How can I solve this find the digits problem more rigorously? 
Adjoin to the digits $739$ three more digits so that the resulting number $739 \text{_ _ _}$ is divisible by $6, 7, 8$, and $9$.

I can do some quick guess and check as well as some little tricks with $9$, $7$, $6$, and $8$ to arrive to an answer, but how could I show this with more rigor (like more concrete number theory)?
Answers: $739368$ and $739872$
 A: HINT
Let indicate 
$$n=739xyz=7\cdot10^5+3\cdot10^4+9\cdot10^3+x\cdot10^2+y\cdot10+z$$
and consider


*

*$n\equiv 0 \pmod 6 \implies 4+4x+4y+z\equiv 0 \pmod 6\implies 4x+4y+z=2+6k$

*$n\equiv 0 \pmod 7 \implies 3+2x+3y+z \equiv0 \pmod 7 \implies 2x+3y+z=4+7h$

*$n\equiv 0 \pmod 8\implies 4x+2y+z\equiv 0 \pmod 8 \implies 4x+2y+z=8s$

*$n\equiv 0 \pmod 9 \implies 1+x+y+z \equiv 0 \pmod 9\implies x+y+z=8+9t$

A: Let the number be $[739ABC]$. 
$$\begin{array}
&\hline
&  &739000&[ABC]&[739ABC]&\text{Remarks}\\
\hline
\mod 9 && 1 &8 & 0& A+B+C=9z+8=8,17,26&\cdots (1)\\
\mod 8 && 0 & 0 & 0& [ABC]=8n&\cdots (2)\\
\mod 7 && 3 & 4 & 0&[ABC]=7m+4&\cdots (3)\\
\hline
& & & & &(m,n,z\in\mathbb Z; A+B+C<27)\\
\hline
\end{array}$$
Divisibility by $8$ and $9$ also ensures by $6$, so no need to check further.
Working through$^*$ the above gives
$$[ABC]=368\;\text{or}\; 872$$

*More details
Combining $(2),(3)$ gives $7m=8n-4=4(2n-1)=4\cdot 7\cdot (2q-1)$, i.e. $m=4(2q-1)$. 
Hence $[ABC]=4+7m=8(7q-3) \; (q\in\mathbb Z)$,
i.e. $[ABC]=32,88,144,\cdots,984$.   
Combining with $(1)$ gives 
$9a=7q-4\; (a\in \mathbb Z)$, which only works for $q=7,16$, giving 
$[ABC]=368\text{ or } 872$.
A: Say your digits are $a,b,c$. Since it is divisible by 6,7,8,9 must be divisible by 7, 8, 9 (then divisibility by 6 is implied).


*

*divisibility by 9 implies $7+3+9+a+b+c$ is divisible by 9

*divisibility by 8 implies $100a+10b+c$ is divisible by 8, and also that $c$ is even


Can you derive criterion from divisibility by 7? 
A: A number is divisible by $6,7,8$ and $9$ if and only if it’s divisible by the least common multiple of $6,7,8$ and $9$, which is $504$. So you need to find those numbers between $739000$ and $739999$ that are divisible by (i.e., integer multiples of) $504$. If you divide $739000$ by $504$, you get a remainder of $136$, so the first multiple of $504$ larger than $739000$ is $504-136 = 368$ greater than $739000$, which is $739368$. And $739368+504=739872$ is the next multiple of $504$ after that and still fits the requirement.
