The number 142857.

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The number 142857 is quite interesting.

Check this out:

142857·1 = 142857
142857·2 = 285714
142857·3 = 428571
142857·4 = 571428
142857·5 = 714285
142857·6 = 857142

Those answers contains the exact same numbers in the same order.
The only difference is which number starts.

142857·7 = 999999

And, if you take the first 3 numbers from answers 1-6, and add them
to the other three numbers, you get:

142 + 857 = 999
285 + 714 = 999
428 + 571 = 999
571 + 428 = 999
714 + 285 = 999
857 + 142 = 999

If you divide the numbers in 3 parts, you get:

14 + 28 + 57 = 99
28 + 57 + 14 = 99
42 + 85 + 71 = 99 + 99
57 + 14 + 28 = 99
71 + 42 + 85 = 99 + 99
85 + 71 + 42 = 99 + 99

If we take number 142857, and put it in a triangle like this,
we get very interesting results:

1·7 + 3 = 10
14·7 + 2 = 100
142·7 + 6 = 1000
1428·7 + 4 = 10000
14285·7 + 5 = 100000
142857·7 + 1 = 1000000
1428571·7 + 3 = 10000000
14285714·7 + 2 = 100000000
142857142·7 + 6 = 1000000000
1428571428·7 + 4 = 10000000000
14285714285·7 + 5 = 100000000000
142857142857·7 + 1 = 1000000000000
1428571428571·7 + 3 = 10000000000000
14285714285714·7 + 2 = 100000000000000
142857142857142·7 + 6 = 1000000000000000
1428571428571428·7 + 4 = 10000000000000000
14285714285714285·7 + 5 = 100000000000000000
142857142857142857·7 + 1 = 1000000000000000000

And here's one more thing:
If you divide the number 142857 with 9, you get 15873,
something that gives you a quite funny result when you multiply with
any number that be divided by seven and remain whole:

142857 · (7 · 1) / 9 = 111111
142857 · (7 · 2) / 9 = 222222
142857 · (7 · 3) / 9 = 333333
142857 · (7 · 4) / 9 = 444444
142857 · (7 · 5) / 9 = 555555
142857 · (7 · 6) / 9 = 666666
142857 · (7 · 7) / 9 = 777777
142857 · (7 · 8) / 9 = 888888
142857 · (7 · 9) / 9 = 999999

So, what did you guys think of this nice number?

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You left out
1 + 4 + 2 + 8 + 5 + 7 = 9 + 9 + 9
which works for the same reason as the other groupings.

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Hm...It seems that I missed that. Confused

Does anyone know if there is another number like this?

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See what you can do with
0588235294117647 (yes, there's a leading zero)
multiplied by 1 through 17.

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9 is a magic number
When I was a youth, my Daddy told me that 9 is a mgic number. As I got older I learned that this was truer than I realized.

Take any number and reverse the digits; then subtract the smaller from the larger. e.g. 86453 - 35468 = 50985. The result will ALWAYS be divisible by 9.

Take any number and add up the digits. If this sum results in more than 1 digit, repeat the process. The single digit result is equal to the remainder of the original number when divided by 9.

e.g., 5,736,423: 5+7+3+6+4+2+3 = 30; 3+0 = 3;

5,736,423/9 = 637,380 + 3/9

The interesting number you're playing with in your original post is a factor of 9 ....

Maybe 9 is more 'magical' than I thought ...

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Re: 9 is a magic number

pjnbarb wrote:

When I was a youth, my Daddy told me that 9 is a mgic number. As I got older I learned that this was truer than I realized.

Take any number and reverse the digits; then subtract the smaller from the larger. e.g. 86453 - 35468 = 50985. The result will ALWAYS be divisible by 9.

Take any number and add up the digits. If this sum results in more than 1 digit, repeat the process. The single digit result is equal to the remainder of the original number when divided by 9.

e.g., 5,736,423: 5+7+3+6+4+2+3 = 30; 3+0 = 3;

5,736,423/9 = 637,380 + 3/9

The interesting number you're playing with in your original post is a factor of 9 ....

Maybe 9 is more 'magical' than I thought ...

