BR Maths Corner1

 BRFite
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Re: BR Maths Corner1
just wondering if any BRFites are working/have worked on the mathematics of viscous fluid flows?? I am writing a paper and need some input/ advice
Re: BR Maths Corner1
Okay  A little quiet here.. Let me post here an old problem...
When I was in High school (or 1st year in college) one of the Jain Guru asked me (I was considered good in Math) an old problem which old Jain Mathematicians worked on.
The numbers:
11
111
1111
11111
etc...
They knew 11 was prime, and wanted to see if any such other number is prime.
First , simple to prove, any number where '1' is not being repeated prime number of times is not prime. (so all numbers like
1111, 111111, etc are not prime .
(Above is very easy to prove)
They knew, 111 is not prime (111 divisible by 3)
11111 is not prime (divisible by 41)
and also 1111111 is not prime (divisible by 239)
What I was asked, (and told that old Jain Mathematician did not know the answer to)
is '1' '11 times that is 11111111111
is prime or not?
So assume, you don't have wiki/internet to search and do not have a computer to do bruteforce search..(But allowed to use a calculator) can you answer that question.
(I always tell my sons  In our days we used bamboo slide rules and log tables
It is an open ended question  Any method which makes searching for a factor or (proving the number is prime or not) a little bit easier would be nice...

I will give the answer (to what is the factor  though now any computer/calculator will do it in a second  not to mention, few mouse clicks in google) , and comments (if there is any interest)
I enjoyed the challenge then, learned quite some math in the process, and did provide the answer to the Jain Guru, who was impressed and wished me that I always will enjoy math.. I still do.
(Of course, you can also deal with 1111111111111 or 1111111111111111 or other such numbers)
When I was in High school (or 1st year in college) one of the Jain Guru asked me (I was considered good in Math) an old problem which old Jain Mathematicians worked on.
The numbers:
11
111
1111
11111
etc...
They knew 11 was prime, and wanted to see if any such other number is prime.
First , simple to prove, any number where '1' is not being repeated prime number of times is not prime. (so all numbers like
1111, 111111, etc are not prime .
(Above is very easy to prove)
They knew, 111 is not prime (111 divisible by 3)
11111 is not prime (divisible by 41)
and also 1111111 is not prime (divisible by 239)
What I was asked, (and told that old Jain Mathematician did not know the answer to)
is '1' '11 times that is 11111111111
is prime or not?
So assume, you don't have wiki/internet to search and do not have a computer to do bruteforce search..(But allowed to use a calculator) can you answer that question.
(I always tell my sons  In our days we used bamboo slide rules and log tables
It is an open ended question  Any method which makes searching for a factor or (proving the number is prime or not) a little bit easier would be nice...

I will give the answer (to what is the factor  though now any computer/calculator will do it in a second  not to mention, few mouse clicks in google) , and comments (if there is any interest)
I enjoyed the challenge then, learned quite some math in the process, and did provide the answer to the Jain Guru, who was impressed and wished me that I always will enjoy math.. I still do.
(Of course, you can also deal with 1111111111111 or 1111111111111111 or other such numbers)
Re: BR Maths Corner1
AmberG ji,
It is an elementary but interesting problem.I cant figure out why 1111...1 say 35 times(not a multiple of 3) is not prime.Much less your problem.Any hints.Request some tutoring help.
It is an elementary but interesting problem.I cant figure out why 1111...1 say 35 times(not a multiple of 3) is not prime.Much less your problem.Any hints.Request some tutoring help.

 BRF Oldie
 Posts: 12410
 Joined: 19 Nov 2008 03:25
Re: BR Maths Corner1
Its a GP. So any nontrivial odd number of significant digits will end up as nonprime.

 BRF Oldie
 Posts: 7979
 Joined: 31 May 2004 11:31
 Location: The rings around Uranus.
Re: BR Maths Corner1
I tried a bruteforce approach, but that only took me so far. Alas, my madrassa math skills are limited. The key here is trying to understand repeating digits. From Amberji we know that:
11  2 digits, prime
111  3 digits, nonprime, factors 3 & 37
1111  4 digits, nonprime, factors 11 & 101
1111111  7 digits, nonprime, factors I don't know
11111111111  11 digits, based on information above I assume nonprime
Please do tell us how to figure out the algorithm simple enough for us Zaid Hamid types?
11  2 digits, prime
111  3 digits, nonprime, factors 3 & 37
1111  4 digits, nonprime, factors 11 & 101
1111111  7 digits, nonprime, factors I don't know
11111111111  11 digits, based on information above I assume nonprime
Please do tell us how to figure out the algorithm simple enough for us Zaid Hamid types?
Re: BR Maths Corner1
krishnapremi wrote:AmberG ji,
It is an elementary but interesting problem.I cant figure out why 1111...1 say 35 times(not a multiple of 3) is not prime.Much less your problem.Any hints.Request some tutoring help.
Because 111....1 (35 times) is divisible by 11111
(and also by 1111111 )
((Knowing 35 is divisible by 5)_
Easy to check ...
(11111,11111,11111,11111,11111,11111,11111 = 11111 * 1 00001 00001 00001 00001 00001 00001)
(This is same as saying that (a^n1) is divisible by (a1) )
( Note that (1111...35 times is same as (10^35 1 )/9 and if a=10^5 and n = 7 you get the same thing)

 BRF Oldie
 Posts: 12410
 Joined: 19 Nov 2008 03:25
Re: BR Maths Corner1
Note that the series can be represented as a GP:
1+a+a^2+....+a^n, where a=10.
So the number is S=(a^{n+1}1)/(a1).
Note that if you have n=odd, implies n+1=even=2k+1 say where k<n.
if n+1i seven, this implies the numerator can be factored at least once (a^k1)(a^k+1).
Repeat this until you reach k=2 or k odd.
If k is odd, a^k1 is divisible by a1 which means S has nontrivial factors.
If k=2,
then also a^21 is divisible by a1=9.
1+a+a^2+....+a^n, where a=10.
So the number is S=(a^{n+1}1)/(a1).
Note that if you have n=odd, implies n+1=even=2k+1 say where k<n.
if n+1i seven, this implies the numerator can be factored at least once (a^k1)(a^k+1).
Repeat this until you reach k=2 or k odd.
If k is odd, a^k1 is divisible by a1 which means S has nontrivial factors.
If k=2,
then also a^21 is divisible by a1=9.

 BRF Oldie
 Posts: 7979
 Joined: 31 May 2004 11:31
 Location: The rings around Uranus.
Re: BR Maths Corner1
Ok, so:
111...1 (11 times) is represented by (10^11  1)/9, then what are the actual factors since 11 is not divisible?
111...1 (11 times) is represented by (10^11  1)/9, then what are the actual factors since 11 is not divisible?

