**WARNING - MATHS AHEAD!**

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.**

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.

How about base-12?

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.

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