This is an easy problem i guess..but don't know why I haven't made much headway in it. Would appreciate a clear solution t this :).

Find the number of natural numbers strictly not greater than 10000 such that 2^n-n^2 is divisible by 7.

I analyzed using mod upto 20 and found 5 cases positive but after that I found there was no consistent pattern and left it.

I would kindly request one and all to come forward and solve this. Thanks

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## Comments

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TopNewestHi Krishna, here is a solution: Make a table with \(n, 2^n, n^2\) as the headings. Then evaluate these values in mod 7 from n=1 to n=21. When \(2^n \bmod 7 = n^2 \bmod 7\), then \(2^n-n^2\) must be divisible by 7.

You find that \(2^n\) cycles (2,4,1) every three, \(n^2\) cycles (1,4,2,2,4,1,0) every seven. Hence the values of \((2^n-n^2) \bmod 7\) cycle every 21. In every cycle, you find that \(2^n \bmod 7 = n^2 \bmod 7\) 6 times, hence from 1 to 10000 this would happen 10000/21*6 = 2856 times, plus the two more times at n=9998, 10000, hence the answer is 2860.

Hopefully my working was correct, but if it is isn't, I hope you know how to solve it now :)

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But can you prove why?

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Hi! In what way did you approach the problem? Was it not by this method?.... And dude, you must tell me about the Mathcounts too.. how to improve in those kind of problems and al...

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Hi Finn and Krishna, the cycles of three and seven repeat every 21 because 21 is the LCM of 3 and 7. In fact, the LCM of any two coprime integers is their product.

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\[1, 2, 3, 1, 2, 3, \dots\]

is a pattern without knowing the next term? You're just making an educated guess. :P

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This is exactly what I did but I left it in the middle :/..... Thanks a lot @Michael Ng

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Exactly. I did uptil 20 and found 5 cases positive. After a chain till 21, yes there are 6 positive cases. And does this completely prove that if there are two cycles( here- n^2 and 2^n) you have to multiply them to find something like a net cyclicity ( here we multiplied 3 and 7 to get 21).....???

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I'm going to rewrite it with \(\LaTeX\) so I can get a better feel for the problem. :D

\[7 \mid 2^n-n^2\]

And just for clarification, is \(n\) less than \(10000\) or is the expression less than \(10000\)?

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@Daniel Liu @Finn Hulse

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Wow this is an excellent problem. I'll have to take a look at it later though, I have to go to school. :D

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Finn, once you come back don't forget to solve it.@Daniel Liu YOU TOO :p.. And both of you, please tell me how to solve problems in Mathcounts with expertise like you. I couldn't solve even problems like these - " find the number of numbers that divide 6^2014 but not 6^2013." Hoping that you would help :)

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\[2015^2-2014^2=4029 \times 1=4029\]

to get the answer. @Daniel Liu I think I did this right, right? Anyways that's just the number of factors of \(6^{2014}\) minus the number of factors of \(6^{2013}\) so really not too hard. :D

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To summarize, the numbers hat divide \(6^{2014}\) but not \(6^{2013}\) is the number of factors of \(6^{2014}\) minus the number of factors of \(6^{2013}\).

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@Finn Hulse @Daniel Liu - Sorry for tagging you people repeatedly... I wanted to know whether you guys get coached for solving these types of problems..or else how do you do so?

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Also, search up "Competition Math for Middle School" by Batterson. That's also a good book.

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here.

AoPS ship their books worldwide. You can see their booksHowever, international shipping does cost a hefty amount (about 40 USD) so you might want to see if you can get a copy of the book somewhere else.

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@Daniel Chiu for advice just like you but now it's reversed! But as far as advice goes, keep a running tab on what you DON'T know, and try to learn it. Get really down and dirty with one topic, and make it perfect. For example, I just finished with tangent spheres/sectors and stuff dealing with circles. :D

I dunno usually I fail but I've gotten exponentially better. It used to be me askingLog in to reply

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@Finn Hulse - Don't tell me that you have school in May also. Then when would you have vacations? Life in India totally sucks... :( :/

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NO! I just spent an hour writing a huge solution, and then deleted one of my other comments and my solution disappeared! @Silas Hundt I'm so mad! :O

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Sorry @Krishna Ar! :O

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