 BRF Oldie
 Posts: 12410
 Joined: 19 Nov 2008 03:25
Re: BR Maths Corner1
No that will be (10^{11+1}1)/(101).
So (10^121)/9 =(10^61)(10^6+1)/9 =(10^31)(10^3+1)(10^6+1)/9
But 10^31=(101)(10^2+10+1).
So this give sthe factors as (10^2+10+1), (10^3+1), (10^6+1)
So (10^121)/9 =(10^61)(10^6+1)/9 =(10^31)(10^3+1)(10^6+1)/9
But 10^31=(101)(10^2+10+1).
So this give sthe factors as (10^2+10+1), (10^3+1), (10^6+1)
Re: BR Maths Corner1
brihaspati wrote:Note that the series can be represented as a GP:
1+a+a^2+....+a^n, where a=10.
So the number is S=(a^{n+1}1)/(a1).
.
Actually even without worrying about GP you just see:
11111... n times = (99999...n times)/9 = (10^n1)/(101)
which is similar to saying "a=10 and S=(a^{n+1}1)/(a1)
(Except there is a typo and S should be (a^n1)/(a1) (because your GP goes only up to 1+a+a^2 ...a^(n1) to get 111.. n times)
So this give sthe factors as (10^2+10+1), (10^3+1), (10^6+1)
(Does not work, as there was a small error in previous steps..you need to factor (10^111) and NOT (10^121) , (apart from trivial factor of 9)
HTH
(Added later: Just to see that one can check that ((10^6) + 1) * ((10^3) + 1) * 111 = 111 111 111 111 (twelve 1's ).
Any way any factors of 11111111111 yet? (or is it prime)?
Last edited by Amber G. on 23 Dec 2009 08:46, edited 2 times in total.
Re: BR Maths Corner1
brihaspati wrote:Note that if you have n=odd, implies n+1=even=2k+1 say where k<n.
Bji,nitpick again.For accuracy sake.
AmberG ji,
Thanks for educating this madrassa math type.

