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When looking at mathematical sequences some might find cycles in these
sequences only interesting when whole numbers return again and again
after a fixed number of terms (the loop length).
In practice, however, such sequences are the exceptions.
What sequences usually show is a digital cycle in which only the
final digits are repeated in loops of, say, 6, 8, 12 or 20 terms long,
dependent on the numeral system used.
In sequences in which the numbers grow longer the number of final digits
which become part of a cycle will also grow larger.
In a binary system, for instance, a 3-term final one-digit
cycle will then be superceded by a 6-term final two-digit
cycle, this one in turn by a 12-term final three-digit cycle, this one in
turn by a 24-term final four-digit cycle, this one in turn by a 48-term
final five-digit cycle, and so on and so forth.
Of course, the five-digit cycle was there from the very beginning, but it
will take some time to detect it.
Now, almost everyone familiar with numbers in the base-10 or 'denary'
system will know that a number of which the final digit is 0 is divisible
by 10, a number of which the final digit is 5 is divisible by 5 and that
an even number (ending in 0, 2, 4, 6, or 8) is divisible by 2.
These facts demonstrate that single final digits do or can contain
information, regardless of the total length of the number in digits.
But what other information about divisibility do or can final digits
contain, not only in the denary system, but also in numeral systems with a
smaller base number?
The following table should give a general answer to this question:
| FACTUAL INFORMATION CONTAINED
IN FINAL DIGIT(S) |
| |
|
WHOLE NUMBER DIVISIBLE BY |
| SYSTEM |
|
2 | |
3 | |
4 | |
5 | |
6 | |
7 | |
8 | |
9 | |
10 |
| Base 10 |
|
0, 2, 4,
6, 8 |
|
-- |
|
00, 04,
..., 96 |
|
0, 5 |
|
-- |
|
-- |
|
000, 008,
..., 992 |
|
-- |
|
0 |
| Base 9 |
|
-- |
|
0 |
|
-- |
|
-- |
|
-- |
|
-- |
|
-- |
|
0 |
|
-- |
| Base 8 |
|
0, 2,
4, 6 |
|
-- |
|
0, 4 |
|
-- |
|
-- |
|
-- |
|
0 |
|
-- |
|
-- |
| Base 7 |
|
-- |
|
-- |
|
-- |
|
-- |
|
-- |
|
0 |
|
-- |
|
-- |
|
-- |
| Base 6 |
|
0, 2, 4 |
|
0, 3 |
|
-- |
|
-- |
|
0 |
|
-- |
|
-- |
|
-- |
|
-- |
| Base 5 |
|
-- |
|
-- |
|
-- |
|
0 |
|
-- |
|
-- |
|
-- |
|
-- |
|
-- |
| Base 4 |
|
0, 2 |
|
-- |
|
0 |
|
-- |
|
-- |
|
-- |
|
00 |
|
-- |
|
-- |
| Base 3 |
|
-- |
|
0 |
|
-- |
|
-- |
|
-- |
|
-- |
|
-- |
|
00 |
|
-- |
| Base 2 |
|
0 |
|
-- |
|
00 |
|
-- |
|
-- |
|
-- |
|
000 |
|
-- |
|
-- |
|
Note: in the base-10 system any number of which the last two digits
interpreted as a number in itself (4, 8, 12, 16 until 96) are divisible by
4 is itself divisible by 4, because 100 and multiples of 100 are divisible
by 4; and any number of which the last three digits interpreted as a
number in itself (8, 16, 24 until 96, and 104 until 992) are divisible by
8 is itself divisible by 8, because 1000 and multiples of 1000 are
divisible by 8.
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You may wonder why one cannot even see whether a number is odd or even in
the base-9, the base-7, the base-5 and the base-3 systems, the odd-numbered
systems.
To explain why we should first have a closer look at the final one-digit
cycles in the sequences which are used in the place-value notations of
numbers.
They are of the following form, in which b is the base number:
xn = bn, n≥0, b0 = 1,
b1 = b
The positional notation itself turns the sequence into a finite series
with a separate coefficient for each power term, while putting the largest
quantity on the left, the smaller ones in descending order on the right:
s = a0bn + a1bn-1+...+
an-1b1 + anb0
As usual the value for n=0 in the sequence may or may not play a role in
the detection of cycles.
