How many positive integers \(n\) satisfy the condition \( 2^n + n \mid 8^n + n\)?

How many positive integers \(n\) satisfy the condition \( 2^n + n \mid 8^n + n\)?

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TopNewest@Calvin Lin- could you please give an entire solution to this? – Sanjib Kumar Tripathy · 6 months ago

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i did it already. please post the entire solution.i am quite...disturbed by this probelem – Sanjib Kumar Tripathy · 6 months ago

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@Calvin Lin-i really have no idea how to go further. Could you please post a solution to this? – Sanjib Kumar Tripathy · 6 months ago

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So we were able to remove the \( 8^n \) and replaced it with a \( n \times 4 ^n \). How can we remove the \( 4^n \) now to further simplify the statement? – Calvin Lin Staff · 6 months ago

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@Calvin Lin-so sorry! it is 2^n+n| 2^n(2^n+1)(2^n-1) – Sanjib Kumar Tripathy · 6 months ago

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– Calvin Lin Staff · 6 months ago

That's a good start. Actually, \( 8^n\) is "more complicated" than \(n\). How can we "eliminate exponents" to only obtain polynomials?Log in to reply

@Calvin Lin-sir, please wait because i need to get the answer today. I dnt have anyone here else to help me out. – Sanjib Kumar Tripathy · 6 months ago

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@Calvin Lin-2^n+n|(2^n-1)(2^n +n ) – Sanjib Kumar Tripathy · 6 months ago

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This doesn't quite look right to me. I now do not know what you are doing. Please list out your steps carefully, and do not just jump to the final step. – Calvin Lin Staff · 6 months ago

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@Calvin Lin-i did it already, multiplied and subtracted to to get (2^n-1)(2^n +n )|2^n+n. all i dnt know is how to get a polynomial out of it – Sanjib Kumar Tripathy · 6 months ago

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– Calvin Lin Staff · 6 months ago

Are you sure the RHS is 2^n+n? That seems to be the LHS. At this point in time, I'm not certain what you're doing.Log in to reply

i did it already, multiplied and subtracted to to get (2^n-1)(2^n+n )|2^n+n. all i dnt know is how to get a polynomial out of it – Sanjib Kumar Tripathy · 6 months ago

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@Calvin Lin-can you make me understand how to get a polynomial from this? – Sanjib Kumar Tripathy · 6 months ago

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IE We have a \( 2^{2n} ( \times n) \) term. How can we get rid of it? – Calvin Lin Staff · 6 months ago

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(2^n-1)(2^n-1)|2^n+n – Sanjib Kumar Tripathy · 6 months, 1 week ago

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(2^2n)n-n|2^n+n – Sanjib Kumar Tripathy · 6 months, 1 week ago

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– Calvin Lin Staff · 6 months, 1 week ago

Great. So once again, " \( 2^{2n} \times n \) is more complicated than \( 2^n \)". How can we simplify further?Log in to reply

i tried that too. I multiplied 2^2n to 2^n+n and subtracted 8^n+n from that and proceeded as above. What i dont get is what to do next. – Sanjib Kumar Tripathy · 6 months, 1 week ago

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– Calvin Lin Staff · 6 months, 1 week ago

What is the equation that you get? How can we simplify it further?Log in to reply

@CalvinLin-this was my approach. If 2^n+n devides 8^n+n then it devides 8^n+n-2^n+n.I wanted to eliminate n. Now i arrive at finding 2^n+n devides (2^n)(2^n+1)(2n-1). Here i am stuck to find the answer – Sanjib Kumar Tripathy · 6 months, 1 week ago

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– Calvin Lin Staff · 6 months, 1 week ago

That's a good start. Actually, \( 8^n\) is "more complicated" than \(n\). How can we "eliminate exponents" to only obtain polynomials?Log in to reply

i tried dat way too. I believe i am leaving a minute detail. Can ya please send me a solution? – Sanjib Kumar Tripathy · 6 months, 1 week ago

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– Calvin Lin Staff · 6 months, 1 week ago

You can write up your solution and obtain feedback from the community on how to improve it.Log in to reply

through brute force 1,2 come. but what canbe a logical way to do this? – Sanjib Kumar Tripathy · 6 months, 1 week ago

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I tried. cant get a clue to start – Sanjib Kumar Tripathy · 6 months, 1 week ago

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Diophantine equation, which approach can we use? – Calvin Lin Staff · 6 months, 1 week ago

Since we have aLog in to reply

What have you tried? Where are you stuck? – Calvin Lin Staff · 6 months, 1 week ago

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