<s>[INST] Middle school number theory Find at least three numbers that satisfy all three conditions:
(1) there is a remainder of $1$ when the number is divided by $2$;
(2) there is a remainder of $2$ when the number is divided by $3$;
(3) there is a remainder of $3$ when the number is divided by $4$.
Since the LCM($2, 3, 4$) is $12$, all of them can divide $12$ without giving a remainder.
So, $12 - 1=11$ gives 
a remainder of $1$ when divided by $2$,
a remainder of $2$ when divided by $3$,
a remainder of $3$ when divided by $4$.
Similarly $12*2-1=23$
and $12*3-1=35$
How would I explain to middle school students the "pulling out of a hat" of 
first the $LCM-1$,
and secondly the $LCM*(n)-1$ parts?
 [/INST] The simple answer could just be write out the sets; i.e, 
$S_1=\{...,-3,\color{red}{-1},1,3,5,7,9,\color{red}{11},13,15,17,19,21,\color{red}{23},25...\}$
$S_2=\{...,-4,\color{red}{-1},2,5,8,\color{red}{11},14,17,20,\color{red}{23},26,...\}$
$S_3=\{...,-5,\color{red}{-1},3,7,\color{red}{11},15,19,\color{red}{23},27,...\}$.
As noted in the other answer, immediately we see that $-1,11$, and $23$ satisfy the conditions.  From there, we can note that perhaps every $12$th number also satisfies.  Then note that $LCM(2,3,4)=12$.  Finally construct your formula as $LCM\cdot{n}-1$.
I tried not to use any modular arithmetic since they are middle schoolers, so I hope this helps.
$\mathbf{EDIT}$  After thinking about it, the last thing i would do is verify that the number $12n-1$ leaves the appropriate remainders when divided by $2,3,$ and $4$.  Convert $12n-1$ to $12n+11$ for arithmetic's sake and show
$$\frac{12n+11}{2}=6n+5+\frac{1}{2}$$
$$\frac{12n+11}{3}=4n+3+\frac{2}{3}$$
$$\frac{12n+11}{4}=3n+2+\frac{3}{4}$$
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