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

\frac{2010}{2011} < \frac{x}{y} < \frac{2011}{2012} then compute the

minimum value for x + y and the

values of x and y which achieves

this minimum

Commented by mr W last updated on 04/Nov/21

\frac{x}{y} = \frac{4021}{4023}

Commented by HongKing last updated on 04/Nov/21

how my dear S er solution please

Commented by Rasheed.Sindhi last updated on 04/Nov/21

S i r m r W, w h e n y o u h a v e s o m e t i m e

You can't use 'macro parameter character #' in math mode

A c t u a l l y m y a n s w e r {d o e s n}^{'} t m a t c h

t h e a n s w e r o f t h e q u e s t i o n e r \dots

T h a n k s i n a d v a n c e s i r!

Commented by mr W last updated on 04/Nov/21

i h a v e c h e c k e d . y o u r a n s w e r i s c o r r e c t .

Commented by Rasheed.Sindhi last updated on 05/Nov/21

THANKS A LOT SIR!

Answered by mr W last updated on 04/Nov/21

x m u s t f u l f i l l

⌈ \frac{2011 x}{2010} ⌉ - ⌊ \frac{2012 x}{2011} ⌋ ⩾ 2

w e g e t x_{m i n} = 4021

y_{m i n} = ⌊ \frac{2012 \times 4021}{2011} ⌋ + 1 = 4023

{(x + y)}_{m i n} = 4021 + 4023 = 8044

\frac{x}{y} = \frac{4021}{4023}, \frac{6031}{6034}, \frac{6032}{6035}, \frac{8041}{8045}, \dots

Commented by Rasheed.Sindhi last updated on 06/Nov/21

\frac{2010}{2011} < \frac{x}{y} < \frac{2011}{2012}

S i r i n t h i s c a s e :

x_{m i n} = 2010 + 2011 = 4021

y_{m i n} = 2011 + 2012 = 4023

I s i t a c o i n c i d e n c e

o r g e n e r a l l y i f \frac{a}{b} < \frac{x}{y} < \frac{c}{d}

t h e n x_{m i n} = a + b & y_{m i n} = c + d

a n d {(x + y)}_{m i n} = a + b + c + d ?

Commented by mr W last updated on 06/Nov/21

a v e r y n i c e t h o u g h t s i r!

b u t i t h i n k h e r e {i t}^{'} s j u s t a c o i n c i d e n c e .

w i t h x = a + c a n d y = b + d

i t f u l f i l l s i n d e e d

\frac{a}{b} < \frac{x}{y} < \frac{c}{d}

b u t x = a + c {m u s t n}^{'} t b e x_{m i n} a n d

y = b + d {m u s t n}^{'} t b e y_{m i n} . f o r e x a m p l e

w h e n g c d (a + c, b + d) \neq 1 .

Commented by mr W last updated on 06/Nov/21

a n e x a m p l e

\frac{13}{75} < \frac{x}{y} < \frac{9}{50}

x_{m i n} = 3, y_{m i n} = 17

\frac{13}{75} < \frac{3}{17} < \frac{9}{50}

Commented by Rasheed.Sindhi last updated on 06/Nov/21

ㄒ卄卂几Ҝ丂爪尺山丂丨尺!