find all such numbers: if we make its last digit, say k, as its first digit, the number becomes k times large as before. (□□...□k)→(k□□...□)=k×(□□...□k)


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Question Number 103888 by mr W last updated on 18/Jul/20

$f i n d a l l s u c h n u m b e r s :$ $i f w e m a k e i t s l a s t d i g i t, s a y k, a s i t s$ $f i r s t d i g i t, t h e n u m b e r b e c o m e s k$ $t i m e s l a r g e a s b e f o r e .$ $(◻ ◻ . . . ◻ k) \to (k ◻ ◻ . . . ◻) = k \times (◻ ◻ . . . ◻ k)$
	Answered by MAB last updated on 18/Jul/20
	$x=Σ_(i=0) ^n 10^i a_i where a_i ∈{0,1...9} and a_0 =k x′=10^n k+Σ_(i=0) ^(n−1) 10^i a_(i+1) =kx hence 10^n k+Σ_(i=0) ^(n−1) 10^i a_(i+1) =k^2 +Σ_(i=0) ^(n−1) 10^(i+1) a_(i+1) (10^n −k)k=9Σ_(i=0) ^(n−1) 10^i a_(i+1) (10^n −k)k=9kx 10^n −k=9x k=1 and x=111...1$
	$x = \underset{i = 0}{\sum^{n}} 10^{i} a_{i} w h e r e a_{i} \in {0, 1 . . . 9} a n d a_{0} = k$ $x^{'} = 10^{n} k + \underset{i = 0}{\sum^{n - 1}} 10^{i} a_{i + 1} = k x$ $h e n c e$ $10^{n} k + \underset{i = 0}{\sum^{n - 1}} 10^{i} a_{i + 1} = k^{2} + \underset{i = 0}{\sum^{n - 1}} 10^{i + 1} a_{i + 1}$ $(10^{n} - k) k = 9 \underset{i = 0}{\sum^{n - 1}} 10^{i} a_{i + 1}$ $(10^{n} - k) k = 9 k x$ $10^{n} - k = 9 x$ $k = 1 a n d x = 111 . . . 1$
	Commented by mr W last updated on 18/Jul/20

	$t h a n k s s i r!$
	Answered by mr W last updated on 18/Jul/20

	$l e t x = \underset{↽ n d i g i t s ⇁}{◻ ◻ . . . ◻ ◻} ⩽ 10^{n} - 1$ $N = \underset{↽ n d i g i t s ⇁}{◻ ◻ . . . ◻ ◻} k = 10 x + k$ $k \underset{↽ n d i g i t s ⇁}{◻ ◻ . . . ◻ ◻} = 10^{n} k + x$ $10^{n} k + x = k (10 x + k)$ $\Rightarrow x = \frac{k (10^{n} - k)}{10 k - 1} ⩽ 10^{n} - 1$ $c a s e k = 1 :$ $x = \frac{10^{n} - 1}{9} = \underset{↽ n d i g i t s ⇁}{111 . . 111}$ $x = \frac{10^{n} - 1}{9} = \underset{↽ n d i g i t s ⇁}{111 . . 111}$ $\Rightarrow N = \underset{↽ n d i g i t s ⇁}{111 . . 1111} w i t h n \in N$ $c a s e k = 2 :$ $x = \frac{2 (10^{n} - 2)}{19}$ $w e c a n f i n d$ $n = 18 m - 1 w i t h m \in N$ $e x a m p l e n = 17 :$ $x = 10526315789473784$ $\Rightarrow N = 105263157894737842$ $c h e c k :$ $2 \times 105263157894737842$ $= 210526315789473784$ $e x a m p l e n = 35 :$ $x = 10526315789473784210526315789473784$ $\Rightarrow N = 105263157894737842105263157894737842$ $c h e c k :$ $2 \times 105263157894737842105263157894737842$ $= 210526315789473784210526315789473784$ $(t o b e c o n t i n u e d)$
	Commented by mr W last updated on 18/Jul/20

	$c a s e k = 3 :$ $h e r e {w e}^{'} l l t r y a n o t h e r w a y t o f i n d$ $t h e n u m b e r x .$ $n o t e : i n f o l l o w i n g t h e d i g i t s o f t h e$ $n u m b e r a r e d i s p l a y e d i n v e r s e l y, i . e .$ $t h e l a s t d i g i t i s t h e l e f t m o s t .$ $39731427155698602685728443013$ $\times 3 97314271556986026857284430139$ $∣\leftarrow l a s t d i g i t f i r s t d i g i t \to∣$ $\Rightarrow x = 103448275862068965517241379$ $\Rightarrow N = 1034482758620689655172413793$ $c h e c k :$ $3 \times 1034482758620689655172413793$ $= 3103448275862068965517241379$ $w e s e e t h e d i g i t s o f n u m b e r N m a y b e$ $r e p e a t e d a n y t i m e s . s o w e h a v e$ $i n f i n i t e n u m b e r s w h i c h s a t i s f y t h e$ $r e q u i r e m e n t .$
	Commented by mr W last updated on 18/Jul/20

	$c a s e k = 4 :$ $u s i n g t h e s a m e m e t h o d a s b e f o r e,$ $4652014$ $\times 4 6520146$ $\Rightarrow x = 10256$ $\Rightarrow N = 102564$ $c h e c k :$ $4 \times 102564$ $= 410256$ $s i m i l a r l y t h e d i g i t s o f N m a y b e$ $r e p e a t e d a n y t i m e s . f o r e x a m p l e$ $N = 102564102564102564$
	Commented by mr W last updated on 18/Jul/20

	$c a s e k = 5 :$ $x = \frac{k (10^{n} - k)}{10 k - 1} = \frac{5 (10^{n} - 5)}{7 \times 7}$ $i t s e e m s t o h a v e n o s o l u t i o n .$
	Commented by mr W last updated on 19/Jul/20

	$c a s e k = 6 :$ $x = \frac{k (10^{n} - k)}{10 k - 1} = \frac{6 (10^{n} - 6)}{59}$ $i t s e e m s t o h a v e n o s o l u t i o n .$
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