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

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dionisos
Posts: 96
Joined: Sat Aug 29, 2015 11:03 am

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

Philosophy Explorer wrote: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 three-digit 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 three-digit 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 three-digit 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
What you try to demonstrate is true, but your demonstration, even if it is a good try, is wrong.
You demonstrate that 9 divide the difference, of anagram of the form (abc, cab).
but anagram of 3 digits, could also be of the form (abc, abc), (abc, acb), (abc, bca), (abc, bac),(abc, cba).

The rest of your demonstration is wrong for the same kind of reason.
In final, you only demonstrate that 9 divide the difference of anagram of a particular form.

dionisos
Posts: 96
Joined: Sat Aug 29, 2015 11:03 am

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

Philosophy Explorer wrote:....................The Completed Extended Proof

(this post should be read together with the first two posts of this thread)

First keep in mind that (x^p)^1/p = x, p not equal to 0. This will help you to understand what follows.

The difference between the power of two anagram numbers with p = 0 is trivial. This difference always equals 0 (since the two expressions becomes equal to 1 and the difference between the ones is 0) and 0 divided by 9 is always 0, a trivial result.

I am asserting the theorem is always true for p being any natural number. In fact I'm asserting that the theorem is true for any real number p (except for p = 0), a bonus.

Let's say we have y = (cab)^p and x = (abc)^p where again a, b and c can take on any values from 0 to 9. Let's raise each expression to the 1/p power (again p not equal to 0). Now we have y^1/p = (cab) and x^1/p = (abc). Taking their difference, we have (cab) - (abc). Comparing with the first post, this is already proven to be divisible by nine and if we replace 1/p by n (where n is allowed to be any real number except 0) and then replace n by p (since p is a dummy variable), this proves this part of the theorem for three-digit numbers and by following the same steps as shown in the second post, this completes the extended proof for powers of anagram numbers of any number with unspecified number of digits).

PhilX
This demonstration is also wrong, for many reasons.
You could see that 21^(1/2)-12^(1/2) is not a whole number, and then 9 don’t divide it.

Also, numbers of the form a/b are rational numbers, which is a subset of real numbers.

Here is a demonstration of what you wanted to demonstrate:
a and b anagrams numbers, and p>0 a whole number.
a^p-b^p = (a^(p-1)+b^(p-1))(a-b)
9 divide (a-b) (because it is the difference of two anagrams numbers).
then 9 divide (a^(p-1)+b^(p-1))(a-b)=a^p-b^p

You could see you was near with your last demonstration with p=2^n.
Instead of dividing p by 2 in each iteration, until i get (a-b), i just removed 1 to p.(what give me a one iteration step, and this work for each p).
And like you put it, the case for p=0 is straightforward.

Philosophy Explorer
Posts: 5621
Joined: Sun Aug 31, 2014 7:39 am

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

dionisos wrote:
Philosophy Explorer wrote:....................The Completed Extended Proof

(this post should be read together with the first two posts of this thread)

First keep in mind that (x^p)^1/p = x, p not equal to 0. This will help you to understand what follows.

The difference between the power of two anagram numbers with p = 0 is trivial. This difference always equals 0 (since the two expressions becomes equal to 1 and the difference between the ones is 0) and 0 divided by 9 is always 0, a trivial result.

I am asserting the theorem is always true for p being any natural number. In fact I'm asserting that the theorem is true for any real number p (except for p = 0), a bonus.

Let's say we have y = (cab)^p and x = (abc)^p where again a, b and c can take on any values from 0 to 9. Let's raise each expression to the 1/p power (again p not equal to 0). Now we have y^1/p = (cab) and x^1/p = (abc). Taking their difference, we have (cab) - (abc). Comparing with the first post, this is already proven to be divisible by nine and if we replace 1/p by n (where n is allowed to be any real number except 0) and then replace n by p (since p is a dummy variable), this proves this part of the theorem for three-digit numbers and by following the same steps as shown in the second post, this completes the extended proof for powers of anagram numbers of any number with unspecified number of digits).

PhilX
This demonstration is also wrong, for many reasons.
You could see that 21^(1/2)-12^(1/2) is not a whole number, and then 9 don’t divide it.