 BRF Oldie
 Posts: 7979
 Joined: 31 May 2004 11:31
 Location: The rings around Uranus.
Re: BR Maths Corner1
Amberji,
I couldn't bear to look anymore and did find the factors, so 11111111111 is not a prime, but I did not look up the algorithm. Please do tell us the algorithm.
Thanks.
I couldn't bear to look anymore and did find the factors, so 11111111111 is not a prime, but I did not look up the algorithm. Please do tell us the algorithm.
Thanks.
Re: BR Maths Corner1
Math is Fun in the sense that one finds inspiration from what seems to be unrelated problems.
For example see my problem about "ankan" (Prob 2 )
from http://forums.bharatrakshak.com/viewtopic.php?f=24&t=4201&p=519411&hilit=ankan#p519411
And solutions related to that. (what if we looking for ankan=1 (actually all 1's)
Here are some related problems which old Jain (and other Hindu) Mathematician worked: (And yes, there may be some relationships with my problem with proving that '11111...' may be prime or what kind of factors that have.
Now let me state that at first blush it may appear, completely unrelated problem, (but there is relationship to above (1111...'s) problem))
So let us
****** (changing gears)
We all know, if a number is rational (that is, can be written as (p/q) where p and q both are integers and q is not zero) it can be written as repeating decimals.
So, now consider a prime n , greater than 5,
what is the pattern of decimals notation of 1/n ?
Specially what is the length of the cycle?
For example: (Let me do the work for first few primes for you ..
1/7 = .142857 142857 142857 ... ( repeats after 6, or cycle lenght =6)
1/11 = .09 09 09 (cycle = 2)
1/13 = .076923 076923 076923 ... (cycle = 6)
or 1/17 = .0588235294117647 0588235294117647 ... (cycle = 16)
1/19 =.052631578947368421 052631578947368421 ... (Cycle = 18)
or take some larger prime
1/41 = .02439 02439 02439 024390 ... (cycle = 5)
or another one..
1/53 = .0188679245283 0188679.... ( cycle = 13)
As I said before, it is open ended question so any comments are welcome... Can anything be said about the relationship between cycle length and n
Happy Holidays!
P.S yes, 11111111111 is not prime, and is = 21649 * (513239) ... But more of that later...
For example see my problem about "ankan" (Prob 2 )
from http://forums.bharatrakshak.com/viewtopic.php?f=24&t=4201&p=519411&hilit=ankan#p519411
And solutions related to that. (what if we looking for ankan=1 (actually all 1's)
Here are some related problems which old Jain (and other Hindu) Mathematician worked: (And yes, there may be some relationships with my problem with proving that '11111...' may be prime or what kind of factors that have.
Now let me state that at first blush it may appear, completely unrelated problem, (but there is relationship to above (1111...'s) problem))
So let us
****** (changing gears)
We all know, if a number is rational (that is, can be written as (p/q) where p and q both are integers and q is not zero) it can be written as repeating decimals.
So, now consider a prime n , greater than 5,
what is the pattern of decimals notation of 1/n ?
Specially what is the length of the cycle?
For example: (Let me do the work for first few primes for you ..
1/7 = .142857 142857 142857 ... ( repeats after 6, or cycle lenght =6)
1/11 = .09 09 09 (cycle = 2)
1/13 = .076923 076923 076923 ... (cycle = 6)
or 1/17 = .0588235294117647 0588235294117647 ... (cycle = 16)
1/19 =.052631578947368421 052631578947368421 ... (Cycle = 18)
or take some larger prime
1/41 = .02439 02439 02439 024390 ... (cycle = 5)
or another one..
1/53 = .0188679245283 0188679.... ( cycle = 13)
As I said before, it is open ended question so any comments are welcome... Can anything be said about the relationship between cycle length and n
Happy Holidays!
P.S yes, 11111111111 is not prime, and is = 21649 * (513239) ... But more of that later...
Re: BR Maths Corner1
Happy New year.
Here is a beautiful geometry theorem which most people I know have not heard before. (The theorem was first published in
19??, which is really surprising because one would guess that such a simple theorem should have been known long ago)
Take any triangle, trisect all angles, the triangle which is formed inside (see below for clarification) will always be an equilateral triangle. Can you prove it?
(ABC is any triangle, P,Q and R are three points inside such as, Angles RAB,=RAQ=QAC  each 1/3 or BAC  , similarly RBA=RBP=PBC and BCP=PCQ=QCA prove PQR is equilateral triangle )
Here is a beautiful geometry theorem which most people I know have not heard before. (The theorem was first published in
19??, which is really surprising because one would guess that such a simple theorem should have been known long ago)
Take any triangle, trisect all angles, the triangle which is formed inside (see below for clarification) will always be an equilateral triangle. Can you prove it?
(ABC is any triangle, P,Q and R are three points inside such as, Angles RAB,=RAQ=QAC  each 1/3 or BAC  , similarly RBA=RBP=PBC and BCP=PCQ=QCA prove PQR is equilateral triangle )
Re: BR Maths Corner1
Ok First some disclaimers. I am not a guru in anything significant least of all math.. So this math problem is going to be trivial for the BR folks that tend to hang around in this neighbhood. However I got into this becuase my kid is in middle school and I was looking for a way to encourage his intrest in the subject.
This problem seems doable for us layman dads, moms or kids as the case might be...
Got this problem from the here http://www.moems.org/zinger.htm
The first ten numbers in a sequence are
1, 2, 2, 3, 3, 3, 4, 4, 4, 4, ....
What is the 500th number in the sequence?
I will be honest.. It took me at least a couple of hours to solve this problem and had to research some formulas first..
This problem seems doable for us layman dads, moms or kids as the case might be...
Got this problem from the here http://www.moems.org/zinger.htm
The first ten numbers in a sequence are
1, 2, 2, 3, 3, 3, 4, 4, 4, 4, ....
What is the 500th number in the sequence?
I will be honest.. It took me at least a couple of hours to solve this problem and had to research some formulas first..
Re: BR Maths Corner1
Jayram wrote:
What is the 500th number in the sequence?
.
Thirty two
Re: BR Maths Corner1
1, 2, 2, 3, 3, 3, 4, 4, 4, 4, ....What is the 500th number in the sequence?
It's a classic and the answer could be ANYTHING! for example the logic "MY 500th term is 37 is as valid as any other. Point is given any finite amount of terms in the series, guessing the next number is more of a 'puzzle' than math.
So unless some one tells the pattern explicitly , that is , 1 is repeated once, then 2 is repeated twice, and 3 is repeated thrice...etc (number n is repeated n times).. the problem is poorly worded. (I know, that kind of problems do appear but any serious mathematician will find the problem poorly stated... and I belive it ought not appear in serious math competition so I am a little surprised that it appeared in Moen  well I might write to them
Having said that, and assuming the obvious pattern etc, if one knows that sum of 1+2+3...n is n(n+1)/2 all one has to do is to take square root of 2*500 to get the right number.
One way to think is to write them as:
1
22
333
4444
......
........
nnnnnnnnnn
and upside down
nnnnnnnnn
.......
.......
4444
333
22
1
Combine them, (that is write them side by side  so first line is 1nnnnnnn, second line is 22.. and last line is nnnnnn..1) you get a neat (approx) square with n on each side with 2*500 numbers, so answer should be near square root of 1000 (since the square is not exactly nxn but (n+1)xn the answer may be 1 this way or that but that can easily checked)
.
Re: BR Maths Corner1
Amber G. wrote:1, 2, 2, 3, 3, 3, 4, 4, 4, 4, ....What is the 500th number in the sequence?
It's a classic and the answer could be ANYTHING! for example the logic "MY 500th term is 37 is as valid as any other. Point is given any finite amount of terms in the series, guessing the next number is more of a 'puzzle' than math.
So unless some one tells the pattern explicitly , that is , 1 is repeated once, then 2 is repeated twice, and 3 is repeated thrice...etc (number n is repeated n times).. the problem is poorly worded. (I know, that kind of problems do appear but any serious mathematician will find the problem poorly stated... and I belive it ought not appear in serious math competition so I am a little surprised that it appeared in Moen  well I might write to them
Having said that, and assuming the obvious pattern etc, if one knows that sum of 1+2+3...n is n(n+1)/2 all one has to do is to take square root of 2*500 to get the right number.
