If ABCDE*4 = EDCBA. ABCDE,EDCBA are five digit numbers. A,B,C,D and E are all natural numbers ( 1-9) without repeating any natural number. Why is it deducible that E cannot be three?
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$\begingroup$ If E=3, then A=2. $\endgroup$– Geoffrey TrangJul 20 at 20:19
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$\begingroup$ Why don’t you ask for the number ABCDE? There is a unique solution. $\endgroup$– Herbert KociembaJul 21 at 15:18
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2 Answers
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This feels a bit trivial, but since both OP and the first answer (now deleted) seem to be having trouble seeing the obvious:
If ABCDE * 4 = EDCBA, inspection of the first digits reveals that E ≥ A * 4. (Inequality is there because of a possible carry.)
Since A is at least 1, E is at least 4.
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The only possible values for A is 1 or 2, because else the result of multiplication would have 6 digits. But since 4*E mod 10 = A, A has to be even. So A must be 2 and E must be 8 or 9.
Moreover B has to be 1 or 2 else there would be a carry when multiplying and E would be 9 which is impossible because 4 * 9 mod 10 = 6 and not 2. Since the numbers do no repeat we have B=1. So we have 21xy8 * 4=8yx12. From 4*(10y+8) mod 100 =12 we can conclude y=2 or y=7, since the numbers do no repeat we have y=7. From 21x78 * 4=87x12 it finally follows x=9.
21978*4=87912
So asking for that number would make the problem much more interesting compared to the given version.
