mathmari
Gold Member
MHB
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Hey! 
Question 1: We consider $\frac{2n-1}{n+7}$. For which $n$ is this term an integer? I have done the following:
We set $n+7=m \Rightarrow n=m-7$.
Then we get $$\frac{2n-1}{n+7}=\frac{2(m-7)-1}{(m-7)+7}=\frac{2m-15}{m}$$ So $m$ has to be a divisor of $15$, i.e. $m\in \{1,3,5,15\}$, therefore $n\in \{-6, \ -4, \ -2, \ 8\}$.
Question 2: Calculate $12673^{37}\pmod 5$. I have done the following:
From Euler's theorem we have $x^4\equiv 1\pmod 5$.
Then we get \begin{align*}12673^{9\cdot 4+1}\pmod 5&\equiv \left (12673^{4}\right )^9\cdot 12673 \pmod 5\\ & \equiv 1^9\cdot 12673 \pmod 5\\ & \equiv 12673 \pmod 5\\ & \equiv \left (2534\cdot 5+3\right )\pmod 5\\ & \equiv 3\pmod 5\end{align*}
Is everything correct and complete? :unsure:
Question 1: We consider $\frac{2n-1}{n+7}$. For which $n$ is this term an integer? I have done the following:
We set $n+7=m \Rightarrow n=m-7$.
Then we get $$\frac{2n-1}{n+7}=\frac{2(m-7)-1}{(m-7)+7}=\frac{2m-15}{m}$$ So $m$ has to be a divisor of $15$, i.e. $m\in \{1,3,5,15\}$, therefore $n\in \{-6, \ -4, \ -2, \ 8\}$.
Question 2: Calculate $12673^{37}\pmod 5$. I have done the following:
From Euler's theorem we have $x^4\equiv 1\pmod 5$.
Then we get \begin{align*}12673^{9\cdot 4+1}\pmod 5&\equiv \left (12673^{4}\right )^9\cdot 12673 \pmod 5\\ & \equiv 1^9\cdot 12673 \pmod 5\\ & \equiv 12673 \pmod 5\\ & \equiv \left (2534\cdot 5+3\right )\pmod 5\\ & \equiv 3\pmod 5\end{align*}
Is everything correct and complete? :unsure: