I dont know how to start with this questions.It is an olympiad sum.
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2Hint: consider, more generally, $3^{2^n}-1$. Induction should help. – Wojowu Sep 10 '16 at 14:05
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This question has shown up before in different guises. It is a common exercise in elementary number theory and/or group theory that powers of $3$ cover exactly one half of the odd residue classes modulo $2^k$ for any $k>2$. Equivalently, the order of $3$ modulo $2^k$ is $2^{k-2}$. I am not sure about the existence of an exact duplicate though. – Jyrki Lahtonen Sep 10 '16 at 14:18
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1But I am a bit disappointed about the way you phrased this question. If you are preparing for a math contest you will need to be prepared to face problems where you don't know how to start. Then you need to do some testing to get an idea of how the land lies. To that end you need to identify some features of the question that may or may not be relevant, and check if they might help. Here the fact that the exponent is a power of two is such a feature. So check what happens when the exponent is a smaller power of two. You will then observe what most of the answerers are getting at. – Jyrki Lahtonen Sep 10 '16 at 14:30
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See for example here. Looks like I considered this a should-be-a-duplicate already four years ago :-) – Jyrki Lahtonen Sep 10 '16 at 14:40
4 Answers
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Hint: note that $3^{4096}-1$ is a difference of squares. Factor it and consider the factors.
Ross Millikan
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HINT:
Observe that the highest power of $2$ that divides $3^2-1$ is $3$
Now $3^{2n}=(1+8)^n\equiv1\pmod8\implies3^{2n}+1\equiv2\pmod8\equiv2\pmod4$
So, the highest power of $2$ that divides $3^{2n}+1$ is $1$ for integer $n\ge0$
So, the highest power of $2$ that divides $3^{(2^2)}-1=(3^2-1)(3^2+1)$ will be $3+1$
So, the highest power of $2$ that divides $3^{(2^3)}-1=(3^2-1)(3^2+1)(3^4+1)$ will be $3+1+1$
Observe that $4096=2^{12}$
lab bhattacharjee
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In the first line there is ambiguity, you mean $2^3$, I thought it was 3. – P Vanchinathan Sep 10 '16 at 14:47
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@PVanchinathan, I wrote power , not the exact number. Sorry for the confusion. – lab bhattacharjee Sep 10 '16 at 14:48
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HINT.-$4096=2^{12}$ and $$3^{2^{12}}-1= (2+1)^{2^{12}}-1=\sum_{k=0}^{2^{12}-1}\binom{2^{12}}{k}$$
Piquito
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Using Binomial expansion for $3^{4096}$, we get
$$3^{4096} - 1 = (1 + 2)^{4096} - 1 = (1 + \binom{4096}{1}\cdot2 + \binom{4096}{2}\cdot 2^2 + ... ) - 1 = \binom{4096}{1}\cdot 2 + \binom{4096}{2}\cdot 2^2 + \dots$$
Therefore, we can say that $2^{13}$ divides $3^{4096} - 1$ making $13$ as the highest positive integer.
Ramil
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