9 is a magical number for a reason. Let me introduce all of you to the wonderful definition and use of "congruence". We say x is congruent to y modulo n if n divides x-y (alternatively x is congruent to y if x modulo n = y modulo n). The wonderful thing about congruence are the following theorems:
1. a is congruent to a modulo any n.
2. if a is congruent to b modulo n, and b is congruent to c modulo n, then a is congruent to c modulo n.
3. a is congruent to b modulo n if and only if b is congruent to a modulo n.

4. if a is congruent to x modulo n, and b is congruent to y modulo n, then a+b is congruent to x+y modulo n, and a*b is congruent to x*y modulo n.

Theorems 1 through 3 are very easy to prove. I'll prove 4:

a+b-x-y=a-x+b-y and both a-x and b-y are divisible by n, therefore n divides a+b-(x+y) and therefore a+b is congruent to x+y modulo n.

a*b-x*y=(a-x+x)*(b-y+y)-x*y=(a-x)(b-y)+(a-x)y+(b-y)x+x*y-x*y=(a-x)(b-y)+(a-x)y+(b-y)x
which is obviously divisible by n. Therefore a*b is congruent to x*y modulo n.

Now... Why is this so great? Because now you can easily see, for example, that (5^20)-4 is divisible by 7.
Why?
Because 5 is congruent to -2 modulo 7.
(-2)^5=-32 which is congruent to 3 modulo 7.
(-2)^20 is therefore congruent to 3^4 modulo 7, which is 81 and is congruent to 4 modulo 7. Minus 4, this is congruent to 0 modulo 7, which means that it is divisible by 7.

Let's get back to why 9 is such a special number. That's because we have ten fingers, and we therefore count on a decimal basis. Meaning that each digit represents a power of 10. 10, however, is congruent to 1 modulo 9, and that is why 9 is important.

Lets take the example of taking a number, then flipping its digits and then subtracting. Why does that work? (why is it divisible by 9)

The first number is (10^26)a+(10^25)b+...+z (add more letters if need be). This is congruent to a+b+...+z modulo 9. We subtract a+10*b+...+(10^26)z from the original number. However, that number too is congruent to a+b+...+z, and therefore by subtracting it from the original number we receive a number that is congruent to 0 modulo 9. In other words it is divisible by 9.

Number theorists use congruence a lot. In fact the major theorem that involves congruence is the following:
If p is a prime then for every number a that is not divisible by p, there is a number b such that a*b is congruent to 1 modulo p.

I will not prove this theorem now because it isn't as simple as the things I have already proven and I'm very tired right now. If you'll ask me to, I'll prove it.

Hope I've tickled your imagination with the possibilities of number theory,

Rara Avis.

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The square root of

12345678987654321

is 111111111 (nine digits).

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I found out the 142857 when I was doing my division tables as a kid. I used tricks to remember odd fractions and this number is one of the ones I remember.

1/7= 142857 repeating
2/7= 285714 repeating
3/7= 428571 repeating
4/7= 571428 repeating
5/7= 714285 repeating
6/7= 857142 repeating

You start with 142857 and for each numerator, you take the next highest number.

When I used to tutor kids, I would show them this trick. I never realized how much more was capabile with this number. . .

Too freaking weird.

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Re: 9 is a magic number

Rara Avis wrote:

9 is a magical number for a reason. Let me introduce all of you to the wonderful definition and use of "congruence". We say x is congruent to y modulo n if n divides x-y (alternatively x is congruent to y if x modulo n = y modulo n). The wonderful thing about congruence are the following theorems:
1. a is congruent to a modulo any n.
2. if a is congruent to b modulo n, and b is congruent to c modulo n, then a is congruent to c modulo n.
3. a is congruent to b modulo n if and only if b is congruent to a modulo n.

4. if a is congruent to x modulo n, and b is congruent to y modulo n, then a+b is congruent to x+y modulo n, and a*b is congruent to x*y modulo n.

Theorems 1 through 3 are very easy to prove. I'll prove 4:

a+b-x-y=a-x+b-y and both a-x and b-y are divisible by n, therefore n divides a+b-(x+y) and therefore a+b is congruent to x+y modulo n.

a*b-x*y=(a-x+x)*(b-y+y)-x*y=(a-x)(b-y)+(a-x)y+(b-y)x+x*y-x*y=(a-x)(b-y)+(a-x)y+(b-y)x
which is obviously divisible by n. Therefore a*b is congruent to x*y modulo n.

