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If $ABCDE\times4 = EDCBA$. Here $A, B, C, D, E$ represent distinct nonzero digits and $ABCDE$ and $EDCBA$ are five digit numbers. Why can't E be three?

IraeVid
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    Please show your own effort to solve the question. It is the way it works. Also, providing some further content to the question may make it have a better quality. Please also consult https://math.meta.stackexchange.com/questions/9959/how-to-ask-a-good-question on "how to ask a good question". As it stays, we have an "answer request", this is not a site of the shape "here is my question, give me an answer". Instead, some fair share of the effort is encouraged, showing what you know makes an answer easier to type and to focus on the essence. – dan_fulea Jul 20 '23 at 19:22
  • what have you tried? Write some of your ideas about solution of the question. – Lion Heart Jul 20 '23 at 19:40
  • You can’t not find the answer if you try. And it’s not quite trivial but close what E is. – gnasher729 Jul 20 '23 at 20:22
  • You don't need to identify the digits to be able to answer the question. In fact, you can answer it after only doing some preliminary investigation into $E$ and one other digit. – Paul Sinclair Jul 21 '23 at 21:03
  • $E \ge 4A.$ If $E = 3$ then $A = 0$ but then $E\ne 3$ – user317176 Jul 25 '23 at 07:22

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Assume $E=3$, then $ABCD3 * 4 = A'B'C'D'2$, so for that to equal $EDCBA$, $A$ would need to be 2. But if $A$ is 2, the first digit of $ABCDE*4$ is greater or equal to 8, which is a contradiction to it being $EDCBA$, with $E=3$ as the first digit.

watertrainer
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