An Odd Numbers' Puzzle from an Ancient Egyptian Binary Puzzle.
Hossam Aboulfotouh
Today is January 7; and many people likes 7 and any thing that includes or starts with 7. I therefore would like to introduce the following note that seems it created a puzzle to me.
160 is the total sum of the following sequential 10 odd numbers;
7, 9, 11, 13, 15, 17, 19, 21, 23, and 25.
They form a progression series, where 16 is its hidden average; both 9 and 25 are terms/parts of this series; and the relation between these three numbers were said to be uncomprehended before the days of Pythagoras. The difference "2" between any two terms equals 1/8 of their average 16.
If one multiplied each of its 10 terms by 1/16, the results would be the 10 terms of the series in the solution of the ancient Egyptian puzzle that were numbers Problem#64 in the so-called "Rhind Mathematical Papyrus", i.e., in the papyrus, the total sum of the ten terms is 10. Its smallest term was (7 * 1/16) = 1/4+1/8+1/16 and its largest term was (25 * 1/16) = 1 + 1/2+ 1/16. See the 10 terms of RMPp#64 via this link:
http://www.math.buffalo.edu/mad/Ancient-Africa/mad_ancient_egypt_algebra.html
My note here is that, in the spectrum of odd numbers, the total sum of the series of numbers from 7 to 25 equal 160, i.e., it equals 16 * 10 = 2Ex^4 * 10Ex^1.
One could rewrite it in the form [2Ex^n * 10Ex^(n-3)], were n = 4 for the above series.
The next series of odd numbers that its total sum comply with the condition [2Ex^n * 10Ex^(n-3)] is the series of odd numbers from 7 to 505, where its sum is 64000 = (2Ex^6 * 10Ex^3). It’s the total sum of 250 odd numbers.
That is, 505 is the number that comes after 25. Using the normal ways of calculation in Excel spread sheet, I was not able to find the number(s) that come(s) after 505, because the numbers become bigger and bigger.
My question here, is there any way (an equation or so) to identify the intervals for the coming but working odd numbers? Taking into consideration that the sum of the numbers missed the interval of 3200 = 2Ex^5*10Ex^2, the case might be false at other intervals too.
Or if you can get those numbers using your own computer program, is it possible to write them here?
Best regards.
Hossam Aboulfotouh
LudovicoVan
> Greetings and Happy New Year to the Moderators, the List Members and the Readers.
Hello there: you are welcome, though maybe note that this is a Usenet
group, and not even a moderated one, so beware... :)
> Today is January 7; and many people likes 7 and any thing that includes or starts with 7. I therefore would like to introduce the following note that seems it created a puzzle to me.
>
> 160 is the total sum of the following sequential 10 odd numbers;
>
> 7, 9, 11, 13, 15, 17, 19, 21, 23, and 25.
>
> They form a progression series, where 16 is its hidden average; both 9 and 25 are terms/parts of this series; and the relation between these three numbers were said to be uncomprehended before the days of Pythagoras. The difference "2" between any two terms equals 1/8 of their average 16.
>
> If one multiplied each of its 10 terms by 1/16, the results would be the 10 terms of the series in the solution of the ancient Egyptian puzzle that were numbers Problem#64 in the so-called "Rhind Mathematical Papyrus", i.e., in the papyrus, the total sum of the ten terms is 10. Its smallest term was (7 * 1/16) = 1/4+1/8+1/16 and its largest term was (25 * 1/16) = 1 + 1/2+ 1/16. See the 10 terms of RMPp#64 via this link:http://www.math.buffalo.edu/mad/Ancient-Africa/mad_ancient_egypt_alge...
>
> My note here is that, in the spectrum of odd numbers, the total sum of the series of numbers from 7 to 25 equal 160, i.e., it equals 16 * 10 = 2Ex^4 * 10Ex^1.
>
> One could rewrite it in the form [2Ex^n * 10Ex^(n-3)], were n = 4 for the above series.
