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.


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

  • 3
    $\begingroup$ 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

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.


The largest Tuesday number is...

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

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 2
    $\begingroup$ Notice a number may be cyclic without being Tuesday. Tuesday is a stronger requirement. $\endgroup$ – Fimpellizieri Sep 29 '15 at 16:45

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.

  • 1
    $\begingroup$ 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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