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Question Number 192142 by mehdee42 last updated on 09/May/23

$Q u e s t i o n$ $l e t x =< a_{n} a_{n - 1} . . . a_{1} a_{0} > \in N; a_{0} \neq 0 &$ $y =< a_{n} a_{n - 1} . . . a_{1} > \in N b e$ $t w o n a t u r a l n u m b e r s$ $s u c h t h a t \frac{x}{y} \in N$ $f i n d t h e n u m b e r ‘ ‘ x^{″} ?$
	Answered by deleteduser1 last updated on 09/May/23
	$For a 2-digit number,x;possible values: 99,88,77,66,55,48,44,39,36,33,28,26,...,22,19,...,11 Now,for an n-digit number(n atleast 3) (where a_0 ≠0) (x/y)=((10x)/(x−a_0 ))=((10(x−a_0 )+10a_0 )/(x−a_0 ))=10+((10a_0 )/(x−a_0 )) ⇒x−a_0 ∣10a_0 ⇒x−a_0 ≤10a_0 ⇒11a_0 ≥x But this is impossible since max{11a_0 }=99 which is not atleast a 3-digit number. ⇒Only 2-digit solutions exist.$
	$F o r a 2 - d i g i t n u m b e r, x; p o s s i b l e v a l u e s :$ $99, 88, 77, 66, 55, 48, 44, 39, 36, 33, 28, 26, . . ., 22, 19, . . ., 11$ $N o w, f o r a n n - d i g i t n u m b e r (n a t l e a s t 3)$ $(w h e r e a_{0} \neq 0)$ $\frac{x}{y} = \frac{10 x}{x - a_{0}} = \frac{10 (x - a_{0}) + 10 a_{0}}{x - a_{0}} = 10 + \frac{10 a_{0}}{x - a_{0}}$ $\Rightarrow x - a_{0} ∣ 10 a_{0} \Rightarrow x - a_{0} ⩽ 10 a_{0}$ $\Rightarrow 11 a_{0} ⩾ x$ $B u t t h i s i s i m p o s s i b l e s i n c e m a x {11 a_{0}} = 99$ $w h i c h i s n o t a t l e a s t a 3 - d i g i t n u m b e r .$ $\Rightarrow O n l y 2 - d i g i t s o l u t i o n s e x i s t .$
	Commented by mehdee42 last updated on 09/May/23

	$i t i s v e r y b e a u t i f u l s o l u t i o n .$ $i n a d d i t i o n a c c o r d i n g t o t h e c o n d i t i o n ‘ ‘ a_{0} \neq 0^{″}$ $x \neq 10, 20, . . ., 90$
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