>
> The next series of odd numbers that its total sum comply with the condition [2Ex^n * 10Ex^(n-3)] is the series of odd numbers from 7 to 505, where its sum is 64000 = (2Ex^6 * 10Ex^3). It’s the total sum of 250 odd numbers.
>
> That is, 505 is the number that comes after 25. Using the normal ways of calculation in Excel spread sheet, I was not able to find the number(s) that come(s) after 505, because the numbers become bigger and bigger.
>
> My question here, is there any way (an equation or so) to identify the intervals for the coming but working odd numbers? Taking into consideration that the sum of the numbers missed the interval of 3200 = 2Ex^5*10Ex^2, the case might be false at other intervals too.
>
> Or if you can get those numbers using your own computer program, is it possible to write them here?
(Note that my notation is tentative:)
Since the sum of the first n odd naturals is:
Sum_{i=1 to n} (2^i - 1) = n^2
the problem becomes solving the following equation, with n>m and p>=3:
Sum_{i=1 to n} (2^i - 1) - Sum_{j=1 to m} (2^j - 1) = 2^p * 10^(p-3)
i.e.:
n^2 - m^2 = 2^p * 10^(p-3)
Now (and if I have made no mistakes above), I wouldn't know if that
has a closed-form solution, but by inspection with a spreadsheet I've
found that there is one solution for p=3, and multiple solutions (an
increasing number of them for increasing p) for each p>3 at least up
to p=7 (after that I couldn't check).
For instance, these are valid solutions (the first one gets in order):
p=3: n=3, m=1 ==> 3 + 5 = 8
p=4: n=13, m=3 ==> 7+...+25 = 160
p=5: n=57, m=7 ==> 15+...+113 = 3200
p=6: n=253, m=3 ==> 7+..+505 = 64000
p=7: n=1137, m=113 ==> 227+...+2273 = 1280000
-LV
Hossam Aboulfotouh
> * 10^1.
> >
> > One could rewrite it in the form [2^n *
> 10^(n-3)], were n = 4 for the above series.
> >
> > The next series of odd numbers that its total sum
> series of odd numbers from 7 to 505, where its sum is
> odd numbers.
> >
> > That is, 505 is the number that comes after 25.
> Using the normal ways of calculation in Excel spread
> sheet, I was not able to find the number(s) that
> come(s) after 505, because the numbers become bigger
> and bigger.
> >
> > My question here, is there any way (an equation or
> so) to identify the intervals for the coming but
> working odd numbers? Taking into consideration that
> the sum of the numbers missed the interval of 3200 =
> intervals too.
> >
> > Or if you can get those numbers using your own
> computer program, is it possible to write them here?
>
Reply of LudovicoVan on Jan. 7, 2011, 11:22PM
Greetings LudovicoVan;
Thank you very much for your valuable reply; in fact I was not aware about the equation for getting the sum of the odd numbers; I found its explanation/proof in one of the pages of Dr. Math: http://mathforum.org/library/drmath/view/56866.html
Thanks for your proposed equation: [n^2 - m^2 = 2^p * 10^(p-3)]
I have noticed that you have used the term m^2 in the above equation as variable; i.e., the first term in the series that comply with the condition of the total sum is diverse and 10 do not divide the number of terms in some of the found series of odd numbers. In your results, the series: [3 + 5 = 8], [15+...+113 = 3200], and [227+...+2273 = 1280000] are of this type and comprise the diverse-case in both the "first term" in the series and the "modular divider" of its number of terms, if the later is an intrinsic property in the found series. I was not expect that this case exists; it is very interesting that you have shown they are there.
My original question was to find the other type that all of them start with number 7, i.e. m^2 = 9 in the above equation. In my opinion, this case makes 10 the "modular divider" of the number of terms in all the found series; following the proposed condition by the author of Ancient Egyptian problem#64 in the so-called "Rhind Mathematical Papyrus". That is, I would like to find the other series of odd numbers that comply with the following (based on your suggestion):
[n^2 -(9) = 2^p * 10^(p-3)]
Best regards.
