Alice and Bob play the following game. They alternately select distinct nonzero digits from $1$ to $9$, until they have chosen seven such digits. Consider the resulting seven-digit number by joining the digits in the order selected, with the seventh digit appearing last (i.e. $\overline{A_1B_2A_3B_4A_6B_6A_7}$). Alice wins if and only if the resulting number is the last seven decimal digits of some perfect seventh power. Which player has the winning strategy?

I have tried this but got nothing useful.


Wins Alice because she has the possibility of choosing the last digit and is enough to choose something coprime with 10, say 1. Then the remaining number seen as an element of $\mathbb{Z}/10^7\mathbb{Z}$ is an invertible element of that ring. And the group of unit,G, of $\mathbb{Z}/10^7\mathbb{Z}$ has $\phi(10^7)$ elements that is a number coprime with 7. Thus consider the map $G \to G: g \to g^7$, is a group homomorphism, and is injective(the order of an element in the kernel has to divide two coprime numbers, thus it has to be 1) so is surjective. So in G every element is a 7-th power, particularly the one resulting by the choice of Alice as last digit( namely, anyone between 1,3,7,9), but this is exactly what makes Alice a winner.

  • $\begingroup$ It's too concise, can you write a bit more detail please? $\endgroup$ – user45220 Jul 23 '14 at 19:57
  • $\begingroup$ @Calvin Lin: i'm sorry that you can't get the explanation , but what i wrote is true. If the starting player choose as last digit, say, 1, then you get a number of 7 digit wich is asked to be a 7-th perfect power in $\mathbb{Z}/10^7\mathbb{Z}$, but since the element is invertible in that ring you can observe that the group of units of that ring has a number of element coprime with 7, so taking 7-th power is an automorphsim of that group, and so every invertible element is a seventh power in that ring $\endgroup$ – res Jul 24 '14 at 10:54
  • $\begingroup$ @CalvinLin: now tell me if the one in my comment can be clear even to you. It would be a good test of clarity $\endgroup$ – res Jul 24 '14 at 11:55
  • $\begingroup$ It makes sense to me (except that Alice is a 'she'). But I think your proof depends on 7 being a primitive root mod $10^7$, which you haven't shown. $\endgroup$ – TonyK Jul 24 '14 at 12:40
  • $\begingroup$ @TonyK: no i think you're wrong. Suppose G is a finite abelian group with 7 coprime with |G|. Then consider the group omomorphism $g \to g^7$. Is injective thanks to Lagrange theorem(an element in the kernel would have order that divides two coprime numbers so it has to be 1) and thus surjective(being G finite). So is a bjection, and thus every element is a 7-th power. Indeed it is true for every finite group, and for any number coprime with |G|, instead of 7. $\endgroup$ – res Jul 24 '14 at 12:46

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