BR Maths Corner1
Re: BR Maths Corner1
Few comments wrt IMO 2017.. while we wait for results..
 Problem 1 is fairly easy, if anyone likes to try it here (or ask that to a bright student).
(Basically the problem is 
 start with a number (positive integer > 1)
 get the next number by adding 3 to it.. except (see next line)
 but if the number is a square .. take the square root.
(This way you will get increasing numbers except once in a while you fall back  for example if you start with 91, you get 91, 94, 97, 100 go down and get 10, 13, 16  go down get 4, go down get 2, 5, 8, .. etc.)
Your job is to find a *all* starting numbers such that , the series becomes a loop..
***
I think, because of this easy problem, there will be quite a few bronze medals awarded ..even countries which do not get any medals usually may get a bronzemedals this year.
I expect every contestant (or nearly everyone) of India will get some kind of medal this year.
(My paanwallah says that there are golds for Sutanay and Ananth )
Paanwallah also says China has not done that well as it generally does..Korea may have done better)
(Results will be announced tomorrow (Friday evening) after Jury meeting)
 Problem 1 is fairly easy, if anyone likes to try it here (or ask that to a bright student).
(Basically the problem is 
 start with a number (positive integer > 1)
 get the next number by adding 3 to it.. except (see next line)
 but if the number is a square .. take the square root.
(This way you will get increasing numbers except once in a while you fall back  for example if you start with 91, you get 91, 94, 97, 100 go down and get 10, 13, 16  go down get 4, go down get 2, 5, 8, .. etc.)
Your job is to find a *all* starting numbers such that , the series becomes a loop..
***
I think, because of this easy problem, there will be quite a few bronze medals awarded ..even countries which do not get any medals usually may get a bronzemedals this year.
I expect every contestant (or nearly everyone) of India will get some kind of medal this year.
(My paanwallah says that there are golds for Sutanay and Ananth )
Paanwallah also says China has not done that well as it generally does..Korea may have done better)
(Results will be announced tomorrow (Friday evening) after Jury meeting)

 BRFite
 Posts: 1114
 Joined: 29 Mar 2017 06:37
Re: BR Maths Corner1
UBji,
Did you write this letter? While LIGO detection can still be questioned but questioning black holes is another thing. How does one explain the increased speed of star movements around center of milky way?
Did you write this letter? While LIGO detection can still be questioned but questioning black holes is another thing. How does one explain the increased speed of star movements around center of milky way?
Re: BR Maths Corner1
Interestingly Hafez Assad (Bashar alAssad's son) is taking part in International Math Olympiad this year. He is representing Syria.
Since the news is on some newspapers already I think it is not unsportsmanlike, or breaking privacy issues to mention it here.
(Actually the son himself has spoken to Brazil's Globo newspaper giving away his privacy  I don't understand why he drew attention on himself).
Since the news is on some newspapers already I think it is not unsportsmanlike, or breaking privacy issues to mention it here.
(Actually the son himself has spoken to Brazil's Globo newspaper giving away his privacy  I don't understand why he drew attention on himself).
Re: BR Maths Corner1
^^^ There is some time before the results are declared.... There is a jury meeting .. some haggling (where the cutoff for medals ought to be decided ) and even discussion on how much "partial credit" one must get.. results may be there later today (Rio time).
The IMO is strange .. All students in India and US have gotten full credit for P1 and P4..
All got 0 in P3 .. (and many 0's in P6)
Korea has done exceptionally well.. China not so..
Here are partial results for India  (? = known score but can not be shown)
Prob 1 (All got 7) (7 is perfect score)
Prob 2 (3,3,4,?,4,4) = Very good.
Prob 3 (All got 0)
Prob 4 (All got 7)
Prob 5 (??????)
Prob 6 (All got 0 except IND 3 got 3) (Hope IND3 Ananat gets gold  depends on cutoff )
IN all very good performance. Congrats.
These scores are similar to USA where everyone got perfect score in P1, and P4, 0 in P3, almost 0 in P6.
Very few (I head of only one Australian who got perfect 7) got any point in P3.. Most got 0.
Funny part is P1 and P4 gives you 14..
Very little (a point or more) more gives you a bronze (so everyone is getting a bronze or better)
A point or 2 more will give you silver..
I think "?" in P5 ( even decent partial credit) may get gold..
The IMO is strange .. All students in India and US have gotten full credit for P1 and P4..
All got 0 in P3 .. (and many 0's in P6)
Korea has done exceptionally well.. China not so..
Here are partial results for India  (? = known score but can not be shown)
Prob 1 (All got 7) (7 is perfect score)
Prob 2 (3,3,4,?,4,4) = Very good.
Prob 3 (All got 0)
Prob 4 (All got 7)
Prob 5 (??????)
Prob 6 (All got 0 except IND 3 got 3) (Hope IND3 Ananat gets gold  depends on cutoff )
IN all very good performance. Congrats.
These scores are similar to USA where everyone got perfect score in P1, and P4, 0 in P3, almost 0 in P6.
Very few (I head of only one Australian who got perfect 7) got any point in P3.. Most got 0.
Funny part is P1 and P4 gives you 14..
Very little (a point or more) more gives you a bronze (so everyone is getting a bronze or better)
A point or 2 more will give you silver..
I think "?" in P5 ( even decent partial credit) may get gold..
Last edited by Amber G. on 22 Jul 2017 02:48, edited 1 time in total.
