Proving that the difference between any two anagram numbers is always a multiple of nine

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Proving that the difference between any two anagram numbers is always a multiple of nine
First, in regards to Rilx, he was wrong on several accounts. First he did the algebra wrong. Then he left out some steps. Then he indicated that an infinity of digits would be involved which isn't true (and I'm going to prove it).
To start this off, let's suppose we have these two threedigit numbers: abc and cab where a, b and c can take on any values from 0 to 9 (this is sufficient to guarantee we're covering any threedigit number that you like in base 10). Now rewrite those anagram numbers as follows:
100a + 10b + c
100c + 10a + b
Taking the difference, we get:
100(a  c) + 10(b  a) + (c  b)
Next, we break down this expression as follows:
(a  c) + 99(a  c) + (b  a) + 9(b  a) + (c  b)
Rearranging terms, we get:
(a  c) + (b  a) + (c  b) + 99(a  c) + 9(c  b)
Cancelling like terms, we're left with:
99(a  c) + 9(c  b)
This proves that my theorem works for any threedigit anagram numbers as the last expression is evenly divisible by nine.
In my next post I will extend this proof to any two anagram numbers with any amount of digits.
PhilX
To start this off, let's suppose we have these two threedigit numbers: abc and cab where a, b and c can take on any values from 0 to 9 (this is sufficient to guarantee we're covering any threedigit number that you like in base 10). Now rewrite those anagram numbers as follows:
100a + 10b + c
100c + 10a + b
Taking the difference, we get:
100(a  c) + 10(b  a) + (c  b)
Next, we break down this expression as follows:
(a  c) + 99(a  c) + (b  a) + 9(b  a) + (c  b)
Rearranging terms, we get:
(a  c) + (b  a) + (c  b) + 99(a  c) + 9(c  b)
Cancelling like terms, we're left with:
99(a  c) + 9(c  b)
This proves that my theorem works for any threedigit anagram numbers as the last expression is evenly divisible by nine.
In my next post I will extend this proof to any two anagram numbers with any amount of digits.
PhilX
Last edited by Philosophy Explorer on Mon Aug 17, 2015 12:19 pm, edited 1 time in total.

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Re: Proving that the difference between any two anagram numbers is always a multiple of nine
To extend the proof I will use mathematical induction.
I will use the following expressions: abc... and cab..., for the anagram numbers with equal, but indeterminate number of digits where the numbers are at least three digits apiece and each digit can take on values from 0 to 9.
First step: Assume this is true for threedigit numbers (which is already proven in the post just above).
Second step: assume this is true for two anagram numbers with any number of digits. Then it must be shown that for any two anagram numbers with n digits, that this implies it must also be true for two anagram numbers with n + 1 digits derived from the anagram numbers with n digits.
To make this clearer, let's say we have abcd and abdc where we assume their difference is a multiple of nine. The next digit up we have abcde and abdce. Now we can break down those numbers into 10abcd + e and 10abdc + e. Taking their difference, we get 10(abcd  abdc) + e  e and we're left with 10(abcd  abdc) which, by assumption, what's inside the parentheses is divisible by nine and the difference between abcde and abdce must also be divisible by nine by implication. It's also obvious this method extends to anagram numbers of any size.
What about twodigit numbers? They're trivial and you can easily figure out their difference is also divisible by nine.
This completes the proof.
PhilX
I will use the following expressions: abc... and cab..., for the anagram numbers with equal, but indeterminate number of digits where the numbers are at least three digits apiece and each digit can take on values from 0 to 9.
First step: Assume this is true for threedigit numbers (which is already proven in the post just above).
Second step: assume this is true for two anagram numbers with any number of digits. Then it must be shown that for any two anagram numbers with n digits, that this implies it must also be true for two anagram numbers with n + 1 digits derived from the anagram numbers with n digits.
To make this clearer, let's say we have abcd and abdc where we assume their difference is a multiple of nine. The next digit up we have abcde and abdce. Now we can break down those numbers into 10abcd + e and 10abdc + e. Taking their difference, we get 10(abcd  abdc) + e  e and we're left with 10(abcd  abdc) which, by assumption, what's inside the parentheses is divisible by nine and the difference between abcde and abdce must also be divisible by nine by implication. It's also obvious this method extends to anagram numbers of any size.
What about twodigit numbers? They're trivial and you can easily figure out their difference is also divisible by nine.
This completes the proof.
PhilX