In this case the value 1 for n=0 is fully acceptable for b=3, b=7 and b=9,
but it distorts the picture for the other base numbers and we shall leave
it out.
We shall also leave out here the power sequence of a base-number raised to
a power n and expressed in itself.
The sequence of the denary 10 raised to the power 1, 2, 3, 4 et cetera,
which gives 10, 100, 1000, 10000, et cetera in the denary system, is such
a rather uninteresting example.
The minimum requirement for the recognition of cyclicity is two full
cycles, or perhaps even two cycles plus the first term of the third one.
Therefore, we should not stop before having seen the first nine terms of
these base-number sequences:
| BASIC POWER
SEQUENCES |
| |
|
BASE NUMBER |
| N |
|
2 | |
3 | |
4 | |
5 | |
6 | |
7 | |
8 | |
9 |
| 1 |
|
2 |
|
3 |
|
4 |
|
5 |
|
6 |
|
7 |
|
8 |
|
9 |
| 2 |
|
4 |
|
9 |
|
16 |
|
25 |
|
36 |
|
49 |
|
64 |
|
81 |
| 3 |
|
8 |
|
27 |
|
64 |
|
125 |
|
216 |
|
343 |
|
512 |
|
729 |
| 4 |
|
16 |
|
81 |
|
256 |
|
625 |
|
1296 |
|
2401 |
|
4096 |
|
6561 |
| 5 |
|
32 |
|
243 |
|
1024 |
|
3125 |
|
7776 |
|
16807 |
|
32768 |
|
59049 |
| 6 |
|
64 |
|
729 |
|
4096 |
|
15625 |
|
46656 |
|
117649 |
|
262144 |
|
531441 |
| 7 |
|
128 |
|
2187 |
|
16384 |
|
78125 |
|
279936 |
|
823543 |
|
2097152 |
|
4782969 |
| 8 |
|
256 |
|
6561 |
|
65536 |
|
390625 |
|
1679616 |
|
5764801 |
|
16777216 |
|
43046721 |
| 9 |
|
512 |
|
19683 |
|
262144 |
|
1953125 |
|
10077696 |
|
40353607 |
|
134217728 |
|
387420489 |
|
Note: all numbers are in denary notation.
Sets of one, two or four numbers between horizontal lines and with their
last digits in red denote second final one-digit cycles as the
first set of the same length does not yet prove anything by itself.
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The above table clearly shows that the powers of the odd base numbers are
always odd themselves.
The sum of an even number of odd numbers is even, but the sum of an odd
number of odd numbers remains odd.
So the final digit of an unknown number longer than one digit does not tell
us anything in this respect, that is, if it is expressed in a numeral
system with an odd base number.
Since the digital cyclicity of a sequence provides us with the final digit
or digits only (sometimes also with an initial digit) and not with the
entire number, we cannot apply a whole lot of divisibility rules either
(such as that an integer is divisible by 7 if the number obtained by
multiplying its last 'digit' by 2 and subtracting the result from the
number formed by the remaining digits can be evenly divided by 7).
And then, one cannot afford to apply a rule which makes use of one or more
final digits to a number in a different numeral system than the rule is
made for.
Thus, the number 5 is 101 in the binary system and therefore a
number ending in (0)101 in the binary system may have the looks of
a denary number ending in 5, and may seduce us into believing that it is
divisible by 5 or 101.
But, as as matter of fact, a number ending in 101 can only in a
minority of cases be divided by 101.
It can only be divided by 101 if the entire binary number consists
of nothing else than sets of digits which interpreted as numbers are
themselves divisible by 101, such as 0, 101,
1010 and 1111, for then
s = 101·1 + A·10^11 +
A·10^B + A·10^B + ... in which A refers to
the place for a number, represented by one or more digits, which is
divisible by 101 and B to the place for the number of digits
following the digits of A
(in other words, denary
s = 5·1 + A·8 + A·2^B + A·2^B + ... in
which A is a place for a multiple of 5 and B the place for a number equal
to or larger than 3, the length of 5 in the binary system).