Re: BR Maths Corner1
As the things stand  we don't know the medal counts but ... ranking ..
At present it looks like ..
Republic of Korea
Singapore
Taiwan
People's Republic of China
Bulgaria
Vietnam
Greece
Czech Republic
Russian Federation
Japan
Georgia
Romania
Mexico
Belarus
At present it looks like ..
Republic of Korea
Singapore
Taiwan
People's Republic of China
Bulgaria
Vietnam
Greece
Czech Republic
Russian Federation
Japan
Georgia
Romania
Mexico
Belarus
Last edited by Amber G. on 22 Jul 2017 05:13, edited 1 time in total.
Re: BR Maths Corner1
Thanks. This is a nice site, one can also find all the past results, IMO problems and statistics..
Re: BR Maths Corner1
Amber G. wrote:KJo wrote:Have any of you guys looked into Srinivasa Ramanujan's work? I would like to read a technical book on his discoveries some day.
There is lot of discussion about Ramanujan's work (and life) in this thread.. I have given some links for his notebooks and other works. In fact there are many problems from his in dhaga  simplified in layman's terms, I have posted here with solutions and insights.. just do a search and that may be of your interest.
There are many very good references, such as:
An overview of his notebooks
Hardy's book "A Mathematician's Apology" is classic.
Another classic is by Hardy,Aiyangar, Seshu Aiyar, Wilson  "Collected Papers of Srinivasa Ramanujan"
There are, of course, many more!
Amber ji, thanks for your reply and the information. I will take a look!
Re: BR Maths Corner1
Okay  Final results of IMO may be up around 9PM (Rio time) FWIW the final standing..since grading has been done (except for a few cases scores of few problems_for_few_students are still not public and so estimates are taken) so this may change a little) . Medals cutoff still to be made public later.. In all, I think this contest, as I said earlier is little weird.
KOR
VNM
CHN
USA
JPN
IRN
HEL
TWN
GEO
BLR
UKR
SGP
CZE
THA
NLD
BGR
RUS
(The official site generally updates the result right away after the final Jury verdict  official results  so it may become available in a few hours)
KOR
VNM
CHN
USA
JPN
IRN
HEL
TWN
GEO
BLR
UKR
SGP
CZE
THA
NLD
BGR
RUS
(The official site generally updates the result right away after the final Jury verdict  official results  so it may become available in a few hours)
Re: BR Maths Corner1
Okay  results have been released on the official site!
India  3 Bronze and 3 HM  oh well (rough)
(Two people did not get 7 on prob 4 as I expected  and 14 was a point below cutoff)
(Sadly 18 was not good enough for silver  Anant's "?" turned out to be 0 so no silver  I was even hoping for a gold  for him)
Ranking  Korea, China, Vietnam, US, Iran ..
Korea got 6 gold!
US  3 gold, 3 silver... (Ankan did get a gold  Congrats)
(India ranked 52  a number which really doesn't mean much)
India  3 Bronze and 3 HM  oh well (rough)
(Two people did not get 7 on prob 4 as I expected  and 14 was a point below cutoff)
(Sadly 18 was not good enough for silver  Anant's "?" turned out to be 0 so no silver  I was even hoping for a gold  for him)
Ranking  Korea, China, Vietnam, US, Iran ..
Korea got 6 gold!
US  3 gold, 3 silver... (Ankan did get a gold  Congrats)
(India ranked 52  a number which really doesn't mean much)
Re: BR Maths Corner1
India always seems to underwhelm at these competitions. What is reason for the same?
I dont believe lack of funds, developing country etc since Vietnam (and even places like Thailand, Iran etc) ALWAYS seems to come within top 10 despite seemingly worse infra and focus than ours
I dont believe lack of funds, developing country etc since Vietnam (and even places like Thailand, Iran etc) ALWAYS seems to come within top 10 despite seemingly worse infra and focus than ours
Re: BR Maths Corner1
^The entire education system needs a total overhaul in curriculum, methodology, and quality of teaching. Good news is that so much resource is now freely available in electronic form that lack of quality teaching can be largely overcome by using such resources. Make them widely available and incorporate them into teaching. Assign homeworks and grade them, practice is what makes perfect.
Re: BR Maths Corner1
^^^About 1015 years ago I knew then teamleader/coach of India's team and have had some interesting discussions. (Also interesting is there have been many indian origin students in US team who have done quite well representing US  I have been involved a little with US teams). I may put some thoughts here..but it may generate some "debate" which admins may not like ..
Meanwhile, some may like this discussion..
Why is India's performance at the International Math Olympiad so poor?
Meanwhile, some may like this discussion..
Why is India's performance at the International Math Olympiad so poor?
Re: BR Maths Corner1
Thanks for the link. Some interesting points.
Re: BR Maths Corner1
Two Indian Gold medallists  but neither representing India !
Iran continues to do well...while India is behind even countries like Bangladesh, Saudi Arabia and Mexico !
The govt needs to take this up at a national scale and start some program to entice the best to take this on...it is becoming an embarrassment for the country.
Iran continues to do well...while India is behind even countries like Bangladesh, Saudi Arabia and Mexico !
The govt needs to take this up at a national scale and start some program to entice the best to take this on...it is becoming an embarrassment for the country.
Re: BR Maths Corner1
As said before, this Olympiad was weird, strangest I have seen in decades  as many will agree. (More of this later).
But this is a very tough break for Indian team.
