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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

$S i r m r W, w h e n y o u h a v e s o m e t i m e$ $You can't use 'macro parameter character #' in math mode$ $A c t u a l l y m y a n s w e r {d o e s n}^{'} t m a t c h$ $t h e a n s w e r o f t h e q u e s t i o n e r . . .$ $T h a n k s i n a d v a n c e s i r!$
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

	$x m u s t f u l f i l l$ $⌈ \frac{2011 x}{2010} ⌉ - ⌊ \frac{2012 x}{2011} ⌋ ⩾ 2$ $w e g e t x_{m i n} = 4021$ $y_{m i n} = ⌊ \frac{2012 \times 4021}{2011} ⌋ + 1 = 4023$ ${(x + y)}_{m i n} = 4021 + 4023 = 8044$ $\frac{x}{y} = \frac{4021}{4023}, \frac{6031}{6034}, \frac{6032}{6035}, \frac{8041}{8045}, . . .$
	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!
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