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Question Number 198563 by mr W last updated on 22/Oct/23
find all numbers (with any number  of digits) satisfying  (abcd...xyz)×2=(zyx...dcba)
$${find}\:{all}\:{numbers}\:\left({with}\:{any}\:{number}\right. \\ $$$$\left.{of}\:{digits}\right)\:{satisfying} \\ $$$$\left(\underline{{abcd}…{xyz}}\right)×\mathrm{2}=\left(\underline{{zyx}…{dcba}}\right) \\ $$
Answered by AST 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..
$$\mathrm{2}{z}\equiv{a}\left({mod}\:\mathrm{10}\right);{a}\leqslant\mathrm{4}\Rightarrow{a}\in\left\{\mathrm{0},\mathrm{2},\mathrm{4}\right\};{a}\geqslant{z} \\ $$$${a}=\mathrm{0}\Rightarrow{z}=\mathrm{0};{question}\:{remains}\:{the}\:{same} \\ $$$${a}=\mathrm{2}\Rightarrow{z}\overset{\mathrm{5}} {\equiv}\mathrm{1}\Rightarrow{z}=\mathrm{6}\Rightarrow\mathrm{2}{b}..{y}\mathrm{6}×\mathrm{2}=\mathrm{6}{b}…{y}\mathrm{2}{X} \\ $$$${a}=\mathrm{4}\Rightarrow{z}\equiv\mathrm{2}\left({mod}\:\mathrm{5}\right)\Rightarrow{z}=\mathrm{7}\Rightarrow\left(\mathrm{4}{b}…{y}\mathrm{7}\right)×\mathrm{2}=\left(\mathrm{7}{y}…{b}\mathrm{4}\right){X} \\ $$$$ \\ $$$${No}\:{solution}..\: \\ $$
Commented by mr W last updated on 25/Oct/23
i agree. thanks!
$${i}\:{agree}.\:{thanks}! \\ $$

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