All three bronze medalists missed silver by just one point. (They got 18 points, 19 would have gotten them silver)
(I do not understand  they were all in top 25%  traditionally good enough for silver  the jury discussion for cutoff went on for a long time  I do wish Indian team leader (and jury members) would have advocated 18 for this cutoff (many were thinking it might be as low as 17)
Similarly Aditya missed bronze by 2 points only! (Many were thinking 14 or 15 would be cutoff for bronze, but it was 16). He did not get any points in 4 of the problems  (i would have hoped that he should have written anything  even partial steps to eke out a point or two.. in one of the four problems  I generally advise kids to write something even the rough work down  after all you have hours)
 Similarly the three medalist who did not get a single point in problem 3,5,6 ..should have tried to write something  may be even some rough outline on how they were thinking to attack the problems)
(As I said before, I was hoping Anant who had a "?" (points contested) in P5 would have gotten at least a point or two in P5; that would have given him silver. USA's Ankan who almost had identical scores in other problems, his "?" in P5 became 7 and he ended up winning a gold. I was hoping that Anant's medal was different .. as I said before I was even hoping for a gold for him)
(Tarush and Sutanay did not get a perfect score in relatively easy problem 4  a disappointment (as they thought that they have done well in that problem, and in my estimates I have given them 7 in those problems)  this may be a simple case of not knowing how to write answers in mathematically rigorous way and training may help)
But this is a very tough break for Indian team.
All three bronze medalists missed silver by just one point. (They got 18 points, 19 would have gotten them silver)
(I do not understand  they were all in top 25%  traditionally good enough for silver  the jury discussion for cutoff went on for a long time  I do wish Indian team leader (and jury members) would have advocated 18 for this cutoff (many were thinking it might be as low as 17)
Similarly Aditya missed bronze by 2 points only! (Many were thinking 14 or 15 would be cutoff for bronze, but it was 16). He did not get any points in 4 of the problems  (i would have hoped that he should have written anything  even partial steps to eke out a point or two.. in one of the four problems  I generally advise kids to write something even the rough work down  after all you have hours)
 Similarly the three medalist who did not get a single point in problem 3,5,6 ..should have tried to write something  may be even some rough outline on how they were thinking to attack the problems)
(As I said before, I was hoping Anant who had a "?" (points contested) in P5 would have gotten at least a point or two in P5; that would have given him silver. USA's Ankan who almost had identical scores in other problems, his "?" in P5 became 7 and he ended up winning a gold. I was hoping that Anant's medal was different .. as I said before I was even hoping for a gold for him)
(Tarush and Sutanay did not get a perfect score in relatively easy problem 4  a disappointment (as they thought that they have done well in that problem, and in my estimates I have given them 7 in those problems)  this may be a simple case of not knowing how to write answers in mathematically rigorous way and training may help)
Re: BR Maths Corner1
Wondering if anyone here found this interesting .. or found an interesting solution...
.. All indian (and USA) students got this problem right and scored a perfect score..
Only loop one gets is 3,6,9...3,6,9 (Any number divisible by 3 will finally end up with this loop (sqrt(9)=3) .. All other numbers will never loop.
Intuitively it is easy as one checks for 2,3,4 etc.. (what else you are supposed to do in an exam  test for some small numbers and see if there is a pattern and you can prove the result). Writing a proof is what they were looking for.
.. All indian (and USA) students got this problem right and scored a perfect score..
Amber G. wrote:Few comments wrt IMO 2017.. while we wait for results..
 Problem 1 is fairly easy, if anyone likes to try it here (or ask that to a bright student).
(Basically the problem is 
 start with a number (positive integer > 1)
 get the next number by adding 3 to it.. except (see next line)
 but if the number is a square .. take the square root.
(This way you will get increasing numbers except once in a while you fall back  for example if you start with 91, you get 91, 94, 97, 100 go down and get 10, 13, 16  go down get 4, go down get 2, 5, 8, .. etc.)
Your job is to find a *all* starting numbers such that , the series becomes a loop..
***
Only loop one gets is 3,6,9...3,6,9 (Any number divisible by 3 will finally end up with this loop (sqrt(9)=3) .. All other numbers will never loop.
Intuitively it is easy as one checks for 2,3,4 etc.. (what else you are supposed to do in an exam  test for some small numbers and see if there is a pattern and you can prove the result). Writing a proof is what they were looking for.
Re: BR Maths Corner1
To day is Word Maths Day: See if you can prove this from Ramajujan's problem:
(One of the relatively easy problem)
(One of the relatively easy problem)
Re: BR Maths Corner1
Amber G. wrote:To day is Word Maths Day: See if you can prove this from Ramajujan's problem:
(One of the relatively easy problem)
Are we allowed to use limits? Some form the Squeeze theorem can be used, if my intuition is correct. I worked on it a little bit; for ~ 30 minutes. I am not sure whether I making it too complicated.
I tried the usual "Gaussian"/surd tricks which did no work. Looks like the proof is a little deeper and might take longer than the half an hour I devoted to the solution. Of course, my half an hour is a lifetime for some who have seen aims liar problems before, to whit Prof. Bruce Berendt.
Last edited by Vayutuvan on 17 Oct 2017 06:08, edited 1 time in total.
Re: BR Maths Corner1
Arjun wrote:Two Indian Gold medallists  but neither representing India !
Iran continues to do well...while India is behind even countries like Bangladesh, Saudi Arabia and Mexico !
