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Here is a Olympiad Problem and i have a solution for it already , please tell me know if i will get full marks for my solution or not (i think my solution is short than official solution)? You can also post your alternative solutions ^_^

Let x=0.$a_1$$a_2$$a3$... where $a_i$ is 1 if nos. of positive divisors of $i$ is even and 0 if they are odd .Prove that x is irrational .

My solution is : $a_i$ has odd nos of divisors iff i is a perfect square else it has even nos of divisors . Thus , $a_i$ is 0 iff $i$ is perfect square else it is 1. Also a nos is rational if in its decimal representation , there is periodicity in it. But difference/gap between two perfect square goes on increasing , ie gap between two 0s after decimal goes on increasing and thus periodicity is not possible . Hence x is irrational .

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I like it! I'm not sure if your solution here is an outline or the full proof that you are going to submit. For submission, you should rigorously justify the remarks that you have made here. –  Euler....IS_ALIVE Feb 3 '13 at 1:50
@MaggiIggam: You would need to justify the remarks about the number of divisors, and you would also need to justify that the gap between any two perfect squares is strictly increasing. No matter how trivial these may seem, they're part of a full solution. –  Clayton Feb 3 '13 at 2:07
There are some minor imprecisions. If a number is rational, then its decimal representation is ultimately periodic. (Your version states the result in the other direction, true but not what we need here.) Also, what you intend to say is that the gap between consecutive squares is increasing. It is a good argument overall. However, it might be a very early question in the contest, and those are often marked in a quite picky way. The imprecisions may well bring you down substantially. –  André Nicolas Feb 3 '13 at 2:16
@Clayton: I am not so sure about justifying, for example, the remark about the number of divisors. A contest kid knows this sort of thing. As long as the quoted result is stated correctly, one would not necessarily expect a proof. –  André Nicolas Feb 3 '13 at 2:38
@AndréNicolas: I've not participated in too many contests ($2$ I think), so I wasn't sure what would be typical to know versus not being rigorous enough. Thanks for the tip! –  Clayton Feb 3 '13 at 2:57

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