The table below should illustrate this for the first thirty-three
multiples of 101:
| SEQUENCE OF MULTIPLES OF
5 OR 101 |
| N |
|
BASE 10 |
|
BASE 2 |
|
N |
|
BASE 10 |
|
BASE 2 |
| 0 |
|
0 |
|
0 |
|
16 |
|
80 |
|
1010000 |
| 1 |
|
5 |
|
101 |
|
17 |
|
85 |
|
1010101 |
| 2 |
|
10 |
|
1010 |
|
18 |
|
90 |
|
1011010 |
| 3 |
|
15 |
|
1111 |
|
19 |
|
95 |
|
1011111 |
| 4 |
|
20 |
|
10100 |
|
20 |
|
100 |
|
1100100 |
| 5 |
|
25 |
|
11001 |
|
21 |
|
105 |
|
1101001 |
| 6 |
|
30 |
|
11110 |
|
22 |
|
110 |
|
1101110 |
| 7 |
|
35 |
|
100011 |
|
23 |
|
115 |
|
1110011 |
| 8 |
|
40 |
|
101000 |
|
24 |
|
120 |
|
1111000 |
| 9 |
|
45 |
|
101101 |
|
25 |
|
125 |
|
1111101 |
| 10 |
|
50 |
|
110010 |
|
26 |
|
130 |
|
10000010 |
| 11 |
|
55 |
|
110111 |
|
27 |
|
135 |
|
10000111 |
| 12 |
|
60 |
|
111100 |
|
28 |
|
140 |
|
10001100 |
| 13 |
|
65 |
|
1000001 |
|
29 |
|
145 |
|
10010001 |
| 14 |
|
70 |
|
1000110 |
|
30 |
|
150 |
|
10010110 |
| 15 |
|
75 |
|
1001011 |
|
31 |
|
155 |
|
10011011 |
|
|
|
|
|
|
32 |
|
160 |
|
10100000 |
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Note: the sets of two, four, eight and sixteen numbers between horizontal
lines with their last digits in red denote full second final one-, two-,
three- and four-digit cycles.
From n=20 to n=32 the last two digits of the senary number are shown in
blue, because this is only the beginning of a (second) twenty-term final
two-digit cycle.
For the same reason the last five digits of the binary number are shown in
blue for n=32, because we may assume (without further empirical proof)
that this is the beginning of a (second) thirty-two-term final five-digit
cycle.
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We may not be able to derive factual information about the
divisibility by 101 of a binary number ending in 101, yet we
can say something about the probability of such a number being
divisible by 101.
In general that probability is 1/5=20% for all numbers, so it might be
interesting if that percentage deviates for numbers ending in 101.
Of course, if we had to know in practice whether a number can be evenly
divided by 5, we would do better by sticking to the denary system, but that
is not the point here.
The point here is that the last one or more digits of a number contain
factual and/or statistical information.
And how much or of what kind will depend on the numeral system.
If we know the length of the number which ends in 101, or if we are
able to come up with an estimate of its length which is accurate enough, we
can calculate the chance of its being divisible by 101.
For a number of 9 digits long, for example, the first 6 digits are the
variable part.
The number of variations with 0 and 1 is then 2^6, but the numbers
themselves will not start with 0, so the number of possible variations is
only half of that: 2^5=32.
In the multiples-of-5 sequence table above we find 6 binary six-digit
numbers which are divisible by 101, so the probability of a binary
9-digit number ending in 101 and being divisible by 101 is
6/32=18.8%, which is significantly smaller than the general average.
Here are the results for binary positive integers from 3 to 10 digits
long:
PROBABILITY OF
DIVISIBILITY OF
A BINARY INTEGER ENDING IN 101 |
LENGTH
in digits |
|
DIVISIBILITY BY 101
probability (with explanation) |
| 3 |
|
100% |
(101 itself) |
| 4 |
|
0% |
(0/1) |
| 5 |
|
0% |
(0/2) |
| 6 |
|
25% |
(1/4) |
| 7 |
|
25% |
(2/8) |
| 8 |
|
25% |
(4/16) |
| 9 |
|
18.8% |
(6/32) |
| 10 |
|
20.3% |
(13/64) |
One or more final digits can sometimes contain all the information we are
interested in, but there is no guarantee whatsoever.
