BR Maths Corner1
Re: BR Maths Corner1
The International Math Olympiads 2018 are currently in process.
Good luck to Alll, specially Indian and US teams.
US Team:
 James Lin (Contestant 1)(19 years old, Gold in IMO 2017)
• Participation at IMO: 2017 (G)
 Adam Ardeishar (Contestant 2)(16 years old)
 Mihir Anand Singhal (Contestant 3)(17 Years old)
 Andrew GuAndrew Gu (Contestant 4) (17 years old), Gold in IMO 2017)
 Michael RenMichael Re ((Contestant 5) (18 years old
 Vincent HuangVincent Huan (Contestant 6)(17 years old, Silver in 2017)
• Participation at IMO: 2017 (S)
Hi, and Good luck to PoShen Loh (Leader)(A well know mathematics family, both brothers and sister has won gold medals for US in past  Time passes fast have known him for 20+ years)
Indian Team:
 Sutanay Bhattacharya (Contestant 1) (17 years old, Bronze in 2016, HM in 2017)
 Spandan GhoshSpandan Ghosh (Contestant 2)(16 years old)
 Amit Kumar Mallik (Contestant 3)( 16 years old)
 Anant Mudgal (Contestant 4) (17 years old, 2015 HM, 2016 Bronze, 2017 Bronze)
 Pulkit Sinha (Contestant 5) ( 18 years old)
 Pranjal Srivastava (Contestant 6) (14 Years old)
Good luck to Alll, specially Indian and US teams.
US Team:
 James Lin (Contestant 1)(19 years old, Gold in IMO 2017)
• Participation at IMO: 2017 (G)
 Adam Ardeishar (Contestant 2)(16 years old)
 Mihir Anand Singhal (Contestant 3)(17 Years old)
 Andrew GuAndrew Gu (Contestant 4) (17 years old), Gold in IMO 2017)
 Michael RenMichael Re ((Contestant 5) (18 years old
 Vincent HuangVincent Huan (Contestant 6)(17 years old, Silver in 2017)
• Participation at IMO: 2017 (S)
Hi, and Good luck to PoShen Loh (Leader)(A well know mathematics family, both brothers and sister has won gold medals for US in past  Time passes fast have known him for 20+ years)
Indian Team:
 Sutanay Bhattacharya (Contestant 1) (17 years old, Bronze in 2016, HM in 2017)
 Spandan GhoshSpandan Ghosh (Contestant 2)(16 years old)
 Amit Kumar Mallik (Contestant 3)( 16 years old)
 Anant Mudgal (Contestant 4) (17 years old, 2015 HM, 2016 Bronze, 2017 Bronze)
 Pulkit Sinha (Contestant 5) ( 18 years old)
 Pranjal Srivastava (Contestant 6) (14 Years old)
Re: BR Maths Corner1
IMO contest is just over.. Rumors (per students) for the first day (yesterday)
Problem 1 All 6 got it
Problem 2 1 perfect, 1 near perfect, 1 half, 3 no.
Problem 3 (None got it).
*** Official results and problems will be out shortly (in a day or so) and I will post the results here.
I think US will beat China this year and will come first.
Problem 1 All 6 got it
Problem 2 1 perfect, 1 near perfect, 1 half, 3 no.
Problem 3 (None got it).
*** Official results and problems will be out shortly (in a day or so) and I will post the results here.
I think US will beat China this year and will come first.
Last edited by Amber G. on 09 Jul 2018 23:03, edited 2 times in total.
Re: BR Maths Corner1
The third problem of International Math olympiad happens to be an old problem, I am sure I saw it in Scientific American's Gardenr's column. (My father was interested in such problems so I have happy memories of playing with such patterns. Of course, I had lots of fun introducing Pascal's triangle (and magic squares) to my own kids and nephews/nieces.) I though the India team would have been able to do it, as this kind of patters have been studied in India long time ago.
Here is the problem, try your hand on it . (feel free to post solution or effort).
>>> Problem from day 1 of 2018 IMO:
An antiPascal triangle is an equilateral triangular array of numbers such that, except for the numbers in the top row, each number is the absolute value of the difference of the two numbers immediately above it.
For example, the following is an antiPascal triangle with four rows which contains *every* integer from *1* to *10*
Does there exist an antiPascal triangle with 2018 rows which contains every integer from *1* to *1 + 2 + 3 + ...+ 2018*?
Here is the problem, try your hand on it . (feel free to post solution or effort).
>>> Problem from day 1 of 2018 IMO:
An antiPascal triangle is an equilateral triangular array of numbers such that, except for the numbers in the top row, each number is the absolute value of the difference of the two numbers immediately above it.
For example, the following is an antiPascal triangle with four rows which contains *every* integer from *1* to *10*
Code: Select all
6 10 1 8
4 9 7
5 2
3
Does there exist an antiPascal triangle with 2018 rows which contains every integer from *1* to *1 + 2 + 3 + ...+ 2018*?
Re: BR Maths Corner1
Added later: The official results have been posted. (https://www.imoofficial.org/results.aspx ) But I am keeping my original message. It was fun to predict/guess while partial results were known.
**** Original message ***
IMO 2018 
Based on partial results (and information from coaches/students, rumors)..
USA has done excellent ..looks like they (and China) are #1 and #2..(** Official results, USA 1, Russia 2, China 3)
at least two people in USA have gotten perfect score in at least 5 of the six problems. Impressive.
(Edited later: James Lin did get perfect score in *all* the problems and Mihir Anand Singhal got 40/42)
(Official results will be out shortly, but partial results are available to participants/coaches)
India  I hoped veteran Anant to get a gold this time but it does not look like so. Newest and youngest Pranjal (14 years and I think youngest in history from India) has done well (but I was hoping more) (*** edited later *** India ranked 28)
Any way from partial results I think/hope  India has chances to get 3 silvers and 12 bronzes Fingers crossed (** edited later *** India did get 3 silver and 2 bronze)
Congrats to all.
**** Original message ***
IMO 2018 
Based on partial results (and information from coaches/students, rumors)..
USA has done excellent ..looks like they (and China) are #1 and #2..(** Official results, USA 1, Russia 2, China 3)
at least two people in USA have gotten perfect score in at least 5 of the six problems. Impressive.
