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 $739872$
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HINT
Let indicate
$$n=739xyz=7\cdot10^5+3\cdot10^4+9\cdot10^3+x\cdot10^2+y\cdot10+z$$
and consider
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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).
Can you derive criterion from divisibility by 7?
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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 $[ABC]=368\text{ or } 872$.
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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.
Well, some quick eliminating.
If a number is divisible by $9$ and $8$ then it is divisible by $3$ and $2$ and therefore $6$. So we don't have to worry about $6$. If we can find a number that is divisible by $8$ and $9$, it will be divisible by $6$.
$8|1000$ so $8$ divides $739000$ so we just have to find $abc$ to be divisible by $8$.
Using the "sum of digits have same remainder with divided by $9$". $739000 \equiv 7+ 3 + 9 \equiv 1 \mod 8$. So $739000 + abc \equiv 1 + abc \mod 9$.
So we need $abc \equiv -1 \equiv 8 \mod 9$.
And $739000 \equiv 7000000 + 39000 \mod 7 \equiv 35000 +4000 \mod 7 \equiv 3500 + 500 \equiv 490 + 10 \equiv 7 + 3\equiv 3 \mod 7$. So $739000 + abc\equiv 3 + abc \mod 7$. So we need $abc \equiv - 3 \equiv 4 \mod 7$.
So we need $abc \equiv 0 \mod 8$ and $abc \equiv 4 \mod 7$ and $abc \equiv 8 \mod 9$.
We can use the chinese remainder theorem.
To get $abc \equiv 4 \mod 7$ and $abc \equiv 0 \mod 8$ we must find $m = 0 + 8k = 4 + 7j \mod 7*8$. By trial (trial being comparing $4,11,18,25,32,39,46,53$ to $0, 8, 16, 24, 32, 40, 48$) and error (error being $4,11,18,25,39,46,53,0, 8, 16, 24, 40, 48$ don't work) we find that $32$ is the correct answer.
By chinese remainder theorem $abc \equiv 32 \mod 56$ is the only solution mod $56$.
Bear with me: $56 \equiv 2 \mod 9$ (by the "sum of digits of a multiple of $9$ is a multiple of $9$") and $32 \equiv 5 \mod 9$ (ditto) so $32 +k*56 \equiv 5 + k*2 \mod 9$. So $32 + 6*56 = 368\equiv 8 \mod 9$ and $368$ is a) divisible by $8$, b) has remainder $4$ when divided by $7$ and c) has remainder $8$ when divided by $9$.
To find other numbers where this is true we must add $7*8*9 = 504$.
So the two numbers are $739368$ and $739872$