How can I solve this find the digits problem more rigorously?

HINT

Let indicate

$$n=739xyz=7\cdot10^5+3\cdot10^4+9\cdot10^3+x\cdot10^2+y\cdot10+z$$

and consider

  • $n\equiv 0 \pmod 6 \implies 4+4x+4y+z\equiv 0 \pmod 6\implies 4x+4y+z=2+6k$
  • $n\equiv 0 \pmod 7 \implies 3+2x+3y+z \equiv0 \pmod 7 \implies 2x+3y+z=4+7h$
  • $n\equiv 0 \pmod 8\implies 4x+2y+z\equiv 0 \pmod 8 \implies 4x+2y+z=8s$
  • $n\equiv 0 \pmod 9 \implies 1+x+y+z \equiv 0 \pmod 9\implies x+y+z=8+9t$

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 $739872$