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

Question Number 56301 by naka3546 last updated on 13/Mar/19

M i n i m u m v a l u e o f b t h a t s a t i s f y t h e

f o l l o w i n g i n e q u a l i t y

\frac{53}{201} < \frac{a}{b} < \frac{4}{15} i s \dots

Commented by naka3546 last updated on 13/Mar/19

a, b a r e p o s i t i v e i n t e g e r s .

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Mar/19

201 = 3 \times 67 15 = 3 \times 5

\frac{53}{201} = \frac{265}{1005} a n d \frac{4}{15} = \frac{4 \times 67}{15 \times 67} = \frac{268}{1005}

\frac{a}{b} = \frac{266}{1005} \leftarrow f i r s t a n s w e r

\frac{a}{b} = \frac{267}{1005} \leftarrow s e c o n d a n s w e r

s o b = 1005

Commented by mr W last updated on 14/Mar/19

w i t h m i n i n m u m b : \frac{9}{34}

t h e n e x t a r e \frac{13}{49}, \frac{14}{53}, \frac{17}{64}, \frac{19}{72} \dots .

Commented by naka3546 last updated on 14/Mar/19

h o w t o g e t i t, s i r .

Commented by MJS last updated on 14/Mar/19

\frac{53}{201} < \frac{a}{b} < \frac{4}{15}

\frac{53}{201} b < a < \frac{4}{15} b

b \neq 15 n \land b \neq 201 n

⌈ \frac{53}{201} b ⌉ > ⌊ \frac{4}{15} b ⌋ xor ⌈ \frac{53}{201} b ⌉ = ⌊ \frac{4}{15} b ⌋ = a

now we must try

Commented by Kunal12588 last updated on 13/Mar/19

s i r \frac{53}{201} < \frac{33}{125} < \frac{4}{15}

{d o n}^{'} t k n o w t h e r e c a n b e s m a l l e r a n s w e r s .

Commented by naka3546 last updated on 14/Mar/19

\frac{9}{34} m a y b e, I {d o n}^{'} t k n o w a b o u t i t .

Answered by mr W last updated on 15/Mar/19

t h i s i s m y m e t h o d :

\frac{53}{201} < \frac{a}{b} < \frac{4}{15}

\Rightarrow \frac{201}{53} > \frac{b}{a} > \frac{15}{4}

\Rightarrow \frac{15 a}{4} < b < \frac{201 a}{53}

\Rightarrow ⌊ \frac{15 a}{4} ⌋ + 1 ⩽ b ⩽ ⌈ \frac{201 a}{53} ⌉ - 1

f o r a g i v e n v a l u e o f a w e c a l c u l a t e

⌊ \frac{15 a}{4} ⌋ + 1 a n d ⌈ \frac{201 a}{53} ⌉ - 1 .

i f ⌊ \frac{15 a}{4} ⌋ + 1 ⩽ ⌈ \frac{201 a}{53} ⌉ - 1, t h e n b e x i s t s,

a n d b = ⌊ \frac{15 a}{4} ⌋ + 1 \dots ⌈ \frac{201 a}{53} ⌉ - 1 .

i f w e s t a r t w i t h a = 1, 2, 3, \dots {w e}^{'} l l g e t a l l

p o s s i b l e (i n f i n i t e) s o l u t i o n s :

a = 9 \Rightarrow b = 34

a = 13 \Rightarrow b = 49

a = 14 \Rightarrow b = 53

a = 17 \Rightarrow b = 64

a = 19 \Rightarrow b = 72

a = 21 \Rightarrow b = 79

a = 22 \Rightarrow b = 83

a = 23 \Rightarrow b = 87

a = 24 \Rightarrow b = 91

a = 25 \Rightarrow b = 94

a = 29 \Rightarrow b = 109

a = 30 \Rightarrow b = 113

a = 31 \Rightarrow b = 117

a = 32 \Rightarrow b = 121

a = 33 \Rightarrow b = 124, 125

a = 35 \Rightarrow b = 132

\dots \dots

a = 117 \Rightarrow b = 439, 440, 442, 443

\dots \dots

Commented by mr W last updated on 15/Mar/19

i s t h e r e a n e a s i e r w a y t h a n t h i s ?

Commented by MJS last updated on 16/Mar/19

no .

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