Now... Why is this so great? Because now you can easily see, for example, that (5^20)-4 is divisible by 7.
Why?
Because 5 is congruent to -2 modulo 7.
(-2)^5=-32 which is congruent to 3 modulo 7.
(-2)^20 is therefore congruent to 3^4 modulo 7, which is 81 and is congruent to 4 modulo 7. Minus 4, this is congruent to 0 modulo 7, which means that it is divisible by 7.

Let's get back to why 9 is such a special number. That's because we have ten fingers, and we therefore count on a decimal basis. Meaning that each digit represents a power of 10. 10, however, is congruent to 1 modulo 9, and that is why 9 is important.

Lets take the example of taking a number, then flipping its digits and then subtracting. Why does that work? (why is it divisible by 9)

The first number is (10^26)a+(10^25)b+...+z (add more letters if need be). This is congruent to a+b+...+z modulo 9. We subtract a+10*b+...+(10^26)z from the original number. However, that number too is congruent to a+b+...+z, and therefore by subtracting it from the original number we receive a number that is congruent to 0 modulo 9. In other words it is divisible by 9.

Number theorists use congruence a lot. In fact the major theorem that involves congruence is the following:
If p is a prime then for every number a that is not divisible by p, there is a number b such that a*b is congruent to 1 modulo p.

I will not prove this theorem now because it isn't as simple as the things I have already proven and I'm very tired right now. If you'll ask me to, I'll prove it.

Hope I've tickled your imagination with the possibilities of number theory,

Rara Avis.

To be honest, it's the first thing I've ever seen in modular arithmatic. I'd never heard the expression before I read your post.

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what exactly is meant by the term, "modulo"?

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By "a is congruent to b modulo n" I mean n divides a-b.
The classic meaning of modulo is as in "b modulo n", which means "the number between 0 and n-1 that you get from the leftovers of when you divide b by n."
Thus "a is congruent to b modulo n" means "a modulo n = b modulo n".

It's quite an amazing field and a lot of number theory is based on it. There is one theorem in particular that number theorists use a lot:

If n and a are coprime (meaning that there aren't any primes that divide both n and a) then (a^phi(n)) is congruent to 1 modulo n (by phi(n) I mean the function that takes n and gives you back the number of numbers between 1 and n-1 which are coprime to n. In particular phi(p) when p is a prime is equal to p-1.)

Fermat proved this theorem for primes and his theorem as known as "Fermat's little theorem". Those who like to tease call it "Little Fermat's theorem".

Proving this theorem isn't very hard. I can probably write it down in one or two posts, and it involves very interesting ideas (plus the beginning of my favorite mathematical topic: "Abstract Algebra"). However, it would take a bit of my time. I would gladly prove it, however, if there is demand. Is there?

Glad to be of service,

Rara Avis.

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Re: The number 142857.

Canoy wrote:

The number 142857 is quite interesting.
So, what did you guys think of this nice number?

1·7 + 3 = 10
14·7 + 2 = 100
142·7 + 6 = 1000
1428·7 + 4 = 10000
14285·7 + 5 = 100000
142857·7 + 1 = 1000000
1428571·7 + 3 = 10000000
14285714·7 + 2 = 100000000
142857142·7 + 6 = 1000000000
1428571428·7 + 4 = 10000000000
14285714285·7 + 5 = 100000000000
142857142857·7 + 1 = 1000000000000
1428571428571·7 + 3 = 10000000000000
14285714285714·7 + 2 = 100000000000000
142857142857142·7 + 6 = 1000000000000000
1428571428571428·7 + 4 = 10000000000000000
14285714285714285·7 + 5 = 100000000000000000
142857142857142857·7 + 1 = 1000000000000000000

This little excerpt of your post I think is where the ability for this number to do what it does comes from. Like Rara stated the modulus of the equation is what gives rise to the recurring pattern. The recurring pattern in this case is really the numbers 3,2,6,4,5,1. These are the modulus or left overs if you performed long division on 1/7. Since 1 is less then seven you pop in a decimal place and put in an imaginery zero so you divide 10 by 7 and get 1 with 3 left over thus mod 3. You then divide 30 by 7 and get 4 mod 2. You then divide 20 by 7 and get 2 mod 6. Etc.

