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A positive integer n (without leading zeros) has the property that shifting the rightmost digit of n to the left end doubles the number.
Examples: 1->1, 1234->4123, 2020->202

What is the smallest n with this property?

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    $\begingroup$ Very nice question. $\endgroup$
    – hexomino
    Aug 18 '20 at 17:01
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    $\begingroup$ This is a way cooler question than I first thought it was :) $\endgroup$ Aug 19 '20 at 4:55
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    $\begingroup$ This has been discussed in a youtube video: youtube.com/watch?v=1lHDCAIsyb8 $\endgroup$ Aug 19 '20 at 7:41
  • $\begingroup$ Wow I did not expect such complexity from such a simple question! Well done. $\endgroup$ Aug 19 '20 at 9:54
  • $\begingroup$ @NilayGhosh actually I prefer the answer given in the video. Very nice and easy to understand. $\endgroup$ Aug 19 '20 at 9:58
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I think the answer is

$$N = 20 \left(\frac{10^{17} -2}{19}\right) + 2$$

Proof

Suppose we write our original number as $$N = a_n 10^n + a_{n-1}10^{n-1} +\ldots + a_0 = \displaystyle \sum_{j=0}^n a_j 10^j$$ Then the equation described in the problem is $$ 2 \displaystyle \sum_{j=0}^n a_j 10^j = a_0 10^n + \displaystyle \sum_{j=1}^n a_j 10^{j-1}$$ Rearranging gives $$ \displaystyle \sum_{j=1}^n a_j ((2 \times 10^j) - 10^{j-1}) = a_0 (10^n - 2)$$ which means that $$ 19 \displaystyle \sum_{j=1}^n a_j 10^{j-1} = a_0 (10^n -2)$$ Now notice that the left hand side is divisible by $19$ so the right hand side must be also but since $a_0$ is coprime to $19$, this means that $10^n - 2$ is divisible by $19$. Therefore, we are looking for the smallest power of $10$ which is congruent to $2$ modulo $19$.

Going through powers of $10$ modulo $19$ gives $10, 5, 12, 6, 3, 11, 15, 17, 18, 9, 14, 7, 13, 16, 8, 4, 2, \ldots$.
Hence, the smallest power of $10$ that works is $10^{17}$. Plugging this into our equation gives $$ \displaystyle \sum_{j=1}^{17} a_j 10^{j-1} = a_0 \frac{10^{17} -2}{19}$$ Clearly, we cannot pick $a_0=1$ as the right-hand side will have too few digits, but if we pick $a_0=2$ (to achieve the minimum) then it looks safe that we will have a $17$-digit number on the right-hand side and we can just pick the rest of the $a_j$ appropriately on the left.

This means that the smallest $N$ which works must be $$N = 20 \left(\frac{10^{17} -2}{19}\right) + 2$$

Computer check

Working it out with a computer it seems the value for $N$ above is $105263157894736842$ and doubling this gives $210526315789473684$ so this does indeed work.

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    $\begingroup$ This is a beautiful proof, I love it. $\endgroup$
    – Sciborg
    Aug 18 '20 at 17:17
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    $\begingroup$ 'rot13("Lbh zvffrq n punapr gb nccyl yvggyr Srezng urer. (10k2 = 1 zbq 19, urapr 10^(19-2) = 2 zbq 19).")' Apart from that, nicely done! $\endgroup$ Aug 18 '20 at 18:00
  • $\begingroup$ Hats off @hexomino. Incredible $\endgroup$
    – DrD
    Aug 18 '20 at 18:02
  • $\begingroup$ @Paul :rot13(V guvax vg vf erdhverq gb tb guebhtu nyy gur cbjref bs 10 zbq 19 nf qbar ol urkbzvab. Yvggyr Srezng cebivqrf n fbyhgvba, ohg abg arprffnevyl gur fznyyrfg (cevzvgvir ebbg). Sbe rknzcyr: jvgu 10^13 = 10 zbq 13 (yvggyr Srezng) jr trg 10^(13-2) = 4 zbq 13, ohg jr nyfb unir 10^5 = 4 zbq 13) $\endgroup$
    – ThomasL
    Aug 18 '20 at 19:08
  • $\begingroup$ @ThomasL rot13("Tvir lbh gung. Fgvyy, fubjvat gung 2 vf cevzvgvir vf nf cnvayrff nf Cuv_18(2) = 64 - 8 + 1 = 0 zbq 19 (Cuv_18(K)=K^6-K^3+K vf 18gu plpybgbzvp cbyl). V\'q fnl gung unaqvyl orngf tbvat guebhtu cbjref bs 10 zbq 19.") $\endgroup$ Aug 18 '20 at 20:21

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