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I start solving this by trying $2k_1+1=3k_2+2$, but it doesn't make any sense, because I still have to use $4k_3+3$.

If I sum all three equations together, I get $3n=(2k_1+3k_2+4k_3)+6$. This may be, but how can I determine the factors $k_1, k_2 $and$\ k_3$?

thunder
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    Hint: you can rewrite this as $n\equiv -1 \pmod {2,3,4}$. Though, in truth, the numbers here are so small that trial and error is probably just as fast. – lulu Mar 27 '20 at 19:56
  • Sorry, high school student here. Is there another way solving this without using modular arithmetic? – thunder Mar 27 '20 at 20:03
  • Sure, trial and error. Really, that just takes a few seconds. But...you don't need modular arithmetic to follow my hint. All I said is that your conditions can be rewritten to say that $n$ is, simultaneously, $1$ less than a multiple or each of $2$, $3$, and $4$. – lulu Mar 27 '20 at 20:20

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First you have to solve the syatem: $$\left\{\begin{matrix} 2x+1=3y+2 \\ 3y+2=4z+3 \end{matrix}\right.$$ The sultions are: $$x=6k+5 \vee y=4k+3 \vee z=3k+2, k\in Z$$ We have to find the value of $k$ for which $12k+11$ is prime, so $k=0$. We can conclude that the smallest $n$ is $11$.

Matteo
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