It is the fact that when you divide 50 by 7 you get 7 mod 1 that brings you back to 10 divided by 7 which causes the repetition of the chain to start all over again.

There are lots of other numbers that exhibit the same properties as 142857. They can all be obtained by locating a prime number that when 1 is divided by this number the number of decimal places before the start of the repeating modulus is equal to the original prime number minus 1.

ie: 1/19 = .052631578947368421 which is 18 decimal place and then repeats.

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Re: 9 is a magic number

Rara Avis wrote:

pjnbarb wrote:
When I was a youth, my Daddy told me that 9 is a mgic number. As I got older I learned that this was truer than I realized.

Take any number and reverse the digits; then subtract the smaller from the larger. e.g. 86453 - 35468 = 50985. The result will ALWAYS be divisible by 9.

Take any number and add up the digits. If this sum results in more than 1 digit, repeat the process. The single digit result is equal to the remainder of the original number when divided by 9.

e.g., 5,736,423: 5+7+3+6+4+2+3 = 30; 3+0 = 3;

5,736,423/9 = 637,380 + 3/9

The interesting number you're playing with in your original post is a factor of 9 ....

Maybe 9 is more 'magical' than I thought ...

9 is a magical number for a reason. Let me introduce all of you to the wonderful definition and use of "congruence". We say x is congruent to y modulo n if n divides x-y (alternatively x is congruent to y if x modulo n = y modulo n). The wonderful thing about congruence are the following theorems:
1. a is congruent to a modulo any n.
2. if a is congruent to b modulo n, and b is congruent to c modulo n, then a is congruent to c modulo n.
3. a is congruent to b modulo n if and only if b is congruent to a modulo n.

4. if a is congruent to x modulo n, and b is congruent to y modulo n, then a+b is congruent to x+y modulo n, and a*b is congruent to x*y modulo n.

Theorems 1 through 3 are very easy to prove. I'll prove 4:

a+b-x-y=a-x+b-y and both a-x and b-y are divisible by n, therefore n divides a+b-(x+y) and therefore a+b is congruent to x+y modulo n.

a*b-x*y=(a-x+x)*(b-y+y)-x*y=(a-x)(b-y)+(a-x)y+(b-y)x+x*y-x*y=(a-x)(b-y)+(a-x)y+(b-y)x
which is obviously divisible by n. Therefore a*b is congruent to x*y modulo n.

Now... Why is this so great? Because now you can easily see, for example, that (5^20)-4 is divisible by 7.
Why?
Because 5 is congruent to -2 modulo 7.
(-2)^5=-32 which is congruent to 3 modulo 7.
(-2)^20 is therefore congruent to 3^4 modulo 7, which is 81 and is congruent to 4 modulo 7. Minus 4, this is congruent to 0 modulo 7, which means that it is divisible by 7.

Let's get back to why 9 is such a special number. That's because we have ten fingers, and we therefore count on a decimal basis. Meaning that each digit represents a power of 10. 10, however, is congruent to 1 modulo 9, and that is why 9 is important.

Lets take the example of taking a number, then flipping its digits and then subtracting. Why does that work? (why is it divisible by 9)

The first number is (10^26)a+(10^25)b+...+z (add more letters if need be). This is congruent to a+b+...+z modulo 9. We subtract a+10*b+...+(10^26)z from the original number. However, that number too is congruent to a+b+...+z, and therefore by subtracting it from the original number we receive a number that is congruent to 0 modulo 9. In other words it is divisible by 9.

Number theorists use congruence a lot. In fact the major theorem that involves congruence is the following:
If p is a prime then for every number a that is not divisible by p, there is a number b such that a*b is congruent to 1 modulo p.

I will not prove this theorem now because it isn't as simple as the things I have already proven and I'm very tired right now. If you'll ask me to, I'll prove it.

Hope I've tickled your imagination with the possibilities of number theory,

Rara Avis.

