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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)
findallnumbers(withanynumberofdigits)satisfying(abcdxyz)×2=(zyxdcba)
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..
2za(mod10);a4a{0,2,4};aza=0z=0;questionremainsthesamea=2z51z=62b..y6×2=6by2Xa=4z2(mod5)z=7(4by7)×2=(7yb4)XNosolution..
Commented by mr W last updated on 25/Oct/23
i agree. thanks!
iagree.thanks!

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