Hossam Aboulfotouh
LudovicoVan
Of course, that should read:
Sum_{i=1 to n} (2*i - 1) = n^2
Same below:
> > the problem becomes solving the following equation,
> > with n>m and p>=3:
>
> > Sum_{i=1 to n} (2^i - 1) - Sum_{j=1 to m} (2^j - 1)
> > 1) = 2^p * 10^(p-3)
>
> > i.e.:
>
> > n^2 - m^2 = 2^p * 10^(p-3)
>
> > Now (and if I have made no mistakes above), I
> > wouldn't know if that
> > has a closed-form solution, but by inspection with a
> > spreadsheet I've
> > found that there is one solution for p=3, and
> > multiple solutions (an
> > increasing number of them for increasing p) for each
> > p>3 at least up
> > to p=7 (after that I couldn't check).
>
> > For instance, these are valid solutions (the first
> > one gets in order):
>
> > p=3: n=3, m=1 ==> 3 + 5 = 8
>
> > p=4: n=13, m=3 ==> 7+...+25 = 160
>
> > p=5: n=57, m=7 ==> 15+...+113 = 3200
>
> > p=6: n=253, m=3 ==> 7+..+505 = 64000
>
> > p=7: n=1137, m=113 ==> 227+...+2273 = 1280000
>
> Greetings LudovicoVan;
>
> Thank you very much for your valuable reply; in fact I was not aware about the equation for getting the sum of the odd numbers; I found its explanation/proof in one of the pages of Dr. Math:http://mathforum.org/library/drmath/view/56866.html
>
> Thanks for your proposed equation: [n^2 - m^2 = 2^p * 10^(p-3)]
>
> I have noticed that you have used the term m^2 in the above equation as variable; i.e., the first term in the series that comply with the condition of the total sum is diverse and 10 do not divide the number of terms in some of the found series of odd numbers. In your results, the series: [3 + 5 = 8], [15+...+113 = 3200], and [227+...+2273 = 1280000] are of this type and comprise the diverse-case in both the "first term" in the series and the "modular divider" of its number of terms, if the later is an intrinsic property in the found series. I was not expect that this case exists; it is very interesting that you have shown they are there.
>
> My original question was to find the other type that all of them start with number 7, i.e. m^2 = 9 in the above equation. In my opinion, this case makes 10 the "modular divider" of the number of terms in all the found series; following the proposed condition by the author of Ancient Egyptian problem#64 in the so-called "Rhind Mathematical Papyrus". That is, I would like to find the other series of odd numbers that comply with the following (based on your suggestion):
>
> [n^2 -(9) = 2^p * 10^(p-3)]
Yes, I didn't have the problem statement clear.
I have tried with a spreadsheet for p up to 8, and the only solutions
I could find are the ones you already know for p=4 and p=6.
But I'd guess the above equation can be treated some way, for instance
this transformation:
n^2 - 9 = 2^p * 10^(p-3)
n^2 = 2^p * 10^(p-3) + 9
n = sqrt( 2^p * 10^(p-3) + 9 )
so the problem, I guess, becomes that of finding when the argument of
the sqrt is a perfect square. But I cannot help that far -- and I do
not guarantee for the correctness of the above anyway... :)
-LV
OwlHoot
>
> [..]
>
> n^2 - m^2 = 2^p * 10^(p-3)
If p > 0 the LHS is even, so that n +/- m are each even. So for p >= 3
the general solution can be expressed as:
(n - m)/2 = 2^a 5^c
(n + m)/2 = 2^b 5^d
or equivalently:
m = 2^b 5^d - 2^a 5^c
n = 2^b 5^d + 2^a 5^c
where a, b, c, d are non-negative integers with:
a + b = 2 p - 5
c + d = p - 3
Cheers
John Ramsden
Milo Gardner
Your suggestion
"> > 160 is the total sum of the following sequential 10
> odd numbers;
> >
> > 7, 9, 11, 13, 15, 17, 19, 21, 23, and 25.