Arjun ji, since you hold that the IQs must be high and within the same band for people of Indian origin, whether in India or elsewhere, shouldn't that be enough to get equivalent scores?
Re: BR Maths Corner1
My 0.0002 cent to this discussion. I personally believe all people around the world are statistically the same. 80% donkeys, 10% mediocre, 10% really smart. Successful nations and societies are able to identify these 10% really smart folks.cultivate them and then magic happens. In India a large number of kids don't even get a decent education (we loose a lot of smart people this way), forget things like love for Maths and Science. The few who get some decent education are mostly driven by hunger pangs and usually opt for the Kota system of education, which is based more on pattern matching and recall factor.
I feel the primary responsibility for cultivating your child's interest fall on the parents. We should actively encourage if we can, our wards to follow their dreams and not the rat race. Also it is high time govt starts spending some quality money on school education.The Govt seems to have abdicated it's primary responsibility on education to the private sector. and that is SAD.
I feel the primary responsibility for cultivating your child's interest fall on the parents. We should actively encourage if we can, our wards to follow their dreams and not the rat race. Also it is high time govt starts spending some quality money on school education.The Govt seems to have abdicated it's primary responsibility on education to the private sector. and that is SAD.
Re: BR Maths Corner1
Amber G. wrote:Wondering if anyone here found this interesting .. or found an interesting solution...
.. All indian (and USA) students got this problem right and scored a perfect score..

 Problem 1 is fairly easy, if anyone likes to try it here (or ask that to a bright student).
[b]Your job is to find a *all* starting numbers such that , the series becomes a loop..
I got 1 (trivial case) and {n n = 0 (mod 3)}
I will take a shot at proving that there are no more.
Last edited by Vayutuvan on 18 Oct 2017 04:10, edited 1 time in total.

 BRFite
 Posts: 448
 Joined: 07 Jul 2017 20:50
Re: BR Maths Corner1
See if you can prove this from Ramajujan's problem:
Nice. Easily proven using induction once you rewrite the infinitely nested function in the form: f(k+1) = g(k)
Re: BR Maths Corner1
Edited later: Missed the above post (was posted while I was posting). Nice Hint..(More details)
Hi  The problem is from Ramanujan published in Journal of Indian Mathematical Society  so if interested look it up.
Of course, it is from a general class of problems..
For solution of (which gives this kind of surds..)
(f(x))^2 = ax +(n+a)^2+xf(x+n)
gives a solution f(x)=x+n+a
(Which is of course, very easy to check) but the above gives that kind of problems...
***
**** Stop here if you just want a hint *** Reading more will give away the solution..
For example, In a particular case (easy to verify)
We start (1+x)^2 = 1+2x+x^2 = 1+x(x+2).
of write this as (1+x) = sqrt (1 + x (x+2)) when x=2 we get 3=sqrt(1+2(2+2))
Now for write 2+2 = 1+3 and apply the same formula...
we get 3 = sqrt (1+2*sqrt(1+3*5))
Now do the same for 5..
****
In other words ... write 3 = sqrt(1+2*4))
write 4 = sqrt(1+3*5)
Write 5 as sqrt (1+4*6)
and so on ... and you have the original problem solved.. (see how easy math can be )
Vayutuvan wrote:Amber G. wrote:To day is Word Maths Day: See if you can prove this from Ramajujan's problem:
(One of the relatively easy problem)
Are we allowed to use limits? Some form the Squeeze theorem can be used, if my intuition is correct. I worked on it a little bit; for ~ 30 minutes. I am not sure whether I making it too complicated.
I tried the usual "Gaussian"/surd tricks which did no work. Looks like the proof is a little deeper and might take longer than the half an hour I devoted to the solution. Of course, my half an hour is a lifetime for some who have seen aims liar problems before, to whit Prof. Bruce Berendt.
Hi  The problem is from Ramanujan published in Journal of Indian Mathematical Society  so if interested look it up.
Of course, it is from a general class of problems..
For solution of (which gives this kind of surds..)
(f(x))^2 = ax +(n+a)^2+xf(x+n)
gives a solution f(x)=x+n+a
(Which is of course, very easy to check) but the above gives that kind of problems...
***
**** Stop here if you just want a hint *** Reading more will give away the solution..
For example, In a particular case (easy to verify)
We start (1+x)^2 = 1+2x+x^2 = 1+x(x+2).
of write this as (1+x) = sqrt (1 + x (x+2)) when x=2 we get 3=sqrt(1+2(2+2))
Now for write 2+2 = 1+3 and apply the same formula...
we get 3 = sqrt (1+2*sqrt(1+3*5))
Now do the same for 5..
****
In other words ... write 3 = sqrt(1+2*4))
write 4 = sqrt(1+3*5)
Write 5 as sqrt (1+4*6)
and so on ... and you have the original problem solved.. (see how easy math can be )
Re: BR Maths Corner1
Vayutuvan wrote:Amber G. wrote:Wondering if anyone here found this interesting .. or found an interesting solution...
.. All indian (and USA) students got this problem right and scored a perfect score..

 Problem 1 is fairly easy, if anyone likes to try it here (or ask that to a bright student).
[b]Your job is to find a *all* starting numbers such that , the series becomes a loop..
I got 1 (trivial case) and {n: n = 0 (mod 3)}
I will take a shot at proving that there are no more.
Yes, The *only* case is (3,6,9). There is no other solution.
Hint: Very easy to prove no solution for numbers like (3k+2).
