Polydivisible Number/Examples/381,654,729

Theorem
The integer $381 \, 654 \, 729$ is the only polydivisible number which is pandigital in the sense of excluding zero.

Proof
First it is demonstrated that indeed $381 \, 654 \, 729$ has this property:

It remains to be demonstrated that it is the only such number.

Let $\sqbrk {abcdefghi}$ be a polydivisible number which is pandigital.

By Divisibility by 5:
 * $e = 5$

By Divisibility by 2:
 * $b, d, f, h$ are even.

Hence:
 * $a, c, g, i$ are odd.

By Divisibility by 3:
 * $a + b + c$, $d + 5 + f$, $g + h + i$ are divisible by $3$.

By Divisibility by 8:
 * $\sqbrk {fgh}$ is divisible by $8$.

Because $f$ is even:
 * $\sqbrk {f00}$ is divisible by $8$.

We must therefore have:
 * $\sqbrk {gh}$ divisible by $8$.

Thus:
 * $\sqbrk {gh}$ must be one of $16, 32, 72, 96$.

Because $i$ is odd and $g + h + i$ are divisible by $3$:
 * $\sqbrk {ghi}$ must be one of $321, 327, 723, 729, 963$.

By Divisibility by 4:
 * $\sqbrk {cd}$ is divisible by $4$.

Because $c$ is odd:
 * $d$ must be $2$ or $6$.

Suppose $h = 6$.

Then $\sqbrk {ghi} = 963$ and $d = 2$.

Since $d + 5 + f$ is divisible by $3$ and $f$ is even:
 * $f = 8$

The remaining even number $b$ must be $4$.

So:
 * $\sqbrk {abcdefghi} = \sqbrk {a4c258963}$

Thus the only remaining options for $a$ and $c$ are $1$ and $7$.

However, we find that neither $1472589$ nor $7412689$ is divisible by $7$.

Therefore we must have:
 * $h = 2$

It follow therefore that:
 * $d = 6$

Because $d + 5 + f$ is divisible by $3$, and $f$ is even:
 * $f = 4$

The remaining even number $b$ must be $8$.

So:
 * $\sqbrk {abcdefghi} = \sqbrk {a8c654g2i}$

Since $a + 8 + c$ is divisible by $3$:
 * $\sqbrk {a8c}$ must be one of $183, 189, 381, 387, 783, 789, 981, 987$.

Suppose $g = 3$.

Then:
 * $\sqbrk {a8c}$ must be one of $189, 789, 981, 987$.

Inspection of the possible numbers reveals that none of $1896543$, $7896543$, $9816543$, $9876543$ is divisible by $7$.

Suppose $g = 7$.

Then:
 * $\sqbrk {a8c}$ must be one of $183, 189, 381, 981$.

Among $1836547$, $1896547$, $3816547$, $9816547$, only $3816547$ is divisible by $7$.

The results follow.