(Edited later: James Lin did get perfect score in *all* the problems and Mihir Anand Singhal got 40/42)
(Official results will be out shortly, but partial results are available to participants/coaches)
India  I hoped veteran Anant to get a gold this time but it does not look like so. Newest and youngest Pranjal (14 years and I think youngest in history from India) has done well (but I was hoping more) (*** edited later *** India ranked 28)
Any way from partial results I think/hope  India has chances to get 3 silvers and 12 bronzes Fingers crossed (** edited later *** India did get 3 silver and 2 bronze)
Congrats to all.
Re: BR Maths Corner1
Congratulations to U.S. team on their 1st place victory at the 59th International Mathematical Olympiad in Romania!
(5 golds, 1 Silver, 1 perfect score, Wow!)
https://pbs.twimg.com/media/Dh6eKflXcAAhp2Z.jpg
Congratulations to Indian team. (3 Silvers, 2 Bronze, 1 HM).
I was watching it for last few days, few friends were participating  and it was fun to guess/predict results based on partial scores etc.
I was hoping for Gold for Anant and Pranjal (14 years and I think youngest in history from India)
Full Official results are now posted on IMO site.
https://www.imoofficial.org/results.aspx
(5 golds, 1 Silver, 1 perfect score, Wow!)
https://pbs.twimg.com/media/Dh6eKflXcAAhp2Z.jpg
Congratulations to Indian team. (3 Silvers, 2 Bronze, 1 HM).
I was watching it for last few days, few friends were participating  and it was fun to guess/predict results based on partial scores etc.
I was hoping for Gold for Anant and Pranjal (14 years and I think youngest in history from India)
Full Official results are now posted on IMO site.
https://www.imoofficial.org/results.aspx
Re: BR Maths Corner1
^^^ For the record ( Name, score (out of 42) and Medal)
IND1 Sutanay Bhattacharya 21 Bronze
IND2 Spandan Ghosh 21 Bronze
IND3 Amit Kumar Malik 10 HM
IND4 Anant Mudgal} 26 Silver
IND5 Pulkit Sinha 26 Silver
IND6 Pranjal Srivastava 28 Silver
(I am specially impressed by Pranjal  He is youngest  had 4 perfect scores (Prob 3 and Prob 6 were missed)  and method used were nice).. I am sure he is going to do very well in the next olympiad and math in general.
***
Prob 3, (which I posted above) turned out to be rather difficult. None of Indian team solved it. Only 2 from USA did it.
I was kind of surprised as I posted above, when I checked, the problem has indeed came in Scientific American. (This is probably 3040 years late, but I knew how to solve it then ..)
Problem is posted above. (With the hint that solution is in SA one can do internet search)..
(Hint: Answer, is one can not find a solution for N>5 (so not solution for 2018))
One solution posted in SA came from 4th grader (for N=6)indeed quite simple.
IND1 Sutanay Bhattacharya 21 Bronze
IND2 Spandan Ghosh 21 Bronze
IND3 Amit Kumar Malik 10 HM
IND4 Anant Mudgal} 26 Silver
IND5 Pulkit Sinha 26 Silver
IND6 Pranjal Srivastava 28 Silver
(I am specially impressed by Pranjal  He is youngest  had 4 perfect scores (Prob 3 and Prob 6 were missed)  and method used were nice).. I am sure he is going to do very well in the next olympiad and math in general.
***
Prob 3, (which I posted above) turned out to be rather difficult. None of Indian team solved it. Only 2 from USA did it.
I was kind of surprised as I posted above, when I checked, the problem has indeed came in Scientific American. (This is probably 3040 years late, but I knew how to solve it then ..)
Problem is posted above. (With the hint that solution is in SA one can do internet search)..
(Hint: Answer, is one can not find a solution for N>5 (so not solution for 2018))
One solution posted in SA came from 4th grader (for N=6)indeed quite simple.
Re: BR Maths Corner1
Also, (since I don't see any news in MSM  specially Indian news papers  some more data,,
Cutoffs
Bronze: 16
Silver: 25
Gold: 31
Ranking
1. USA
2. Russia
3. China
4. Ukraine
5. Thailand
6. Taiwan
7. Korea
8. Singapore
9. Poland
10. Indonesia..
....
28 India
Team results at : http://www.imoofficial.org/year_country_r.aspx?year=2018&column=total&order=desc
Cutoffs
Bronze: 16
Silver: 25
Gold: 31
Ranking
1. USA
2. Russia
3. China
4. Ukraine
5. Thailand
6. Taiwan
7. Korea
8. Singapore
9. Poland
10. Indonesia..
....
28 India
Team results at : http://www.imoofficial.org/year_country_r.aspx?year=2018&column=total&order=desc
Re: BR Maths Corner1
Congratulations to Akshay Venkatesh for winning the Fields Medal along with, Peter Scholze, Alessio Figalli and Caucher Birkar.
Born in New Delhi, raised in Austraila, and like Manjul another one from Princeton.
Akshay has won SASTRA Ramanujan Prize and his work in Number theory makes him true prodigy like Ramanujan  has done lot of work. In 2010, as I mentioned in brf dhaga, he was an invited speaker at the International Congress of Mathematicians (Hyderabad), BTW I predicted his and few other's from India's name then as no Indian has won the Fields prize then. Manjul Bhargava did not win in 2010 but won in 2014  first Indian origin. (and BTW Modi recruited him as participant in GIAN). Akshay won this year. (Kiran Kedalya is the only one I missed, 2 out of 3 is not bad!!)
I remember when Modi made his first trip to Australia, he told how both Austraila and India are proud of Akshay.
(He mentioned it along with common bond like Cricket between India/Austraila greats  How many world leaders even know about such things?  I mean how many PM would even know that greatest Australian mathematician was born in India).
Again congratulations.
Born in New Delhi, raised in Austraila, and like Manjul another one from Princeton.
Akshay has won SASTRA Ramanujan Prize and his work in Number theory makes him true prodigy like Ramanujan  has done lot of work. In 2010, as I mentioned in brf dhaga, he was an invited speaker at the International Congress of Mathematicians (Hyderabad), BTW I predicted his and few other's from India's name then as no Indian has won the Fields prize then. Manjul Bhargava did not win in 2010 but won in 2014  first Indian origin. (and BTW Modi recruited him as participant in GIAN). Akshay won this year. (Kiran Kedalya is the only one I missed, 2 out of 3 is not bad!!)