One way to think is to write them as:
1
22
333
4444
......
........
nnnnnnnnnn
and upside down
nnnnnnnnn
.......
.......
4444
333
22
1
Combine them, (that is write them side by side  so first line is 1nnnnnnn, second line is 22.. and last line is nnnnnn..1) you get a neat (approx) square with n on each side with 2*500 numbers, so answer should be near square root of 1000 (since the square is not exactly nxn but (n+1)xn the answer may be 1 this way or that but that can easily checked)
.
I went the same route...my answer is above. Hopefully that is the right answer.
Re: BR Maths Corner1
Amber G. wrote:1, 2, 2, 3, 3, 3, 4, 4, 4, 4, ....What is the 500th number in the sequence?
It's a classic and the answer could be ANYTHING! for example the logic "MY 500th term is 37 is as valid as any other. Point is given any finite amount of terms in the series, guessing the next number is more of a 'puzzle' than math.
So unless some one tells the pattern explicitly , that is , 1 is repeated once, then 2 is repeated twice, and 3 is repeated thrice...etc (number n is repeated n times).. the problem is poorly worded. (I know, that kind of problems do appear but any serious mathematician will find the problem poorly stated... and I belive it ought not appear in serious math competition so I am a little surprised that it appeared in Moen  well I might write to them
I see your point precision counts especially when wording a test question!!
Having said that, and assuming the obvious pattern etc, if one knows that sum of 1+2+3...n is n(n+1)/2 all one has to do is to take square root of 2*500 to get the right number.
This is what I used to figure this out. I did have to make the leap to consider the numbers as clumps of 1+2+3 and not get distracted by the repetion which can throw one off from applying the classic addition formula.
So I got the same answer as Viv except I took much longer than him.
One way to think is to write them as:
Combine them, (that is write them side by side  so first line is 1nnnnnnn, second line is 22.. and last line is nnnnnn..1) you get a neat (approx) square with n on each side with 2*500 numbers, so answer should be near square root of 1000 (since the square is not exactly nxn but (n+1)xn the answer may be 1 this way or that but that can easily checked)
.
Thanks for the learning. Graphically good visualization technique. Will add it to my armour when it comes to teaching time..
Re: BR Maths Corner1
While on the topic of MOEMS this Jewesh kid we know on my sons soccer team while in 4th grade took the Math Olympiad for school kids and was one of only 209 kids worldwide to get a perfect score of 25. This when he was in 4th grade. He was double promoted to 6th grade and promptly topped the school district scores for this years class. So he has now earned an automatic spot on an elite (national winners etc..) middle school team Science Olympiad team as well. My son hopes to get admitted into that school next year. You need to be invited based on your test scores etc.. This is all public school system BTW.
Re: BR Maths Corner1
Jayram wrote:[
[
This is what I used to figure this out. I did have to make the leap to consider the numbers as clumps of 1+2+3 and not get distracted by the repetion which can throw one off from applying the classic addition formula.
So I got the same answer as Viv except I took much longer than him.
I turned it to n^2 + n 1000 = 0
This was approx same as doing sqrt(1000) i.e. applied quadratic formula which simplified to sqrt(4001)/2. Then tried with summation of first 31 terms and found it has to be 32. No geometric visualization for me
Re: BR Maths Corner1
Okay  some comments on numbers like "11111...." and which one are primes etc... (See the problem posted a few posts above)
******
As ankan problem showed, and a few post above (here) It was noted: (Let me repeat it for clarity)
For any prime (>5) n, if you write (1/n) as repeating decimal, and if the cycle of this is m
Then (easy to prove)
11111...m times is divisible by n
So given (see previous post)
One can deduct: 111111 is divisible by 7 (and not prime)
(111111 is also divisible by 13, 1111111111111111 (16 times) is divisible by 17 etc...
This also means:
11111 is divisible by 41 (or 11111 is not prime)
and 1111111111111 (13 times) is not prime either, (it is divisible by 53)
Neat!
( If one happened to know, from the tables  for example
1/21649= .00004619151 00004619151 00004619151 (cycle = 11)
One could have said 11111111111 is not prime but divisible by 21649)
********
Another neat property, which is very important (First proved by Euler it in 18th century, but it seem old Chinese mathematicians knew it ..I have not seen any reference)
the (if n is prime )cycle (of 1/n) it self divides (n1).
(one can see, for example, for a case of 41, the cycle is 5 and (411) is divisible by 5.
This gives 2 nice ways to check if a number is prime..
(For all n>5)
1  to check '1111...n times' is prime or not, (n is prime here) , you have to check divisibility by (2kn+1) *only*
This means, if you want to check '11111', just check if it is divisible by (10k+1 oe 11,21,31,41 ..etc only)
So for example to 'check' '11111111111' one need not try *all* primes less than sqrt(11111111111) but only of the type
23,45,67... etc only.
2. One need not work in decimal system, for example in binary... if n is prime, it means (2^n2) has to be divisible by n. Good thing is one can calculate 2^n very fast (by repeated squaring) if one finds 2^n2 is not divisible by n one can say n is not prime. This is how I was able to (without computer programing but checking 2^(1111111110) mod 11111111111) find out that 11111111111 is not prime.
******
As ankan problem showed, and a few post above (here) It was noted: (Let me repeat it for clarity)
For any prime (>5) n, if you write (1/n) as repeating decimal, and if the cycle of this is m
Then (easy to prove)
11111...m times is divisible by n
So given (see previous post)
1/7 = .142857 142857 142857 ... ( repeats after 6, or cycle lenght =6)
1/11 = .09 09 09 (cycle = 2)
1/13 = .076923 076923 076923 ... (cycle = 6)
or 1/17 = .0588235294117647 0588235294117647 ... (cycle = 16)
1/19 =.052631578947368421 052631578947368421 ... (Cycle = 18)
or take some larger prime
1/41 = .02439 02439 02439 024390 ... (cycle = 5)
or another one..
1/53 = .0188679245283 0188679.... ( cycle = 13)
One can deduct: 111111 is divisible by 7 (and not prime)
(111111 is also divisible by 13, 1111111111111111 (16 times) is divisible by 17 etc...
This also means:
11111 is divisible by 41 (or 11111 is not prime)
and 1111111111111 (13 times) is not prime either, (it is divisible by 53)
Neat!
( If one happened to know, from the tables  for example
1/21649= .00004619151 00004619151 00004619151 (cycle = 11)
One could have said 11111111111 is not prime but divisible by 21649)
********
Another neat property, which is very important (First proved by Euler it in 18th century, but it seem old Chinese mathematicians knew it ..I have not seen any reference)
the (if n is prime )cycle (of 1/n) it self divides (n1).
(one can see, for example, for a case of 41, the cycle is 5 and (411) is divisible by 5.
This gives 2 nice ways to check if a number is prime..
(For all n>5)
1  to check '1111...n times' is prime or not, (n is prime here) , you have to check divisibility by (2kn+1) *only*
This means, if you want to check '11111', just check if it is divisible by (10k+1 oe 11,21,31,41 ..etc only)
So for example to 'check' '11111111111' one need not try *all* primes less than sqrt(11111111111) but only of the type
23,45,67... etc only.
2. One need not work in decimal system, for example in binary... if n is prime, it means (2^n2) has to be divisible by n. Good thing is one can calculate 2^n very fast (by repeated squaring) if one finds 2^n2 is not divisible by n one can say n is not prime. This is how I was able to (without computer programing but checking 2^(1111111110) mod 11111111111) find out that 11111111111 is not prime.
Re: BR Maths Corner1
BTW, apart from 11 only primes of that type are: (All known only in 20th century)
1111111111111111111 (19 TIMES)
11111111111111111111111 (23 TIMES)
Also 317 times and 1031 times too.
(Other 49081 times 86453 times are most likely prime....and none other less than 100,000 or so  which were found in last few years only
(Indeed they are very rare)
1111111111111111111 (19 TIMES)
11111111111111111111111 (23 TIMES)
Also 317 times and 1031 times too.
(Other 49081 times 86453 times are most likely prime....and none other less than 100,000 or so  which were found in last few years only
(Indeed they are very rare)

