ONTC Final Exam

Discussion on Bangladesh National Math Camp
Epshita32
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Re: ONTC Final Exam

Unread post by Epshita32 » Wed Sep 02, 2015 10:23 am

In p3 , let's have two odd numbers - 2n+1 and 2p+1 . Their sum = 2n+2p+2 = 2(n+p+1) . Suppose 2n+2p+2 is divisible by 2 , but not 4 . So n+p+1 is not divisible by 2 . The difference = 2n-2p = 2(n-p) . n+p+1 is not divisible by 2 . So n+p must be divisible by 2 . That can be possible in two ways . Either both n and p are even or both of them are odd . If both of them are odd , then n-p = odd-odd = even . Or if both of them are even , n-p = even - even = even . Either way n-p is an even and therefore divisible by 2 . Let n-p = 2*k . Difference = 2*(n-p) = 2*2*k =4*k . So 2*(n-p) is divisible by 4 . Is this correct ? :idea:

Nirjhor
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Re: ONTC Final Exam

Unread post by Nirjhor » Wed Sep 02, 2015 1:47 pm

Epshita32 wrote:In p3 , let's have two odd numbers - 2n+1 and 2p+1 . Their sum = 2n+2p+2 = 2(n+p+1) . Suppose 2n+2p+2 is divisible by 2 , but not 4 . So n+p+1 is not divisible by 2 . The difference = 2n-2p = 2(n-p) . n+p+1 is not divisible by 2 . So n+p must be divisible by 2 . That can be possible in two ways . Either both n and p are even or both of them are odd . If both of them are odd , then n-p = odd-odd = even . Or if both of them are even , n-p = even - even = even . Either way n-p is an even and therefore divisible by 2 . Let n-p = 2*k . Difference = 2*(n-p) = 2*2*k =4*k . So 2*(n-p) is divisible by 4 . Is this correct ? :idea:
Yes.
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.


Revive the IMO marathon.

Epshita32
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Re: ONTC Final Exam

Unread post by Epshita32 » Wed Sep 02, 2015 3:44 pm

In p8 , the number is irrational . Cause prime numbers never end . Can anyone give me the proof of why they never end ? :?

rah4927
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Re: ONTC Final Exam

Unread post by rah4927 » Wed Sep 02, 2015 6:52 pm

Epshita32 wrote:In p8 , the number is irrational . Cause prime numbers never end . Can anyone give me the proof of why they never end ? :?
The criterion for the irrationality of a number isn't that there are an infinite number of digits after the decimal. Take an easy example, like $\dfrac{1}{3}=.333\cdots$.

The actual condition for (ir)rationality is different. A number is rational if and only some sequence of digits repeat continuously again and again and again....

You can learn about rationality and irrationality from many resources, even The Secondary Textbook of Bangladesh.

So what you actually have to show in P8 is that if all the prime numbers are all written down in a row, then no sequence of digits can repeat continuously. A good way to show that is by considering what the last few digits of prime numbers can be. But my solution doesn't use elementary tools, and so I consider my solution to be a bad one.

Epshita32
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Re: ONTC Final Exam

Unread post by Epshita32 » Wed Sep 02, 2015 7:14 pm

Rah4927 , I know about what you wrote . Never ending was my first step . Disorder in sequence would have been the next . I just kind of couldn't do it and confused me well . Well , and thank you .

rah4927
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Re: ONTC Final Exam

Unread post by rah4927 » Wed Sep 02, 2015 7:55 pm

Epshita32 wrote:Rah4927 , I know about what you wrote . Never ending was my first step . Disorder in sequence would have been the next . I just kind of couldn't do it and confused me well . Well , and thank you .
To get a feel for how you should deal with this type of problem, take a look here-(http://math.stackexchange.com/questions ... irrational). It might feel a little complicated in spite of the fact that it isn't.

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Masum
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Re: ONTC Final Exam

Unread post by Masum » Wed Sep 02, 2015 8:49 pm

আমাদের ক্লাস এইটের বইয়ে পৌনপুনিক নিয়ে কিছু আলোচনা ছিল। ওখান থেকে, এই লেমাটা প্রমাণ করার চেষ্টা কর আগেঃ
একটা সংখ্যা মূলদ হয় যদি ও কেবল যদি দশমিকের পরের(যদি কিছু থাকে) অংশে কোন পর্যায়ক্রমিক ধারা থাকে। যেমন, ১৫.২৪২২৪২২৪২২...
খেয়াল কর যে, এখানে পর্যায়টা নির্দিষ্ট। মানে যত কিছুই কর, এটা পরিবর্তন হবে না। এই পেরেকটাকে অবশ্য একটা শক্তিশালি হাতুড়ি দিয়ে শায়েস্তা করা লাগে।
One one thing is neutral in the universe, that is $0$.

rah4927
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Re: ONTC Final Exam

Unread post by rah4927 » Wed Sep 02, 2015 10:22 pm

Masum wrote:আমাদের ক্লাস এইটের বইয়ে পৌনপুনিক নিয়ে কিছু আলোচনা ছিল। ওখান থেকে, এই লেমাটা প্রমাণ করার চেষ্টা কর আগেঃ
একটা সংখ্যা মূলদ হয় যদি ও কেবল যদি দশমিকের পরের(যদি কিছু থাকে) অংশে কোন পর্যায়ক্রমিক ধারা থাকে। যেমন, ১৫.২৪২২৪২২৪২২...
খেয়াল কর যে, এখানে পর্যায়টা নির্দিষ্ট। মানে যত কিছুই কর, এটা পরিবর্তন হবে না। এই পেরেকটাকে অবশ্য একটা শক্তিশালি হাতুড়ি দিয়ে শায়েস্তা করা লাগে।
Is there no elementary solution to this problem? I have been trying to work with the last few digits of primes but I can't seem to solve this problem without resorting to sledgehammers.

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Masum
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Re: ONTC Final Exam

Unread post by Masum » Wed Sep 02, 2015 11:48 pm

Yes, there is an elementary solution. But it uses Cyclotomic Polynomial.
One one thing is neutral in the universe, that is $0$.

rah4927
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Re: ONTC Final Exam

Unread post by rah4927 » Thu Sep 03, 2015 12:11 am

Masum wrote:Yes, there is an elementary solution. But it uses Cyclotomic Polynomial.
Hmm, my solution was to use Drichlet to show that the final digits of a prime can be $11\cdots 1$ and that the number of $1$'s can be increased arbitrarily high and so no sequence of integers can ever repeat(unless all the primes at one point become repunit primes but that's not possible by Bertrand's Postulate, and also because there are infinitely many primes of the form $4k+1$, which has an elementary proof.I forgot to mention this in my solution, damn it!).

I am guessing the solution using cyclo is to first prove a special case of Drichlet (the $p\equiv 1\pmod n$ case) and then basically apply the same idea. But I was talking about something more elementary, because it just seems obvious that the decimal can't be recurring.
Last edited by rah4927 on Thu Sep 03, 2015 10:58 am, edited 1 time in total.

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