Let $k, n \in \mathbb{Z}$.
How can I find a solution for this problem?
Also, where are some resources to solve a similar problem? I came across this question when attempting to find a solution to $2^n \equiv 0 \mod 12$.
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3Can $3$ divide the left hand side? – Randall Feb 24 '19 at 20:04
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To solve a similar problem, review divisibility and prime number decomposition, i.e., $12=2^2\cdot 3$ and $2^n=2\cdots 2$, with $12\mid 2^n$. – Dietrich Burde Feb 24 '19 at 20:26
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@DietrichBurde In other words, never, since $2^n$ doesn't contain any $3$'s. – Jossie Calderon Feb 24 '19 at 20:31
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Yes, exactly. And "resources" are just texts on elementary number theory for this. They all deal with divisibility and prime numbers. – Dietrich Burde Feb 24 '19 at 20:40
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If $$2^n=12k$$ then $$2^{n-2}=3k$$ $$2^{n-2}\equiv 0 \mod{3}$$ $$(-1)^{n}\equiv 0 \mod{3}$$ $$0\in\{-1,1\}$$ Which is not true - contradicts the original equation.
Peter Foreman
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1Well I just proved why $3$ does not divide $2^n$ if it wasn't obvious to the OP. – Peter Foreman Feb 24 '19 at 20:29