LTE flavoured divisibility
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Find all natural numbers $n$ such that $7^n$ divides $9^n-1$.
- Fatin Farhan
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Re: LTE flavoured divisibility
$$7^n \mid 9^n-1 \Rightarrow 7 \mid 9^n-1$$.
But $$7 \nmid 9^n-1$$.
As $$n \geq 1$$ there is no such $$n$$.
But $$7 \nmid 9^n-1$$.
As $$n \geq 1$$ there is no such $$n$$.
"The box said 'Requires Windows XP or better'. So I installed L$$i$$nux...:p"
Re: LTE flavoured divisibility
This is wrong. Smallest counterexample: \(n=3\). In fact \(9^n-1\equiv 2^n-1~(\bmod~7)\). And \(2^{3m}=8^m\equiv 1~(\bmod~7)\). So all multiples of \(3\) works (for the case of \(7\)).Fatin Farhan wrote:But $$7 \nmid 9^n-1$$.
As $$n \geq 1$$ there is no such $$n$$.
Fun fact: \(7\mid 2^n-1\) is IMO 1964 Problem 1.
- 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.
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
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- Posts:107
- Joined:Sun Dec 12, 2010 10:46 am
Re: LTE flavoured divisibility
Fatin, how did you guess that $7 \nmid (9^n-1)$?