Six-digit split decision

In my country this problem is written as $\overline{abcdef} ⋮ (\overline{abc}\times \overline{def})$ so here is my solution: $\overline{abcdef} ⋮ (\overline{abc}\times \overline{def})\Rightarrow {\begin{cases}\overline{abcdef}⋮\overline{abc}\\\overline{abcdef}⋮\overline{def}\end{cases}}$

$\Rightarrow {\begin{cases}(\overline{abc000}+\overline{def})⋮\overline{abc}\\(\overline{abc000}+\overline{def})⋮\overline{def}\end{cases}}$ $\Rightarrow {\begin{cases}\overline{def}⋮\overline{abc}\\\overline{abc000}⋮\overline{def}\end{cases}}$ . This is because $\overline{abc000}\div \overline{abc}=1000$ and $\overline{def}\div \overline{def}=1$ .

Because $\overline{def}⋮\overline{abc}$ , let $\overline{def}=k \times \overline{abc}$ ( $k$ is a positive integer). We know that $\overline{abc}\ge 100$ and $\overline{def}\le 999$ , this implies that $k<10$ or $k \le 9$ .

Because $\frac{\overline{abc000}}{\overline{def}}\in N$ , then $\frac{\overline{abc}\times 1000}{k \times \overline{abc}}\in N$ or $\frac{1000}{k}\in N$ $\Rightarrow 1000 ⋮k$ . However $k \le 9$ as stated above, we have $k\in \left\{1;2;4;5;8\right\}$ satisfies both condition. There are three cases:

Case 1: $k=1$ or $\overline{def}=\overline{abc}$ . We know that $\overline{abc}=100a+10b+c$ and same goes for $\overline{def}$ .

We have $\overline{abcdef}=\overline{abc000}+\overline{abc}=100000a+10000b+1000c+100a+10b+c=1001(100a+10b+c)$ .

Also note that $\overline{abc}\times \overline{def}=\overline{abc}\times \overline{abc}=\left(100a+10b+c\right)^2$ .

Because $\frac{\overline{abcdef}}{\overline{abc}\times \overline{def}}\in N$ (implied from the problem), $\frac{1001(100a+10b+c)}{(100a+10b+c)^2}\in N$ or $\frac{1001}{100a+10b+c}\in N$ , then $1001⋮\overline{abc}$ . The only three-digit divisor of 1001 is 143, satisfies $\overline{abc}\ge 100$ , so $\overline{abc}=143$ .

So the first six-digit number that satisfies all the conditions is 143143 (correct because $143143⋮(143 \times 143)$ ), but we will continue to see if there are more numbers.

Case 2: $k=2$ or $\overline{def}=2 \times \overline{abc}$ .

We have $\overline{abcdef}=\overline{abc000}+\overline{def}=\overline{abc}\times 1000 + 2 \times \overline{abc}=1002\times \overline{abc}$ .

Also note that $\overline{abc}\times \overline{def}=\overline{abc}\times 2 \times \overline{abc}=(\overline{abc})^2 \times 2$ .

Because $\frac{\overline{abcdef}}{\overline{abc}\times \overline{def}}\in N$ , $\frac{1002\times \overline{abc}}{(\overline{abc})^2 \times 2}\in N$ , or $\frac{501}{\overline{abc}}\in N$ , then $501⋮\overline{abc}$ . There are two three-digit divisors of 501, they are 167 and 501, but if $\overline{abc}=501$ then $\overline{def}=1002$ , this is incorrect. However if $\overline{abc}=167$ then we know that the result is 167334 , this matches the condition from the problem, so we move on to the final case.

Case 3: $k=5$ or $\overline{def}=5 \times \overline{abc}$ .

We have $\overline{abcdef}=\overline{abc000}+\overline{def}=\overline{abc}\times 1000 + 5 \times \overline{abc}=1005\times \overline{abc}$ .

Also note that $\overline{abc}\times \overline{def}=\overline{abc}\times 5 \times \overline{abc}=(\overline{abc})^2 \times 5$ . Because $\frac{\overline{abcdef}}{\overline{abc}\times \overline{def}}\in N$ , $\frac{1005\times \overline{abc}}{(\overline{abc})^2 \times 5}\in N$ , or $\frac{201}{\overline{abc}}\in N$ , then $201⋮\overline{abc}$ . The only three-digit divisor of 201 is 201, this will make $\overline{def}=5 \times \overline{abc}=201 \times 5 =1005$ , which is not a three-digit number. So case 3 has no solutions.

EDIT: As stated above, $1000⋮k$ and $k \le 9$ , but I forgot to mention two more cases of $k$ , in this case $k=4$ and $k=8$ will make it 5 cases total. However, both $k=4$ and $k=8$ has no solutions (you can try to prove it yourself, similar to case 3), so luckily the result doesn't change.

We conclude that the largest six-digit number satisfies the condition as stated in the problem is $\boxed{167334}$ .