> >
> > They form a progression series, where 16 is its
> hidden average; both 9 and 25 are terms/parts of this
> series; and the relation between these three numbers
> were said to be uncomprehended before the days of
> Pythagoras. The difference "2" between any two terms
> equals 1/8 of their average 16."
is interesting. Progressions of an arithmetic and geometric nature were well known before the time of Pythagoras.
Concerning Ahmes and the Kahun Papyrus scribe a formula allowed the finding any one of four arithmetic progression parameters, knowing the three parameters, were discussed in RMP 40, 64 and the KP per:
" ... the KP scribe used formula 1.0:
(1/2)d(n-1) + S/n = Xn (formula 1.0)
with,
d = differential, n = number of terms in the series, S = sum of the series, Xn = largest term in the series allowed three(of the four) parameters: d, n, S and Xn, to algebraically find the fourth parameter.
When n was odd, x (n/2) = S/n,
and x 1 + xn = x2 + x(n -1) = x3 + x(n -2) = ... = x(n/2) = S/n "
Egyptian scribes also looked for unity sums. The unity sum that you cited may be found in a hieratic text ... but until a sum near the type you have suggested ... I'll continue to discuss the number theory that Ahmes discussed per:
http://ahmespapyrus.blogspot.com/2009/01/ahmes-papyrus-new-and-old.html
and a third n/p conversion method discussed by the unity
sum (1)= 53/53 = 2/53 + 3/53 + 5/53 + 15/53 + 28/53
in RMP 36.
Clearly the number theory used by Pythagoras, Greeks, Hellenes, Coptics and Arabs created general purpose n/p tables. One set of tables was left in the 400 AD Akhmim Papyrus that cited n/3, n/5, n/7, ...m n/33 tables .., with the n/17 and n/19 tables being of particular interest.
Thank you for ponder ancient number theory, and a few of the skills that ancient scribes may have possessed.
Best Regards,
Milo Gardner
Hossam Aboulfotouh
If n^2= 64009, and m^2=9
What are the values of your suggested [a, b, c, & d] for the above numbers. And what is the advantage(s) of your solution; in my opinion, using the suggestion of LudovicoVan [n^2 - m^2 = 2^p * 10^(p-3)] is easier.
Also, if you have applied your suggestion numerically, did you find other series of odd numbers that comply with the above equation that either starts with 7 or other numbers.
Regards
Hossam Aboulfotouh
Hossam Aboulfotouh
Greetings Milo,
Away from the classical understanding of, and talks on, the mathematical hieratic text of the ancient Egyptians, in this thread I see the issue from different perspective.
In my opinion, the essence of the series of the following 10 odd numbers: 7, 9, 11, 13, 15, 17, 19, 21, 23, and 25, is its inherent intrinsic property that it is somehow hints to the so-called the Pythagorean idea for finding the hypotenuse of the right angled triangle that its sides are 3 and 4.
If you applied the same ancient Egyptian reckoning process used in RMPp64 on the above series. That is, by using a problem statement that says "a perimeter of a circle equal 160 and divided into 10 parts, were the difference between each of them is 1/8 of their average; find the terms of the series?"
If you followed the same process of solution of RMPp64, you will find out at the end that the way to get the largest number of the series is to add 9 to 16 and get 25. Taking into consideration that each of the terms of the series in RMPp64 equal 1/16 of its counterpart in the sequence of the above series of odd numbers.
Regards
Hossam Aboulfotouh
http://fotouh.netfirms.com
Milo Gardner
The context of hieratic math often was the equal distribution of one or more products as wages. Equal was defined by proportion and arithmetic progressions per:
http://ahmespapyrus.blogspot.com/2009/01/ahmes-papyrus-new-and-old.html
and a dozen other blogs, and internet sites.