 BRFite
 Posts: 348
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 Location: Pandora
Re: BR Maths Corner1
Hi all. This is my first post after 5 years of lurking. The urge to post as soon as you sign up  did you get that when you signed up? The obvious choice  go to BENIS, bash the Bakis and get laughs / streetcred. Decided to leave that for my 2nd post, and focus on a really interesting math concept.
If you read Carl Sagan's book "Contact", the epilogue of the book shows the existence of God on a computer monitor. The computer in this case was calculating the digits of "pi" to find a sign of intelligent design, and after a few trillion digits the digits became a series of 1 and 0, which on a raster display would be represented as a circle...hence proved that the universe has a "creator" who coded the shape of a circle into pi. I thought to myself "that must be a few thousand dots, all lined up as a circle  so very unlikely that pi would have a sequence of several thousand digits that follow this pattern, OK I buy Sagan's logic that this proves the existence of God".
Then I came across this site : http://www.angio.net/pi/piquery
It allows you to define a string, say 123456, and search the first 200 million digits of pi. Almost everyone's birthday can be found  in just the first 200 million digits! Mine is around the 303,000th digit.
Now, if we take the first trillion or quadrillion digits of pi, would we not find a whole lot more? For example, the entire text of the Bible, or the Encyclopedia Britannica?
For that matter, instead of reading this post, you can go to digit 10^345+72980, and read the next 2000 digits to get my post  it's all there!
Am I reading something wrongly here? If pi is an infinite random series, then by definition every finite sequence that you care to define MUST be there somewhere in pi.
If you read Carl Sagan's book "Contact", the epilogue of the book shows the existence of God on a computer monitor. The computer in this case was calculating the digits of "pi" to find a sign of intelligent design, and after a few trillion digits the digits became a series of 1 and 0, which on a raster display would be represented as a circle...hence proved that the universe has a "creator" who coded the shape of a circle into pi. I thought to myself "that must be a few thousand dots, all lined up as a circle  so very unlikely that pi would have a sequence of several thousand digits that follow this pattern, OK I buy Sagan's logic that this proves the existence of God".
Then I came across this site : http://www.angio.net/pi/piquery
It allows you to define a string, say 123456, and search the first 200 million digits of pi. Almost everyone's birthday can be found  in just the first 200 million digits! Mine is around the 303,000th digit.
Now, if we take the first trillion or quadrillion digits of pi, would we not find a whole lot more? For example, the entire text of the Bible, or the Encyclopedia Britannica?
For that matter, instead of reading this post, you can go to digit 10^345+72980, and read the next 2000 digits to get my post  it's all there!
Am I reading something wrongly here? If pi is an infinite random series, then by definition every finite sequence that you care to define MUST be there somewhere in pi.