It also depends on the numeral system whether the number presented in
that system gives us the information we want in a slower or in a faster
way.
Thus, the quaternary system is a very handy system for seeing whether a
number is divisible by 2, 4 or 8 -- much handier than the
denary system.
Yet, when it is divisibility by 3 and by 9 we are after (something for
which we need the whole number in both systems) the denary system
distinguishes between the sum of the (numbers formed by the single) digits
being divisible by 3 and the sum of the 'digits' being divisible by 9.
In the former case the number is indeed divisible by 3, in the latter case
by 9 as well.
In the quaternary system, however, there is only one such divisibility
rule: if the sum of the (numbers formed by the single) digits is divisible
by 3, the number itself may be divisible by 3, 6 or 9 (when we confine
ourselves to the first ten positive integers).
You may check the sums of the base-4 digits in the following table:
FIRST THIRTY-THREE
QUATERNARY INTEGERS
DIVISIBLE BY 3 (3), 12 (6) OR 21
(9) |
| N3 |
N6 |
N9 |
|
BASE 10 |
|
BASE 4 |
|
N3 |
N6 |
N9 |
|
BASE 10 |
|
BASE 4 |
| 0 |
0 |
0 |
|
0 |
|
0 |
|
16 |
8 |
|
|
48 |
|
300 |
| 1 |
|
|
|
3 |
|
3 |
|
17 |
|
|
|
51 |
|
303 |
| 2 |
1 |
|
|
6 |
|
12 |
|
18 |
9 |
6 |
|
54 |
|
312 |
| 3 |
|
1 |
|
9 |
|
21 |
|
19 |
|
|
|
57 |
|
321 |
| 4 |
2 |
|
|
12 |
|
30 |
|
20 |
10 |
|
|
60 |
|
330 |
| 5 |
|
|
|
15 |
|
33 |
|
21 |
|
7 |
|
63 |
|
333 |
| 6 |
3 |
2 |
|
18 |
|
102 |
|
22 |
11 |
|
|
66 |
|
1002 |
| 7 |
|
|
|
21 |
|
111 |
|
23 |
|
|
|
69 |
|
1011 |
| 8 |
4 |
|
|
24 |
|
120 |
|
24 |
12 |
8 |
|
72 |
|
1020 |
| 9 |
|
3 |
|
27 |
|
123 |
|
25 |
|
|
|
75 |
|
1023 |
| 10 |
5 |
|
|
30 |
|
132 |
|
26 |
13 |
|
|
78 |
|
1032 |
| 11 |
|
|
|
33 |
|
201 |
|
27 |
|
9 |
|
81 |
|
1101 |
| 12 |
6 |
4 |
|
36 |
|
210 |
|
28 |
14 |
|
|
84 |
|
1110 |
| 13 |
|
|
|
39 |
|
213 |
|
29 |
|
|
|
87 |
|
1113 |
| 14 |
7 |
|
|
42 |
|
222 |
|
30 |
15 |
10 |
|
90 |
|
1122 |
| 15 |
|
5 |
|
45 |
|
231 |
|
31 |
|
|
|
93 |
|
1131 |
|
|
|
|
|
|
|
|
32 |
16 |
|
|
96 |
|
1200 |
Divisibility by 6 means divisibility by 3 AND divisibility
by 2, and the latter feature is in the quaternary system only a
question of the last digit being 0 or 2.
So, if the last digit in the table above is 0 or 2, the
quaternary number is divisible by 6.
But
i* cannot find or think of
a similar rule which distinguishes quaternary integers which are divisible
by 9 from the other quaternary integers which are divisible by 3.
Here, i will have to leave the initiative to the reader.
The information contained in final digits may be a mixed bag, it is not
an empty bag.
67.MNW
| * |
The first-person singular pronoun is
spelled with a small i, as i do not consider myself a Supreme
Being or anything else of that Ilk. |
|