Also, numbers of the form a/b are rational numbers, which is a subset of real numbers.

Here is a demonstration of what you wanted to demonstrate:
a and b anagrams numbers, and p>0 a whole number.
a^p-b^p = (a^(p-1)+b^(p-1))(a-b)
9 divide (a-b) (because it is the difference of two anagrams numbers).
then 9 divide (a^(p-1)+b^(p-1))(a-b)=a^p-b^p

You could see you was near with your last demonstration with p=2^n.
Instead of dividing p by 2 in each iteration, until i get (a-b), i just removed 1 to p.(what give me a one iteration step, and this work for each p).
And like you put it, the case for p=0 is straightforward.
Two things dionisos:

First you wrote 21^(1/2)-12^(1/2) which is the wrong number. You should have written
(21^2)^1/2 - (12^2)^1/2 which is the right form of the number I am discussing. So
(21^2)^1/2 = (21)^2/2 = 21^1 = 21 and
(12^2)^1/2 = (12)^2/2 = 12^1 = 12. The difference between 21 and 12 is 9 which is evenly divisible by 9. Apparently you're having a problem with the notation. Just reread what I wrote in the first sentence.

If you're restricting a and b to be whole numbers, then yes, a/b is a rational number (b not equal to 0). But I'm not restricting a and b to be whole numbers. I'm letting them be any real number (except b can't be equal to 0).

Looking over your counterexample, you are correct because when you're taking the square root of 12 and 21, the difference isn't divisible by 9 as the decimal expressions beyond the decimal point won't cancel when you take their difference (but look at anagram numbers 441 and 144, their difference is 297 which is evenly divisible by 9; also look at their square roots 21 and 12 and the difference between them is 9 which is of course divisible by 9).

Your counterexample forced me to reconsider the situation so I'm amending my theorem to restricting p to be just any natural number which will take care of the problem.

Are we in agreement now?

PhilX

dionisos
Posts: 96
Joined: Sat Aug 29, 2015 11:03 am

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

Just a note: You should write (21^2)^(1/2) and not (21^2)^1/2, because the '^' has priority otherwise.

The problem is not only about the fact that p is a real or a whole number, i can only find counter example when it is not a whole number, but your demonstration is still false otherwise. (even if what you try to demonstrate is true).

let y = (cab)^p and let x = (abc)^p (ok)
then y^(1/p) = (cab) and x^(1/p) = (abc) (ok)
then y^(1/p) - x^(1/p) = (cab) - (abc) (ok)
then 9 divide y^(1/p) - x^(1/p) (ok)
let n=1/p (ok)
then 9 divide y^n - x^n (ok)
let p = n (not ok, you have already defined p)
Look, if you need this last step, and you can’t use n directly, it is only because you need to use "y = (cab)^p", but you can’t use it, if you don’t use the same p.
And in your demonstration, p have no need to be a whole number, if it was right, it should be right for any real number p.
Then my counter example should finish to convince you that your demonstration is wrong, for p a whole number or not.

Philosophy Explorer
Posts: 5621
Joined: Sun Aug 31, 2014 7:39 am

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

I'm still working on hammering out the complete proof.

I was looking over the binomial theorem which suggests this is always true (unproven so far).

If we have two anagram numbers, x and y, then:

y^2•x - x^2•y is always divisible by 9.

Check this out on your calculator.

PhilX

dionisos
Posts: 96
Joined: Sat Aug 29, 2015 11:03 am

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

Philosophy Explorer wrote:I'm still working on hammering out the complete proof.

I was looking over the binomial theorem which suggests this is always true (unproven so far).

If we have two anagram numbers, x and y, then:

y^2•x - x^2•y is always divisible by 9.

Check this out on your calculator.

PhilX
Yes, i don’t know if you saw it, but i demonstrated the general case here : viewtopic.php?f=26&t=16596&start=45#p221433

Philosophy Explorer
Posts: 5621
Joined: Sun Aug 31, 2014 7:39 am

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

dionisos wrote:
Philosophy Explorer wrote:....................The Completed Extended Proof

(this post should be read together with the first two posts of this thread)

First keep in mind that (x^p)^1/p = x, p not equal to 0. This will help you to understand what follows.