Your point of view tends to be modern in scope per:
"Away from the classical understanding of, and talks on, the mathematical hieratic text of the ancient Egyptians, in this thread I see the issue from different perspective.
In my opinion, the essence of the series of the following 10 odd numbers: 7, 9, 11, 13, 15, 17, 19, 21, 23, and 25, is its inherent intrinsic property that it is somehow hints to the so-called the Pythagorean idea for finding the hypotenuse of the right angled triangle that its sides are 3 and 4.
If you applied the same ancient Egyptian reckoning process used in RMPp64 on the above series. That is, by using a problem statement that says a perimeter of a circle equal 160 and divided into 10 parts, were the difference between each of them is 1/8 of their average; find the terms of the series?
If you followed the same process of solution of RMPp64, you will find out at the end that the way to get the largest number of the series is to add 9 to 16 and get 25. Taking into consideration that each of the terms of the series in RMPp64 equal 1/16 of its counterpart in the sequence of the above series of odd numbers."
Neither of our points of view are classical in scope. My view reports that well defined scribal mathematics actually used by scribes to solve ancient problems.
Your point of view takes an aspect of an ancient scribal problem and overlays a modern situation exposing a possible ancient math skill.
For myself I prefer a large set of direction connections to scribal math. RMP 64, for example, is decoded by documenting its inner arithmetic progression workings to RMP 40 and the Kahun Papyrus, as well as the number theory of the 2/n table and its two additional n/p conversion rules.
Your point of view provides speculation that modern math proposes additional aspects of an ancient math problem without actually revealing the ancient math context, and the number theory used by scribes, placing a modern cart before the ancient horse.
Best Regards,
Milo Gardner
Hossam Aboulfotouh
>
> > Greetings Hossam,
> >
> > Your suggestion
> >
> > "> > 160 is the total sum of the following
> sequential
> > > odd numbers;
> > > >
> on, the mathematical hieratic text of the ancient
> Egyptians, in this thread I see the issue from
> different perspective.
>
> In my opinion, the essence of the series of the
> following 10 odd numbers: 7, 9, 11, 13, 15, 17, 19,
> 21, 23, and 25, is its inherent intrinsic property
> that it is somehow hints to the so-called the
> Pythagorean idea for finding the hypotenuse of the
> right angled triangle that its sides are 3 and 4.
>
> If you applied the same ancient Egyptian reckoning
> process used in RMPp64 on the above series. That is,
> by using a problem statement that says "a perimeter
> of a circle equal 160 and divided into 10 parts, were
> the difference between each of them is 1/8 of their
> average; find the terms of the series?"
>
> If you followed the same process of solution of
> RMPp64, you will find out at the end that the way to
> get the largest number of the series is to add 9 to
> 16 and get 25. Taking into consideration that each of
> the terms of the series in RMPp64 equal 1/16 of its
> counterpart in the sequence of the above series of
> odd numbers.
>
On Jan 11, 2011 10:13 AM , Milo Gardner wrote
Milo,
Once again, the title of the thread is "An Odd Numbers' Puzzle from an Ancient Egyptian Binary Puzzle" i.e., we are speaking about new puzzle form an ancient Egyptian puzzle, where the solution of the 1st series of the new puzzle comply with the Pythagorean law of the right angled triangle.
In order to observe the similarity, solve the following:
A perimeter of a circle equal 160 units and divided into 10 parts, where the difference between each of them is 1/8 of their average; find the terms of the series?"
Based on the solution of RMPp64, the solution of the above is as follows.
The number of terms 10 minus 1 gives 9
The difference is 1/8 of 16, which equals 2
1/2 of the difference equals 1
9 times 1 gives 9
9 plus the average 16 gives the largest term 25
The terms of the series are: 7, 9, 11, 13, 15, 17, 19, 21, 23, and 25. Total 160.
Finlay, and this is the core objective of the thread, what are the other series of odd numbers that starts with 7 and comply with this equation [n^2 - 9 = 2^p * 10^(p-3)] based on the suggestion of LV.
