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I and Ron were talking casually over coffee. Just then a thought sprung into my mind; could there exist numbers such that if their digits are reversed, for example $123\to321$, the new number is exactly double the original?

NOTE: I am talking about integers excluding 0 (Thanks! Bass)

Ron replied, "Of course there are!"

After two seconds of silence,

Ron added, "But the smallest such number is 18 digits long."

When I asked him how did he know, he assured me that it was nothing more than 4th grade maths.

How did Ron deduce this?

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    $\begingroup$ And mention if you are using base 10. "Digit" on its own is meaningless. $\endgroup$ – Martin Argerami Jan 8 '18 at 11:23
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    $\begingroup$ @MartinArgerami "Digit" on its own means base 10 in standard usage. If other bases are envisioned then this should be clearly stated. $\endgroup$ – John Coleman Jan 8 '18 at 13:35
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    $\begingroup$ @JohnColeman I agree, and yet an answer that claims “Since there are no rules on which base you use…” has 12 votes at the moment. So specifying the base in the question would indeed have been good IMO. $\endgroup$ – ShreevatsaR Jan 8 '18 at 18:34
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    $\begingroup$ Did anyone else click on this because they thought it was a question about Harry Potter? $\endgroup$ – Obie 2.0 Jan 9 '18 at 8:35
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    $\begingroup$ @RickvanOsta How does the existence of non-standard usage refute a claim about standard usage? When used without qualification in a context where other bases are not under consideration the phrase "3-digit number" means a number between one hundred and a nine hundred ninety nine. There are, of course, other bases out there, and for such number systems there is a nonstandard usage whereby "digit" denotes a number between 0 and the base minus 1. In the hexadecimal system "C" is a "digit", but it is clearly not standard English usage to refer to "C" as a digit. $\endgroup$ – John Coleman Jan 9 '18 at 12:46
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We know two things about the number:

A: If you reverse it and divide by 2, you get the original number
B: If you multiply it by 2 and reverse it, you get the original number again.

Deducing from there,

* From A, the first digit must be even, otherwise the reversed number wouldn't be divisible by 2
* From B, the first digit must be < 5, otherwise the doubled number would have too many digits (a number with leading zeroes would be ill-formed, so we exclude those straight away)
So the first digit must be either 2 or 4.
* From B, the last digit must be either twice the first digit (if the second digit <5), or one more than that (if the second digit >= 5). (We can omit the usual "modulo 10" bit here, because our numbers are smaller than 5)
* If the first digit is 2, then the last digit must be 4 or 5. However, using A, we get a number ending in 2, divided by 2; the result must end in either 6 or 1. So 2 cannot be the first digit.
* If the first digit is 4, the last digit must be either 8 or 9. Again using A, a number ending in 4 divided by 2 will end in either 2 or 7. So 4 cannot be the first digit either

It would seem that Ron must have used the single-phase Stetson-Milliner method, because

apart from the trivial 0, such a number doesn't exist at all, and Ron totally pulled his answer out of a hat.

(Hope I didn't make any mistakes there; tried to double-check, but it's difficult to proofread one's own logic because one becomes blind to any flaws.)

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    $\begingroup$ I agree with your statement A. But if a number shows this property(say N), will 2N also show this property? $\endgroup$ – prog_SAHIL Jan 8 '18 at 10:55
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    $\begingroup$ @prog_sahil, Count again. That number doubles when you shift the last digit to the beginning, and you stole it from a youtube video. $\endgroup$ – Bass Jan 8 '18 at 11:00
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    $\begingroup$ But this puzzle is a lot different from that one. $\endgroup$ – prog_SAHIL Jan 8 '18 at 11:33
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    $\begingroup$ Yes, it is. Somehow, that answer also has 18 digits, and somehow, you thought it solved this puzzle. There are very few scenarios where those coincidences are likely. Also, I happened to find the uncredited source of your recent river-crossing problem, identical even up to the protagonist's name. These comments are probably not the right place to sort this out. $\endgroup$ – Bass Jan 8 '18 at 12:33
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    $\begingroup$ "From B, the first digit must be < 5, otherwise the doubled number would have too many digits" is not necessarily true. Take, for example, 75: (75 * 2 = 150) → (051 = 51) $\endgroup$ – Vaelus Jan 8 '18 at 18:08
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Since there are no rules on which base you use:

If you were to do this in base-2, or binary, the number 01 (1) would become 10 (2), which doubles it's value. :D

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  • $\begingroup$ The stated example limits to at least Base 4 so yours is excluded. $\endgroup$ – user2617804 Jan 8 '18 at 13:16
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    $\begingroup$ And the example did not meets the "double" specification, so he can use whatever base he want. $\endgroup$ – Alix Eisenhardt Jan 8 '18 at 15:45

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