The difference between the power of two anagram numbers with p = 0 is trivial. This difference always equals 0 (since the two expressions becomes equal to 1 and the difference between the ones is 0) and 0 divided by 9 is always 0, a trivial result.

I am asserting the theorem is always true for p being any natural number. In fact I'm asserting that the theorem is true for any real number p (except for p = 0), a bonus.

Let's say we have y = (cab)^p and x = (abc)^p where again a, b and c can take on any values from 0 to 9. Let's raise each expression to the 1/p power (again p not equal to 0). Now we have y^1/p = (cab) and x^1/p = (abc). Taking their difference, we have (cab) - (abc). Comparing with the first post, this is already proven to be divisible by nine and if we replace 1/p by n (where n is allowed to be any real number except 0) and then replace n by p (since p is a dummy variable), this proves this part of the theorem for three-digit numbers and by following the same steps as shown in the second post, this completes the extended proof for powers of anagram numbers of any number with unspecified number of digits).

PhilX
This demonstration is also wrong, for many reasons.
You could see that 21^(1/2)-12^(1/2) is not a whole number, and then 9 don’t divide it.

Also, numbers of the form a/b are rational numbers, which is a subset of real numbers.

Here is a demonstration of what you wanted to demonstrate:
a and b anagrams numbers, and p>0 a whole number.
a^p-b^p = (a^(p-1)+b^(p-1))(a-b)
9 divide (a-b) (because it is the difference of two anagrams numbers).
then 9 divide (a^(p-1)+b^(p-1))(a-b)=a^p-b^p

You could see you was near with your last demonstration with p=2^n.
Instead of dividing p by 2 in each iteration, until i get (a-b), i just removed 1 to p.(what give me a one iteration step, and this work for each p).
And like you put it, the case for p=0 is straightforward.
What about the terms a•b^(p - 1) and -b•a^(p - 1) which results from multiplying through by (a - b)? Can you prove in general their difference is also divisible by 9? (I was already aware of this situation)

We have more work to do and maybe the binomial theorem may not be the best way to complete the proof (but it has suggested that the difference between those terms always being divisible by 9 may be true).

Quite a problem.

PhilX

dionisos
Posts: 96
Joined: Sat Aug 29, 2015 11:03 am

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

Yes, i did a mistake on it.

Philosophy Explorer
Posts: 5621
Joined: Sun Aug 31, 2014 7:39 am

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

dionisos wrote:Yes, i did a mistake on it.
Don't worry about the mistake. Even math greats such as Euler make mistakes. The important thing is to learn from our mistakes and we have gained some new possible knowledge (namely the difference between those terms seems to be also divisible by nine).

PhilX

dionisos
Posts: 96
Joined: Sat Aug 29, 2015 11:03 am

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

I don’t worry, it was only to inform you that i saw it, but thank you I wrote a rapid python program, and it found no counter example, i will try to find a demonstration with you then.

Code: Select all

``````#!/bin/python

def get_anagram(x):
str_x = str(x).zfill(3)
str_ana = str_x + str_x + str_x
return int(str_ana)

for a in range(100, 500):
b = get_anagram(a)
for p in range(2, 6):
if (pow(a, p) - pow(b, p))%9 == 0:
print("ok for:" + str(a) + "/" + str(b) + "/" + str(p))
else:
print("counter")
print(str(a) + "/" + str(p))
exit(1)
``````

Scott Mayers
Posts: 1638
Joined: Wed Jul 08, 2015 1:53 am

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

I'd join in but my head is still in the Monty Hall problem at present. I see that 'weights' to which we apply upon things make it harder to be certain about our math without extreme care to the details.

dionisos
Posts: 96
Joined: Sat Aug 29, 2015 11:03 am

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

ok, i got it right this time:

P(n) = "9 divide (a^n - b^n), and 9 divide (a^(n-1) - b^(n-1))"

if (a, b) are anagrams, then
9 divide (a - b)
a^2-b^2=(a + b)(a - b)
then 9 divide (a^2-b^2)
then p(2)

If P(n), then:
9 divide a^n - b^n and
9 divide a^(n-1) - b^(n-1)
a^(n+1)-b^(n+1) = (a^n - b^n)(a + b) - ba^n + ab^n = (a^n - b^n)(a + b) - ab(a^(n-1) - b^(n-1))
9 divide (a^n - b^n)(a + b)
and
9 divide ab(a^(n-1) - b^(n-1))
then
9 divide (a^n - b^n)(a + b) - ab(a^(n-1) - b^(n-1))
then 9 divide a^(n+1)-b^(n+1)
and 9 divide a^n - b^n
then P(n+1)

P(2) and (P(n)⇒P(n+1)) then ∀n>2, P(n)
Then 9 divide a^n - b^n.

