eg 217 = 2 + 1 + 7 = 10 = 1 + 0 = 1

32961 = 3 + 2 + 9 + 6 + 1 = 21 = 2 + 1 = 3

incidentally adding 9 to a number leaves it unchanged (also it is divisible by 3 and 9)

you can see from the above that what were base 10 numbers are reduced to base 9

with the exception of zero itself

Ranks showing numbers multiplied together

so for example if we multiplied 27 by 31 these have ranks of 9 and 4 respectively

0___ 1 2 3 4 5 6 7 8 9 .... <- numbers from 1 to 9

1___ 1 2 3 4 5 6 7 8 9.... <- ranks obtained by multiplication

2___ 2 4 6 8 1 3 5 7 9.... <- ditto

3___ 3 6 9 3 6 9 3 6 9

4___ 4 8 3 7 2 6 1 5 9

5___ 5 1 6 2 7 3 8 4 9

6___ 6 3 9 6 3 9 6 3 9

7___ 7 5 3 1 8 6 4 2 9

8___ 8 7 6 5 4 3 2 1 9

9___ 9 9 9 9 9 9 9 9 9

the result of mutiplying 27 * 31 = 837 = 8 + 3 + 7 = 18 = 9 as predicted in the table above

There are many symetries and inverse symetries of the first 8 numbers in this table ..

also multiplying by 9 always results in 9

Note that the squares of numbers (i.e the main diagonal left to right)

are limited to the following values

1 2 3 4 5 6 7 8 9 <- number to be squared base 9

1 4 9 7 7 9 4 1 9 <- result base 9

so we can say that 2345 = 2 + 3 + 4 + 5 = 14 = 5

cannot be the square of a whole number(integer)

whereas 256 = 4 could be (and is), 259 = 7 could be (and isnt)

Note also that these properties dont just apply in number base 10

but in all bases .. its the first and unstated law of numbers)

OK now lets try and do something usefull with the idea

Lets try deducing the square root of 961 using this system :-

Using a base 9 system .. i.e adding up the numbers

we get 961 = 9+6+1 = 16 = 1+6 = 7

we could just have easily have divided by 9

the remainder being 7

[ 7 = 961(mod 9) is how its usually written]

the sums of squares in a base 9 system

as we have already seen can be tabulated as :-

1 2 3 4 5 6 7 8 9 <- number to be squared base 9

1 4 9 7 7 9 4 1 9 <- result base 9

so 961 = 7 means that it must have come from

numbers of rank 4 or 5 i.e. 9k+4 or 9k+5

the likely numbers

are 9k+4 .... 4, 13, 22, 31, 40, 49, 58 etc

and 9k+5 .... 5, 14, 23, 32, 41, 50, 59 etc

I could take a guess since I know that 20 squared is about 400

and that 40 squared is about 1600

that the number must be one of 31 or 32

I could also try something a bit more systematic

In this example it is like looking at the at digit

of 961 and seeing what 2 numbers could make it

This is the same as using a number base 10 in the same way

that we used the 9 base above

can be tabulated as :-

1 2 3 4 5 6 7 8 9 10 <- number to be squared base 10

1 4 9 6 5 6 9 4 1 0 <- result base 10

961 base 10 = 1 so it can only result from a rank 1 or 9 number

the likely numbers

are 10k+1 .... 1, 11, 21, 31, 41, 51, 61 etc

and 10k+9 .... 9, 19, 29, 39, 49, 59, 69 etc

In this case there are 4 commonalities with the whole set

above these are 31, 41, 49 and 59

If I applied the same logic as before 20 being too small and 40

too large then there is patently only one solution

This has worked quite well for 961 as its a small number

but with a number such as 96100 using a base 9 and 10 system

would give a lot more feasible solutions located near 310

Can you think of a way that you might reduce the number of likely

possibilities .. there are several.

Cubic equations are somewhat trickier to solve

If we look at a tabulation of cubes base 9 we find

1 2 3 4 5 6 7 8 9 <- number to be cubed base 9

1 8 9 1 8 9 1 8 9 <- result base 9

which is a suprisingly limited set of results

[multipes of 9 +or- 1 in a sequence + - = + - ..]

If we look at the sum of 2 cubes

then we find the possible outcomes are

1+8=9, 1+1=2, 8+8=7, 9+anything stays the same

i.e. the outcomes are 1,2,7,8,9

Thats a very limited set of results

You might try to apply this to Fermats Last theorem

A very interesting subject in analysis is the nature

of rational [can be made from a ratio of numbers eg 1/3]

and irrational numbers [eg the square root of 2]

Rational numbers have a reciprocating sequence

eg 1/7 the sequence is 142857 you will notice that

this is 1 shorter than the length of the divisor 7

and that the rank of the sequence is 9

If we limit ourselfes to prime numbers

[nos that can only be divided by 1 and themself]

we find that some numbers have shorter sequences

such as 41 = 02439:

and these are always an integer division

of 1 less than the divisor 41 -1 = 40 in this case

[would you like to hazard a guess as to wether this

is true in number bases other than 10]

In many cases the resulting sequences are reflexive

about their mid point eg 1/13 = 076923:

where this is the case ... 076 and 923

the sums of the corresponding digits is 9

0+9=9 7+2=9 6+3=9

These features are a result of the relationship

of the number with the base