What mathematical term describes numbers whose component digits form
at least one algebraic equation?

For instance, 6,294 can be expressed as 6^2 = 9*4, but 1,137 cannot be
similarly "balanced".

How about 11 - 3 > 7

I should have clarified that the digits are to be used individually,
not combined into new numbers.

Also, the only parameter that makes the question meaningful is =.

I don't know whether or not there's a name for this phenomenon, but
there is a game based upon it:

"If you have ever looked at a set of numbers and seen a pattern or an
equation hidden in the numerals, then you will instantly be hooked by
BEDMAS."

http://www.alagan.com/

1,137   surely:  1 = 1^37  also: 1 = 1 mod 37

frde,

you arent allowed to combine digits, but your on the right track...but
1=1^3^7 is fine.


...i would like to se a set of the numbers that fit in this category.
i love doing problems like this but theres always the question of what
math operations can be used.  If there is a term for these numbers
there must be a formal definition.  If this definition exists it would
finally explain what operations count.  I would feel cheap if i solved
1132 by saying the integral of 1 from 1 to 3 was 2. 1=1^3^2 feels a
little better.

if there isnt a name yet there should be (balanced is nice).  And i
think +,-,*,/,^ (and one = sign) should suffice as operators.

the answer to your question is 'yes'.  there is a name for these
numbers and you just made it.  Now all you have to do is publish a
paper on 'em.  :P

?fractl,

GA has this amazing effect when I sit down to ask a question--my mind
blanks beyond belief.

And so:
1. 1,137 was a pretty sorry example of a non-balanced number. I think
7,784 and 910 are better (correct me if I'm wrong again)
2. You're absolutely right about the operators issue (+,-,*,/,^ and one =)
3. I would venture to guess that, much as with prime numbers, someone,
somewhere would be interested to find the largest non-balanced number.
How soon can we get a supercomputer up and running?

qna,

784 and 910 are 'unbalanced' using the 6 operators we agreed on.
The largest unbalanced number wont take a supercomputer...its most likely
not that large...PROVING that its the largest may require a bit of time
(and creativity).

I came up with a few laws regarding the set of balanced numbers that may
help you determine who belongs in the set can anyone come up with a few more?.


A NUMBER IS BALANCED IF:
1. Any group of digits that form a balanced number may be considered a 'zero'
2. Any group of digits that form a balanced number may be considered a 'one'
3. If a number contains at least two 'zeros'.
4. If a number begins with a 'one' (or a 'zero' 'one') and contains a second 
   'one'or 'zero' not next to the first.


let A,B... be arbitrary numbers of any length:

ex 1.

    Given a balanced number AB one can assume A=B.
    if this is the case A-B = 0
    if A = 13 and B=740,  1+3 = 7-4+0 so AB is balanced.
    XABY can then be X(A-B)Y or X0Y

ex 2.

    Given a balanced number AB one can assume A=B.
    if this is the case A/B = 0
    if A = 13 and B=740,  1+3 = 7-4+0 so AB is balanced.
    XABY can then be X(A/B)Y or X1Y
    in the case where A=B=0 one can not divide 0/0 to get one.
    consider 0^0 instead, to get the same result.

ex 3.

    The number A0B0C is balanced because A*0=B*0*C

ex 4.

    The number 1A1B is balanced because 1^A=1^B
    The number 1A0B is balanced because 1=A^(0*B)
    Likewise 01A1B is balanced because 0+1^A=1^B
    and 01A0B is balanced because 0+1=A^(0+B)




I'm making a program on my calculator to find the first balanced #s
i should have a set of them for you sometime today.

*wheeee this is fun!*

fractl,

I'm glad you find this subject as fascinating as I do.

My original basis for balanced numbers was that the order of the 6
operators be entirely flexible. For example, 33,333 balances as 3 ? 3
= (3 ? 3) * 3. Similarly, 242 becomes -2 + 4 = 2. Please factor these
changes into your calculations.

To add to your list of laws, a number WITH AT LEAST TWO DIGITS is
balanced if it contains:

1. Four or more digits, with one number (e.g., A,AAA)
2. Four digits, with two numbers (e.g., A,BAB or A,BBA)
3. Three digits, with two equivalent and the third a zero (e.g., A0A, AA0)
4. An even number of digits, with the first and second half containing
equivalent digits in any order (e.g., ABC,ABC or CAB,BCA).

Concerning the largest non-balanced (unbalanced?) number, I suggest we
first refine the laws to successfully exclude numbers above a certain
number of digits.

I was not allowing a negative sign in my program I interpret a -x as
0-x and i dont think the defiition of a negative is the same as a
minus sign.  is 3345 okay because 3*-3+4=-5?  Parenthesis are
inportant, though and should be used however needed (sadly my program
wasnt advanced enough to do that).
Regardless...senior TI-92 says the first 34 numbers are:

1=1, 2=2, 3=3, 4=4, 5=5, 6=6, 7=7, 8=8, 9=9, 1*0=0, 1+0=1, 1-1=0,
1*1=1, 1+1=2, 1=1^3, 1=1^4, 1=1^5, 1=1^6, 1=1^7, 1=1^8, 1=1^9, 1=2^0,
1^2=1, 1*2=2, 1+2=3, 1=3^0, 1^3=1, 1=3-2, 1*3=3, 1+3=4, 1=4^0, 1^4=1,
1=4-3, 1*4=4

I calculated the 1st 1,250 but i dont have a way to paste them online atm.
I'm going to teach it how to use parenthesis before i try to go
further, no point in generating an incomplete list.

From what I saw the frequency was pretty consistant (around 50%) ...
ill see what happens when the numbers get larger.

I like your rules.  I wonder if we can get a complete list.

i looked on the The On-Line Encyclopedia of Integer Sequences [
http://www.research.att.com/~njas/sequences/ ] and these numbers arent
there.

-fractl

Qna, I have good news and bad news...

The Bad News:

Sadly your term 'balanced' already exists.  Using the same database I
mentioned in my last comments I found the following:
"Name: Balanced numbers: numbers n such that phi(n) (A000010) divides
              sigma(n)"
This site also makes mention of it:
http://www.geocities.com/bob_kraus_2000/PUZZLES_DIGITAL.html

I really liked balanced, but perhaps a new name is in
order...'equation generating numbers'? Until a new name is found I?ll
keep using ?balanced? and ?unbalanced? though

There are many more rules I have come up with and it's becoming clear
to me that there will be no way to define the whole set of balanced
numbers using a realistic number of rules like the ones we've come up
with.

I have taught my calculator to consider parenthesis and discovered it
wont be able to compute past 4 digits.
Using @ to represent +,-,*,/,^ i found the following equation forms:

2 DIGITS:
A=B

3 DIGITS:
A@B = C
A = B@C

4 DIGITS:
A@(B@C) = D
(A@B)@C = D
A@B = C@D
A = B@(C@D)
A = (B@C)@D

There are 20 ways to write this for 5 digits, considering the 3
operators there are 2500 equations to check for EACH 5 digit number.
Even a computer would be chugging through the 6 digit numbers so
finding the largest unbalanced number may be a challenge after all (if
there is a largest).

The Good News:

I'm not giving up!
I?m working on finding the properties of numbers using only +, only -
only * etc. and I'll try to build toward the balanced set.
It's been a while since I had a problem like this to work on...I hope
you're having fun as well.

-Fractl

P.S. ?inportant? in my last post should be ?iMportant?.  I have no
idea why I forgot how to spell that.

fractl,

I think you're on the right track, but I'd like to come clean on a few issues:

1. I have no idea what type of "calculator" you're referring to
2. I'm a little rusty on my advanced math--it's been 10 years since
high school. So bear with me...
3. My sixth sense tells me the largest unbalanced number (if there is
one, as you rightly point out) contains far more digits than we've
discussed so far
4. For the record, my interest in balanced numbers took shape around
the age of 12, attempting to balance numbers on car licence plates.

Ideas on how to move forward:
1. We should plan proactively. Skipping directly to calculations may
be tempting, but won't be productive without deciding on methods first
2. We need to solicit advice from people who have experience with this
sort of thing. Going back to my initial posts, the closest similarlity
I've ever detected is to prime numbers, and you know how complicated
those can become
3. Perhaps we ought to move the discussion to a forum outside GA, one
that can get our creative juices flowing.

In the meantime, don't worry about the name. We can settle on that later...

* similarity, not similarlity *

P.S. How about "in-built" numbers?

Or may be we can call unbalanced numbers "nonbers"? :)

In-built is fine with me (?nonbers? is great...?out-built? would sound
weird) Although names aren?t important, we need a some terminology to
use.  Your initial question asked for a mathematical term so I felt I
needed to inform you ?balanced? couldn?t fly.

I have a modest TI graphing calculator collection; I computed the
first 500 in-built numbers using a TI-92+
(http://education.ti.com/images/products/graphing/92pbig.jpg) and the
next 200 using a Voyage 200
(http://education.ti.com/us/product/tech/v200/features/features.html).
 I would be lost doing math without them.

I don't know how much 'advanced math' we'll run into but I'm pretty
comfortable with it (I?m a 3rd year math major).  If we get to a
roadblock we could always post our question here (you seem to have 8
dollars left).

This does seem similar to the issues people encounter with primes just
as finding non-primes are simple and rule based (no evens, no mult. of
three etc) finding in-built numbers can be done with a rule system
too.  Primes seem tied to the nonbers (HEHE...we need to keep that!)
the only way to determine one is by brute force.  Perhaps, like
primes, we may discover ways to anticipate when the next will occur. 
Finding a formula for this will most likely be ugly if not impossible,
finding the largest remains to be seen, finding proof that the set of
nonbers is finite (or infinite) may be the first step to finding the
other two.

Although balanced numbers are new to me I have been obsessed with
license plate patterns for a while (I can determine the area of the
state a plate is from by its suffix and format).  In the obligatory
list of initials following my senior yearbook quote I included many
common plate suffixes from my area (4SI, GT, 0V0...) My games usually
consisted of making sentences out of the letters ([965·DGO] => 965
Doubtfully Gracious Ocelots).

If you know of a better place to go to discuss this I'd be glad to (we
seem to be the only ones on GA still on this topic).

As I was working on the + only, - only systems today I realized that
the ?^? is the odd man out. The ?+? is complemented by ?-?, ?*? by
?/?...if we include ?^? we should also have roots (e.g. 3^4=81..the
4?81(the 4th root of 81) =3) it doesn?t have as simple a symbol as ?^?
but it should be considered.  If not that we should only use the 4
basic operators (once we know enough about the basic four series we
could make a jump to include more).

Since this series relies so much on non-natural math, that is, things
like base 10 digits, order of operations, 5/6 meaning .833 and not 1.2
all of these things are human influenced and, as such, make it less
likely for the set to have a nice equation.
The best way to begin looking at this problem is:
1. Data collection (get the sets of numbers that belong in in-built
systems of various rules [+system, +-system, */system +-*/^system
etc.])
2. Eliminating the man made factor (what do the sets look like in
binary, make new symbols (A->B = A-B ...A<-B = B-A) etc.)
3. Build from simple sets to the +-*/^ set.
4. Definitely get others involved.

I suggest you familiarize yourself with sigma notation, and product
notations (if you aren't too up on that) I think they may be useful
tools in figuring this out.  Here are some links that may help you
with these concepts:

http://www.math.ilstu.edu/day/courses/old/305/contentsummationnotation.html
http://www.purplemath.com/modules/numbbase.htm

the next 2000 not 200 using the V200....not that it matters in the least.

I?m approaching the in-built series using only the basic 4 operators.
I have made a table breaking down each digit into the two-digit
combinations that can produce it (e.g. 1 can be made by 2-1, 3/3, 0+1
etc.)
There are 30 ways to make a 0 using 2 digits, 22 ways to make a 1 and
between 13-17 for the rest.

My logic here is that I can take an in-built number (99 for example)
and, using the table, expand it to create new in-built numbers (918,
090, 729).
Basically if AB is in-built (A having 1 digit, B having 2) and A=B
then any in-built number containing an A would still be in-built when
A is replaced by B. So, given AB=693 (6=9-3), if I know that 862 is
in-built then 8932 is as well.
I generated a reverse table, listing all 100 2-digit numbers and the 1
digit numbers they can make.  There are 19 numbers that cannot make a
digit using the four basic operators
(28,29,37,38,39,46,47,48,49,56,57,58,59,67,68,69,78,79,89).  A number
like 46789 or 56789 that only contain these numbers is sure to be a
nonber, if my logic is correct.

