20 responses so far ↓

• 1 Samira // Nov 14, 2007 at 12:07 am

  hey guys…there’s little error in the diagram when I’m explaining the rows of three circles. The rows of three circles include the rows along the rim of the circle, and the THREE circle-rows going out from the center of the circle, not TWO. Sorry about the confusion!

• 2 Ali - (York, UK) // Nov 14, 2007 at 2:18 am

  Hi Samira

  First of all, thank you for being so supportive and also for giving me an A+. It encourages me to solve maths problems now even more.
  Then,
  this weeks solusion is:

  1—18—3
  19,17,15,9
  2—16—4—8—10
  14,12,13,7
  6—11—5

  It’s image can be seen here:
  http://i9.tinypic.com/80lpvmh.jpg

  I spent 5.5 hours to solve this problem and just did it. There are some hints that could help like;
  1+2+…+19=190
  &
  22 X 12 = 264
  So,
  264 - 190 = 74
  There are 6 numbers which are repeated 3 times in 264 and just one number repeating 6 times. Hence, we need to add 7 numbers each of which is added two times(as we already added each one once in 190), and add one of them 3 more times to satisfy that extra 74.
  If we add up numbers 1 to 7 and multiply it by 2, the result is 56;
  74-56=18
  18/3=6
  This shows that the number in the middle cannot be mor than 6.
  Also for 18 and 19 we have;
  1+19+2=22
  1+18+3=22
  These are the only ways to use 18 and 19 in the hexagon. and for 17 we have;
  1+17+4=22
  2+17+3=22
  These are the only ways for 17 to be used. By playing with them we’ll get that the number in the centre circle must be either of 1,2,3 or 4.
  if we choose 4 which is closer to 6(that we found above), we need to change some of our edge numbers to satisfy the sum of 62 rather than 56. Then we can have;
  74-62=12
  12/3=4 (which means 4 in the centre)
  As we need to use 1,2,3 and 4, so there must be a change on 5,6 and 7 and we should add 3 in any way to them and then the remaining digits are our other 3 edges.
  Writing down the all posibilities and simplify them by crossing out the ones with two same numbers (e.g 6,6,9) and keeping just one of the repeated ones(e.g. 6,7,8), there are only 3 posibilities:
  5,6,10
  5,7,9
  6,7,8
  Checking these three sets of edge numbers we end up with only 5,6,10 being accepted.
  The rest is straightforward.

  and,
  Happy thanks giving!

  [IMG]http://i9.tinypic.com/80lpvmh.jpg[/IMG]

• 3 Kamran // Nov 15, 2007 at 9:12 am

  Would you please say my name in your program so I won’t feel you are ignoring me. I am bad in math but that doesn’t make me a bad person.

• 4 Samira // Nov 15, 2007 at 11:43 am

  Ali, that was impressive! If you want, now try it with 23…it works!

  Kamran, of course you’re not a bad person for not being good in math! I hope you’re learning more math from these problems!

• 5 Iraj // Nov 16, 2007 at 12:09 am

  I have no Idea about your show Samira? but I love math.
  I am getting ready for Putnam as well, I don’t know if you are familiar with that or not though!

  (I didn’t get the last problem you raised?)

• 6 Samira // Nov 16, 2007 at 6:11 pm

  Hi Iraj. I love the putnam! I took that exam when I was in undergrad. Now THOSE problems are difficult. It’s in December, right?

• 7 Ali - (York, UK) // Nov 17, 2007 at 4:50 am

  Hi Samira

  That was a good challenge! and to be honest, I was a bit lucky to find the answer.

  Here’s the answer for hexagon-23 :

  1–19–3
  18,16,14,12
  4–13–6–9–8
  17,15,7,5
  2–11–10

  and it’s image is here:

  http://i9.tinypic.com/8dzo8lw.jpg

  I’m sure you don’t want me to explain it.
  I loved your challenge and thanks for involving me.

  and…
  Kamran!
  mentioning your name in the show does not make you either a good person or a good mathematician. Being good in anything is totally up to you and how serious you take it.
  Always try to become valuable rather than famous.

• 8 Kamran // Nov 19, 2007 at 8:54 am

  Samira, here’s my message to you:
  010110010110111101110101001000000110000101110010011001010010000001110010011001010110000101101100011011000111100100100000011100000111001001100101011101000111010001111001

• 9 Ali - (York, UK) // Nov 19, 2007 at 4:18 pm

  Samira!
  On behalf of you I’m gonna reply Kamran’s message.
  Kamran! Here’s your answer:
  596F7526172065207265616C6C7920707265747479

  Now! This is your challenge!
  Can you find any relation between your message and it’s answer?!

• 10 Samira // Nov 19, 2007 at 10:54 pm

  hmm…I believe that’s the hexadecimal equivalent to Kamran’s binary message! Am I right?!

  You know there are 10 kinds of people in the world…Those who can think in binary, and those who can’t

• 11 Ali - (York, UK) // Nov 20, 2007 at 10:58 am

  Yup!
  I knew you know the answer. That was for Kamran to stop complaining and start to think.
  But anyway, you’re right.
  and your “10 kinds of people” expression was so so nice! I loved it and gonna use it!
  thanks

• 12 Rob W // Nov 20, 2007 at 4:53 pm

  I tried to use your binary joke at a party and everyone looked at me like I was crazy!

• 13 Ali - (York, UK) // Nov 21, 2007 at 1:53 am

  Don’t worry! That’s because they were from the group who CAN’T.

