[ *July 10, 2014* ]

Plea for interpreting ninedigitals __with a leading zero__ as pandigitals !

by 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.