Using this reverse table I can test a number without needing to use
every operation. I had mentioned how slow the calculations were to
test numbers of only five digits...below is a 6 digit number proven by
hand to be a nonber. Each level reduces two digits into one, the top
line should be interpreted as follows:
(((3+5)/8)*9)-2=5
((8/8)*9)-2=5
(1*9)-2=5
9-2=5
7=5
FALSE
Only numbers that end with a double (11,22,33,44...) are in-built. 
Yes, it?s a long process...but this would have taken over 20000 lines
if done the old method.  Once this information is entered into a
program computation will be about 200 times faster.  I haven?t yet
come up with a proof that this method works, however.  I have tried it
out on a few numbers and it seems to be correct, but I plan on
developing a proof that there are no numbers that this method misses.


358925??88925??1925??925??75
      ?      ?     ?    ??97
      ?      ?     ??175??75
      ?      ?     ?    ??85
      ?      ?     ?    ??12
      ?      ?     ??197??97
      ?      ?           ??92
      ?      ??0925??025??25
      ?      ?     ?    ??05
      ?      ?     ?    ??07
      ?      ?     ??925??75
      ?      ?     ?    ??97
      ?      ?     ??075??75
      ?      ?     ?    ??05
      ?      ?     ?    ??02
      ?      ?     ??097??97
      ?      ?           ??07
      ?      ?           ??02
      ?      ??8875??175??75
      ?      ?     ?    ??85
      ?      ?     ?    ??12
      ?      ?     ??075??75
      ?      ?     ?    ??05
      ?      ?     ?    ??02
      ?      ?     ??815??75
      ?      ?     ?    ??85
      ?      ?     ?    ??95
      ?      ?     ?    ??85
      ?      ?     ?    ??86
      ?      ?     ??882??12
      ?      ?           ??02
      ?      ?           ??84
      ?      ?           ??86
      ?      ??8897??197??97
      ?             ?    ??92
      ?             ??097??97
      ?             ?    ??07
      ?             ?    ??02
      ?             ??882??12
      ?                   ??02
      ?                   ??84
      ?                   ??86
      ??35875??8875??175??75
      ?      ?     ?    ??85
      ?      ?     ?    ??12
      ?      ?     ??075??75
      ?      ?     ?    ??05
      ?      ?     ?    ??02
      ?      ?     ??815??75
      ?      ?     ?    ??85
      ?      ?     ?    ??95
      ?      ?     ?    ??85
      ?      ?     ?    ??86
      ?      ?     ??882??12
      ?      ?           ??02
      ?      ?           ??84
      ?      ?           ??86
      ?      ??3515??815??75
      ?      ?     ?    ??85 
      ?      ?     ?    ??95 
      ?      ?     ?    ??85   
      ?      ?     ?    ??86
      ?      ?     ??345??75 
      ?      ?     ?    ??79 
      ?      ?     ??355??85 
      ?      ?     ?    ??81  
      ?      ?     ?    ??80   
      ?      ?     ??365??95   
      ?      ?     ?    ??31
      ?      ?     ??355??85 
      ?      ?     ?    ??81  
      ?      ?     ?    ??80
      ?      ?     ??356??86   
      ?      ??3582??882??12
      ?             ?    ??02 
      ?             ?    ??84
      ?             ?    ??86
      ?             ??354??84
      ?             ?    ??31 
      ?             ?    ??39 
      ?             ??356??86 
      ??35875??8897??197??97 
              ?     ?    ??92 
              ?     ??097??97 
              ?     ?    ??07 
              ?     ?    ??02 
              ?     ??882??12   
              ?           ??02 
              ?           ??84
              ?           ??86 
              ??3582??882??12
                     ?    ??02 
                     ?    ??84
                     ?    ??86
                     ??354??84
                     ?    ??31 
                     ?    ??39 
                     ??356??86 
                           
                           
My battle plan is as follows:

Prove/disprove the breakdown method
Can all in-built numbers be made by expanding digits (as mentioned above)
Can nonbers be produced in a similar way?
Use this method to derive a formula for the amount of in-built numbers
or nonbers of n digits
Is there an infinite amount of nonbers?
If not, what is the largest?
Finally...expand these methods to in-built sets of all kinds
(including ones with ^, mod, !, ? etc.)
                           