 BRFite
 Posts: 348
 Joined: 21 Jan 2010 19:24
 Location: Pandora
Re: BR Maths Corner1
Jayram wrote:Ok First some disclaimers. I am not a guru in anything significant least of all math.. So this math problem is going to be trivial for the BR folks that tend to hang around in this neighbhood. However I got into this becuase my kid is in middle school and I was looking for a way to encourage his intrest in the subject.
This problem seems doable for us layman dads, moms or kids as the case might be...
Got this problem from the here http://www.moems.org/zinger.htm
The first ten numbers in a sequence are
1, 2, 2, 3, 3, 3, 4, 4, 4, 4, ....
What is the 500th number in the sequence?
I will be honest.. It took me at least a couple of hours to solve this problem and had to research some formulas first..
Jayram, this is equivalent to asking "in the series 1,2,3,4... where does the cumulative sum exceed 500"?
Re: BR Maths Corner1
Amber G. wrote:the (if n is prime )cycle (of 1/n) it self divides (n1).
(one can see, for example, for a case of 41, the cycle is 5 and (411) is divisible by 5.
This gives 2 nice ways to check if a number is prime..
(For all n>5)
1  to check '1111...n times' is prime or not, (n is prime here) , you have to check divisibility by (2kn+1) *only*
This means, if you want to check '11111', just check if it is divisible by (10k+1 oe 11,21,31,41 ..etc only)
So for example to 'check' '11111111111' one need not try *all* primes less than sqrt(11111111111) but only of the type
23,45,67... etc only.
One need not work in decimal system, for example in binary... if n is prime, it means (2^n1) has to be divisible by n.
Amberji,
1)I am not able to make sense of above statement on 2^n1..
2)How on earth can one guess about 1/21649 ie cycle length? The problem though interesting,does not seem 'natural' as far as approach goes.
3)Ankan is not necessary for the problem,only cycle length
I will revert back regarding cycle length for general n and n prime.
Re: BR Maths Corner1
krishnapremi wrote:One need not work in decimal system, for example in binary... if n is prime, it means (2^n1) has to be divisible by n.
Amberji, I have a couple of questions about that statement myself.
1. What does being prime have to do with working in binary or decimal. A number is prime whether you represent it in binary, decimal, octal, hex, base36, roman numerals, cuniform writing or whatever. Am I missing something here.
2. Say n is 11. 2^11  1 is 2047, which is definitely not divisible by 11. But 11 is prime.
Re: BR Maths Corner1
ArmenT wrote:krishnapremi wrote:One need not work in decimal system, for example in binary... if n is prime, it means (2^n1) has to be divisible by n.
Amberji, I have a couple of questions about that statement myself.
1. What does being prime have to do with working in binary or decimal. A number is prime whether you represent it in binary, decimal, octal, hex, base36, roman numerals, cuniform writing or whatever. Am I missing something here.
2. Say n is 11. 2^11  1 is 2047, which is definitely not divisible by 11. But 11 is prime.
Few comments.
With respect to 2, what you should have 2^112 (base is 2) which is divisible by 11, and so is
(3^113) or (4^114) or (10^1110) ... all are divisible by 11.
Of course, if a number is prime or not, does not depend on the base ...
but in base (say b) (In standard number theory it is called 'witness' is b ) if a number n is prime then (b^n  n) is divisible by n
The reverse is not always true. That is if (b^n  n ) is divisible by n , n may not be a prime, but most likely a prime.
To just give an idea, among first 25 trillion numbers, there are about 1 trillion prime numbers, but there are only about 20,000 of them are 'pseduo primes' when one uses witness 2. ( only 1 in 50 million  false results if you use base 2)
Point is it is MUCH faster to calculate if 2^n2 is divisible by n or not then to try factoring n.
If n , is say a 100 digit number, a computer will take billions of billions of billions times .. the age of the universe ... to try all numbers less than 10^50 as divisors.. while the other test will take less than a second.
The problem, here is, of course, one can not be 100% sure, but if you use even a few witness your chances of being wrong would be VERY small (say less than 1 in trillions) .. and this method is good enough to find "primes" for industrial grade cryptography.
All this, of course, is standard number theory ....
Re: BR Maths Corner1
krishnapremi wrote:
One need not work in decimal system, for example in binary... if n is prime, it means (2^n1) has to be divisible by n.
Amberji,
1)I am not able to make sense of above statement on 2^n1..
2)How on earth can one guess about 1/21649 ie cycle length? The problem though interesting,does not seem 'natural' as far as approach goes.
3)Ankan is not necessary for the problem,only cycle length
I will revert back regarding cycle length for general n and n prime.[/quote]
Krisnapremiji  we have a typo, correct statement is:
If n is a prime ===> 2^n 2 is divisible by n
or for that matter for any b, ( b^n  b) is divisible by n
2. One did not "guess" it. Old mathematicians, just like they had multiplication tables, planetary positions tables etc, had complied those things ..There are references in Math History that some mathematicians had those cycle lengths for first 100,000 numbers or so.
2b ) what one has, that for example, for '11111111111' , one need not check *all* numbers from 1 and up, only numbers like 22k+1 (21649 is in fact 984*22+1) , that shortens the work by a factor of 10 or more.
3) Ankan (or similar theorem) is *necessary* to prove that the method works.
Regards.
Re: BR Maths Corner1
Fidel Guevara:
When I actually looked at the next 2000 characters of pi's decimal digits starting from 10 ^ 345 72,980 I found the following: I found these somewhat random digits which, when translated into utf8 code, produced:
Don't know if the above is totally random or not!  Since the translation is not perfect, I can do not know if it proves or disproves God's existance ...
Anyway:
With respect to that, there are two classic problems
(these problems could be hard unless you know the solution, but they have appeared in High School level math competitions )
Given any sequence of digits, say '20100128' (todays date) prove that:
Q1  There exists a n such that first 8 digits of 2^n are '20100128'
Q2  There exists a m such that m! starts with '20100128'
(Here m! means 1*2*3*4*5.......*m)
When I actually looked at the next 2000 characters of pi's decimal digits starting from 10 ^ 345 72,980 I found the following: I found these somewhat random digits which, when translated into utf8 code, produced:
Fidel Guevara wrote: हाय सब. यह मेरी पहली पोस्ट है गुप्त के 5 साल बाद. के रूप में जल्द ही पोस्ट के रूप में आप साइन अप करने से आग्रह करता हूं  तुम हो कि जब आप साइन अप किया है? स्पष्ट विकल्प  BENIS जाना, Bakis पार्टी की योजना बनाई और हंसते हुए कहते हैं मिल / सड़कcred. का फैसला किया जाने के लिए है कि मेरी 2nd पद के लिए, और ध्यान एक बहुत दिलचस्प गणित अवधारणा पर.
यदि आप कार्ल Sagan है पुस्तक "सम्पर्क" पढ़ें, पुस्तक के उपसंहार एक कंप्यूटर मॉनीटर पर भगवान के अस्तित्व को दर्शाता है. इस मामले में कंप्यूटर "ढोंगी" के अंकों की गणना था बुद्धिमान डिजाइन का संकेत मिल जाए, और उसके बाद कुछ खरब अंक अंक 1 और 0 है, जो एक रेखापुंज प्रदर्शन पर एक चक्र के रूप में प्रतिनिधित्व किया जाएगा .. की एक श्रृंखला बन गया . इसलिए साबित कर दिया कि ब्रह्मांड एक "निर्माता" जो अनुकरणीय में एक वृत्त के आकार कोडित है. मैं खुद के लिए सोचा "है कि कुछ हजार बिंदु होगा, सब एक चक्र के रूप में लाइन में खड़ा  बहुत बहुत संभावना नहीं है कि गड़बड़ी कई हजार अंक का एक दृश्य है कि इस पद्धति का पालन करना होगा, ठीक है मैं Sagan तर्क है कि यह भगवान के अस्तित्व को साबित करता खरीद ".
तब मैं इस साइट भर में आया: http://www.angio.net/pi/piquery
यह तुम एक स्ट्रिंग निर्धारित करने की अनुमति देता है, 123456 कहते हैं, और गड़बड़ी के पहले 200 मिलियन अंक खोज. लगभग सभी के जन्मदिन सिर्फ पहले 200 मिलियन अंक में  हो पाया जा सकता! मेरा 303.000 अंक वीं के आसपास है.
अब, अगर हम गड़बड़ी की पहली खरब या quadrillion अंक ले, हम पाते हैं कि नहीं? होगा एक पूरी बहुत अधिक उदाहरण के लिए, बाइबिल, या ब्रिटैनिका विश्वकोश के पूरे पाठ को?
कि क्या बात है, बजाय इस पोस्ट को पढ़ने के लिए, आप अंक 10 ^ 345 72,980 कर सकते हैं, और अगले 2000 अंक को पढ़ने के लिए अपने पद मिल  यह सब वहाँ है!
क्या मैं कुछ पढ़ने के गलत तरीके से यहाँ? यदि pi एक अनंत यादृच्छिक श्रृंखला, परिभाषा तक सीमित है तो हर दृश्य है कि आप को परिभाषित करने के लिए जरूरी वहाँ pi में कहीं परवाह है.
Don't know if the above is totally random or not!  Since the translation is not perfect, I can do not know if it proves or disproves God's existance ...
Anyway:
With respect to that, there are two classic problems
(these problems could be hard unless you know the solution, but they have appeared in High School level math competitions )
Given any sequence of digits, say '20100128' (todays date) prove that:
Q1  There exists a n such that first 8 digits of 2^n are '20100128'
Q2  There exists a m such that m! starts with '20100128'
(Here m! means 1*2*3*4*5.......*m)
Re: BR Maths Corner1
Amberji,
Thanks.I still have doubts.
1)x=1/7=0.142857... then 10^6x=142857.142857..., 999999/7=142857
implies 111111*9/7=142857, implies 7 divides 111111.This argument goes through for other numbers as well.So why ankan?
2)p divides a^n a does not need ankan.
3)Please excuse my obsession.Is it 'natural' to check for 984*22+1,from 1 to 984.
4)What is the proof for the result?cycle length of 1/p divides p1 5)Can you throw some light on cycle length for gen.1/n.
I seek your indulgence because i want to 'see' this problem through
Thanks.I still have doubts.
1)x=1/7=0.142857... then 10^6x=142857.142857..., 999999/7=142857
implies 111111*9/7=142857, implies 7 divides 111111.This argument goes through for other numbers as well.So why ankan?
2)p divides a^n a does not need ankan.
3)Please excuse my obsession.Is it 'natural' to check for 984*22+1,from 1 to 984.
4)What is the proof for the result?cycle length of 1/p divides p1 5)Can you throw some light on cycle length for gen.1/n.
I seek your indulgence because i want to 'see' this problem through
Last edited by svenkat on 30 Jan 2010 07:51, edited 1 time in total.

