find all numbers (with any number of digits) satisfying (abcd...xyz)×2=(zyx...dcba)


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Question Number 198563 by mr W last updated on 22/Oct/23

$f i n d a l l n u m b e r s (w i t h a n y n u m b e r$ $o f d i g i t s) s a t i s f y i n g$ $(\underset{―}{a b c d . . . x y z}) \times 2 = (\underset{―}{z y x . . . d c b a})$
	Answered by deleteduser1 last updated on 25/Oct/23
	$2z≡a(mod 10);a≤4⇒a∈{0,2,4};a≥z a=0⇒z=0;question remains the same a=2⇒z≡^5 1⇒z=6⇒2b..y6×2=6b...y2X a=4⇒z≡2(mod 5)⇒z=7⇒(4b...y7)×2=(7y...b4)X No solution..$
	$2 z \equiv a (m o d 10); a ⩽ 4 \Rightarrow a \in {0, 2, 4}; a ⩾ z$ $a = 0 \Rightarrow z = 0; q u e s t i o n r e m a i n s t h e s a m e$ $a = 2 \Rightarrow z \overset{5}{\equiv} 1 \Rightarrow z = 6 \Rightarrow 2 b . . y 6 \times 2 = 6 b . . . y 2 X$ $a = 4 \Rightarrow z \equiv 2 (m o d 5) \Rightarrow z = 7 \Rightarrow (4 b . . . y 7) \times 2 = (7 y . . . b 4) X$ $N o s o l u t i o n . .$
	Commented by mr W last updated on 25/Oct/23

	$i a g r e e . t h a n k s!$
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