Q1 ∴ x=<1a_1 a_2 ...a_n >∈N & y=<a_1 a_2 ...a_n 1>∈N if y=3x then , find the smallest value of x Q2 ∴ with the above conditions ,what other values can be placed besides the number ♮ 1 ε


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Question Number 192172 by mehdee42 last updated on 10/May/23

$Q 1 ∴ x =< 1 a_{1} a_{2} . . . a_{n} >\in N & y =< a_{1} a_{2} . . . a_{n} 1 >\in N$ $i f y = 3 x t h e n, f i n d t h e s m a l l e s t$ $v a l u e o f x$ $Q 2 ∴ w i t h t h e a b o v e c o n d i t i o n s, w h a t o t h e r v a l u e s$ $c a n b e p l a c e d b e s i d e s t h e n u m b e r ♮ 1 ε$
	Answered by AST last updated on 10/May/23

	$1) x = 10^{n} + \frac{y - 1}{10}; y = 3 x \Rightarrow x = 10^{n} + \frac{3 x - 1}{10}$ $\Rightarrow 10 x = 10^{n + 1} + 3 x - 1 \Rightarrow 7 x = 10^{n + 1} - 1$ $10^{n + 1} - 1 \equiv 0 (m o d 7) \Rightarrow 3^{n + 1} \equiv 1 (m o d 7) \Rightarrow 3^{n} \overset{7}{\equiv} 5$ ${o r d}_{3} (7) = 6; 3^{5} \overset{7}{\equiv} 5; ϕ (7) = 6 \Rightarrow n = 6 k + 5; m i n (n) = 5$ $\Rightarrow m i n (x) = \frac{10^{6} - 1}{7} = 142857$ $2) x = p 10^{n} + \frac{3 x - p}{10} \Rightarrow 10 x = p 10^{n + 1} + 3 x - p$ $\Rightarrow 7 x = p (10^{n + 1} - 1)$ $T h e f i r s t d i g i t o f \frac{p (10^{n + 1} - 1)}{7} s h o u l d b e p$ $T h i s i s o n l y t r u e f o r p = 1, 2$ $\Rightarrow p = 1 a n d 2 a l w a y s w o r k w h e n n = 6 k + 5 .$ $H e n c e, o n l y 2 c a n b e p l a c e d b e s i d e s t h e n u m b e r$ $♮ 1 ε .$
	Commented bymehdee42 last updated on 10/May/23

	$v e r y n i c e S i r$ $f o r p = 2; t h e l o w e s t v a l u e$ $x = 285714$
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