If you can find any in-built numbers that cannot be ?built? as I
mentioned above or ones that don?t break down to a true statement tell
me.  That would disprove my method.
I realize this requires a lot of work (I wish I had a way to send you
the tables I made) but a problem like this needs time.  If my method
is correct...then I am SURE that 358925 is a nonber! Have you ever
been sure that a 6+ digit long number cannot be solved by doing the
calculations by hand (I know that I always worry I?ve overlooked
something)?
I have seen these problems asked in many places:
This is the only link that comes to mind, though:
http://answers.google.com/answers/threadview?id=404057

From what I know there is not a definitive method of solving these
problems other than creative thinking.
Do you know of any sites that make mention of this?  I?d like to see
if there is a better method out the-re.

-Fractl                         
 P.S. if you need any clarification about my method (I realize it
could be explained better) just ask.

Also, it may be useful to set the equation equal to 0. Ex. given a
number N that can be divided into parts A and B instead of trying to
prove that A = B ir may be more fruitful to prove that A-B = 0

Hi, I've only had a chance to quickly skim through your commments but
this seems like a very interesting topic. Just a few ideas that popped
into my head while reading.

My guess would be that there would not be a formula for finding
balanced or unbalanced numbers. This seems like something that you
would use a genetic algorithm for which basically means an educated
guess.

I know I keep coming back to this, but how exactly are prime numbers
calculated on supercomputers?

Also, if the largest known prime, at 7,235,733 digits, is
2^24,036,583-1, what range of nonbers can be expected?

Again, going forward, methodology is key.

yay...more comments!!

Grigri,
As to your idea of setting the numbers to zero?while it does work I
don?t see any advantage to that approach. 6745 is
in-built?6-7=4-5?using your method would force parenthesis (a notation
which ive come to loathe since beginning this project) into the
equation?(6-7)-(4-5)=0?the method works?but it seems a bit more
cumbersome.
This train of thought did lead me to an idea, though.  Rather than
stating the equations as true or false values, I can use the A-B=0
method to look for data on how far off numbers are from being equal.

910 is a nonber.
9-1?0 is the closest the equation gets to being equal ((9-1)-0=8 off)

789 is also a nonber
7?8/9 is the closest this gets to being equal.  Since the difference
there (6 1/9 is smaller than the one for 910 one could say it?s less
of a nonber.  Theres a chance that this may be useful in analyzing the
set of in-builts and nonbers.  It certainly raises some questions,
though:
What happens to the nonber difference rating as more digits are added?
What would the mean difference be if all possible false equations are
considered (not just the closest to 0)?would it look different for
nonbers than for in-built numbers?
What if, instead of A-B=0, I used A/B=1.  How would that affect the
data mentioned above?

I disagree with the comment saying that there is no formula for
this?not because I have any evidence to the contrary, but because I am
convinced there has to be a formula for everything.  If anything we
just lack the language to express it.  I will concede that there may
not be a practical formula, or even an obtainable formula.  I won?t be
satisfied until i'm able to prove or disprove the possibility, though.

Qna,
 To the best of my knowledge a prime is computed by determining that
the number in question can not be evenly divided by any of the prime
numbers less than the square root of the number itself.
So to test 47 we divide by 2, 3 and 5
By definition a prime is a number that has no factors other than one
and itself.  One can eliminate the non-primes because any multiple of
a non prime would have a smaller prime component.  Checking all the
numbers from 2 to the number in question is not necessary, once you
hit the square root you?ve checked them all.  For example, with
15?once you find that 3 is a factor (less than the root) you also know
that 5 is as well (above the root).
The other trick they have is to find patterns to predict large primes.
 The one you sited is a Mersenne prime (of the form Mn=2^(n-1) ) there
are other prime predicting formulae?none proven and none perfect?but
it?s an infinitely better strategy than just picking big numbers
haphazardly.  I?d like to note here that out of the first 24036583
Mersenne numbers only 41 were found to be prime.  Those are, sadly,
good odds when looking for huge primes.  I hope to develop some kind
of way to anticipate large nonbers (the table helps quite a bit).