Ok I'll give 'er a go

1) a is congruent to a mod n for all n (is this just applicable to positive integers?)

n|(a-a) for all n or n divides zero fo all real positive integers.

2) if a congruent to b mod n and b congruent to c mod n, then a congruent to c mod n

n|(a-b) & n|(b-c)

There are integers k & j st kn=(a-b) & jn=(b-c), so (a-b)+(b-c)=(a-c)=kn+jn=n(k+j). Therefore, n|(a-c) and a is congruent to c mod n

3) a congruent to b mod n iff b is congruent to a mod n

if a congruent to b mod n then n|(a-b)
if b congruent to a mod n then n|(b-a)
but |(a-b)|=|(b-a)| and if n|(a-b) then n|(b-a)
the other way around is also true.

Ok, the last thrm stab

If p is a prime then for every number a that is not divisible by p, there is a number b such that a*b is congruent to 1 modulo p.

assume there's a number b (is b prime to p?) st p|(ab-1)

there exist two integers s & t st a=sp+t where 0<=t<s

and s' & t' st b=s'p+t' where 0<=t'<s'

(ab)=(sp+t)(s'p+t')=ss'p^2+st'p+s'tp+tt'

therefore tt'=(ab)mod p

where t<p and t'<p

there exits two integers s" & t" st tt'=s"p+t" and t"=(tt')mod p=1 or t"=1

I think I'm lost, but I'll continue to muddle blindly

if b is arbitrary and t"=1 assume that s"=1 and tt"=p+1 which is an even number (p is odd, p not =2). then I'm still lost and getting a headache.

I'll look at this again a little later.

Nevertheless, I can see how this would be a handy tool in the box--thanks

Rap

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I'm glad to see you are enjoying the new tool I gave you. I'll start proving stuff:

gcd(a,b) means the greatest common divisor between a and b. For example gcd(12,8)=4.

Theorem (that is very basic in number theory): For any two natural (a number is called natural if it is a positive integer) numbers n and m, there are integers a and b such that an+bm=gcd(n,m).

Proof:
Lets look at the set (collection) of all the elements xn+ym for all integers x and y such that xn+ym is a natural number. Lets take the smallest number in the set, an+bm. We want to show that an+bm=gcd(n,m). First we will show that any element of the set is divisible by an+bm. Take an arbitrary element from the set: xn+ym, and divide it by an+bm with remainder:
xn+ym=q(an+bm)+r such that 0<=r<an+bm (because if the remainder is bigger than an+bm then you add one to q and you subtract an+bm from r. You repeat this process until 0<=r<an+bm).
Therefore r=(x-qa)n+(y-bq)m. Assume (we will get a contradiction) that r>0. Then that means that r is in the set (because x-qa and y-bq are integers). But r<an+bm, which is a contradiction to the fact that an+bm is the smallest element of the set. Therefore r=0. Which means xn+ym=q(an+bm) which means that any arbitrary element from the set is divisible by an+bm.
Both n and m are in the set and are therefore divisible by an+bm. Now we want to show that an+bm is the largest number with that property. That is equivalent to saying that for any number, t, such that both n and m are divisible by t, an+bm is divisible by t. That is obviously correct because if n=tk and m=tl, then an+bm=atk+btl=t(ak+bl).

QED

Why does this help? Take a prime p, and a number x that is not divisible by p. That means that gcd(p,x)=1 and therefore there are numbers a and b such that ap+bx=1. That means that bx is congruent to 1 modulo p. Tada.

Stick around. I'll post some more later. Please tell me if you like these explanation posts I'm sending.

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Proofs and format
I looked over that proof (had to print it out to use a more contemporary format with fountain pen, paper and ink). Once I did that the proof became obvious.

Moreover, I seem to remember the first part from Euclid as he developed a method to determine greatest common denominator (GCD) in Book 5 (I think). The second part, the extension using the ap+ bx=1, with a&b being integers was a twist that I didn't consider. But once you determine the gcd(p,x)=1, then application of this twist shows that bx has to be congruent to p mod 1.

Sweet

Yes please continue to pose theorems. I may even pose some question.

As for Fermat's Little Theorem (as opposed to his last) I think I need a little further clarification.