I remember when Modi made his first trip to Australia, he told how both Austraila and India are proud of Akshay.
(He mentioned it along with common bond like Cricket between India/Austraila greats  How many world leaders even know about such things?  I mean how many PM would even know that greatest Australian mathematician was born in India).
Again congratulations.
Re: BR Maths Corner1
How about this??? This was 8 years ago: (Link: https://forums.bharatrakshak.com/viewtopic.php?p=923025#p923025
I think I deserve BRF prize!
I think I deserve BRF prize!
Amber G. wrote:Amber G. wrote:Rumors is that Kiran Kedlaya is in for Fields Medal. (Guys you heard it here first!). He is under 40, and is an invited speaker. People may recall I had talked about him in our math thread. He has won many IMO medals ... for those who don't know him, he is at present a prof at MIT.
(Of course, It is a guarded secret, and no one is suppose to know till the announcement is made)
Other I think may be Ngo or may be Lurie or Manjul Bhargava.. Lets us wait and see.
Time is quite near ... Any rumor in Hyderabad?
I'll add another Indian name here: Ashkay Venkatesh
Re: BR Maths Corner1
Courtesy/ #via: @venug
https://aperiodical.com/2018/09/hlfblo ... ypothesis/
HLF Blogs: What is the Riemann Hypothesis?
#Vyoman, #Sky_High
Aha.. so, to answer this question: What would the consequences of a proof be?
"
The most straightforward consequence, if the least mathematically interesting, would be that the person to do it would be awarded one million dollars. "
calling gurus to explain what benefits the world on solving this?
https://aperiodical.com/2018/09/hlfblo ... ypothesis/
HLF Blogs: What is the Riemann Hypothesis?
#Vyoman, #Sky_High
Aha.. so, to answer this question: What would the consequences of a proof be?
"
The most straightforward consequence, if the least mathematically interesting, would be that the person to do it would be awarded one million dollars. "
So, there is no emphatic/direct impact on encryption
#Vyoman, #Sky_High
I guess, we are covered by having higher number of bits to cover up for possible hits  for example: SHA3 1024 bits
venug
The author says there could be as it will make prime number finding easy
venug
2m
#Vyoman, #Sky_High
Perhaps we will take this to BRF math dhaaga. we have couple or profs there
venug
Ok
calling gurus to explain what benefits the world on solving this?
Re: BR Maths Corner1
For those of you curious about the #Riemannhypothesis, this thread is excellent. It explains why skepticism about the “proof” is warranted.
https://twitter.com/stevenstrogatz/stat ... 00066?s=21
Re: BR Maths Corner1
SaiK wrote: <<<.... about
HLF Blogs: What is the Riemann Hypothesis?
 Perhaps we will take this to BRF math dhaaga. we have couple or profs there
calling gurus to explain what benefits the world on solving this?
Few comments:
 We have discussed the Riemann Hypothesis here in BRF at a few times. Some really good explanation can be found <this brf post> or or here . IMO they are easy to read and cover some basic math and importance. I am going to requote those two posts for convenience.
 I got interested in this wonderful path when I was fairly young (highschooler) and saw a series like (1+1/4+1/9+1/16+1/25+...(1/n^2))... and was fascinated that sum of these kind of simple looking serieses come out to be very interesting. For example sum of above series is (pi*pi/6).
The zeta function  for numbers >1 is defined as:
ζ(s) = (1+(1/2)^s + (1/3)^s+(1/4)^s+ .....
for example ζ(2) = pi*pi/6 , ζ(4)= pi^4/90 etc .
(These serieses generally taught only at senior level in UG or Graduate Math level  Though it is easy to explain it earlier._
 I was fortunate enough to learn a lot about these function as I was interested in Ramnujan's work and had a few professors who were leading expert in the field. What is interesting is that this once considered purely mathematical field has lot of application in Physics  et String theory, Nuclear Physics, Statistical Physics etc. For me, the fascinating part was study of nuclear spectra of complex nuclei. There is Mehta, Wigner, Dyson model using Random Matrices where statistical properties behaves very much like zero's of zeta function. (see other posts down).
Among others, a lot of work has been done by Ramanujan and a few other Indian Physicists/Mathematicians in the field.
If Riemann Hypothesis is proven many other math problems will find a solution. Most believe the hypothesis is correct but rigorous proof has not been given yet. Though people, using computers, have checked for very very high numbers and found the hypothesis to be correct.
So what is the hypothesis  (Please also read any standard source ot Wiki for details).
 ζ(s) (Zeta function) for s>1 and real can be defined as above.
 ζ(s) for complex values can be defined as "analytical continuation" and that value can be defined even for negative values or even s=0 or s=1 etc. ( See note **1 on why this is so "controversial" )
One also finds that zeta of a negative even number is zero. (Which again is very strange if one looks at the series method only)
ζ(2) = ζ(4) = ζ(6) = .... = 0
Other than these values there are other values when ζ(s) is zero. These values are called "nontrivial values of zeta's zero)
Defined that way it is found that all "nontrivial zeros of zeta function" lie on one vertical straight line (Real part = 1/2). And their distribution follows a particular form. No one has been able to prove that yet but this has been found to be correct for billions of values tested.
Now if the hypothesis is proven many theorems in math will be proven, hence a lot of interest.
Hope this helps.
Re: BR Maths Corner1
^^^ note **1 from above..
One of "interesting" fact which has fascinated lot of people is:
zeta function when s=0 or negative number  If one does "analytical continuation" makes sense and one finds:
ζ(0) = 1/2
ζ(1) = 1/12
These makes NO sense if one just uses series method. for example
ζ(0) becomes = 1+1+1+1+1+1.......
(How in the world this can be 1/2 !!!)
and ζ(1) = 1+2+3+4+5+6+...
(How in the world this can be 1/12) !!!
*** There is a famous story where Ramanujan talks about 1+2+3+4+...= 1/12 (see that in my post in BRF before)
Point here is in a weird way there is some sense (as we see in string theory or quantum electrodynamics  ) is having those values as what Ramanujan said.
But there is enough in Wiki and other popular media  check out Ramanujan sums etc..
One of "interesting" fact which has fascinated lot of people is:
zeta function when s=0 or negative number  If one does "analytical continuation" makes sense and one finds:
ζ(0) = 1/2
ζ(1) = 1/12
These makes NO sense if one just uses series method. for example
ζ(0) becomes = 1+1+1+1+1+1.......
