Polydivisible Number/Examples/381,654,729

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Theorem

The integer $381 \, 654 \, 729$ is the only polydivisible number which is penholodigital.


Proof

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

\(\ds 3\) \(=\) \(\ds 1 \times 3\)
\(\ds 38\) \(=\) \(\ds 2 \times 19\)
\(\ds 381\) \(=\) \(\ds 3 \times 127\)
\(\ds 3816\) \(=\) \(\ds 4 \times 954\)
\(\ds 38 \, 165\) \(=\) \(\ds 5 \times 7633\)
\(\ds 381 \, 654\) \(=\) \(\ds 6 \times 63 \, 609\)
\(\ds 3 \, 816 \, 547\) \(=\) \(\ds 7 \times 545 \, 221\)
\(\ds 38 \, 165 \, 472\) \(=\) \(\ds 8 \times 4 \, 770 \, 684\)
\(\ds 381 \, 654 \, 729\) \(=\) \(\ds 9 \times 42 \, 406 \, 081\)

$\Box$


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

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

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.

$\blacksquare$


Historical Note

David Wells reports in his $1997$ work Curious and Interesting Numbers, 2nd ed. that this factoid was posted in the The Sunday Times as their brainteaser no. $1040$, by a contributor going by the name Upton, but further details have not been uncovered.


Sources

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