7, 9,…., 25 (total sum = 2^4* 10^1)
7, 9,…., 505 (total sum = 2^6*10^3)
7, 9,…., 362038671 (total sum = 2^15*10^12)
The aim is to find the other series that start with 7.
quasi
wrote:
>Finally, and this is the core objective of the thread, what are the other series of odd numbers that starts with 7 and comply with this equation [n^2 - 9 = 2^p * 10^(p-3)] based on the suggestion of LV.
>
>7, 9,…., 25 (total sum = 2^4* 10^1)
>7, 9,…., 505 (total sum = 2^6*10^3)
The one below is incorrect (it doesn't sum to what you claim).
>7, 9,…., 362038671 (total sum = 2^15*10^12)
>
>The aim is to find the other series that start with 7.
What about
7 + 9 = 2^4 * 10^0
and
7 + 9 + 11 + 13 = 2^2 * 10^1
??
In any case, if you are trying to find pairs n,p of integers
satisfying the Diophantine equation
n^2 - 9 = 2^p * 10^(p-3)
the only solutions are
n = 25, p = 4
n = 505, p = 6
which correspond to the first two sequences you listed above.
Clearly we must have p >= 3, otherwise the RHS is not an integer. By
inspection, p = 4 and p = 6 work, and p = 3 and p = 5 fail.
Suppose there is a solution with p > 6.
Clearly n must be an odd positive integer, so write
n = 2k - 1
where k is a positive integer. Then
(k-2)(k+1) = 2^(2p-5) 5^(p-3)
If k is even then we must have
k-2 = 2^(2p-5)
k+1 = 5^(p-3)
and if k is odd then we must have
k-2 = 5^(p-3)
k+1 = 2^(2p-5)
But the function f(p) = 5^(p-3) - 2^(2p-5) is greater than 3 for all
real p > 6.
Thus for p > 6,
k even implies
3 = (k+1) - (k-2) = f(p) > 3, contradiction
and k odd implies
3 = (k+1) - (k-2) = -f(p) < -3, contradiction,
hence, for p > 6, there are no solutions.
But let's take a step back.
I don't think the equation
n^2 - 9 = 2^p * 10^(p-3)
correctly models what you are seeking.
You want find all positive integers n > 4 such that
7 + 9 + ... + (2n - 1)
has the form 2^a 10^b for some nonnegative integers a,b.
Right?
For now, I'll assume that's your objective.
The equation
7 + 9 + 11 + ... + (2n - 1) = 2^a 10^b
yields
(n+3)(n-3) = 2^a 10^b
so it's clear that n must be odd.
Write n = 2k - 1 where k is a positive integer.
After simplifying, the new equation is
(k+1)(k-2) = 2^c 5^b
where c = a+b-2.
But k+1 and k-2 have no common factor except perhaps 3, and in this
case, a common factor of 3 is impossible (since 2^c 5^b is not a
multiple of 3). Hence k+1 and k-2 are relatively prime.
It follows that either
k-2 = 2^c and k+1 = 5^b
or
k-2 = 5^b and k+1 = 2^c
according as k is even or odd.
Thus your question reduces to finding all pairs b,c of integers
satisfying the Diophantine equation
|5^b - 2^c| = 3
By inspection we find
b = 0, c = 2 => 7 + 9 = 2^4 10^0
b = 1, c = 1 => 7 + 9 + 11 + 13 = 2^2 10^0
b = 1, c = 3 => 7 + 9 + ... + 25 = 2^4 10^1
b = 3, c = 7 => 7 + 9 + ... + 505 = 2^6 10^3
Are there other solutions? Probably not, but these kinds of
Diophantine equations are often not so easy to resolve, especially
when you already know that _some_ solutions exist. In particular, it's
no use trying to show that the equation fails mod m, for some m.