It is not really about anagrams in fact: if k divide (a-b), then k divide (a^n - b^n)
That (a, b) are anagrams just allow for the first step.

Philosophy Explorer
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Joined: Sun Aug 31, 2014 7:39 am

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

........................The Binomial Theorem]

Part of math exploration can lead to accidental discoveries. I was seeking (unsuccessfully) to use the binomial theorem to extend my math proof. What happened instead is it lead to a new conjecture connected to another conjecture.

I was seeking to prove that x^n - y^n, x and y anagram numbers and n any natural number, is divisible by 9 and I thought of trying (x - y)^n which will generate x^n +/- y^n
(depending on whether n is odd or even) plus other terms. It is the other terms that are of interest.

Now (x - y)^n is factorable into (x - y)(x - y)...(x - y)(x - y)
n times which means it is divisible by 9 and x^n - y^n is suspected of being divisible by 9 (conjecture). This implies that the remaining terms, if the conjecture is true, are also divisible by 9, even if n is an even number! (that is, x^n + y^n is divisible by 9).

I'm going to check x^n + y^n for 9 divisibility on my calculator and will let you know what happens. In any case, there may be an infinite number of equations that are divisible by 9 resulting from the binomial theorem.

PhilX

Update: A check with my calculator shows that x^n + y^n doesn't work so only x^n - y^n works on my calculator. I may update this further as I'm double checking something else on my calculator.

2nd update: I copied this from Wiki:

(x-y)^3 = x^3 - 3x^2y + 3xy^2 - y^3.

With the middle terms on the RHS and factoring 3, we have:

3(xy^2 - yx^2)

There is also:

5[(xy^4 + 2x^3•y^2) - (x^4•y + 2x^2•y^3)]

when n is 5.

I have checked several anagrams with the expressions in the parentheses showing divisibility by 9. So with the binomial theorem expansion, I'm now conjecturing that the middle terms on the RHS are divisible by 9 when n is an odd number which checks out on my calculator (I would appreciate someone running a program to verify this on his computer as this is tedious to check out on my calculator).

wtf
Posts: 913
Joined: Tue Sep 08, 2015 11:36 pm

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

Here's the proof that the difference of anagrams is divisible by 9.

First, note that x = x0 + x1 * 10 + x2 * 10^2 + ... + xn * 10^n (writing xn or sometimes x_n to mean "x_subscript_n") and likewise y = y0 + y1 * 10 + y2 * 10^2 + ... + yn * 10^n.

Note that since the x_i's are just a rearrangement of the y_i's, it's certainly the case that sum(all the x_i's) - sum(all the y_i's) = 0. So the sum of all (x_i - y_j) is also zero regardless of the order of the xi's and yi's.

Now note that 10 = 1 (mod 9) and also 10^k = 1 (mod 9) for any natural number k.

Then

x - y = (x0 - y0) + (x1 - y1)*10 + (x2 - y2)*10*2 + ... + (xn - yn)*10^n

= sum(all the xi - yi terms) = 0 (mod 9). This is the same as saying that 9 divides x - y.

End of proof

If you haven't seen modular arithmetic before, it's just the remainder on integer division. So 5 = 3 (mod 2) because they both have the same remainder on division by 2. And 10 = 1 (mod 9) for the same reason.

The other thing to know is that addition and multiplication respect remainders. So the remainder of x plus the remainder of y is the same as the remainder of x + y with respect to any fixed divisor. https://en.wikipedia.org/wiki/Modular_arithmetic

dionisos
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### Re: Proving that the difference between any two anagram numbers is always a multiple of nine

Good proof wtf ### Who is online

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