(How in the world this can be 1/2 !!!)
and ζ(1) = 1+2+3+4+5+6+...
(How in the world this can be 1/12) !!!
*** There is a famous story where Ramanujan talks about 1+2+3+4+...= 1/12 (see that in my post in BRF before)
NO it does not make sense to many, but "1 + 2 + 3 + 4 + ⋯ = −1/12" was presented in one of the Ramanujan's book).. If you don't believe me , check out wiki:
http://en.wikipedia.org/wiki/1_+_2_+_3_+_4_+_%E2%80%A6
(go a few paragraphs below under Heuristics)
Here is letter from Ramanujan to Hardy:
"Dear Sir, I am very much gratified on perusing your letter of the 8th February 1913. I was expecting a reply from you similar to the one which a Mathematics Professor at London wrote asking me to study carefully Bromwich's Infinite Series and not fall into the pitfalls of divergent series. … I told him that the sum of an infinite number of terms of the series: 1 + 2 + 3 + 4 + · · · = −1/12 under my theory. If I tell you this you will at once point out to me the lunatic asylum as my goal. I dilate on this simply to convince you that you will not be able to follow my methods of proof if I indicate the lines on which I proceed in a single letter. …
Point here is in a weird way there is some sense (as we see in string theory or quantum electrodynamics  ) is having those values as what Ramanujan said.
But there is enough in Wiki and other popular media  check out Ramanujan sums etc..
Last edited by Amber G. on 13 Dec 2018 09:22, edited 1 time in total.
Re: BR Maths Corner1
My earlier posts from a few years ago reposted here:
Amber G. wrote:
...This also has been mentioned a few times in this dhaga.. in fact the sum of 1/n^2.., that is
S=1+1/4+1/9+1/16+1/25 ...
is some times not covered in many undergraduate math courses... The sum, of course is, (pi^2/6),
This is what fascinated my son. (You just see squares of 1,2,3.. all natural numbers and wow, suddenly pi creeps in).. he got hooked on Math at very young age and this equation was definitely the main cause ( he remembers this as the cause for getting hooked). He learned calculus and many other fun things trying to understand how pi creeps in the sum. (His interest in math is still there.. he did his PhD in Theoretical Physics, which some say is nothing but math with purpose.. )
****
As I have mentioned before, lot of Ramanujan's work was with zeta function.. (which sum of such kind of a series)..
One of silly but really fun thing is zeta (1) which is (1/12), and in some weird sense
sum of (1+2+3+4+5+...) is (1/12)
(These sums are called "ramanujan sums"
(NO it does not make sense to many, but "1 + 2 + 3 + 4 + ⋯ = −1/12" was presented in one of the Ramanujan's book).. If you don't believe me , check out wiki:
http://en.wikipedia.org/wiki/1_+_2_+_3_+_4_+_%E2%80%A6
(go a few paragraphs below under Heuristics)
Here is letter from Ramanujan to Hardy:"Dear Sir, I am very much gratified on perusing your letter of the 8th February 1913. I was expecting a reply from you similar to the one which a Mathematics Professor at London wrote asking me to study carefully Bromwich's Infinite Series and not fall into the pitfalls of divergent series. … I told him that the sum of an infinite number of terms of the series: 1 + 2 + 3 + 4 + · · · = −1/12 under my theory. If I tell you this you will at once point out to me the lunatic asylum as my goal. I dilate on this simply to convince you that you will not be able to follow my methods of proof if I indicate the lines on which I proceed in a single letter. …
Re: BR Maths Corner1
And this post too ..
Amber G. wrote:Eric Demopheles wrote:http://en.wikipedia.org/wiki/1_+_2_+_3_+_4_+_%E2%80%A6
If I am not mistaken, such things were already considered by Euler.
Not really.
It is true that Euler was one of the first (or a giant) who studied these zeta functions (Many still call them EulerRiemannzeta functions) ( Euler–Riemann zeta function, ζ(s), is a function of a complex variablethat analytically continues the sum of the infinite series (mentioned above). The series converges when the real part of s is greater than 1 (BTW, zeta function plays a very important role, not only in analytic number theory but has applications in physics, statistics.. etc. (For example, random matrices (prof Mehta, Wigner, Dyson etc), just to give one example, uses it heavily)
But the critical point is: as a function of a real argument, was studied by Euler without using complex analysis (It was not developed at that time). Riemann extended the Euler definition to a complex variable, proved its meromorphic continuation and established a relation between zeros ζ(s) ...and the distribution of prime numbers etc..And that is where all the "fun" and "crazy" stuff appears..
Hardy, Littlewood and Ramanujan contributed the major part later...
Re: BR Maths Corner1
For the New Year we have a new Mersenne prime M₈₂ ₅₈₉ ₉₃₃=2⁸² ⁵⁸⁹ ⁹³³  1. It has 24,862,048 digits making it the largest known prime.
By the EulerEuclid theorem, we get a new even perfect number 2⁸² ⁵⁸⁹ ⁹³²(2⁸² ⁵⁸⁹ ⁹³³  1) .
It was discovered via the Great Internet Mersenne Prime Search (GIMPS) by Patrick Laroche, one of the thousands of volunteers using the free GIMPS software.
By the EulerEuclid theorem, we get a new even perfect number 2⁸² ⁵⁸⁹ ⁹³²(2⁸² ⁵⁸⁹ ⁹³³  1) .
It was discovered via the Great Internet Mersenne Prime Search (GIMPS) by Patrick Laroche, one of the thousands of volunteers using the free GIMPS software.
Re: BR Maths Corner1
Speaking of 2019  (From mine and few of my friends social media) Ramanujan would notice that:
2019 = 1^4+2^4+3^4+5^4+6^4
***
2019^(2^(9102))2018 is prime.
If you add all the digits (of the above 213 digit prime number) the number is also prime.
This number (sum of digits of the huge prime number) is also emirp.
What is emirp, you ask..
emirp is a prime number (=prime spelled backwards) which, if you reverse all the digits it still remains prime.
