^{nd}partition (857) by the 1

^{st}(142) we get the quotient 6, with a remainder of 5.

So the
number 6 is 1 less than the cyclic prime involved i.e. 7, while the
remainder 5 is 2 less.

This is a
property which universally holds.

For example
for the next cyclic prime (17) the 1

^{st}and 2^{nd}partitions are
05882352
and 94117647. And dividing 94117647 by 05882352 we obtain the quotient 16 and the
remainder 15!

More
complicated - though in their own way equally exciting patterns - can be seen
through breaking the repeating full cycle of digits into more than 2 equal
partitions.

One
intriguing case arises when we divide into 4 equal partitions. This requires
the repeating sequence of digits of the cyclic prime to be divisible by 4!

Up to
100 the only candidates that qualify are the cyclic primes 17, 29, 61 and 97.

So starting
with 17 we now divide its 16 repeating digits into the 4 equal partitions

0588 2352
9411 7647

Now beginning
with the last we divide 7647, 9411 and 2352 respectively by 0588 finding in each case the
quotient and remainder.

We then get
in the 3 cases respectively

13 and 3

16 and 3

4 and 0

Now these 6
numbers can be represented in general terms by the following grid

a

_{11 }a_{12}
a

_{21 }a_{22}
a

_{31 }a_{32}
The
following set of relationships now hold (which universally apply in all
such cases):

a

_{21}is always one less than the cyclic prime in question. Therefore in this case of the cyclic prime 17,
a

_{21 }is 16.So,

a

_{21 }= 16 (i.e. 1 less than the cyclic prime)
a

_{11 }+ a_{31 }=_{ }17 (i.e. the cyclic prime)
a

_{21 }–_{ }a_{22 }= a_{11 }(i.e. 13)_{}

a

_{31 }= a_{22 }+ 1 (i.e. 4)
a

_{11 }+_{ }a_{12 }= a_{21 }– a_{32 }(i.e. 16)
a

_{31 }–_{ }a_{32 }= a_{12 }+ 1_{ }(i.e. 4)
a

_{12 }+_{ }a_{32 }= a_{22 }(i.e. 4)
I then calculated the corresponding grid of numbers with
reference to the cyclic primes 29, 61 and 97 respectively.

These are

p = 29 p = 61 p = 97

12 7 11 9 22 17

28 16
60 49 96 74

17 9 50 40 75 57

Then as we
can see the value for a

_{21 }is always one less than the cyclic prime in question. It then becomes like a fascinating variation on a Suduko puzzle to fill in the remaining values on the grid using the general relationships I have listed (which always hold).
A number of
possible incorrect variations however are consistent with the relationships given. This suggests that the values are determined in a simultaneous holistic manner!

However
there is another interesting observation in that the ratios of terms in each
row seem to approximate closely with each other. So the closest relationships possible as
between the ratios will correspond therefore to the correct entries in the
grid.

The next
cyclic prime with recurring digits divisible by 4 is 109.

So it would
be interesting bearing in mind what I have said to predict the correct grid
entries without manually calculating the values.

_{ }
A simpler situation
relates to three partitions which requires that the repeating decimal sequence
of the cyclic prime be divisible by 3.

The first case therefore arises with the cyclic prime 7.

Again we break the repeating digits into 3 successive
partitions, i.e.

14 28 57

Then starting with the last we divide 57 and 28 by 14
respectively, listing again the quotient and remainder in each case.

This time we get 4 and 1 in the first case and 2 and 0 in
the second

So listing in a grid with general form

a

_{11 }a_{12}
a

_{21 }a_{22}
we get,

4
1

2
0

Then for the next applicable cyclic prime 19, the
corresponding grid is

7
4

11
6

Then the cyclic prime 61 the grid is

47
10

13 2

What is clear in each case is that the sum of quotients i.e.

a

_{11 }+_{ }a_{21 }= p – 1
Likewise a

_{11 }+_{ }a_{12 }+ a_{22 }= p_{ }– 2.