Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

Recently I came across a myth saying that the division by 7 implies the utmost secret of the universe. The multiplication of life, from 1 to 2, to 4, to some power of 2, can be simplified to repeatitive 1 4 2 8 5 and 7, the figure appearing in 1/7.

Here is how it works:

1 2 4 8 16 32 64 128 256 1024

1

2

4

8

1+6=7

3+2=5

6+4=10, 1+0=1

1+2+8=13, 1+3=4

2+5+6=13, 1+3=4

1+0+2+4=8

So far, as you can tell, only 6 figures (1 2 4 5 7 and 8) out of the 9 non-zero figures appear in this abstracted number out of 2^n when n<=10.

And the method used is simple and commonplace in astrology. There is no deliberate manipulation here.

A further evidence is the Yin and Yang Taoism philosophy in China.We have a strange idiom saying 7 up and 8 down, meaning 7 represents the Yang, and the 8 represents the Yin. Indeed here 7 is the largest odd number out of the 3 odds, and 8 is the largest even number out of the evens. Moreover, if you add 1 and 8, 2 and 7, 4 and 5, you always get the result 9. And that is the utmost power number 9 in Daoism (Taoism), which is the unity of both Yin and Yang.

The question is,

Can someone prove the 6 figures rule applies to all 2^n's ?

**X'(y-Xβ)=0**

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

2048 -> 14 -> 5

4096 -> 19 -> 20 -> 2

8192 -> 20 ->2

...

It seems that you will never get 3, 6 or 9 through powers of 2, which are compounded by some figures out of 1, 2, 4, 5, 7 and 8. (3=1+2, 6=1+5 or 2+4, 9=1+8 or 2+7 or 4+5)

**X'(y-Xβ)=0**

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

Hmm, it is easy to prove why 3, 6 an 9 are missing. If the addition of the figures comes out as multiple of 3, the original number 2^k must be divisible by 3. Of course they will not appear. However the questions still remain why all of the 6 remaining appear in abstractions of 2^n, and why 3, 6 and 9 are missing in all 1/7, 2/7 , 3/7, 4/7, 5/7 and 6/7.

**X'(y-Xβ)=0**

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

and why 3, 6 and 9 are missing in all 1/7, 2/7 , 3/7, 4/7, 5/7 and 6/7.

What do you mean? Where do 1/7 to 6/7 come into play, and why does 3/7 (for example) not count?

It seems like you're dabbling in numerology, which while perhaps fun to some people, is a rather "meaningless" venture. Realize (for example) that your sequence would be completely different if humans had 12 fingers instead of 10, because then we'd be counting in base 12.

And the method used is simple and commonplace in astrology.

That's not a good sign.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

Yes I agree it has something to do with base 10.

But the numerology exists in base 10 in concurrence:

the only decimal figures you get from division by 7 will be these 6:

1 2 4 5 7 and 8

aren't they special?

And the only figures you get from adding numbers of 2^n come to

1 2 4 5 7 or 8

Adding numbers together isn't the privillige of astrology, but mathematicians use this too -

Suppose you see 14532, and you can tell it is divisible by 3 since 1+4+5+3+2=15, and 1+5=6, 6/3=2.

**X'(y-Xβ)=0**

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

It is easy to explain why the remainder of any integers divided by 7 will be 0, 1, 2, 3, 4, 5, and 6.

However it is strange that the decimals (after the decimal point) appearing in N/7, are only 0

or 1, 2, 4, 5, 7 and 8

why crossing out 3, 6 and 9? why these particular three?

Is there a mathematical explanation to this phenomenon?

And is there a mathematical proof to the 2^n case?

**X'(y-Xβ)=0**

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,989

Hi;

Your idea is easier to understand in terms of mods. Reducing a number to a single digit or a digital sum as it is sometimes called is done using mod 9. There it is easier to see why the pattern will repeat.

1,2,4,8, (2 * 8 = 16 mod 9 = 7), (2 * 7=14 mod 9 = 5), (2 *5=10 mod 9 = 1)...

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

**Online**

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

Using spreadsheet, I just discovered not only do 2^n corresponds to 6 figures, but the figures follow a special sequence too!

2^0=1 **1**

2^1=2 **2**

2^2=4 **4**

2^3=8 **8**

2^4=16 =>7 **7**

2^5=32 =>5 **5**

2^6=64 =>10=>1 **1**

2^7=128 =>11=>2 **2**

2^8=256 =>13=>4 **4**

2^9=512 =>8 **8**

2^10=1024 =>7 **7**

2^11=2048 =>14=>5 **5**

2^12=4096 =>19=>10=>1 **1**

2^13=8192 =>20=>2 **2**

2^14=16384 =>22=>4 **4**

2^15=32768 =>26=>8 **8**

2^16=65536 =>25 =>7 **7**

2^17=131072 =>14 =>5 **5**

2^18=262114 =>19 =>1 **1**

2^19=524288 =>29=>11 =>2 **2**

2^20=1048576 =>31 =>4 **4**

2^21=2097152 =>26 =>8 **8**

2^22=4194304 =>25 =>7 **7**

2^23=8388608 =>41 =>5 **5**

2^24=16777216 =>37 =>10=>1 **1**

2^25=33554432 =>29=>11=>2 **2**

2^26=67108864 =>40=>4 **4**

2^27=134217728 =>35 =>8 **8**

2^28=268435456 => 43 =>7 **7**

2^29=536870912 => 41 =>5 **5**

2^30=1073741824 =>37 =>10 **1**

So it is very obvious that the figure sums of powers of 2 endure an ascending 2 4 8 process and then a descending process 7 5 1 and the cycle repeats itself. And this is so called the wave, the cycle, the spiral, the yin & yang, whatever.

Anyone have a clue about the cause?

*Last edited by George,Y (2010-03-01 20:18:24)*

**X'(y-Xβ)=0**

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

bobbym wrote:

Hi;

Your idea is easier to understand in terms of mods. Reducing a number to a single digit or a digital sum as it is sometimes called is done using mod 9. There it is easier to see why the pattern will repeat.

1,2,4,8, (2 * 8 = 16 mod 9 = 7), (2 * 7=14 mod 9 = 5), (2 *5=10 mod 9 = 1)...

Thank you!

**X'(y-Xβ)=0**

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

acceleratingly ascending

1 2 4 8

and then acceleratingly descending

8 7 5 1

**X'(y-Xβ)=0**

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

Taking difference:

1 2 4 8 7 5 1

1 2 4 -1 -2 -4

**X'(y-Xβ)=0**

Offline

**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

u get one of the nos. out of the set {1, 4, 2, 8, 5, 7} if u take any no. out of this set and raise it to any power!!

just {0, 3, 6 9} are out of the set!

eg.

5^2=25=2+5=7

5^3=125=1+2+5=8

and so on...

Its quite obvios from the result of 2^n... Right??

*Last edited by ZHero (2010-03-29 20:42:27)*

If two or more thoughts intersect, there has to be a point!

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

That makes sense.

See it another way, (3a)^n gets trapped in its own loop {3,6,9}

beside 0

{1,2,4,8,7,5}must be seperated from{3,6,9}

The hardest part is the ascending and then descending order of

1 2 4 8 7 5 1

**X'(y-Xβ)=0**

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,989

Hi George,Y;

Again, there is nothing mysterious at all working here, just the residuals of the powers of 2 mod 9.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

**Online**

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

They have also similar cycle puzzles like

1 3 9 7 1

+2+6-2-6

and

2 4 8 6 2

seems swap the last two biggest number in a four number ascending sequence could produce a cycle.

**X'(y-Xβ)=0**

Offline

Pages: **1**