By adding the component parts of numbers we can obtain their rank
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