mathrm{53}}{\mathrm{201}}<\frac{\mathrm{33}}{\mathrm{125}}<\frac{\mathrm{4}}{\mathrm{15}} \\ $$$${don}'{t}\:{know}\:{there}\:{can}\:{be}\:{smaller}\:{answers}. \\ $$
Commented by naka3546 last updated on 14/Mar/19
(9/(34)) may be, I don′t know about it.
$$\frac{\mathrm{9}}{\mathrm{34}}\:\:{may}\:\:{be},\:\:{I}\:\:{don}'{t}\:\:{know}\:{about}\:\:{it}. \\ $$
Answered by mr W last updated on 15/Mar/19
this is my method: ((53)/(201)) < (a/b) < (4/(15)) ⇒ ((201)/(53))>(b/a)>((15)/4) ⇒ ((15a)/4)<b<((201a)/(53)) ⇒⌊((15a)/4)⌋+1≤b≤⌈((201a)/(53))⌉−1 for a given value of a we calculate ⌊((15a)/4)⌋+1 and ⌈((201a)/(53))⌉−1. if ⌊((15a)/4)⌋+1≤⌈((201a)/(53))⌉−1, then b exists, and b=⌊((15a)/4)⌋+1...⌈((201a)/(53))⌉−1. if we start with a=1,2,3,... we′ll get all possible (infinite) solutions: a=9⇒b=34 a=13⇒b=49 a=14⇒b=53 a=17⇒b=64 a=19⇒b=72 a=21⇒b=79 a=22⇒b=83 a=23⇒b=87 a=24⇒b=91 a=25⇒b=94 a=29⇒b=109 a=30⇒b=113 a=31⇒b=117 a=32⇒b=121 a=33⇒b=124, 125 a=35⇒b=132 ...... a=117⇒b=439, 440, 442, 443 ......
$${this}\:{is}\:{my}\:{method}: \\ $$$$\frac{\mathrm{53}}{\mathrm{201}}\:\:<\:\:\frac{{a}}{{b}}\:\:<\:\:\frac{\mathrm{4}}{\mathrm{15}} \\ $$$$\Rightarrow\:\frac{\mathrm{201}}{\mathrm{53}}>\frac{{b}}{{a}}>\frac{\mathrm{15}}{\mathrm{4}} \\ $$$$\Rightarrow\:\frac{\mathrm{15}{a}}{\mathrm{4}}<{b}<\frac{\mathrm{201}{a}}{\mathrm{53}} \\ $$$$\Rightarrow\lfloor\frac{\mathrm{15}{a}}{\mathrm{4}}\rfloor+\mathrm{1}\leqslant{b}\leqslant\lceil\frac{\mathrm{201}{a}}{\mathrm{53}}\rceil−\mathrm{1} \\ $$$$ \\ $$$${for}\:{a}\:{given}\:{value}\:{of}\:{a}\:{we}\:{calculate} \\ $$$$\lfloor\frac{\mathrm{15}{a}}{\mathrm{4}}\rfloor+\mathrm{1}\:{and}\:\lceil\frac{\mathrm{201}{a}}{\mathrm{53}}\rceil−\mathrm{1}. \\ $$$${if}\:\lfloor\frac{\mathrm{15}{a}}{\mathrm{4}}\rfloor+\mathrm{1}\leqslant\lceil\frac{\mathrm{201}{a}}{\mathrm{53}}\rceil−\mathrm{1},\:{then}\:{b}\:{exists}, \\ $$$${and}\:{b}=\lfloor\frac{\mathrm{15}{a}}{\mathrm{4}}\rfloor+\mathrm{1}…\lceil\frac{\mathrm{201}{a}}{\mathrm{53}}\rceil−\mathrm{1}. \\ $$$$ \\ $$$${if}\:{we}\:{start}\:{with}\:{a}=\mathrm{1},\mathrm{2},\mathrm{3},…\:{we}'{ll}\:{get}\:{all} \\ $$$${possible}\:\left({infinite}\right)\:{solutions}: \\ $$$${a}=\mathrm{9}\Rightarrow{b}=\mathrm{34} \\ $$$${a}=\mathrm{13}\Rightarrow{b}=\mathrm{49} \\ $$$${a}=\mathrm{14}\Rightarrow{b}=\mathrm{53} \\ $$$${a}=\mathrm{17}\Rightarrow{b}=\mathrm{64} \\ $$$${a}=\mathrm{19}\Rightarrow{b}=\mathrm{72} \\ $$$${a}=\mathrm{21}\Rightarrow{b}=\mathrm{79} \\ $$$${a}=\mathrm{22}\Rightarrow{b}=\mathrm{83} \\ $$$${a}=\mathrm{23}\Rightarrow{b}=\mathrm{87} \\ $$$${a}=\mathrm{24}\Rightarrow{b}=\mathrm{91} \\ $$$${a}=\mathrm{25}\Rightarrow{b}=\mathrm{94} \\ $$$${a}=\mathrm{29}\Rightarrow{b}=\mathrm{109} \\ $$$${a}=\mathrm{30}\Rightarrow{b}=\mathrm{113} \\ $$$${a}=\mathrm{31}\Rightarrow{b}=\mathrm{117} \\ $$$${a}=\mathrm{32}\Rightarrow{b}=\mathrm{121} \\ $$$${a}=\mathrm{33}\Rightarrow{b}=\mathrm{124},\:\mathrm{125} \\ $$$${a}=\mathrm{35}\Rightarrow{b}=\mathrm{132} \\ $$$$…… \\ $$$${a}=\mathrm{117}\Rightarrow{b}=\mathrm{439},\:\mathrm{440},\:\mathrm{442},\:\mathrm{443} \\ $$$$…… \\ $$
Commented by mr W last updated on 15/Mar/19
is there an easier way than this?
$${is}\:{there}\:{an}\:{easier}\:{way}\:{than}\:{this}? \\ $$
Commented by MJS last updated on 16/Mar/19
no.
$$\mathrm{no}. \\ $$

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