I have absolutely no idea about the range?.it may be infinite, there
may be a largest.  Although there isn?t proof supporting it, it seems
to be the common opinion that there aren?t a limit to the number of
primes.  What exactly did you mean by range?

-Fractl

Anyone else wanna help me with this?

In response to Fractl's comment about there being a range of balanced
numbers there is definately a highest "nonber" since any number 9
digits or more is balanced. That's not to say that the highest
"nonber" is 8 digits long, there may be other limitations that we
haven't found yet but we can be certain that "nonber"  is more than 8
digits long. If me just saying something is true isn't enough evidence
feel free to look at my rudimentary proof. ;)

GIVEN 
a number N of length 9 digits
    -no digits repeat
    -the zero digit isn't present
    - 0*A = 0*B

1. given the restrictions on the digits of N we know that n is made up
of the digits 123456789 in some order.

2. pick 3 digits such that (a -(b+/-c)) = 0 
ex.
(6-(4+2))=0
(3-(1+2))=0

3. this mean that you've created a "zero" for one side of the equation
and you have 6 unique digits remaining to work with.

4. Let's assume for a moment that we have only 5 digits remaining. The
only set of five digits that by itself doesn't create another "zero"
is 13579.

5. Then if we add any other digit to 13579 we will be able to create a "zero".

6. Once we have two "zero's" we know that the number is balanced.

Actually I see now that my proof can be generalized to show that
numbers of 7 digits or higher can't be nonbers.

I dont quite understand your proof.
Why can't digits repeat? 358925 is a nonber and it has two 3's.
910 is also a nonber and it contains a zero.
Could you explain you'r logic in coming up with those statements.

I would try to look for a counterexample but i left my precious tables
at work.  I'll see if I can generate a 9+ digit nonber...im 90% sure
its possible.


You have 10% of my belief, use it well.
Fractl

Fractl: there is a very simple proof that there are an infinite number
of primes.  Here it is:

Assume there are a finite number of primes.  Multiply them all
together.  Now you have a number that is divisible by every single
prime (and is also larger than the largest prime).  Add 1 to this
number, and it is divisible by none of the primes, so it is prime. 
Therefore the assumption that the primes are finite is false.

Also, I don't know anything about how a large number is determined to
be prime, but I do know there are much much better ways than checking
every prime up to the square root of the number.  qma cited the
largest known prime at something over 7 million digits.  Its square
root has over 3 million digits.  Doing one operation for every prime
up to 3 million digits (even if they were all known, which they're
not) is I think a task which all the supercomputers in the world,
working for the age of the universe, would not even begin to make a
dent in.

Yes, for n>=17, the number of primes less than n is strictly greater than 
n/(ln(n)).  Since the natural log of a number with 3 million digits
has around 6 digits, n/ln(n) still has around 3 million digits, and
there is no way to check them all.

Tons of info on primality testing at 

http://mathworld.wolfram.com/topics/PrimalityTesting.html

GAH...I knew that!!! I learned that in 8th grade!! And I have had it
re-taught to me (in some form or another) every 2 or so years
since...but for some reason I keep thinking that the proof of infinite
primes in an unsolved mystery...I must be thinking of Goldbach's
Conjecture or something.  Thank you for setting me straight once
again, I'm sure I'll make the same mistake in a few months :P

A slight correction on your statement of Euclid?s proof (not that I
deserve to talk about it after forgetting it completely):
    The sum of all primes+1 is not necessarily a prime. 
    (2*3*5*7*11*13)+1=30030=59*509
    The proof states that the product+1 will either be prime OR yield
a prime factor that is not contained within the set (like 59 and 509).

I have recently been swamped with work and I don't have enough time to
put 100% (or even 10%) into this problem.  I plan to jump back into
this in mid-late Dec.
If there are any other GA math folks who would like to study this set
with me it would be appreciated.

-Fractl T. Carf  (yeah...racecar isn?t the only palindrome around here...)

fractl: you are thinking of the proof for infinite prime pairs.  :)