(Example 13, 71 etc)
****
2019 is the smallest number that can be written in 6 ways as the sum of the squares of 3 primes:
7² + 11² + 43² = 2019
7² + 17² + 41² = 2019
13² + 13² + 41² = 2019
11² + 23² + 37² = 2019
17² + 19² + 37² = 2019
23² + 23² + 31² = 2019
It's also the sum of all unique perfect powers up to 3⁵ = 243:
1² + 2² + 2³ + 2⁴ + 2⁵ + 2⁶ + 2⁷ + 3² + 3³ + 3⁴ + 3⁵ + 5² + 5³ + 6² + 6³ + 7² + 10² + 11² + 12² + 13² + 14² + 15² = 2019
Happy 2019 to all Brfites!
2019 = 1^4+2^4+3^4+5^4+6^4
***
2019^(2^(9102))2018 is prime.
If you add all the digits (of the above 213 digit prime number) the number is also prime.
This number (sum of digits of the huge prime number) is also emirp.
What is emirp, you ask..
emirp is a prime number (=prime spelled backwards) which, if you reverse all the digits it still remains prime.
(Example 13, 71 etc)
****
2019 is the smallest number that can be written in 6 ways as the sum of the squares of 3 primes:
7² + 11² + 43² = 2019
7² + 17² + 41² = 2019
13² + 13² + 41² = 2019
11² + 23² + 37² = 2019
17² + 19² + 37² = 2019
23² + 23² + 31² = 2019
It's also the sum of all unique perfect powers up to 3⁵ = 243:
1² + 2² + 2³ + 2⁴ + 2⁵ + 2⁶ + 2⁷ + 3² + 3³ + 3⁴ + 3⁵ + 5² + 5³ + 6² + 6³ + 7² + 10² + 11² + 12² + 13² + 14² + 15² = 2019
Happy 2019 to all Brfites!
Re: BR Maths Corner1
Speaking of Ramanujan, here is a problem to test interest and resourcefulness of BRF junta in 2019.
The problem is old but see if you can do it  first without external help but later go ahead use computers, internet, ask/phone a friend/expert . See how long it takes to get a answer in this dhaga.
If you have seen this problem before, please do not put a link for at least a month  Just put the answer (and may be how you did it  used computer or googled or asked a youngster interested in math).
I want to see how resourceful people are here (how fast are their computers or search engines etc).. Waiting to see how many days it takes before the first solution appears.
Problem:
Happy 2019!
The problem is old but see if you can do it  first without external help but later go ahead use computers, internet, ask/phone a friend/expert . See how long it takes to get a answer in this dhaga.
If you have seen this problem before, please do not put a link for at least a month  Just put the answer (and may be how you did it  used computer or googled or asked a youngster interested in math).
I want to see how resourceful people are here (how fast are their computers or search engines etc).. Waiting to see how many days it takes before the first solution appears.
Problem:
It is easy to see that:
(17/21)^3 + (37/21)^3 = 6
Can you find another pair of rational numbers a and b such that a^3+b^3 = 6 [/b]
Happy 2019!
Re: BR Maths Corner1
^^^ Okay quite a few weeks passed since the above post. Wondering if any one has a solution(s)? Do we have interest and good resources in terms of computers, internet sources or other resources here in .
(All we need are two fractions whose cubes add up to 6 .
(All we need are two fractions whose cubes add up to 6 .
Re: BR Maths Corner1
1805723/960540, 2237723/960540
I found it on stackexchange website. Didn't get any elation after finding that answer. Probably because I have no idea what Diophantine equation means.
OT  As per wiki, Brahmaguptha studied this equation. I wonder why he was studying equations like that. It's not like they built some aircraft or computer. what's the deal, Modiji??
I found it on stackexchange website. Didn't get any elation after finding that answer. Probably because I have no idea what Diophantine equation means.
OT  As per wiki, Brahmaguptha studied this equation. I wonder why he was studying equations like that. It's not like they built some aircraft or computer. what's the deal, Modiji??
Re: BR Maths Corner1
syam wrote:1805723/960540, 2237723/960540
Nice.
and ( 2237723/960540)^3(1805723/960540)^3 is indeed 6.
There are infinite numbers of answers, and once you found the method, it is easy to calculate. For example if one wanted both numbers to be positive 
(1498088000358117387964077872464225368637808093957571271237/
1097408669115641639274297227729214734500292503382977739220)
and (1659187585671832817045260251600163696204266708036135112763/
1097408669115641639274297227729214734500292503382977739220)
Will work too.
****
This problem is actually inspired by Ramanujan's work where this pattern is explored. In fact "taxi cab numbers" which have been discussed in brf math dhaga many times.. has close relationship with this.
(Problem is Old  once you know the method, it is not difficult to solve)
Since one month has passed, I please go ahead and put links etc / discuss or put other solutions.
I wonder why he was studying equations like that. It's not like they built some aircraft or computer. what's the deal, Modiji??
Actually this particular branch of math does have *many* practical applications. iPhone's cryptography, and also bitcoin cryptography is actually based on finding a solution of this type of equations.
Re: BR Maths Corner1
In related matter  Interestingly, this is from recent math news  Ramanujan will be proud..
It was thought (but not proven one way or other) that was impossible to write 33 as sum of 3 cubes it was found that
Wow! 33 = 8866128975287528 ^ 3  8778405442862239 ^ 3  2736111468807040 ^ 3
Some theory: It was known that 
All numbers of the form 9k+4 or 9k+5 can not be written as sum of three cubes. (we can prove this)
for rest of the numbers the answer is not known.
Some numbers are easy. For example 29 = 3^3+1^3+1^3
or 1 = 1^3+0^3+0^3
also 1 = 10^3+9^31^3 (Ramanujan!)
Some numbers are not that easy.. and it required a computer to check many answers..
For example 29 is easy but 30 is hard.
***
Till recently only 33 and 42 were two numbers less than 100 which people did not know the answer..
For example see this youtube video; The Uncracked Problem with 33
It was thought (but not proven one way or other) that was impossible to write 33 as sum of 3 cubes it was found that
Wow! 33 = 8866128975287528 ^ 3  8778405442862239 ^ 3  2736111468807040 ^ 3
Some theory: It was known that 
All numbers of the form 9k+4 or 9k+5 can not be written as sum of three cubes. (we can prove this)
for rest of the numbers the answer is not known.
Some numbers are easy. For example 29 = 3^3+1^3+1^3
or 1 = 1^3+0^3+0^3
also 1 = 10^3+9^31^3 (Ramanujan!)