Slightly difficult for numbers like (3k+1)
(If you try just few small numbers, like 2,3,4,5 .. the reason will become clear)
Re: BR Maths Corner1
periaswamy wrote:See if you can prove this from Ramajujan's problem:
Nice. Easily proven using induction once you rewrite the infinitely nested function in the form: f(k+1) = g(k)
Hmm. Easy you say?
Re: BR Maths Corner1
Amber G. wrote:Vayutuvan wrote:
I got 1 (trivial case) and {n: n = 0 (mod 3)}
I will take a shot at proving that there are no more.
Yes, The *only* case is (3,6,9). There is no other solution.
Hint: Very easy to prove no solution for numbers like (3k+2).
Slightly difficult for numbers like (3k+1)
(If you try just few small numbers, like 2,3,4,5 .. the reason will become clear)
I got a little further using the Fundamental Theorem of Arithmetic. But that looks a little too round about. There must be shorter solutions. For example, there is theorem that states that every number can be written in the form m^2. k where k is a radical (?) which of course is squarefree.
This is really embarrassing to say the least. After 30 minutes, I am unable to solve the simplest of the problems.
Re: BR Maths Corner1
Vayutuvan wrote:... some who have seenaims liarsimilar problems before, to whit Prof. Bruce Berendt and his students.
Correcting spelling mistake.
Re: BR Maths Corner1
What's Special About This Number?
Kaprekar Number defined by Mr. Kaprekar (also known as Niven Number)
If you know a distinctive fact about a number not listed here, please email me.
primes graphs digits sums of powers bases
combinatorics powers/polygonal Fibonacci
geometry repdigits algebra perfect/amicable
pandigital matrices divisors games/puzzles
0 is the additive identity.
1 is the multiplicative identity.
2 is the only even prime.
3 is the number of spatial dimensions we live in.
4 is the smallest number of colors sufficient to color all planar maps.
5 is the number of Platonic solids.
6 is the smallest perfect number.
7 is the smallest number of sides of a regular polygon that is not constructible by straightedge and compass.
8 is the largest cube in the Fibonacci sequence.
9 is the maximum number of cubes that are needed to sum to any positive integer.
10 is the base of our number system.
11 is the largest known multiplicative persistence.
12 is the smallest abundant number.
13 is the number of Archimedean solids.
14 is the smallest even number n with no solutions to φ(m) = n.
15 is the smallest composite number n with the property that there is only one group of order n.
16 is the only number of the form xy = yx with x and y being different integers.
17 is the number of wallpaper groups.
18 is the only positive number that is twice the sum of its digits.
19 is the maximum number of 4th powers needed to sum to any number.
...
9999 Is a Kaprekar Number
Kaprekar Number defined by Mr. Kaprekar (also known as Niven Number)

 BRFite
 Posts: 448
 Joined: 07 Jul 2017 20:50
Re: BR Maths Corner1
Vayutuvan wrote:Hmm. Easy you say?
Like most of mathematics, tricksy but once you see the trick...
if you rewrite = 3 as 2+1 and eliminate the square root by squaring both sides. then you see (2+1)^2  1 = 2 sqrt(1+3...)
which becomes (3+1)^2  1 = 3*sqrt(1+4....)
so if you define f(k) as sqrt(1+ (k+1) f(k+1))
so you can generalize to (k+1)^2  1 = k * f(k+1) which lends itself to proof by induction.
Re: BR Maths Corner1
By Vickyiyer
I personally believe all people around the world are statistically the same. 80% donkeys, 10% mediocre, 10% really smart.
All healthy people have same amount of brain some apply it to arts others to science some to sports., It is what you do with your time on earth that matters! If you want to be "XYZ" you can be "XYZ" as long as you dedicate enough time and sincere effort towards the cause.
Re: BR Maths Corner1
periaswamy wrote:Vayutuvan wrote:Hmm. Easy you say?
Like most of mathematics, tricksy but once you see the trick...
if you rewrite = 3 as 2+1 and eliminate the square root by squaring both sides. then you see (2+1)^2  1 = 2 sqrt(1+3...)
which becomes (3+1)^2  1 = 3*sqrt(1+4....)
so if you define f(k) as sqrt(1+ (k+1) f(k+1))
so you can generalize to (k+1)^2  1 = k * f(k+1) which lends itself to proof by induction.
Na na. I was not going to read this. But I scanned the initial part. I will reproduce it now hopefully.
Re: BR Maths Corner1
dup  deleted!
Last edited by Amber G. on 18 Oct 2017 21:50, edited 1 time in total.
Re: BR Maths Corner1
Google is fun !
Also (apart from looking at my post many posts above  also it turns out that this problem and its solution has appeared in mathdhaga many years before). one can use google search do "Ramanujan's infinite radicals" (or something like that)..and the first result you will find is :
this
The second result in google is:
Vayutuvan wrote:
Na na. I was not going to read this. But I scanned the initial part. I will reproduce it now hopefully.