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Re: Proving that the difference between any two anagram numbers is always a multiple of nine
Two things:
1) As a reminder to the chicken man, this isn't spam as I've already placed threads like this in the Lounge area only to see them moved into this section.
2) I have a partial proof that the difference between any powers of anagram numbers is divisible by nine too. I'll work up a more general proof later.
PhilX
1) As a reminder to the chicken man, this isn't spam as I've already placed threads like this in the Lounge area only to see them moved into this section.
2) I have a partial proof that the difference between any powers of anagram numbers is divisible by nine too. I'll work up a more general proof later.
PhilX

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Re: Proving that the difference between any two anagram numbers is always a multiple of nine
This works because given any n number of digits, the absolute difference between the two given the same identical numbers can only differ by a power of ten in each common digit. As such, the powers will always cancel the common digits with a multiple of its tens complement.
I'm not sure how far you've delved into complements but this is used to subtract by using addition of some complement and useful for computers.
Take any number, such as 7. Then its ten's complement is 3, meaning 7 + 3 = 10.
To subtract 7 from any number, use its complement, 3, add, then subtract 10 later. (because 10  3 = 7)
49  7 = 49 + [(10  7)  10] = 49 + (3  10) = 49 + 3  10 = 52  10 = 42
How it relates to this theorem is as follows:
abcd bcda = (1000a  a) + (100b  1000b) + (10c  100c) + (d  10d)
..............= 999a + ( 900b) + ( 90c) + (9d)
Notice how any difference of powers of the base, ten, always MUST leave a difference of 1, 10, 100, ... which always leads with 9, the tens complement of zero, subtract one. [9 + 1 + ( 1) = 10  1 + ( 1)] This becomes the 9's complement in the highest significant digit. This is used better in computers for some architectures.
To subtract 7 in 9's complement, 7 becomes 9  7 = 2. This may seem weird but when using binary, the process is easy:
7(base 10) = 111(base 2)
The [base 2 minus one's] complement of 111(base 2) is 001 [note how this is simply just negating the digits]
Example:
10  7 = 1010(base 2)  111(base 2) = 1010 + 001  1000 = 11
Thus you just negate the digits, add one, and remove the highest significant digit. Cool huh?
I'm not sure how far you've delved into complements but this is used to subtract by using addition of some complement and useful for computers.
Take any number, such as 7. Then its ten's complement is 3, meaning 7 + 3 = 10.
To subtract 7 from any number, use its complement, 3, add, then subtract 10 later. (because 10  3 = 7)
49  7 = 49 + [(10  7)  10] = 49 + (3  10) = 49 + 3  10 = 52  10 = 42
How it relates to this theorem is as follows:
abcd bcda = (1000a  a) + (100b  1000b) + (10c  100c) + (d  10d)
..............= 999a + ( 900b) + ( 90c) + (9d)
Notice how any difference of powers of the base, ten, always MUST leave a difference of 1, 10, 100, ... which always leads with 9, the tens complement of zero, subtract one. [9 + 1 + ( 1) = 10  1 + ( 1)] This becomes the 9's complement in the highest significant digit. This is used better in computers for some architectures.
To subtract 7 in 9's complement, 7 becomes 9  7 = 2. This may seem weird but when using binary, the process is easy:
7(base 10) = 111(base 2)
The [base 2 minus one's] complement of 111(base 2) is 001 [note how this is simply just negating the digits]
Example:
10  7 = 1010(base 2)  111(base 2) = 1010 + 001  1000 = 11
Thus you just negate the digits, add one, and remove the highest significant digit. Cool huh?
 Hobbes' Choice
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Re: Proving that the difference between any two anagram numbers is always a multiple of nine
Okay lets try...
Multiples of nine are as follows
9,12, 1B, 24, 2D, 35
so lets try 1235 1+2+3+5=B
how about 3453? 3+4+5+3=F
Nope!
Multiples of nine are as follows.
11, 22, 33, 44, ,55, 66, 77, 88
so lets try 56453= 5+6+4+5+3=27... 2+7=11
Hex doesn't work, Octal does.
Multiples of nine
1001, 10010, 11011, 100100, 101101
so lets try 111111 , 1+1+1+1+1+1=110 1+1+0= 10
Obvious meaningless for binary.
Multiples of nine are as follows
9,12, 1B, 24, 2D, 35
so lets try 1235 1+2+3+5=B
how about 3453? 3+4+5+3=F
Nope!
Multiples of nine are as follows.
11, 22, 33, 44, ,55, 66, 77, 88
so lets try 56453= 5+6+4+5+3=27... 2+7=11
Hex doesn't work, Octal does.
Multiples of nine
1001, 10010, 11011, 100100, 101101
so lets try 111111 , 1+1+1+1+1+1=110 1+1+0= 10
Obvious meaningless for binary.