Some numbers are not that easy.. and it required a computer to check many answers..
For example 29 is easy but 30 is hard.
***
Till recently only 33 and 42 were two numbers less than 100 which people did not know the answer..
For example see this youtube video; The Uncracked Problem with 33
Re: BR Maths Corner1
On this day let me bring you an exciting story....
Anyway I got this email from a math friend who was working on some old Ramajnujan problem regarding Heegner theorem. She found an interesting result. The result is not published yet so you are reading it first hand here.
Let me present it here  she found that amazingly
(ln(12^3(231^21)^3+744)/pi)^2 = 163
Here ln = natural logarithm (that is base e)
Pi = 3.14159265....
etc..
(If you don't believe it, just use a calculator / google to check it out)
For example here: <link>
Of course, the result is rather surprising. Pi is an irrational number. And natural logarithms of even ordinary numbers are almost always irrational so finding an integer result of this rather complicated expression is VERY remarkable. Naturally it has baffled all those mathematicians who were on this email list. Once the result is published I think it will cause a major revolution in number theory.
I am sure, my math friends know that 163 is largest Heegner number  that is this is the largest number where quadratic imaginary number fields admit unique factorisation in their ring of integers.
Anyway I got this email from a math friend who was working on some old Ramajnujan problem regarding Heegner theorem. She found an interesting result. The result is not published yet so you are reading it first hand here.
Let me present it here  she found that amazingly
(ln(12^3(231^21)^3+744)/pi)^2 = 163
Here ln = natural logarithm (that is base e)
Pi = 3.14159265....
etc..
(If you don't believe it, just use a calculator / google to check it out)
For example here: <link>
Of course, the result is rather surprising. Pi is an irrational number. And natural logarithms of even ordinary numbers are almost always irrational so finding an integer result of this rather complicated expression is VERY remarkable. Naturally it has baffled all those mathematicians who were on this email list. Once the result is published I think it will cause a major revolution in number theory.
I am sure, my math friends know that 163 is largest Heegner number  that is this is the largest number where quadratic imaginary number fields admit unique factorisation in their ring of integers.

 BRF Oldie
 Posts: 3528
 Joined: 29 Mar 2017 06:37
Re: BR Maths Corner1
^^amber sir, what is the significance of these heegner numbers in the real world or the result in real world?
Re: BR Maths Corner1
Arjunji 
First  Like a few earlier times  my post was posted on April 1  so a few points..
The result of (ln(12^3(231^21)^3+744)/pi)^2 is about 163.000000000000000000000000000023216777942453... but most calculators will round it of to 163.. and many may fall for the April fool joke..
(A similar version was posted in Scientific American in 1975  I still have that copy  which fooled a few)
***
What is true, remarkably is that this was discovered by Ramanjuan who studied a whole class of numbers where these kind of results come VERY close to an integer.
BTW this is based on a constant known as Ramanjuan Constant (see <link>
There are quite a few such remarkable results  one very well known  e^(pi)  pi is almost 20.
First  Like a few earlier times  my post was posted on April 1  so a few points..
The result of (ln(12^3(231^21)^3+744)/pi)^2 is about 163.000000000000000000000000000023216777942453... but most calculators will round it of to 163.. and many may fall for the April fool joke..
(A similar version was posted in Scientific American in 1975  I still have that copy  which fooled a few)
***
What is true, remarkably is that this was discovered by Ramanjuan who studied a whole class of numbers where these kind of results come VERY close to an integer.
BTW this is based on a constant known as Ramanjuan Constant (see <link>
There are quite a few such remarkable results  one very well known  e^(pi)  pi is almost 20.

 BRF Oldie
 Posts: 3528
 Joined: 29 Mar 2017 06:37
Re: BR Maths Corner1
Amber G. wrote:Arjunji 
First  Like a few earlier times  my post was posted on April 1  so a few points..
1. you completely got me. Your posts are always of serious nature so i dismissed the april 1 thing, even though i saw it later
2. I did see on wiki that the results astonishingly close to integers
3. what i was curious is that what do these numbers mean in real world. That's my question these days. Where do they find application in real world. That was one reason why i found myself disconnected with mathematics during my later years. unfortunately I can't see the video in office and at home I have to make my kid learn no.s and the bolbol gol matol what is mass of black hole
Re: BR Maths Corner1
..what do these numbers mean in real world.
Very good question! That is what many want to figure it out.
Some times people think it is a mere "coincidence" or merely a beautiful result but again it may be have a deep meaning..
What I do know, or learned in last 50 years, is many of these strange results do find practical use  in fields where no one imagined before.
100 years or so ago, Electrical Engineers started using "imaginary numbers", say in measuring impedance and they found a "practical use"..
Now these kind of class of numbers are having applications in cryptography etc..

 BRF Oldie
 Posts: 3528
 Joined: 29 Mar 2017 06:37
Re: BR Maths Corner1
not that it helps a beginner like me but as a meme based on new golmal says like "samajh nahin aaaya but sunke acha laga".
BTW yes you are right. I developed a disdain for mathematics as intellectual orgasms only to realize it later.
1. Einstein had mathematics worked out by Riemann
2. What i have heard today is that String theory to stuck because mathematics is not fully developed and relies on approx solutions
3. Some of the works by Ramanujan is being used in black holes, heard from Neil degrasse Tyson or Michio kaku in one of talk shows
What might be a good idea is to expand these threads with few intermediate areas and take topics by topics
BTW yes you are right. I developed a disdain for mathematics as intellectual orgasms only to realize it later.
1. Einstein had mathematics worked out by Riemann
2. What i have heard today is that String theory to stuck because mathematics is not fully developed and relies on approx solutions
3. Some of the works by Ramanujan is being used in black holes, heard from Neil degrasse Tyson or Michio kaku in one of talk shows
What might be a good idea is to expand these threads with few intermediate areas and take topics by topics
Re: BR Maths Corner1
Do not know where to post this about Kollam/Rangoli art and Mathematics
https://www.atlasobscura.com/articles/i ... artkolam
This reminds me very strongly of the small and beautiful monograph titled Symmetry by my mathematical idol Hermann Weyl. This monograph is very old and explores similar themes of symmetry in Art and Architecture. Roger Penrose who is famous for his work on Singularity theorems in General Relativity has also written a lot on Symmetry to be found in Art.
https://press.princeton.edu/titles/865.html
^^ This is a beautiful book for lay people and has influenced me profoundly when I was young.
https://www.atlasobscura.com/articles/i ... artkolam
This reminds me very strongly of the small and beautiful monograph titled Symmetry by my mathematical idol Hermann Weyl. This monograph is very old and explores similar themes of symmetry in Art and Architecture. Roger Penrose who is famous for his work on Singularity theorems in General Relativity has also written a lot on Symmetry to be found in Art.
https://press.princeton.edu/titles/865.html
^^ This is a beautiful book for lay people and has influenced me profoundly when I was young.
