[ July 10, 2014 ]
Plea for interpreting ninedigitals with a leading zero as pandigitals
Heine Wanderlust (email)

Let a pandigital be any number consisting of each of the ten digits 0 - 9 once and once only
in any order. This definition allows a pandigital to have a lead zero, provided of course it
is the only zero. In that position it is quantitatively dormant, but on multiplication yielding
another pandigital it may acquire quantitative significance if shifted to another position
in the digitstring.

(Can you have a zeroless pandigital? Yes, but then you can have - say - a threeless pandigital
too for that matter: 245678901. Multiply that by 4 and you get another one).

For p = 0123456789, p x the multipliers 1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 20, 22, 23, 25,
26, 31, 32, 34, 35, 40, 41, 43, 44, 50, 52, 53, 61, 62, 70, 71, 80 are all also pandigitals,
(After that p x 81 has repeated digits, and p x 82 has more than 10 digits.)

We can see that to get these pandigitals:

The multiplier must be co-prime to Base – 1, in this case 9. (We could instead say the digits
of the multiplier cannot sum to a multiple of 3). This means when Base – 1 is a prime,
as in base 12 for example, then the equivalent of p (0123456789AB) yields a pandigital
when multiplied by each of the numbers 1 to A, or Base – 2.

The digits of the multiplier must sum to less than Base – 1.

A circular shift on a pandigital retains its multiplicative properties, subject to
the multiple not exceeding 10 digits. For example a circular shift on p,
2345678901 x 4, is also a pandigital.

I certainly am not criticising anybody in particular about lack of clarity on the issue
of ninedigitals or zeroless pandigitals. Rather, I was making the possibly novel
and controversial point that so-called nine digit pandigitals, like 987654321,
are really ten digits long since there's a concealed or commonly omitted 0
at the beginning. This means amongst other things that the quantitatively smallest
pandigital (which I call p) is 0123456789, not 1023456789,
and that a circular shift on p beginning with say 4 goes 4567890123.

0123456789 + 9876543210 = 9999999999

So the smallest and the largest pandigitals sum to the largest possible ten place number.

In fact any pandigital must have an inverse in 9999999999 that is also a pandigital,
but not always in this neat reverse order. For example the inverse of 1026753849
is 8973246150. But another pandigital (which I've already mentioned), 2839450617,
has the inverse 7160549382. The rule seems to be that the differences between
each successive digit must form a palindrome for this inverse reverse to happen.