Also (apart from looking at my post many posts above  also it turns out that this problem and its solution has appeared in mathdhaga many years before). one can use google search do "Ramanujan's infinite radicals" (or something like that)..and the first result you will find is :
this
The second result in google is:
Re: BR Maths Corner1
(Full problem is at: https://forums.bharatrakshak.com/viewtopic.php?p=2211496#p2211496
Here is one solution  If people are still curious. (Solution does NOT require any high math, just elemantry math and logic)
 One key part is, no square is 2 mod 3 (trivial to prove, (3k+1)^2, and (3k+2)^2 both when divided by 3 will leave a remainder 1)
Hence if you start with number like 2 (or 5 or 8 etc) the numbers 2,5,8,11,... will always keep growing, no loop back
(None of the number could be a square)
 Another part is to think about "induction"., for any "large" k, solution  if loop back takes place  will behave same as some number less than k (as long sqrt(k) is less than (k3) etc).. unless sqrt(k) is 2 mod 3.
So basically you have to check for starting numbers 2,3,4,5 only. This is because for any larger number (n3)^2 will be > n . (All other will either come to these numbers or will get to a number like 3k+2 and then go for ever) (
Number (2 and 5) are 3k+2 type ==> No solution)
Number 3 will give (3,6,9,..3,6,9..) ==> looping solution.
Number 4 goes to sqrt(4)=2 ===> No solution
so.
Any number like 3k will eventually fall back to 3 and the loop will be (3,6,9).. All other numbers will finally end up as 3k+2 and will grow ever from that point.
(This is why math is beautiful! .
Amber G. wrote:Vayutuvan wrote:
I got 1 (trivial case) and {n: n = 0 (mod 3)}
I will take a shot at proving that there are no more.
Yes, The *only* case is (3,6,9). There is no other solution.
Hint: Very easy to prove no solution for numbers like (3k+2).
Slightly difficult for numbers like (3k+1)
(If you try just few small numbers, like 2,3,4,5 .. the reason will become clear)
Here is one solution  If people are still curious. (Solution does NOT require any high math, just elemantry math and logic)
 One key part is, no square is 2 mod 3 (trivial to prove, (3k+1)^2, and (3k+2)^2 both when divided by 3 will leave a remainder 1)
Hence if you start with number like 2 (or 5 or 8 etc) the numbers 2,5,8,11,... will always keep growing, no loop back
(None of the number could be a square)
 Another part is to think about "induction"., for any "large" k, solution  if loop back takes place  will behave same as some number less than k (as long sqrt(k) is less than (k3) etc).. unless sqrt(k) is 2 mod 3.
So basically you have to check for starting numbers 2,3,4,5 only. This is because for any larger number (n3)^2 will be > n . (All other will either come to these numbers or will get to a number like 3k+2 and then go for ever) (
Number (2 and 5) are 3k+2 type ==> No solution)
Number 3 will give (3,6,9,..3,6,9..) ==> looping solution.
Number 4 goes to sqrt(4)=2 ===> No solution
so.
Any number like 3k will eventually fall back to 3 and the loop will be (3,6,9).. All other numbers will finally end up as 3k+2 and will grow ever from that point.
(This is why math is beautiful! .
Re: BR Maths Corner1
I have a question about math history, I had once had a disagreement on some blog (the author was muslim, possibly paki) where I said a number system and decimal system which we use today had originated in India and he had dismissed this saying only a very primitive system had been formed in India and greatly improved and modernized by Islamic civilization. Is this true, and if so what were the additional changes made by the them?
Similarly, I've often heard the phrase that Arabs (meaning Islamic scholars) invented algebra which I think isn't really true since equations were around before that in other civilizations (wikipedia has section on equation solving in other civilizations). Again, what was their main change which you could call proper algebra which didn't exist in other civilizations before ? or was it just incremental discoveries in some specific area with algebra systems already existing and this confusion being because it was from them it was introduced to postmedieval west (like the label arabic numerals)?
Somebody had mentioned long back that Indian texts had been translated into arabic and their authors do cite Indian works. How much of their work has evidence of being totally new and how much was sourced from earlier Greek, Persian, Indian, Roman (I don't really see much about that last one)? What parts can attributed clearly to Indian contribution (such as numerals and counting system)?
Similarly, I've often heard the phrase that Arabs (meaning Islamic scholars) invented algebra which I think isn't really true since equations were around before that in other civilizations (wikipedia has section on equation solving in other civilizations). Again, what was their main change which you could call proper algebra which didn't exist in other civilizations before ? or was it just incremental discoveries in some specific area with algebra systems already existing and this confusion being because it was from them it was introduced to postmedieval west (like the label arabic numerals)?
Somebody had mentioned long back that Indian texts had been translated into arabic and their authors do cite Indian works. How much of their work has evidence of being totally new and how much was sourced from earlier Greek, Persian, Indian, Roman (I don't really see much about that last one)? What parts can attributed clearly to Indian contribution (such as numerals and counting system)?
Re: BR Maths Corner1
Shwetank,
It is very hard to pin down who invented (for strict constructionists)/discovered (all others) what. What is recorded is that Al Khwarizmi Al Jabar is considered to be the person from whose name both Algorithm and Algebra are derived. It is not too far fetched in my humble opinion. That said, one cannot dismiss Indian contributions to mathematical foundations, especially the contributions of <i>paaNini</i> and _nyaaya_ for logic. In that sense that blogger is way out of line in his dismissal. Probably he is more of an "eminent historian" rather than a historian of mathematics.
It is very hard to pin down who invented (for strict constructionists)/discovered (all others) what. What is recorded is that Al Khwarizmi Al Jabar is considered to be the person from whose name both Algorithm and Algebra are derived. It is not too far fetched in my humble opinion. That said, one cannot dismiss Indian contributions to mathematical foundations, especially the contributions of <i>paaNini</i> and _nyaaya_ for logic. In that sense that blogger is way out of line in his dismissal. Probably he is more of an "eminent historian" rather than a historian of mathematics.
