In the last two posts of this series, I had
investigated the properties of exponents
in different number systems or bases.
To recap, if you take a number, and
exponentiate it, or raise it to a power,
you can know, at a minimum, the last
digit of the power, because the last digit
of a power recurs in a pattern, based
on the initial number's last digit. If this
is a little unclear, which I suspect it is,
take a look back at the preceding posts.
Now, back to my investigation- I had
found a pattern in base 10, which was
the main reason for starting this series,
and this then led me to find a pattern in
binary, or base 2.
This post is to do the same, if possible,
for base 3 and base 4.
Base 3 is a little more complicated than
binary, as I can't draw an easy parallel
between ternary numbers and decimal
numbers, as I did in the last post with
base 2.
But what I realise now is that the little
connection I made in the last post was
due to the fact that odd and even numbers
rely on their modulus with 2. This may
seem very obvious, but when I was
comparing bases in the last post, it seemed
like a lucky coincidence.
A modulus with 2, or X mod 2, is a
way to write, represent or calculate the
remainder when a number is divided by
2. A mod B, for example, would be the
remainder when A is divided by B.
The last digit of a number in base 3 can
either be 0, 1 or 2. This digit represents
the modulus of the number in base 10,
with respect to 3. For example, the number
11213 is equal to 43 in base 10. Forty-three,
when divided by 3, leaves a remainder of 1.
This makes the task much easier - now we
can evaluate the modulus of a number with
respect to 3, and then relate it to the last
digit of a ternary number. The last digit of
the subsequent exponents, which is the aim
of this exercise, can then be found.
Okay, let's start with the first possibility:
the ternary number ends in 0. This means
that the equivalent decimal number is divisible
by 3. And since any multiple of 3, when
raised to any power, is still a multiple of 3,
the modulus of the powers will also be 0,
and therefore in ternary, the last digit of the
power is also 0.
0 >> 0 (Loop Length of just 1)
Next, let's tackle the numbers ending in one.
They can be represented as (3x +1). Now,
if we expand (3x +1)n using the Binomial
Theorem, then we find that every term, except
for one, will be a multiple of 3. I won't elaborate
on how that is found here. Since the term
which is not a multiple of 3 will determine the
modulus and therefore the last digit of the power,
it is worth examining. In this case, it will
always be 1n. Since this is just 1, the modulus
is 1, and the last digit of the subsequent power,
in base 3, will be 1.
1 >> 1 (Loop Length of just 1)
Finally, we reach the last possibility:
Last Digit = 2. Instead of writing such a
number as (3x +2) in decimal, as was done
in the last case, I will write it as (3x -1).
So (3x -1)n would hinge on the term of (-1)n.
This term fluctuates between 1 and -1,
depending on whether n is odd or even.
Since this alternates, it can be seen that
the last digit of the power goes from 2, to
1, to 2, to 1 again.
2 >> 1 >> 2 >> 1 >> ...
Therefore, for ternary or base 3, the
following table can be drawn up.
investigated the properties of exponents
in different number systems or bases.
To recap, if you take a number, and
exponentiate it, or raise it to a power,
you can know, at a minimum, the last
digit of the power, because the last digit
of a power recurs in a pattern, based
on the initial number's last digit. If this
is a little unclear, which I suspect it is,
take a look back at the preceding posts.
Now, back to my investigation- I had
found a pattern in base 10, which was
the main reason for starting this series,
and this then led me to find a pattern in
binary, or base 2.
This post is to do the same, if possible,
for base 3 and base 4.
Base 3 is a little more complicated than
binary, as I can't draw an easy parallel
between ternary numbers and decimal
numbers, as I did in the last post with
base 2.
But what I realise now is that the little
connection I made in the last post was
due to the fact that odd and even numbers
rely on their modulus with 2. This may
seem very obvious, but when I was
comparing bases in the last post, it seemed
like a lucky coincidence.
A modulus with 2, or X mod 2, is a
way to write, represent or calculate the
remainder when a number is divided by
2. A mod B, for example, would be the
remainder when A is divided by B.
The last digit of a number in base 3 can
either be 0, 1 or 2. This digit represents
the modulus of the number in base 10,
with respect to 3. For example, the number
11213 is equal to 43 in base 10. Forty-three,
when divided by 3, leaves a remainder of 1.
This makes the task much easier - now we
can evaluate the modulus of a number with
respect to 3, and then relate it to the last
digit of a ternary number. The last digit of
the subsequent exponents, which is the aim
of this exercise, can then be found.
Okay, let's start with the first possibility:
the ternary number ends in 0. This means
that the equivalent decimal number is divisible
by 3. And since any multiple of 3, when
raised to any power, is still a multiple of 3,
the modulus of the powers will also be 0,
and therefore in ternary, the last digit of the
power is also 0.
0 >> 0 (Loop Length of just 1)
Next, let's tackle the numbers ending in one.
They can be represented as (3x +1). Now,
if we expand (3x +1)n using the Binomial
Theorem, then we find that every term, except
for one, will be a multiple of 3. I won't elaborate
on how that is found here. Since the term
which is not a multiple of 3 will determine the
modulus and therefore the last digit of the power,
it is worth examining. In this case, it will
always be 1n. Since this is just 1, the modulus
is 1, and the last digit of the subsequent power,
in base 3, will be 1.
1 >> 1 (Loop Length of just 1)
Finally, we reach the last possibility:
Last Digit = 2. Instead of writing such a
number as (3x +2) in decimal, as was done
in the last case, I will write it as (3x -1).
So (3x -1)n would hinge on the term of (-1)n.
This term fluctuates between 1 and -1,
depending on whether n is odd or even.
Since this alternates, it can be seen that
the last digit of the power goes from 2, to
1, to 2, to 1 again.
2 >> 1 >> 2 >> 1 >> ...
Therefore, for ternary or base 3, the
following table can be drawn up.
| Last digit | | | Last digit |
| of initial number | | | of subsequent powers |
| 0 | | | 0 |
| 1 | | | 1 |
| 2 | | | 1, 2, 1,... |
From the above, we can see that the loop lengths
are in 1_1_2. Not so symmetrical. Disappointing,
to say the least. The last two bases I considered
were bases 2 and 10. Both even-numbered bases.
Maybe the pattern only occurs here, because in
both these cases, the pattern in the loop lengths
has been recurring twice. 1-4-4-2-1 and -1- being
the repeating pattern in the loop lengths. In the
next post, I will leave out the intermediate steps
needed in finding out the loop lengths, and instead
leave that to you, the readers, if you so wish. Then,
will it be found whether the pattern found so far
holds true for other bases.
But, of course, that would still only be a very minor
part of a true mathematical investigation. After
finding a few cases that are compliant with a pattern,
the next step is to show that all similar cases also
do the same. I hope to do that in Part 5, perhaps the
final part in this interesting series.
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