Ancient Chinese divination numbers problem

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Lars Bo

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Joined
Oct 5, 2013
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4
Dear Forum
I am working on the translation of an ancient Chinese divination
text.
But I got stuck with a mathematical problem because math was never my
strong point.
About 2000 years ago in ancient China they had a popular method
of divining by numbers produced by counting a bunch of 49 thin sticks in a
certain way.
The method always yielded 6, 7, 8 or 9 which were interpreted
as the divination answer.
However, nearly 1000 years earlier it seems they
used a different method of calculation which always yielded 5, 6, 7 or 8.
I
can understand and use the first method which is described in an ancient text.
My problem is that I can’t figure out how in earlier times they counted to
reach 5, 6, 7 or 8.
Where is the appropriate place to post this question?
Explanations in pdf:
https://www.dropbox.com/s/jk9yuqthe6i065v/Simpel explanation of counting sticks.pdf
https://www.dropbox.com/s/zdgke9s6d8s6djv/Explanation with a little more detail-.pdf
 
Lars Bo said:
Dear Forum
I am working on the translation of an ancient Chinese divination
text.
But I got stuck with a mathematical problem because math was never my
strong point.
About 2000 years ago in ancient China they had a popular method
of divining by numbers produced by counting a bunch of 49 thin sticks in a
certain way.
The method always yielded 6, 7, 8 or 9 which were interpreted
as the divination answer.
However, nearly 1000 years earlier it seems they
used a different method of calculation which always yielded 5, 6, 7 or 8.
I
can understand and use the first method which is described in an ancient text.
My problem is that I can’t figure out how in earlier times they counted to
reach 5, 6, 7 or 8.
Where is the appropriate place to post this question?
Explanations in pdf:
Click to expand...

Mr. Bo, We will not open a pdf file from a stranger. So if you want our assistance regarding this puzzle, you will have to copy and paste the document into this forum.
 
The simple explanation

Divination by counting sticksThe procedure of counting sticks is done three times; the result will always be either 6, 7, 8 or 9:
Step 1.
49 sticks are divided at random in two heaps. One stick is set aside.1
The left-hand heap is then counted out by fours into the right hand, until 1, 2, 3 or 4 sticks remain. These 1, 2, 3 or 4 are discarded.
The sticks now held in the right hand are put back on the table.
The other heap is then taken up in the left hand and counted out by fours, leaving 1, 2, 3 or 4 sticks to be discarded.
The heap on the table is put back together with the heap in the hand.
5 or 9 sticks have been discarded. 40 or 44 sticks remain.
Step 2.
The remaining 40 or 44 are again divided at random in two heaps. One stick is set aside.
The left-hand heap is then counted out by fours into the right hand, until 1, 2, 3 or 4 sticks remain. The remainder is discarded.
The sticks now held in the right hand are put back on the table.
The other heap is then taken up in the left hand and counted out by fours, leaving 1, 2, 3 or 4 sticks to be discarded.
There will now be 40, 36, or 32 sticks left on the table.
Step 3.
These 40, 36 or 32 sticks are again divided into two heaps and one stick set aside.
The left-hand heap is then counted out by fours into the right hand, until 1, 2, 3 or 4 sticks remain. The remainder is discarded.
The sticks now held in the right hand are put back on the table.
The other heap is then taken up in the left hand and counted out by fours, leaving 1, 2, 3 or 4 sticks to be discarded.
36, 32, 28 or 24 sticks remain on the table.
Step 4.
The remaining 36, 32, 28 or 24 sticks are counted out in fours. There will be no remainder.
The result will be 6, 7, 8 or 9.
My question is:
What number of sticks should be used?
And how many should be counted instead of fours to reach 5, 6, 7 or 9?
1 I assume that the single stick which is set aside each time helps to keep track of the three counting operations.
 
A slightly different explanataion

1 The first interim countThe 49 sticks are divided at random and placed on the table in two heaps. One stick, taken from the right-hand heap, is put to the side.
The left-hand heap is then taken in the left hand and counted out by fours into the right hand, until 1, 2, 3 or 4 sticks remain
in the left hand. These 1, 2, 3 or 4 are discarded. The sticks now held in the right hand are put back on the table by themselves.
The right-hand heap is then taken up in the left hand and counted out by fours, leaving 1, 2, 3 or 4 sticks to be discarded. The sticks now in the right hand are put back on the table with the left-hand heap.
(In practice, it is not necessary to count the second heap. Once the discard from the first heap is known, the discard from the second heap can be predicted. If the discard from the first heap is 4, the total discard must be 9; if it is 1, 2 or 3, the total discard must be 5.)
Either 5 or 9 sticks have been discarded, because they can only have occurred as 1 +1 +3,
1 +2+2, 1 +3+1, (each giving a total of 5) or 1 +4+4 (a total of 9). These are now set apart. 44 or 40 sticks remain on the table.
2 The second interim count
The 40 or 44 sticks left lying on the table are now gathered together and randomly divided into two heaps, which are laid side by side. One stick is taken from the right-hand heap and put aside. The rest of the two heaps is counted out by fours in the same manner as the original 49.
(Again it is not necessary to count out the second heap. If the discard from the first heap is 1 or 2, the total discard is bound to be 4; if the first discard is 3 or 4, the total must be 8.)
This time the number of discarded sticks will be 8 or 4, because they must have occurred as 1 +1 +2, 1 +2+1 (each giving a total of 4), 1 +3+4, or 1 +4+3 (each giving a total of 8).
There will now be 40, 36, or 32 sticks left on the table.
3 The third count
The 40, 36 or 32 sticks left in use are divided into two heaps and one stick put aside. Both heaps are told off by fours until 8 or 4 have been discarded between the fingers. 36, 32, 28 or 24 sticks remain on the table.
4 Final step
The remaining 36, 32, 28 or 24 sticks are counted out in fours. There will be no remainder, because 36, 32, 28 and 24 are all divisible by 4. The quotient will be 9, 8, 7 or 6
 
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