Re: BR Maths Corner1
Al Kowarizmi Al Jabar translated Indian books and methods (also tranlsated Greek stuff). There were others too who talked of 'hisabehind'. The basic contributions from Indian sources and their path through the Islamic world are noted in multiple books on mathematics (numerals/decimal system, algebra and trigonometery are all listed there). Nowadays there is contention that Jesuit priests took early Calculus concepts (Kerala school is cited) to Europe and that inspired Descartes and Newton.
btw, the modern numerals are actually modified Indian numerals from Morocco. The rest of Arab world uses different numerals.
btw, the modern numerals are actually modified Indian numerals from Morocco. The rest of Arab world uses different numerals.
Re: BR Maths Corner1
In this dhaga I have talked about Mersenne prime numbers dozens of times. Now we have 50th known Mersenne prime and 50th perfect number!
New Mersenne prime M77232917 found! 2^772329171 is the largest known prime number!
https://www.mersenne.org/primes/press/M77232917.html
New Mersenne prime M77232917 found! 2^772329171 is the largest known prime number!
https://www.mersenne.org/primes/press/M77232917.html
RALEIGH, NC., January 3, 2018  The Great Internet Mersenne Prime Search (GIMPS) has discovered the largest known prime number, 277,232,9171, having 23,249,425 digits. A computer volunteered by Jonathan Pace made the find on December 26, 2017. Jonathan is one of thousands of volunteers using free GIMPS software available at www.mersenne.org/download/.
The new prime number, also known as M77232917, is calculated by multiplying together 77,232,917 twos, and then subtracting one. It is nearly one million digits larger than the previous record prime number, in a special class of extremely rare prime numbers known as Mersenne primes. It is only the 50th known Mersenne prime ever discovered, each increasingly difficult to find. Mersenne primes were named for the French monk Marin Mersenne, who studied these numbers more than 350 years ago. GIMPS, founded in 1996, has discovered the last 16 Mersenne primes. Volunteers download a free program to search for these primes, with a cash award offered to anyone lucky enough to find a new prime. Prof. Chris Caldwell maintains an authoritative web site on the largest known primes, and has an excellent history of Mersenne primes.
The primality proof took six days of nonstop computing on a PC with an Intel i56600 CPU. To prove there were no errors in the prime discovery process, the new prime was independently verified using four different programs on four different hardware configurations.
Aaron Blosser verified it using Prime95 on an Intel Xeon server in 37 hours.
David Stanfill verified it using gpuOwL on an AMD RX Vega 64 GPU in 34 hours.
Andreas Höglund verified the prime using CUDALucas running on NVidia Titan Black GPU in 73 hours.
Ernst Mayer also verified it using his own program Mlucas on 32core Xeon server in 82 hours. Andreas Höglund also confirmed using Mlucas running on an Amazon AWS instance in 65 hours.
Jonathan Pace is a 51year old Electrical Engineer living in Germantown, Tennessee. Perseverance has finally paid off for Jon  he has been hunting for big primes with GIMPS for over 14 years. The discovery is eligible for a $3,000 GIMPS research discovery award.
GIMPS Prime95 client software was developed by founder George Woltman. Scott Kurowski wrote the PrimeNet system software that coordinates GIMPS' computers. Aaron Blosser is now the system administrator, upgrading and maintaining PrimeNet as needed. Volunteers have a chance to earn research discovery awards of $3,000 or $50,000 if their computer discovers a new Mersenne prime. GIMPS' next major goal is to win the $150,000 award administered by the Electronic Frontier Foundation offered for finding a 100 million digit prime number.
Credit for this prime goes not only to Jonathan Pace for running the Prime95 software, Woltman for writing the software, Kurowski and Blosser for their work on the Primenet server, but also the thousands of GIMPS volunteers that sifted through millions of nonprime candidates. In recognition of all the above people, official credit for this discovery goes to "J. Pace, G. Woltman, S. Kurowski, A. Blosser, et al."
About Mersenne.org's Great Internet Mersenne Prime Search
The Great Internet Mersenne Prime Search (GIMPS) was formed in January 1996 by George Woltman to discover new world record size Mersenne primes. In 1997 Scott Kurowski enabled GIMPS to automatically harness the power of thousands of ordinary computers to search for these "needles in a haystack". Most GIMPS members join the search for the thrill of possibly discovering a recordsetting, rare, and historic new Mersenne prime. The search for more Mersenne primes is already under way. There may be smaller, as yet undiscovered Mersenne primes, and there almost certainly are larger Mersenne primes waiting to be found. Anyone with a reasonably powerful PC can join GIMPS and become a big prime hunter, and possibly earn a cash research discovery award. All the necessary software can be downloaded for free at www.mersenne.org/download/. GIMPS is organized as Mersenne Research, Inc., a 501(c)(3) science research charity. Additional information may be found at www.mersenneforum.org and www.mersenne.org; donations are welcome.
For More Information on Mersenne Primes
Prime numbers have long fascinated both amateur and professional mathematicians. An integer greater than one is called a prime number if its only divisors are one and itself. The first prime numbers are 2, 3, 5, 7, 11, etc. For example, the number 10 is not prime because it is divisible by 2 and 5. A Mersenne prime is a prime number of the form 2P1. The first Mersenne primes are 3, 7, 31, and 127 corresponding to P = 2, 3, 5, and 7 respectively. There are now 50 known Mersenne primes.
