For which n is the term an integer & Calculate the equivalence

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Discussion Overview

The discussion revolves around two mathematical questions: the conditions under which the expression $\frac{2n-1}{n+7}$ is an integer for various values of $n$, and the calculation of $12673^{37} \pmod 5$. The scope includes mathematical reasoning and verification of calculations.

Discussion Character

  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant proposes that for the expression $\frac{2n-1}{n+7}$ to be an integer, $n$ must take values derived from the divisors of 15, resulting in $n \in \{-6, -4, -2, 8\}$.
  • Another participant suggests that additional values for $m$ should include -1, -3, -5, and -15, implying that the initial set of values for $n$ may be incomplete.
  • For the second question, multiple participants confirm the correctness of the calculation involving $12673^{37} \pmod 5$, referencing Euler's theorem.

Areas of Agreement / Disagreement

Participants generally agree on the correctness of the second question's solution, while there is disagreement regarding the completeness of the values for $n$ in the first question, with additional values proposed by some participants.

Contextual Notes

The discussion highlights the need for careful consideration of all possible divisors in the first question, indicating that assumptions about the completeness of the solution may be limited.

mathmari
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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:
 
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Question 2 is correct.

In question 1 we need to take -1, -3, -5, -15 as additional values of m
 
kaliprasad said:
Question 2 is correct.

In question 1 we need to take -1, -3, -5, -15 as additional values of m

Ah yes! Except from that everything else is correct and compelete, right?
 
yes
 
kaliprasad said:
yes

Great! Thank you! ☺
 

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