Re: BR Maths Corner1
Amber G. wrote:Also, (since I don't see any news in MSM  specially Indian news papers  some more data,,
Cutoffs
Bronze: 16
Silver: 25
Gold: 31
Ranking
1. USA
2. Russia
3. China
4. Ukraine
5. Thailand
6. Taiwan
7. Korea
8. Singapore
9. Poland
10. Indonesia..
....
28 India
Team results at : http://www.imoofficial.org/year_country_r.aspx?year=2018&column=total&order=desc
Not a good show for India... we must be in the top 3 ideally at the number 1 spot consistently. What is lacking? I mean we are behind Thailand, Taiwan, Indonesia countries with much smaller populations and Indonesia is economically equivalent to richer parts of India.
Re: BR Maths Corner1
Amber G. wrote:..what do these numbers mean in real world.
...
100 years or so ago, Electrical Engineers started using "imaginary numbers", say in measuring impedance and they found a "practical use"...
Following up on AmbgerG ji's reply, something as basic (for us now) as negative numbers were not accepted when they were "discovered"/"invented"
Last edited by Vayutuvan on 19 May 2019 01:31, edited 1 time in total.
Re: BR Maths Corner1
Amber G. wrote: ...
If Riemann Hypothesis is proven many other math problems will find a solution. Most believe the hypothesis is correct but rigorous proof has not been given yet. Though people, using computers, have checked for very very high numbers and found the hypothesis to be correct.
...
No one has been able to prove that yet but this has been found to be correct for billions of values tested.
...
Hope this helps.
Very nice short explanation, AmberG ji.
I want to add on more point here as to why people spend computational resources (sometimes even very valuable supercomputer cycles) for checking the hypothesis/conjecture for as large a value as is possible on the computational infrastructure they have. There are several benefits in carrying out this exercise. It is new (and old) area called computational mathematics.
1. Firstly, if one wants to find a counterexample, then the counterexample has to be larger than what has been checked through computational verification.
2. Since one is computing very large numbers, numbers so large that they cannot be stored in even "long long int" (128 bit integers on Cray, for example), new methods and algorithms have to be/had been developed to do computations with very large number. Mind you, there is no recourse to rounding off because we need to exact arithmetic with integers, one way or the other.
3. Sometimes, the proof of a hypothesis/conjecture can be reduced to proofs of a few cases. Sometimes the number of cases might not be small. This was the case when FCC (Four color conjecture) was proved and changed its status to FCT. The proof was the start of hat one would call "Computational Mathematical" proof.
When Appel and Haken of UIUC announced the proof, they had "1,936 reducible configurations" (special cases). All planar maps, by repeated application of three operations, have been shown to be equivalent one of the 1936 maps. Then a program was written to four colour these 1936 reducible configurations thus proving 4CT or FCT. In 2001, the number of reducible configurations has been decreased to 633 as well improving the algorithm from O(n^4) O(n^2).
On 4CT, there was a lot of controversy as to whether it should be considered a proof. Now it has been accepted that to be a valid proof method.
Many people believe that there is a more elegant short proof of 4CT. Since this a problem that can be stated in simple terms, it would be interesting to see if somebody comes up with a succinct proof. In cases like this, it is a curse to know too much. Most people expect that, if such proof exists, it would most probably be discovered by a bright high school student or an amateur mathematician/hobbyist/enthusiast.
Added later:
From a computational perspective, while the theorem statement is "every planar map is four colorable", finding such a coloring is hard. In fact, it NPcomplete to decide whether a given planar map is 3colorable.
(Technical: I don't know the proof per se. My guess is that 3SAT can be reduced 3coloring of a planar graph, which is equivalent to 3coloring of a planar map)
Re: BR Maths Corner1
tandav wrote:Amber G. wrote:Also, (since I don't see any news in MSM  specially Indian news papers  some more data,,
Cutoffs
Bronze: 16
Silver: 25
Gold: 31
Ranking
1. USA
2. Russia
...
28 India
Team results at : http://www.imoofficial.org/year_country_r.aspx?year=2018&column=total&order=desc
Not a good show for India... we must be in the top 3 ideally at the number 1 spot consistently. What is lacking? I mean we are behind Thailand, Taiwan, Indonesia countries with much smaller populations and Indonesia is economically equivalent to richer parts of India.
Tandavji : FWIW (Subjective and personal perspective)..
I will certainly not call it a "Not a good show for India"..3 Silver and 2 Bronze, and 1 HM, are nothing to be sad about.. in fact quite good. Officially those rankings (country wise) are not serious .. what matters is number of medals. I would have liked if India got a gold or two but I am *very* happy with the result. We are NOT behind, (in my opinion) than Thailand or Taiwan etc...if our measure/aim is to find top quality students for our colleges.
It is not like our neighbors on *many* other countries where most get a zero.. Pakistan, for example has no medals, in fact *none* got more than 9 marks (ditto Sri Lanka or Nepal  where everyone got a zero)... Math Olympiad is fairly difficult exam and competition is quite high caliber. For perspective, most of math prof in India (or US) would not do that well. Difficulty is much higher than say a JEE...
Also for perspective *any one* of the Indian IMO team will be almost certainly be able to get into any top school here in USA.
Re: BR Maths Corner1
Vayutuvan  thanks.