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Re: Proving that the difference between any two anagram numbers is always a multiple of nine
PE has a great proof for base 10 though. I wrote what I did because I recognized how seemingly trivial playing with numbers has actually eventually come into some use as in computers, math, or other areas down the road. I can't figure out how his theorem (or whomever's) may work in practice. But its cool to try proving things like this.

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Re: Proving that the difference between any two anagram numbers is always a multiple of nine
I don't care if my discoveries have any practical use. By nature I'm an explorer. Maybe in time practical uses will be found. I do regard recreational activity as worthwhile in itself for entertainment purposes as well as helping to see the world in a new light (btw Scott, the proof for the sum of digits to nine inspired my proof  I don't know though if I'm original with this. All I can say is it's as new to me as it is to everybody else here and an internet search has turned up nothing regarding my theorem).Scott Mayers wrote:PE has a great proof for base 10 though. I wrote what I did because I recognized how seemingly trivial playing with numbers has actually eventually come into some use as in computers, math, or other areas down the road. I can't figure out how his theorem (or whomever's) may work in practice. But its cool to try proving things like this.
PhilX

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Re: Proving that the difference between any two anagram numbers is always a multiple of nine
If y and x are anagram numbers, I can easily prove that
y^n  x^n will also be divisible by nine when n = 2^k and k is a natural number (making n a power of 2). To make my point clearer, I will choose k = 3. Watch what happens:
We get y^8  x^8. Factor this expression as follows:
(y^4 + x^4)(y^4  x^4). Now factor what's in the second parentheses:
(y^4 + x^4)(y^2 + x^2)(y^2  x^2). Now factor what's in the last parentheses:
(y^4 + x^4)(y^2 + x^2)(y + x)(y  x). Look at what's left in the last parentheses: y  x. We assumed above they're anagram numbers and we've proven the other day that the difference between any two anagram numbers is divisible by nine. This means that y^8  x^8 is also evenly divisible by nine and this scheme will always work when n is a power of 2 which completes this phase of the proof
(it still remains to be proven in general for other cases such as n = 3 or n = 6 which I'm still working on).
PhilX
y^n  x^n will also be divisible by nine when n = 2^k and k is a natural number (making n a power of 2). To make my point clearer, I will choose k = 3. Watch what happens:
We get y^8  x^8. Factor this expression as follows:
(y^4 + x^4)(y^4  x^4). Now factor what's in the second parentheses:
(y^4 + x^4)(y^2 + x^2)(y^2  x^2). Now factor what's in the last parentheses:
(y^4 + x^4)(y^2 + x^2)(y + x)(y  x). Look at what's left in the last parentheses: y  x. We assumed above they're anagram numbers and we've proven the other day that the difference between any two anagram numbers is divisible by nine. This means that y^8  x^8 is also evenly divisible by nine and this scheme will always work when n is a power of 2 which completes this phase of the proof
(it still remains to be proven in general for other cases such as n = 3 or n = 6 which I'm still working on).
PhilX
 Arising_uk
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Re: Proving that the difference between any two anagram numbers is always a multiple of nine
No idea why you brought me into this but since you have I'll remind you that my complaint is with your spammed links and this thread did not start with your usual MO. However, this is not a philosophy of mathematics post but a Mathematics post and as such should not be here but on a Mathematics forum or a recreational maths forum.Philosophy Explorer wrote:...
1) As a reminder to the chicken man, this isn't spam as I've already placed threads like this in the Lounge area only to see them moved into this section.

