I am trying to solve some practice problems and found this one which I can't solve:
Find all the positive integers such that $n+7$ is divisible by $3n-1$
So far I have tried using the usual properties such as $a|b$ then $a|bc$ ; $a|b$ and $a|c$ then $a|bx+cy$ ; etc but I can't get to anything useful. Just by looking, apparently the answers should be 1 and 4 but that's all I can get.
Here on mathstack I was able to find some similar ones but that only have a number (i.e: 7|[whatever expression] but I don't see how to relate them to mine)
If possible, please avoid using congruence for the solution.
Best Answer
While there are definitely approaches with heavier artillery that will work on more broad versions of this problem, sometimes recognising that you get away with using a peashooter is an important skill as well.
First note that for both positive $a \vert b$ means $a \leq b$, so $3n-1 \leq n+7 \implies n\leq 4$ and there are only four cases to check.
For $n=1$, we get $\frac{8}{2}=4$.
For $n=2$, we get $\frac{9}{5}=1.8$.
For $n=3$, we get $\frac{10}{8}=1.25$.
For $n=4$, we get $\frac{11}{11}=1$.
If you also want to check zero, we get $\frac{7}{-1}=-7$. But if you allow negative number inputs, there is more required.