 BRFite
 Posts: 348
 Joined: 21 Jan 2010 19:24
 Location: Pandora
Re: BR Maths Corner1
Amber G. wrote:Fidel Guevara:
When I actually looked at the next 2000 characters of pi's decimal digits starting from 10 ^ 345 72,980 I found the following: I found these somewhat random digits which, when translated into utf8 code, produced:Fidel Guevara wrote: हाय सब. यह मेरी पहली पोस्ट है गुप्त के 5 साल बाद. के रूप में जल्द ही पोस्ट के रूप में आप साइन अप करने से आग्रह करता हूं  तुम हो कि जब आप साइन अप किया है? स्पष्ट विकल्प  BENIS जाना, Bakis पार्टी की योजना बनाई और हंसते हुए कहते हैं मिल / सड़कcred. का फैसला किया जाने के लिए है कि मेरी 2nd पद के लिए, और ध्यान एक बहुत दिलचस्प गणित अवधारणा पर.
यदि आप कार्ल Sagan है पुस्तक "सम्पर्क" पढ़ें, पुस्तक के उपसंहार एक कंप्यूटर मॉनीटर पर भगवान के अस्तित्व को दर्शाता है. इस मामले में कंप्यूटर "ढोंगी" के अंकों की गणना था बुद्धिमान डिजाइन का संकेत मिल जाए, और उसके बाद कुछ खरब अंक अंक 1 और 0 है, जो एक रेखापुंज प्रदर्शन पर एक चक्र के रूप में प्रतिनिधित्व किया जाएगा .. की एक श्रृंखला बन गया . इसलिए साबित कर दिया कि ब्रह्मांड एक "निर्माता" जो अनुकरणीय में एक वृत्त के आकार कोडित है. मैं खुद के लिए सोचा "है कि कुछ हजार बिंदु होगा, सब एक चक्र के रूप में लाइन में खड़ा  बहुत बहुत संभावना नहीं है कि गड़बड़ी कई हजार अंक का एक दृश्य है कि इस पद्धति का पालन करना होगा, ठीक है मैं Sagan तर्क है कि यह भगवान के अस्तित्व को साबित करता खरीद ".
तब मैं इस साइट भर में आया: http://www.angio.net/pi/piquery
यह तुम एक स्ट्रिंग निर्धारित करने की अनुमति देता है, 123456 कहते हैं, और गड़बड़ी के पहले 200 मिलियन अंक खोज. लगभग सभी के जन्मदिन सिर्फ पहले 200 मिलियन अंक में  हो पाया जा सकता! मेरा 303.000 अंक वीं के आसपास है.
अब, अगर हम गड़बड़ी की पहली खरब या quadrillion अंक ले, हम पाते हैं कि नहीं? होगा एक पूरी बहुत अधिक उदाहरण के लिए, बाइबिल, या ब्रिटैनिका विश्वकोश के पूरे पाठ को?
कि क्या बात है, बजाय इस पोस्ट को पढ़ने के लिए, आप अंक 10 ^ 345 72,980 कर सकते हैं, और अगले 2000 अंक को पढ़ने के लिए अपने पद मिल  यह सब वहाँ है!
क्या मैं कुछ पढ़ने के गलत तरीके से यहाँ? यदि pi एक अनंत यादृच्छिक श्रृंखला, परिभाषा तक सीमित है तो हर दृश्य है कि आप को परिभाषित करने के लिए जरूरी वहाँ pi में कहीं परवाह है.
Don't know if the above is totally random or not!  Since the translation is not perfect, I can do not know if it proves or disproves God's existance ...
Anyway:
With respect to that, there are two classic problems
(these problems could be hard unless you know the solution, but they have appeared in High School level math competitions )
Given any sequence of digits, say '20100128' (todays date) prove that:
Q1  There exists a n such that first 8 digits of 2^n are '20100128'
Q2  There exists a m such that m! starts with '20100128'
(Here m! means 1*2*3*4*5.......*m)
Amber, this is a great find! Curious though  which database has pi up to 10^345 digits? Jest kidding!
I'll take a crack at your Q1; at work now, so...
You may find this discussion interesting  blog of the crazy genius Pickover :
http://sprott.physics.wisc.edu/pickover/pimatrix.html
Lots of discussion about pi and what it contains. If you look at pi starting from Graham's Number + 22, and look at the next 1 billion digits, there's a MPEG code for Amber "having fun" with Katrina Kaif!
Re: BR Maths Corner1
Sorry  As noticed before ... My post on jan08 contained a typo:
The original is:
should be
I can no longer edit the post, so I have asked admins to see if they would be kind enough...
Anyway: the correct theorem is:
If n is prime then for any b
(b^n  b) is divisible by n
In case, b and n has no common factors, one can reduce the above to
b^(n1)1 is divisible by n
The theorem, in beauty, is one of my favorite. As one math historian commented something to the effect:
>>>>"Once shown how to prove it, any bright elementary school student can understand the proof. But if asked to prove it, and if the solution is not known, not 1 in a million would be able to find the proof. " >>>>
The theorem, as said before is named after Fermat but actually proven (as far as we know) by Euler. Though others suspected the truth ...
For example Chinese knew the case for 2..
(2^32) is divisible by 3, (2^52) is divisible by 5, or (2^72) is divisible by 7
In fact easy to check (2^n2) is divisible by n when (n = 2,3,5,7,11,13,17.....etc...)
The reverse is not true always , but in most cases it is true..
In fact ( 2^n 2 ) is not divisible by n for almost all nonprime values ...4,6,..15... or any such number ..
(As one guess ...
this is always true : (If ( b^nb) is not divisible by n ==> n is not prime.)
In fact first time you see contradiction is for 341 ... that is (2^3412) is divisible by 341, yet 341 is not a prime.
But this gives a VERY powerful method because, if one finds that (2^n2) is NOT divisible by n then one can tell that n is not prime ... (that was how I was able to prove that 11111111111 is not prime because I could check (2^111111111112) ( I actually checked (2^111111111101) which is does not change a thing) was not divisible by 11111111111 (That I did with nothing more than pen and paper).
not
With a simple program even a 50 (or 1000) digit number, could be proven not prime by this method while if one wants to check all factors it will take more than the age of universe...even with
billion super computer running in parallel...
n
The original is:
2. One need not work in decimal system, for example in binary... if n is prime, it means (2^n1) has to be divisible by n. Good thing is one can calculate 2^n very fast (by repeated squaring) if one finds 2^n1 is not divisible by n one can say n is not prime. This is how I was able to (without computer programing but checking 2^(1111111110) mod 11111111111) find out that 11111111111 is not prime.
should be
2. One need not work in decimal system, for example in binary... if n is prime, it means (2^n2) has to be divisible by n. Good thing is one can calculate 2^n very fast (by repeated squaring) if one finds 2^n[2] is not divisible by n one can say n is not prime. This is how I was able to (without computer programing but checking 2^(1111111110) mod 11111111111) find out that 11111111111 is not prime.
I can no longer edit the post, so I have asked admins to see if they would be kind enough...
Anyway: the correct theorem is:
If n is prime then for any b
(b^n  b) is divisible by n
In case, b and n has no common factors, one can reduce the above to
b^(n1)1 is divisible by n
The theorem, in beauty, is one of my favorite. As one math historian commented something to the effect:
>>>>"Once shown how to prove it, any bright elementary school student can understand the proof. But if asked to prove it, and if the solution is not known, not 1 in a million would be able to find the proof. " >>>>
The theorem, as said before is named after Fermat but actually proven (as far as we know) by Euler. Though others suspected the truth ...
For example Chinese knew the case for 2..
(2^32) is divisible by 3, (2^52) is divisible by 5, or (2^72) is divisible by 7
In fact easy to check (2^n2) is divisible by n when (n = 2,3,5,7,11,13,17.....etc...)
The reverse is not true always , but in most cases it is true..
In fact ( 2^n 2 ) is not divisible by n for almost all nonprime values ...4,6,..15... or any such number ..
(As one guess ...
this is always true : (If ( b^nb) is not divisible by n ==> n is not prime.)
In fact first time you see contradiction is for 341 ... that is (2^3412) is divisible by 341, yet 341 is not a prime.
But this gives a VERY powerful method because, if one finds that (2^n2) is NOT divisible by n then one can tell that n is not prime ... (that was how I was able to prove that 11111111111 is not prime because I could check (2^111111111112) ( I actually checked (2^111111111101) which is does not change a thing) was not divisible by 11111111111 (That I did with nothing more than pen and paper).
not
With a simple program even a 50 (or 1000) digit number, could be proven not prime by this method while if one wants to check all factors it will take more than the age of universe...even with
billion super computer running in parallel...
n
Re: BR Maths Corner1
Trying to answer, hope it more sense:
Yes, this does not require 'ankan' as one can show this as you have shown.
But the proof that, given any prime n (or not prime for that matter, as long as it is not divisible by 2 or 5).. you will find a multile of n which is nothing but all '1's'
(This is what the orginal question of ankan asked to show, .. there I did not have restriction that number is not divisible by 2 or 5 and hence we had  two distinct digist (11111...0..0 for any multiple of n)
What is more relavent here that the cycle length of (1/n) (where n is prime >5) will divide (n1)
The proof of this requires logic like ankan. (The proof of this is not easy, if one does not already know it..)
The proof of this is almost the the same as the proof that cycle length of (1/n) divides (n1).
You don't need ankan  but logic like ankan to prove it (Of course, There are many other methods to prove it too)
If one is writing computer program  (or doing it by hand, and has months of time and/or has a few sisya's to help) the task (to find factors of '11111111111') is some what reduced.. in stead of checking for all numbers ... you need to check only for 23, 45, 67.. etc... So it becomes easier. (Here we know, that if a factor exists it must be of the form (22k+1).
BTW there is, very famous hisory assosiated with the number '1111111111111111111111111111111' in binary or (2^311) or 2147483647 .
(See link http://en.wikipedia.org/wiki/2147483647) which Euler proved , it was prime, (in 1772 !  (and it remained the largest prime known till 1867 according to wiki)  method used by Euler was essentially the same  one only has to try numbers of the form 62k+1 .. actually it can be impoved a little but that will take will need more math ..)
Any number theory book or Internet source can give the proof (It is not trivial  but not too hard either) Here is one source
link form wolfram
or wiki article:
This is essentially same as Fermat's little theorem ... (Check out wiki or any other source)
link
Here are some proofs from wiki:
http://en.wikipedia.org/wiki/Proofs_of_Fermat%27s_little_theorem
Regards.
krishnapremi wrote:Amberji,
1)x=1/7=0.142857... then 10^6x=142857.142857..., 999999/7=142857
implies 111111*9/7=142857, implies 7 divides 111111.This argument goes through for other numbers as well.So why ankan?
Yes, this does not require 'ankan' as one can show this as you have shown.
But the proof that, given any prime n (or not prime for that matter, as long as it is not divisible by 2 or 5).. you will find a multile of n which is nothing but all '1's'
(This is what the orginal question of ankan asked to show, .. there I did not have restriction that number is not divisible by 2 or 5 and hence we had  two distinct digist (11111...0..0 for any multiple of n)
What is more relavent here that the cycle length of (1/n) (where n is prime >5) will divide (n1)
The proof of this requires logic like ankan. (The proof of this is not easy, if one does not already know it..)
2)p divides a^n a does not need ankan.
The proof of this is almost the the same as the proof that cycle length of (1/n) divides (n1).
You don't need ankan  but logic like ankan to prove it (Of course, There are many other methods to prove it too)
3)Please excuse my obsession.Is it 'natural' to check for 984*22+1,from 1 to 984.
If one is writing computer program  (or doing it by hand, and has months of time and/or has a few sisya's to help) the task (to find factors of '11111111111') is some what reduced.. in stead of checking for all numbers ... you need to check only for 23, 45, 67.. etc... So it becomes easier. (Here we know, that if a factor exists it must be of the form (22k+1).
BTW there is, very famous hisory assosiated with the number '1111111111111111111111111111111' in binary or (2^311) or 2147483647 .
(See link http://en.wikipedia.org/wiki/2147483647) which Euler proved , it was prime, (in 1772 !  (and it remained the largest prime known till 1867 according to wiki)  method used by Euler was essentially the same  one only has to try numbers of the form 62k+1 .. actually it can be impoved a little but that will take will need more math ..)
4)What is the proof for the result?cycle length of 1/p divides p1 5)Can you throw some light on cycle length for gen.1/n.
I seek your indulgence because i want to 'see' this problem through
Any number theory book or Internet source can give the proof (It is not trivial  but not too hard either) Here is one source
link form wolfram
or wiki article:
(see few paragraphs deep)http://en.wikipedia.org/wiki/Repeating_decimal
This is essentially same as Fermat's little theorem ... (Check out wiki or any other source)
link
Here are some proofs from wiki:
http://en.wikipedia.org/wiki/Proofs_of_Fermat%27s_little_theorem
Regards.