Mersenne primes have been central to number theory since they were first discussed by Euclid about 350 BC. The man whose name they now bear, the French monk Marin Mersenne (15881648), made a famous conjecture on which values of P would yield a prime. It took 300 years and several important discoveries in mathematics to settle his conjecture.
At present there are few practical uses for this new large prime, prompting some to ask "why search for these large primes"? Those same doubts existed a few decades ago until important cryptography algorithms were developed based on prime numbers. For seven more good reasons to search for large prime numbers, see here.
Previous GIMPS Mersenne prime discoveries were made by members in various countries.
In January 2016, Curtis Cooper et al. discovered the 49th known Mersenne prime in the U.S.
In January 2013, Curtis Cooper et al. discovered the 48th known Mersenne prime in the U.S.
In April 2009, Odd Magnar Strindmo et al. discovered the 47th known Mersenne prime in Norway.
In September 2008, HansMichael Elvenich et al. discovered the 46th known Mersenne prime in Germany.
In August 2008, Edson Smith et al. discovered the 45th known Mersenne prime in the U.S.
In September 2006, Curtis Cooper and Steven Boone et al. discovered the 44th known Mersenne prime in the U.S.
In December 2005, Curtis Cooper and Steven Boone et al. discovered the 43rd known Mersenne prime in the U.S.
In February 2005, Dr. Martin Nowak et al. discovered the 42nd known Mersenne prime in Germany.
In May 2004, Josh Findley et al. discovered the 41st known Mersenne prime in the U.S.
In November 2003, Michael Shafer et al. discovered the 40th known Mersenne prime in the U.S.
In November 2001, Michael Cameron et al. discovered the 39th Mersenne prime in Canada.
In June 1999, Nayan Hajratwala et al. discovered the 38th Mersenne prime in the U.S.
In January 1998, Roland Clarkson et al. discovered the 37th Mersenne prime in the U.S.
In August 1997, Gordon Spence et al. discovered the 36th Mersenne prime in the U.K.
In November 1996, Joel Armengaud et al. discovered the 35th Mersenne prime in France.
Euclid proved that every Mersenne prime generates a perfect number. A perfect number is one whose proper divisors add up to the number itself. The smallest perfect number is 6 = 1 + 2 + 3 and the second perfect number is 28 = 1 + 2 + 4 + 7 + 14. Euler (17071783) proved that all even perfect numbers come from Mersenne primes. The newly discovered perfect number is 277,232,916 x (277,232,9171). This number is over 46 million digits long! It is still unknown if any odd perfect numbers exist.
There is a unique history to the arithmetic algorithms underlying the GIMPS project. The programs that found the recent big Mersenne primes are based on a special algorithm. In the early 1990's, the late Richard Crandall, Apple Distinguished Scientist, discovered ways to double the speed of what are called convolutions  essentially big multiplication operations. The method is applicable not only to prime searching but other aspects of computation. During that work he also patented the Fast Elliptic Encryption system, now owned by Apple Computer, which uses Mersenne primes to quickly encrypt and decrypt messages. George Woltman implemented Crandall's algorithm in assembly language, thereby producing a primesearch program of unprecedented efficiency, and that work led to the successful GIMPS project.
School teachers from elementary through highschool grades have used GIMPS to get their students excited about mathematics. Students who run the free software are contributing to mathematical research. David Stanfill's and Ernst Mayer's verification computations for this discovery was donated by Squirrels LLC (http://www.airsquirrels.com) which services K12 education and other customers.

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Re: BR Maths Corner1
Yayavar wrote:Al Kowarizmi Al Jabar translated Indian books and methods (also tranlsated Greek stuff). There were others too who talked of 'hisabehind'. The basic contributions from Indian sources and their path through the Islamic world are noted in multiple books on mathematics (numerals/decimal system, algebra and trigonometery are all listed there). Nowadays there is contention that Jesuit priests took early Calculus concepts (Kerala school is cited) to Europe and that inspired Descartes and Newton.
btw, the modern numerals are actually modified Indian numerals from Morocco. The rest of Arab world uses different numerals.
A professor in mathematics from Cambridge under whom I studied in the past  literally called the Number system currently used as Hindu numeral system. He was not ashamed to say that ancient Indian mathematics was very well studied and passed to western world by Arabs and that was one of the main drivers of Renaissance and end of dark ages in Europe. Prior to that European mathematics was handicap while using Greek Latin number system which really went to certain length before everything goes haywire in representing numbers.
Having said that Babylonians were quite adept at mathematics and being at cusp of Indian, Chinese and European trade routes they easily borrowed improved things. Arabs are just imposters masking themselves as Babylonians? The modern "Arabs" are Bedouin tribes you had nothing to do with mathematics. A simple question on why there is no scientific discovery in this so called land of mathematical genius is testament of the fact of plagiarism and nothing else. Even in the west most countries were just hunter gatherers and savages and west really only had Romans and Greeks to show something worthwhile. Their knowledge exploded after advent of mass printing which made books on all subjects available to large number of masses and study them widely. Industrial revolution and scientific progress is history from then on. Zero is indeed a mathematical marvel without which rockets could have flown or humans could not have achieved the progress we have now or mathematics, banking, trading and economics would not have progressed as much. I felt proud when he paid tribute to Indian genius who invented zero!!
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