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A Tuesday number is a positive integer $N$ with $d$ digits where the given properties hold true for every positive integer $i$ in the range $0<i<d$

  • $N\times i$ contains the exact same digits (in any order) as $N$
  • Moreover, it should be possible to split $N$ into two parts, such that writing the first half after the second does, in fact, give the value of $N\times i$

Some valid examples of splitting and reordering:

$1371\rightarrow 3711, 7113, 1137, 1371$

$02684\rightarrow 26840, 68402, 84026, 40268, 02684$

Find the largest Tuesday number.

Hint

You probably will require the number to start with a zero.

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    $\begingroup$ the largest one I can think of immediately is 99 $\endgroup$ – Novarg Sep 29 '15 at 14:36
  • $\begingroup$ @Novarg The answer is much larger than that. $\endgroup$ – ghosts_in_the_code Sep 29 '15 at 15:18
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There are some extremely large Tuesday numbers, with billions of digits. Probably there is no bound on how large they can get, though nobody knows how to prove this.

If $n$ is a positive integer, the decimal expansion of $1/n$ is eventually periodic, and the period has at most $n-1$ digits. The period can be exactly $n-1$ digits only if $n$ is a prime number (though this does not happen for every prime). For example, $$ \frac{1}{7}=0.\overline{142857} $$ has period $7-1=6$ (and indeed $7$ is prime).
If the decimal expansion of $1/p$ has period $p-1$, we call $p$ a full reptend prime.

We can obtain a Tuesday number (also called a cyclic number) from one full period of the decimal expansion of the reciprocal of a full reptend prime, and every Tuesday number arises this way. For example, $7$ is a full reptend prime, so $142857$ is a Tuesday number: $$ \begin{array}{ll} 1\cdot142857=142857,&2\cdot 142857=2857\>14,\\ 3\cdot 142857=42857\>1,&4\cdot 142857=57\>1428,\\ 5\cdot 142857=7\>14285,& 6\cdot 142857=857\>142.\\ \end{array} $$

It can be proven that $p$ is a reptend prime if and only if the first $p-1$ powers of $10$ give distinct remainders when divided by $p$ (in this case one says $10$ is a primitive root modulo $p$). Artin's conjecture on primitive roots asserts that every non-square positive integer is a primitive root modulo $p$ for infinitely many primes $p$, and this would imply that there exist arbitrarily large Tuesday numbers.

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The largest Tuesday number is...

unknown. It has been conjectured that there are an infinite number, but not proven. (source)

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  • $\begingroup$ Although I agree that these must be cyclic numbers. It is not clear to me from that link that there are infinite of them. In other words: Is it known whether there is a largest full reptend prime? If it can be proved that there are infinite you should provide it I think $\endgroup$ – Ivo Beckers Sep 29 '15 at 15:06
  • $\begingroup$ @IvoBeckers It appears to be an open conjecture: "In fact, there is no single value of a for which Artin's conjecture is proved." $\endgroup$ – f'' Sep 29 '15 at 15:18
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    $\begingroup$ It does make me wonder the OP's hint. The hint suggests that the solution actually is not a cyclic number because a number with a leading zero can't be cyclic by the usual definition of cyclic I think $\endgroup$ – Ivo Beckers Sep 29 '15 at 15:24
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    $\begingroup$ @ghost I think you typed the number wrong above. It shouldn't have a trailing $0$. (At least, when I multiply $052631578947368421$ by each of the digits from 2 to 18, it meets the criteria above, while your number doesn't.) $\endgroup$ – GentlePurpleRain Sep 29 '15 at 15:54
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    $\begingroup$ Notice a number may be cyclic without being Tuesday. Tuesday is a stronger requirement. $\endgroup$ – Fimpellizieri Sep 29 '15 at 16:45
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If leading zeroes were not allowed, then:

For numbers where $d$ is greater than 11, $N*i$ will have more digits in it than N for the largest value of $i$. The largest cyclic number without a leading zero shown on the Wikipedia page that satisfies the rules of a Tuesday number appears to be 142857. I don't believe there is a higher one as 7 is the largest prime below 10 and 11 doesn't generate a cyclic number.

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    $\begingroup$ This would be correct if leading zeroes weren't allowed. $\endgroup$ – Zandar Sep 29 '15 at 17:15
  • $\begingroup$ @Zandar Yes, I've realised that now, but I can't delete my answer from my phone! $\endgroup$ – Gordon K Sep 29 '15 at 17:16
  • $\begingroup$ @GordonK Leave it undeleted, just add that this would be correct if leading zeroes were disallowed. $\endgroup$ – Rohcana Sep 29 '15 at 17:22
  • $\begingroup$ This answer proves that the accepted answer can't be right! So the OP's solution is probably the right one $\endgroup$ – Ivo Beckers Sep 29 '15 at 21:13
  • $\begingroup$ @Ivo No, the OP's solution was 052631578947368421, which is also too long - it relies on a leading zero. $\endgroup$ – Zandar Sep 29 '15 at 21:25

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