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Re: Proving that the difference between any two anagram numbers is always a multiple of nine
Which I've also explained before.Arising_uk wrote:No idea why you brought me into this but since you have I'll remind you that my complaint is with your spammed links and this thread did not start with your usual MO. However, this is not a philosophy of mathematics post but a Mathematics post and as such should not be here but on a Mathematics forum or a recreational maths forum.Philosophy Explorer wrote:...
1) As a reminder to the chicken man, this isn't spam as I've already placed threads like this in the Lounge area only to see them moved into this section.
PhilX
 vegetariantaxidermy
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Re: Proving that the difference between any two anagram numbers is always a multiple of nine
You could at least explain it more clearly for maths idiots.

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Re: Proving that the difference between any two anagram numbers is always a multiple of nine
What's wrong with using math on this site? It's appropriate as logic and such demonstrations suggest excellent ways to practice good argument skills for philosophy.
 Arising_uk
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Re: Proving that the difference between any two anagram numbers is always a multiple of nine
Nothing, if it's being used to prove a philosophical point but what is being done here is Mathematics and not Philosophy of Mathematics and there are many sites setup just for such things so why post incorrectly, especially in a purported philosophy forum.Scott Mayers wrote:What's wrong with using math on this site? ...
But is it? Whilst Mathematics obviously uses Logic the form is such that unless one understands the Mathematics the logic is wasted, whereas Philosophy has a very suitable tool for demonstrating and practicing good argument skills for Philosophy, in fact it's practically the only thing that is Philosophy's own, Logic and it's variant Symbolic Logic.It's appropriate as logic and such demonstrations suggest excellent ways to practice good argument skills for philosophy.

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Re: Proving that the difference between any two anagram numbers is always a multiple of nine
Well, I follow mathematical logic as well as various other logic and I find it fascinating. Just because you may not follow it doesn't mean that others don't. And it is definitely a part of philosophy too. If you don't understand something, you can always ask too. It makes it interesting to try to find a way to communicate these things for others. I even opened one on the Monty Hall Problem and the Secretary Problem which is easy for anyone to grasp without too much math up front. In fact sometimes others can see errors where a mathematician may miss.Arising_uk wrote:Nothing, if it's being used to prove a philosophical point but what is being done here is Mathematics and not Philosophy of Mathematics and there are many sites setup just for such things so why post incorrectly, especially in a purported philosophy forum.Scott Mayers wrote:What's wrong with using math on this site? ...But is it? Whilst Mathematics obviously uses Logic the form is such that unless one understands the Mathematics the logic is wasted, whereas Philosophy has a very suitable tool for demonstrating and practicing good argument skills for Philosophy, in fact it's practically the only thing that is Philosophy's own, Logic and it's variant Symbolic Logic.It's appropriate as logic and such demonstrations suggest excellent ways to practice good argument skills for philosophy.

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Re: Proving that the difference between any two anagram numbers is always a multiple of nine
I will explain it so that even the chicken man can understand.vegetariantaxidermy wrote:You could at least explain it more clearly for maths idiots.
I'm sure even he can recall that math websites I've posted to had told me the purpose of their websites was for helping out members (i.e. homework problems), not for recreational math. To reinforce what I'm explaining for the benefit of the chicken man, look at the title of this section. It doesn't say philosophy of mathematics, does it? It says LOGIC AND philosophy of mathematics so it just doesn't have to be about philosophy of mathematics which is what I've been doing (the logic part). Now I think even the chicken man realizes that I've been using logic in my math postings and it would be useless for him to persist any further, otherwise I'll just have to pluck out some more of his feathers.
I'm still working on extending the proof to at least where all the powers of anagram numbers, are natural numbers.
PhilX
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