To recap, you have two separate Diophantine equations to be solved in
integers b,c
(1) 5^b - 2^c = 3
(2) 2^c - 5^b = 3
Equation (1) has the known solution
b = 1, c = 1
and equation (2) has the known solutions
b = 0, c = 2
b = 1, c = 3
b = 3, c = 7
If there are any other solutions, that will yield other sequences
satisfying the required conditions, but I doubt that any other
solutions exist.
quasi
Milo Gardner
Stefen, a Romanian colleague applies right triangles to generally convert rational numbers to unit fraction series per
"1/14 = (1/14)*(14/14) = 14/196
196 : 1,2,4,7/,14,....
1/14 = (7 + 4 + 2 + 1)/(2*2*7*7)
= 1/28 + 1/49 + 1/98 + 1/196
How did I find your LCM = 14(not ELCM which is mine) ?
1/196 + 1/49 + 1/28 + 1/98 = (1 + 4 + 7 + 2)/196 = 14/196 = (1/14)*(14/14)
ELCM is LCM which allows me (or egyptians) to wright 1 as a serie of unit fractions.
It's an wonderfull thing I did not learn at school about. I learned from Ahmes ."
much as you discuss RMP 64:
"Once again, the title of the thread is "An Odd Numbers' Puzzle from an Ancient Egyptian Binary Puzzle" i.e., we are speaking about new puzzle form an ancient Egyptian puzzle, where the solution of the 1st series of the new puzzle comply with the Pythagorean law of the right angled triangle.
In order to observe the similarity, solve the following:
A perimeter of a circle equal 160 units and divided into 10 parts, where the difference between each of them is 1/8 of their average; find the terms of the series?"
Based on the solution of RMPp64, the solution of the above is as follows.
The number of terms 10 minus 1 gives 9
The difference is 1/8 of 16, which equals 2
1/2 of the difference equals 1
9 times 1 gives 9
9 plus the average 16 gives the largest term 25
The terms of the series are: 7, 9, 11, 13, 15, 17, 19, 21, 23, and 25. Total 160.
Finlay, and this is the core objective of the thread, what are the other series of odd numbers that starts with 7 and comply with this equation [n^2 - 9 = 2^p * 10^(p-3)] based on the suggestion of LV.
7, 9,?., 25 (total sum = 2^4* 10^1)
7, 9,?., 505 (total sum = 2^6*10^3)
7, 9,?., 362038671 (total sum = 2^15*10^12)
The aim is to find the other series that start with 7."
Ahmes wrote up a beautiful mod 7 problem in RMP 79, within a geometric series, that was passed down from 1650 BCE to the medieval era (1202 AD) almost exactly as originally recorded. The New York Times discussed it last month within the oldest known puzzle:
http://www.nytimes.com/2010/12/07/science/07first.html?_r=2&ref=science
In indirect ways you are correct to stress hieratic mod 7 math issues. Your 'first cut' of RMP 64, as Ahmes wrote it, connects to RMP 79 and the 2/n table by Egyptian unit fraction series methods.
In addition, I see possibilities that Stefan's right triangle method (as cited above) and your proto-Pythagorean triples points of view fit into this ancient story-line.
My historical text point of view pieces ancient math puzzle ideas and methods together by taking an idea here (RMP 64, 40, the 2/n table), there (Kahun Papyrus) and within well documented problems (RMP 79 and the Liber Abaci).
The resultant larger puzzle opens proto-number theory doors that vividly document Ahmes scaled rational numbers n/p by the best LCM m to mn/mp. At this point Ahmes inspected divisors of mp to record the best divisors in red numbers that equaled numerator mn.
By the time of Fibonacci rational numbers n/p were scaled by LCM m in a subtraction context, i.e. (n/p- 1/m) = (mn -p)/mp with (mn -p) set to unity whenever possible, before a right to left unit fraction series was recorded.
Several of the ancient 2050 BCE to 1202 AD doors hint of proto-Pythagorean thinking, as you aptly suggest. One day we will agree across a wider spectrum of math ideas.
Thanks for the discussion.