BTW a better link to St Andrews "Mathematician Biographies and History" is http://www-gap.dcs.st-and.ac.uk/~history/index.html.

Oh and I'm going to explore if there are additional fonts that might be more amenable to mathematic proofs and problems.

Rap

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It's time to introduce all of you to the concept of groups (which is exactly what I'm interested in researching later on in life). A set (collection) with a multiplication table is called a group if:

1. For every a and b in the group G, a*b is also in the group G.

2. a*(b*c)=(a*b)*c

3. There is an element that we shall call 1 such that for every a in G a*1=1*a=a.

4. For every a in G there is an element b in G (which we shall later call a^(-1)) such that a*b=b*a=1

Now notice that * doesn't have to be multiplication. Let me explain. The group you will be most familiar with is the set of all real numbers (except 0) under multiplication. However the set of all real numbers (including 0) under addition (meaning that by * I mean addition) is also a group.

Groups come in all shapes in sizes and there are many finite groups (an infinite number, actually). Notice that we did not demand that a*b=b*a, and therefore it doesn't have to.

The definition of groups isn't very interesting. What is interesting is the amazing way they are explored (especially after what is known as "the isomorphism theorems"). I will only prove some basic theorems about them:

Theorem: Let f be a function from G to G defined as f(a)=g*a for a specific g in G. The theorem says that f is 1-1 (if f(a)=f(b) then a=b) and onto (for every b in G there is an a in G such that f(a)=b).
Proof:
if f(a)=f(b) then g*a=g*b. We will multiply from the left by g^(-1) and we will get a=b. Therefore f is 1-1.

Let b be some arbitrary element in G. f((g^(-1))*b)=b. Therefore f is onto.
QED

Why does this have to do with anything? Because:
Lets look at 1,...,p-1 under regular multiplication and then congruence modulo p. This is a group (the only nontrivial thing to check is that for every a there is a^(-1), but I proved that in the last post).

Lets multiply all of the elements in the group 1*...*p-1. Because of the last theorem that equals f(1)*...*f(p-1)=(g^(p-1))*1*...*p-1 (because in this group a*b=b*a for every a and b).
So we have
(g^(p-1))*1*...*p-1=1*...*p-1
Lets multiply by (1*...*p)^(-1) from the right. We get g^(p-1)=1. Now remember that this "=" is only in the group. When we translate this to the real world we get that g^(p-1) is congruent to 1 modulo p.

Isn't Abstract Algebra the best?

I'll let you figure out why the generalization I mention earlier of Fermat's Little Theorem is correct.

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Re: The number 142857.

nipok wrote:

This little excerpt of your post I think is where the ability for this number to do what it does comes from. Like Rara stated the modulus of the equation is what gives rise to the recurring pattern. The recurring pattern in this case is really the numbers 3,2,6,4,5,1. These are the modulus or left overs if you performed long division on 1/7. Since 1 is less then seven you pop in a decimal place and put in an imaginery zero so you divide 10 by 7 and get 1 with 3 left over thus mod 3. You then divide 30 by 7 and get 4 mod 2. You then divide 20 by 7 and get 2 mod 6. Etc.

It is the fact that when you divide 50 by 7 you get 7 mod 1 that brings you back to 10 divided by 7 which causes the repetition of the chain to start all over again.

There are lots of other numbers that exhibit the same properties as 142857. They can all be obtained by locating a prime number that when 1 is divided by this number the number of decimal places before the start of the repeating modulus is equal to the original prime number minus 1.

ie: 1/19 = .052631578947368421 which is 18 decimal place and then repeats.

Awsome, now I actually have an explination for the qwirky behavior of 1/7. I often wondered how many of these repeating numbers there were, it appears as everything else, infinate. I wonder if you encorporated number theory with an infinate series, if it would add up to a set integer, say for example, they all approach 1/7th or something, or if it approaches some other quirky number.

But then again, I'll leave it to the mathmaticians to solve this problem, the main reason I became an engineer was because I likes the saying "close enough" Smile

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142857
Canoy, Nipok,

I divided the repeating portion of the multipliers on your triangle (326451) by 7 and I get 46635.85714 --- that's as far as my calculator goes but I would bet money the next digit is 2.

What a fun number!

Guido.

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The number 142857.

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