BR Maths Corner-1

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Amber G.
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Re: BR Maths Corner-1

Postby Amber G. » 09 Jul 2018 04:26

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 Po-Shen 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)

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Re: BR Maths Corner-1

Postby Amber G. » 09 Jul 2018 22:49

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.
Last edited by Amber G. on 09 Jul 2018 23:03, edited 2 times in total.

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Re: BR Maths Corner-1

Postby Amber G. » 09 Jul 2018 22:51

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 anti-Pascal 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 anti-Pascal 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 anti-Pascal triangle with 2018 rows which contains every integer from *1* to *1 + 2 + 3 + ...+ 2018*?

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Re: BR Maths Corner-1

Postby Amber G. » 12 Jul 2018 19:29

Added later: The official results have been posted. (https://www.imo-official.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 1-2 bronzes Fingers crossed (** edited later *** India did get 3 silver and 2 bronze)

Congrats to all.

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Re: BR Maths Corner-1

Postby Amber G. » 13 Jul 2018 01:15

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.imo-official.org/results.aspx

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Re: BR Maths Corner-1

Postby Amber G. » 13 Jul 2018 02:08

^^^ 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 30-40 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.

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Re: BR Maths Corner-1

Postby Amber G. » 13 Jul 2018 03:38

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.imo-official.org/year_country_r.aspx?year=2018&column=total&order=desc

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Re: BR Maths Corner-1

Postby Amber G. » 01 Aug 2018 22:56

Congratulations to Akshay Venkatesh for winning the Fields Medal along with, Peter Scholze, Alessio Figalli and Caucher Birkar.
Image

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.

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Re: BR Maths Corner-1

Postby Amber G. » 01 Aug 2018 23:07

How about this??? This was 8 years ago: (Link: https://forums.bharat-rakshak.com/viewtopic.php?p=923025#p923025

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

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Re: BR Maths Corner-1

Postby SaiK » 24 Sep 2018 06:27

Courtesy/ #via: @venug

https://aperiodical.com/2018/09/hlf-blo ... 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?

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Re: BR Maths Corner-1

Postby venug » 25 Sep 2018 04:18

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

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Re: BR Maths Corner-1

Postby Amber G. » 13 Dec 2018 04:00

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 re-quote those two posts for convenience.

- I got interested in this wonderful path when I was fairly young (high-schooler) 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 "non-trivial values of zeta's zero)

Defined that way it is found that all "non-trivial 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.

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Re: BR Maths Corner-1

Postby Amber G. » 13 Dec 2018 04:25

^^^ 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)

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.

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Re: BR Maths Corner-1

Postby Amber G. » 13 Dec 2018 04:33

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) :eek:

(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. …

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Re: BR Maths Corner-1

Postby Amber G. » 13 Dec 2018 04:56

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 Euler-Riemann-zeta 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...



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Re: BR Maths Corner-1

Postby Amber G. » 12 Jan 2019 02:53

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 Euler-Euclid 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.

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Re: BR Maths Corner-1

Postby Amber G. » 12 Jan 2019 02:59

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^(9-1-0-2))-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!

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Re: BR Maths Corner-1

Postby Amber G. » 12 Jan 2019 03:02

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:
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!

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Re: BR Maths Corner-1

Postby Amber G. » 05 Feb 2019 03:35

^^^ 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 :).

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Re: BR Maths Corner-1

Postby syam » 08 Feb 2019 18:09

-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. :oops:

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??

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Re: BR Maths Corner-1

Postby Amber G. » 09 Feb 2019 12:37

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.

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Re: BR Maths Corner-1

Postby Amber G. » 15 Mar 2019 06:08

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+9^3-1^3 (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


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