I got this problem from my son and he picked it from some local math competition. It's fairly simple:

For numbers $k,m,n\in N$ ($k\ge2)$ we know the following:

$$\left(\frac mn\right)^k=0.\overline{x_1x_2...x_9}\tag{1}$$

On the right side we have an infintely repeating sequence of exactly nine digits $x_i\in\{0,1,2,\dots9\}$ ($i=1\dots9)$ and these digits are not necessarily distinct. Find all possible values of expression (1).

The solution seems to be simple: we can replace the repeating sequence of digits with some number $a$:


Relation (1) now becomes:

$$\left(\frac mn\right)^k=\frac{a}{10^9-1}$$


If we assume that $m$ and $n$ are coprime then:


If we are able to find prime factors of $(10^9-1)$ quickly, we are done. True, we can find a few prime factors fairly easily:

$$10^9-1=(10^3)^3-1=(10^3-1)(10^6+10^3+1)=9\times111\times1001001 \\ =3^2\times3\times 37\times3\times333667=3^4\times37\times333667$$

However, the last number (333667) is a tough nut to crack. We can proceed only if we know its factors.

With some help from the computer you can easily find out that 333667 is a prime and the rest of the solution is fairly straightforward.

However, suppose that you are in a real competition - you don't have a computer or a pocket calculator. Factoring 333667 by hand is a time consuming activity and you have other problems to solve as well.

Is there a better approach?

Happy holidays :)

  • $\begingroup$ Well, you need to somehow know at least that $333667$ is squarefree - otherwise, for its factor $p^2$, $1/p^2$ would have purely periodic expansion of length $9$. The problem at this point is actually equivalent to checking if $333667$ is squarefree. $\endgroup$ – Wojowu Jan 2 at 21:14
  • $\begingroup$ @Wojowu Yes, that's the key point. If we know that 333667 is squarefree, we are done. But I have no idea how to prove it quickly. $\endgroup$ – Oldboy Jan 2 at 21:17
  • $\begingroup$ Testing squarefree-ness is a known problem in computational complexity with no known polynomial-time algorithm. This doesn't exactly tell us there is no way to solve this problem easily, since $333667$ might have some magical properties which make it simpler, but I am being a little skeptical here... $\endgroup$ – Wojowu Jan 2 at 21:22
  • $\begingroup$ @JohnDouma A digit can be zero, I have clarified this. $\endgroup$ – Oldboy Jan 2 at 21:28
  • $\begingroup$ There was no need to clarify. I just misunderstood. That's why I deleted the comment. $\endgroup$ – John Douma Jan 2 at 21:30

You want to find whether there exists a prime $p$ such that $p^2\mid n$, where $n=333667$. Suppose that such $p$ exists. Then, we know that $$p\leq \sqrt{n}<578.$$ It is easily seen that $p>11$, so $$13\leq p\leq 577.$$ However, if $p>67$, then $$\frac{n}{p^2}\leq \frac{n}{71^2}<66.$$ Thus, $n$ must have a prime divisor $q<66$ such that $q\mid n$ (noting that $n$ is not a perfect square per TonyK's comment under Ross Millikan's answer). Therefore, $n$ must have a prime divisor that is inclusively between $13$ and $67$: $13$, $17$, $19$, $23$, $29$, $31$, $37$, $41$, $43$, $47$, $53$, $59$, $61$, and $67$. We can easily rule out $37$ as $n-1$ is divisible by $111=3\cdot 37$. This leaves $13$ primes to deal with.

There will be some cumbersome computations. It is not too difficult (but a little bit tedious) to find the square root or the cubic root of $n$ by hand (the cubic root of $n$ is used to obtain $67$ when I say that if $p>67$ then there exists a prime divisor $q<66$). And then you have to divide $n$ by $13$ primes. This is doable, but not very nice.

  • $\begingroup$ A cube root is not good enough because we could have $333667=p^2q$ for $p,q$ prime. But if we show there is no prime factor smaller than the cube root we are done. $\endgroup$ – Ross Millikan Jan 2 at 22:40

To prove $333667$ is squarefree, you just have to show it has no prime factor smaller than $\sqrt[3]{333667} \approx 69$ The small ones can be done by divisibility rules, say $2,3,5,7,11$. That leaves $14$ to try, which is not too bad. You might even know the variants on the classic test for $7$ that you double the last digit and subtract it from the rest of the number. This is based on the fact that $21$ is a multiple of $7$. For $13$ you can note that $39$ is a multiple of $13$ and multiply the last digit by $4$ and add to the rest of the number. For $17$ you can use $51$. That gets you the next few. It would be a few minutes, but if you are quick with arithmetic much less than $10$.

  • 4
    $\begingroup$ Well, you also have to show that $333667$ is not a perfect square. (But that's easy, because a perfect square can't end in $7$.) $\endgroup$ – TonyK Jan 2 at 22:18

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