Re: BR Maths Corner1
https://phys.org/news/201905mathemati ... emann.html
Mathematicians revive abandoned approach to the Riemann Hypothesis
Mathematicians revive abandoned approach to the Riemann Hypothesis
Many ways to approach the Riemann Hypothesis have been proposed during the past 150 years, but none of them have led to conquering the most famous open problem in mathematics. A new paper in the Proceedings of the National Academy of Sciences (PNAS) suggests that one of these old approaches is more practical than previously realized.The paper builds on the work of Johan Jensen and George Pólya, two of the most important mathematicians of the 20th century. It reveals a method to calculate the JensenPólya polynomials—a formulation of the Riemann Hypothesis—not one at a time, but all at once."The beauty of our proof is its simplicity," Ono says. "We don't invent any new techniques or use any new objects in math, but we provide a new view of the Riemann Hypothesis. Any reasonably advanced mathematician can check our proof. It doesn't take an expert in number theory."Although the paper falls short of proving the Riemann Hypothesis, its consequences include previously open assertions which are known to follow from the Riemann Hypothesis, as well as some proofs of conjectures in other fields."."The Riemann Hypothesis is one of seven Millennium Prize Problems, identified by the Clay Mathematics Institute as the most important open problems in mathematics. Each problem carries a $1 million bounty for its solvers.The hypothesis debuted in an 1859 paper by German mathematician Bernhard Riemann. He noticed that the distribution of prime numbers is closely related to the zeros of an analytical function, which came to be called the Riemann zeta function. In mathematical terms, the Riemann Hypothesis is the assertion that all of the nontrivial zeros of the Zeta function have real part ½."His hypothesis is a mouthful, but Riemann's motivation was simple," Ono says. "He wanted to count prime numbers."The hypothesis is a vehicle to understand one of the greatest mysteries in number theory—the pattern underlying prime numbers. Although prime numbers are simple objects defined in elementary math (any number greater than 1 with no positive divisors other than 1 and itself) their distribution remains hidden.The first prime number, 2, is the only even one. The next prime number is 3, but primes do not follow a pattern of every third number. The next is 5, then 7, then 11. As you keep counting upwards, prime numbers rapidly become less frequent."It's well known that there are infinitely many prime numbers, but they become rare, even by the time you get to the 100s," Ono explains. "In fact, out of the first 100,000 numbers, only 9,592 are prime numbers, or roughly 9.5 percent. And they rapidly become rarer from there. The probability of picking a number at random and having it be prime is zero. It almost never happens."
Despite their work, the results don't rule out the possibility that the Riemann Hypothesis is false and the authors believe that a complete proof of the famous conjecture is still far off
Re: BR Maths Corner1
Okay, just for fun  For year 2019.
Given a, b, c are real positive numbers can you prove:
a^2 + b^3 + c^4 + 2019 > 79 (a+b+c)
Given a, b, c are real positive numbers can you prove:
a^2 + b^3 + c^4 + 2019 > 79 (a+b+c)
Re: BR Maths Corner1
Amber G. wrote:Okay, just for fun  For year 2019.
Given a, b, c are real positive numbers can you prove:
a^2 + b^3 + c^4 + 2019 > 79 (a+b+c)
This problem is totally trivial via elementary calculus. Rewrite it as
(a^279a)+(b^379 b)+(c^479c) + babaji (1)
Now notice that if a>79 first term is positive, b> (79)^{1/2} second term is positive and if c> (79)^{1/3} third term is positive so babaji is not needed if a, b, c are big enough so we can restrict a, b, c to finite intervals.
We now examine maxima and minima, say for the c term by differentiation minima is achieved at
4c^379=0, C=(79/4)^{1/3}. Similarly minimum of b term is achieved at B=(79/3)^{1/2} and of a term at, A=79/2. Plug these last numbers in to (1) and if babaji is big then (1) is positive. That is evaluate
(A^279 A)+(B^379B)+(C^479C) and add babaji to show it is positive.
This also tells you the least value of babaji to take.
QED.
Re: BR Maths Corner1
^^^Very Nice.
(Agree that, like many of other "fun" problems I post are fairly easy and are meant to draw interest. The above problem is, as you said trivial using calculus, in fact this was originally meant for bright students in middle/high school and fairly easy even without using calculus.)
****
For example, here is one way using Arithmetic Mean is grater than Geometric Mean property..
Take (a^2) and (79/2)^2 and use AM >= GM
a^2 + (79/2)^2 > 2 * (a) (79/2) >= 79 a
Take (b^3), (79/3)^(3/2), (79/3)^(3/2) as three values and apply AM >= GM
so b^3 + (79/3)^(3/2)+ (79/3)^(3/2) >= 3(b*sqrt(79/3)*sqrt*79/3) = 79 b
and similarly c^4 + (79/4)^(4/3) + (79/4)^(4/3) + (79/4)^(4/3) >= 79 c
Adding above three we get:
a^2+b^3+c^4 + (79/2)+(79/3)^(3/2)+ (79/3)^(3/2)+ (79/4)^(4/3) + (79/4)^(4/3) + (79/4)^(4/3) ) >= 79(a+b+c)
and (79/2)+(79/3)^(3/2)+ (79/3)^(3/2)+ (79/4)^(4/3) + (79/4)^(4/3) + (79/4)^(4/3) is about 1991 definitely less than 2019 . QED
(Agree that, like many of other "fun" problems I post are fairly easy and are meant to draw interest. The above problem is, as you said trivial using calculus, in fact this was originally meant for bright students in middle/high school and fairly easy even without using calculus.)
****
For example, here is one way using Arithmetic Mean is grater than Geometric Mean property..
Take (a^2) and (79/2)^2 and use AM >= GM
a^2 + (79/2)^2 > 2 * (a) (79/2) >= 79 a
Take (b^3), (79/3)^(3/2), (79/3)^(3/2) as three values and apply AM >= GM
so b^3 + (79/3)^(3/2)+ (79/3)^(3/2) >= 3(b*sqrt(79/3)*sqrt*79/3) = 79 b
and similarly c^4 + (79/4)^(4/3) + (79/4)^(4/3) + (79/4)^(4/3) >= 79 c
Adding above three we get:
a^2+b^3+c^4 + (79/2)+(79/3)^(3/2)+ (79/3)^(3/2)+ (79/4)^(4/3) + (79/4)^(4/3) + (79/4)^(4/3) ) >= 79(a+b+c)
and (79/2)+(79/3)^(3/2)+ (79/3)^(3/2)+ (79/4)^(4/3) + (79/4)^(4/3) + (79/4)^(4/3) is about 1991 definitely less than 2019 . QED
Re: BR Maths Corner1
Amber G. wrote:Vayutuvan  thanks.
Welcome sire. Let us revive this dhaga and the fizziks one from fizzling out.
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