 BRF Oldie
 Posts: 9664
 Joined: 19 Nov 2009 03:27

 BRF Oldie
 Posts: 9664
 Joined: 19 Nov 2009 03:27
Re: BR Maths Corner1
The Enemy of My Enemy
By STEVEN STROGATZ
http://opinionator.blogs.nytimes.com/2010/02/14/theenemyofmyenemy/
By STEVEN STROGATZ
http://opinionator.blogs.nytimes.com/2010/02/14/theenemyofmyenemy/
Re: BR Maths Corner1
A review, of a book on ancient and medieval Indian mathematics, by a distinguished modern mathematician David Mumford,Fields medal winner
Did you know that Vedic priests were using the socalled
Pythagorean theorem to construct their firealtars in 800 BCE?; that the differential equationfor the sine function, in finite difference form, was
described by Indian mathematicianastronomers in the fifth century CE?; and that “Gregory’s”series π/4 = 1−1/3 +1/5 −· · · was proven using the
power series for arctangent and, with ingenious summation methods, used to accurately compute π in southwest India in the fourteenth century?
If any of this surprises you, Plofker’s book is for you.
Her book fills a huge gap: a detailed, eminently readable, scholarly survey of the full scope of Indian1 mathematics and astronomy (the two were inseparable in India) from their Vedic beginnings to roughly 1800. There is only one other survey,Datta and Singh’s 1938 History of Hindu Mathematics,recently reprinted but very hard to obtainin the West (I found a copy in a small specialized bookstore in Chennai). They describe in some detail the Indian work in arithmetic and algebra and,supplemented by the equally hard to find Geometry
in Ancient and Medieval India by Sarasvati Amma (1979)
Re: BR Maths Corner1
Excerpts:
In particular,as I mentioned above, one finds here the earliest
explicit statement of “Pythagorean” theorem (so it might arguably be called Baudhayana’s theorem).
Chapter 7 of Plofker’s book is devoted to the crown jewel of Indian mathematics, the work of the Kerala school. Kerala is a narrow fertile strip
between the mountains and the Arabian Sea along the southwest coast of India. Here, in a number of small villages, supported by the Maharaja of
Calicut, an amazing dynasty of mathematicians and astronomers lived and thrived. A large proportion of their results were attributed by later
writers to the founder of this school, Madhava of Sangamagramma, who lived from approximately 1350 to 1425. It seems fair to me to compare him with Newton and Leibniz. The high points of their mathematical work were the discoveries of the power series expansions of arctangent, sine, and cosine.

 BRF Oldie
 Posts: 9664
 Joined: 19 Nov 2009 03:27
Re: BR Maths Corner1
Finding Your Roots
By STEVEN STROGATZ
http://opinionator.blogs.nytimes.com/2010/03/07/findingyourroots/
By STEVEN STROGATZ
http://opinionator.blogs.nytimes.com/2010/03/07/findingyourroots/
Re: BR Maths Corner1
If one could provide some ideas as to how to prove that there is an infinite number of primes.
I would look it up on Google, except I want to be shoved in the right direction without being given the whole proof. Thanks.
I would look it up on Google, except I want to be shoved in the right direction without being given the whole proof. Thanks.
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