Best Regards,
Milo Gardner
Hossam Aboulfotouh
> >
> > > Greetings Hossam,
> > >
> > > Your suggestion
> > >
> > > "> > 160 is the total sum of the following
> > sequential
> > > 10
> > > > odd numbers;
> > > > >
> > > > >
> 19,
> > 21, 23, and 25, is its inherent intrinsic property
> > that it is somehow hints to the so-called the
> > Pythagorean idea for finding the hypotenuse of the
> > right angled triangle that its sides are 3 and 4.
>
> >
> > If you applied the same ancient Egyptian reckoning
> > process used in RMPp64 on the above series. That
> is,
> > by using a problem statement that says "a
> > of a circle equal 160 and divided into 10 parts,
> were
> > the difference between each of them is 1/8 of
> their
> > average; find the terms of the series?"
> >
> 19,
> > 21, 23, and 25, is its inherent intrinsic property
> > that it is somehow hints to the so-called the
> > Pythagorean idea for finding the hypotenuse of the
> > right angled triangle that its sides are 3 and 4.
> >
> > If you applied the same ancient Egyptian reckoning
> > process used in RMPp64 on the above series. That
> is,
> of
> > a circle equal 160 and divided into 10 parts, were
> > the difference between each of them is 1/8 of
> their
> > average; find the terms of the series?
> >
> 7, 9,…., 505 (total sum = 2^6*10^3)
> 7, 9,…., 362038671 (total sum = 2^15*10^12)
>
> The aim is to find the other series that start with
> 7.
>
On Jan 12, 2011 9:02 AM Milo Gardner wrote
Milo,
Your reply is not related to the core objective of this thread.
Solve the new puzzle.
Regards
Hossam Aboulfotouh
Hossam Aboulfotouh
Thank you very much for your extended reply
Based on this equation
n^2 - 9 = 2^p * 10^(p-3)
For p =15, p-3=12
My calculations were proceeded as follows:
2^15=32768
n = SQRT[9+(32768 * 10^12)] = 181019336
Largest term of the series = (181019336 * 2) - 1 = 362038671
That is the series: 7, 9,…. , 362038671 is the third one; however its inner intrinsic property is different from the other two series.
One of these properties is the value of the average, i.e.,
2^4 or 16 is the average of 7, 9, …., 25
2^8 or 256 is the average of 7, 9, …., 505
While the average of the third series equals 181019339
quasi
wrote:
>Based on this equation
>
>n^2 - 9 = 2^p * 10^(p-3)
>
>For p =15, p-3=12
>
>My calculations were proceeded as follows:
>
>2^15=32768
>
>n = SQRT[9+(32768 * 10^12)] = 181019336
No, square it back and you'll see the error.
In fact, 9 + (2^15 * 10^12) is not a perfect square,
Rounded to 4 decimal places,
sqrt(9 + (2^15 * 10^12)) = 181019335.9836
quasi
quasi
Oops -- I meant 181019335.9838
quasi
Hossam Aboulfotouh
Thank you very much; yes you are right, it doesn't.
The cells in my new Excel 7 spread sheet were automatically adjusted in some cells to show numbers without the decimal digits.
In your first reply you said:
" If there are any other solutions, that will yield other sequences
satisfying the required conditions, but I doubt that any other
solutions exist."
The relation between the two averages of the first two series relatively comply with the astronomical distances in our solar system, one is 16 times the other, like the relation between the maiden orbital distance of Mercury and Jupiter, 45 Mkm and 720 Mkm.
2^4 or 16 is the average of 7, 9, …., 25
2^8 or 256 is the average of 7, 9, …., 505
In one of my previous theoretical studies this relation is some harmonic, see my equations via this link:
http://fotouh.netfirms.com/spin-gravity-S2-music-spheres.htm
And this is the reason or the motive to try to find other series of this type, in the spectrum of odd numbers. And I still believe that there should be others, but my computer, even after adjusting its cells' numerical format for 40 round digits, is not able to show correct results after p = 22.