## May 15, 2014

### Horse 1676 - Pandigital Numbers and the Divisibility Test

Following on from the last post about sudoku, a though hit me that every single line and column of a sudoku because they always contain the numbers 1-9, must always tally to 45. This led me to a rather strange sort of conclusion.
Every 9-digit and 10-digit pandigital number (that is, a number that uses all of the digits) must have a digit sum of 45. Also as a result of this, they must all be divisible by 9 since one of the handy divisibility tests is that a the digit sum of any multiple of 9 is also divisible by 9.

Pick any pandigital number:
832,974,156
8 + 3 + 2 + 9 + 7 + 4 + 1 + 5 + 6 = 45
4 + 5 = 9
9 is one less than 10.

Try it yourself. Pick any number you like, multiply it by 9 and see what you get. For example:

11861 * 9 = 106,749
1 + 0 + 6 + 7 + 4 + 9 = 27
2 + 7 = 9
9 is one less than 10 (base-10).

Spin out!

It also stands to reason that any pandigital number either added to or subtracted from another pandigital number must leave a result which is also also divisible by 9 because addition is commutative.

To wit:
527,148,936 - 483,269,517 = 43,879,419
4 + 3 + 8 + 7 + 9 + 4 + 1 + 9 = 45
4 + 5 = 9
9 is one less than 10 (base-10).

Even More spin out!

There must be something even more fundamental going on below the surface; something weirder. I don't believe that there can be anything specifically unique about the number 9 because the divisibility test also works for 3.

229, 167 * 3 = 687,501
6 + 8 + 7 + 5 + 0 + 1 = 27
2 + 7 = 9
9 is one less than 10 (base-10).

9? is 3²

Found you. Thought you could sneak away could you?

For base-10, 10 is 9 + 1. 10 is also 3²+1. The general rule if it exists must be that for a given base n, the divisibility test must work for all n-1's to some power x.

Make the whole thing smaller. Lets pick... I dunno... 6 as the base and add up the digits of a pandigital number in base-6.

251,340
2 + 5 + 1 + 3 + 4 + 0 = 23
2 + 3 = 5
5 is one less than 10 (base-6).

Don't understand? That's probably because you're thinking in base-10 land.
The number 10 is actually: 1*10¹ + 0*10°. 10 in base-10 is one-ten and zero-units.
The number 23 in base 6 is actually: 2*6¹ + 2*6°. 23 in base six is two-sixes and three-units.

We're going to need some new symbols. How about X for ten (because we're Roman) and E for eleven. We've been here before (See Horse 785 (or in base-10 1109))

508,E67,41X,239
5 + 0 + 8 + E + 6 + 7 + 4 + 1 + X + 2 + 3 + 9 = 56
5 + 6 = E
E is one less than 10 (base-12).

Just to make sure of this:
5973 * E = 32033
3 + 2 + 0 + 3 + 3 = E
E is one less than 10 (base-12).

Bringing this all together.

My conjecture, if it isn't already a proven fact is that for any given base-n, the multiples of any given number n-1, must also add up to a multiple of n-1. Likewise since pandigital numbers which use every digit once, always are a multiple of n-1, then they too with have a digit sum of some multiple of n-1.

So then for base-36:
A = 10, B = 11, C = 12, D = 13... until I ran out of normal characters.

One pandigital number at random is
RIC,2HQ,59P,AF4,7OE,GSU,NBJ,M3D,LX1,6TW,Z80,VYK (which by the way is 10,700,559,216,665,253,478,593,063,215,400,573,113,142,093,263,213,829,432,580,898,768,839,245,824 in base-10)

Then the sum of the digits is...
R + I + C + 2 + H + Q + 5 + 9 + P + A + F + 4 + 7 + O + E + G + S + U + N + B + J + M + 3 + D + L + X + 1 + 6 + T + W + Z + 8 + 0 + V + Y = HI
H + I = Z
Z is one less than 10 (base-36).
Or in base 10... 17 + 18 = 35.

AHAH!
Huzzah, huzzah, Have I just won the game of maths? I hope so. My brain hurts.