• 14 Vafa // Nov 21, 2007 at 11:55 pm

  Hi, this is my first visit to this blog and bebin.tv. I am happy to find some other people interested in maths. I guess we could work together. I am currently doing my PHD in category theory and have given several talks and if you like we can share something. unfortunately I do not like filming myself but I can send you my files so you can use them. My last talk can be found here
  http://www.maths.usyd.edu.au/u/AusCat/abstracts/070523vk.html

  Also I can share with you, some of the maths puzzles that I have made like:

  1- In how many ways, you can put one of the numbers 0,1 and 2 in each of the boxes of a 4by 4 table such that the sum of each column and each row is 2

  2- (this one appeared in the 27th annual (2005) university of Maryland, High School Mathematics competition, Part II question 5)

  Problem: There are 2005 people at a meeting. At the end of the meeting, each person who has shaken hands with at most 10 people is given a red T-shirt with the message “I am unfriendly”. Then each person who has shaken hands with people who recieved red-Tshirts is given a blue T-shirt with the message “All of my friends are unfriendly”. (some lucky people might get both red and blue T-shirts, for example, those who shook no one’s hand.) Prove that the number of people who recieved blue T-shirts is less than or equal to the number of people who recieved red T-shirts.

  And finally Keep up the good work.

  Best Wishes

  Vafa

• 15 Vafa // Nov 22, 2007 at 8:19 am

  by the way, since this blog is powered by wordpress, you should be able to write out $latex \LaTeX$ markups which is quite good, let’s test it.

  Let $latex \displystyle k$ be a positive integer, and let $\displaystyle x_1,\,x_2,\,…,\,x_n$ be positive real numbers. prove that:

  $latex \displaystyle \left(\sum_{i=1}^{n}\frac{1}{1+x_i}\right)\left(\sum_{i=1}^{n}x_i\right)\leq\left(\sum_{i=1}n\frac{x_{i}^{k+1}}{1+x_i}\right)\left(\sum_{i=1}^{n}\frac{1}{x_{i}^{k}}\right)$

• 16 Vafa // Nov 22, 2007 at 8:24 am

  How come it does not work here, bt worked on the original wordpress:

  http://mathspassion.wordpress.com/category/latex/

  mayb LaTeX pluggings ae not installed.

• 17 Vafa // Nov 24, 2007 at 4:57 am

  A Mathematician’s Love Letter:

  De-Morgan’s Apartment,
  Binomial Avenue,
  United States of Matrices.

  My Dear Love,

  Yesterday, I was passing by your rectangular house in trigonometric
  lane. There I saw you with your cute circular face, conical nose and
  spherical eyes, standing in your triangular garden. Before seeing you
  my
  heart was a null set, but when a vector of magnitude (likeness) from
  your
  eyes at a deviation of theta radians made a tangent to my heart, it
  differentiated.

  My love for you is a quadratic equation with real roots, which only
  you can solve by making good binary relation with me. The cosine of
  my
  love for you extends to infinity. I promise that I should not resolve
  you
  into partial functions but if I do so, you can integrate me by
  applying
  the limits from zero to infinity.You are as essential to me as an
  element
  of a set. The geometry of my life revolves around your acute
  personality.
  My love, if you do not meet me at parabola restaurant on date 10 at
  sunset, when the sun is making an angle of 160 degrees, my heart
  would be
  like a solved polynomial of degree 10. With love from your higher
  order
  derivatives of maxima and minima, of an unknown function.

  Yours ever loving,
  Pythagoras

• 18 Samira // Nov 24, 2007 at 1:10 pm

  HAHAHA that letter was hilarious! Thanks for sharing, Vafa, and thanks for the comments. Your problems are great! Your abstract sounds very interesting…I wish I was there to here your talk! I’ll look for my abstract for the talk I gave about a year ago and share it with you. I’m starting a new research soon, and if I have questions, I’ll be asking YOU!

  Here’s an article I thought you’d like…it’s in the same category as the love letter…

  A public school teacher was arrested today at John F. Kennedy
  International Airport as he attempted to board a flight while in
  possession of a ruler, a protractor, a set square, a slide rule and a
  calculator.

  At a morning press conference, Attorney General Alberto Gonzales said he
  believes the man is a member of the notorious Al-gebra movement. He did
  not identify the man, who has been charged by the FBI with carrying
  weapons of math instruction.

  “Al-gebra is a problem for us,” Gonzales said. “They desire solutions by
  means and extremes, and sometimes go off on tangents in a search of
  absolute value. They use secret code names like ‘x’ and ‘y’ and refer to
  themselves as ‘ unknowns’, but we have determined they belong to a
  common denominator
  of the axis of medieval with coordinates in every country. As the Greek
  philanderer Isosceles used to say, ‘There are three sides to every
  triangle.”

  When asked to comment on the arrest, President Bush said,
  “If God had wanted us to have better weapons of math instruction, He
  would have given us more fingers and toes.” White House aides told
  reporters they could not recall a more intelligent or profound statement
  by the president.

• 19 Vafa // Nov 24, 2007 at 6:49 pm

  ok, that was good. What is the area of your research? I am a categorist.by the way we had an interesting talk this week in our university and it was about “Cutting it fine: Euclid’s problem of slicing the circle” . The speaker was “Douglas Rogers”. You can find his paper at

  http://www.ics.mq.edu.au/~gerry/slices.pdf

• 20 Vafa // Dec 2, 2007 at 10:11 am

  What happened to your next problem?
  I hope that everything goes fine.

  An Interesting Problem:
  I am offering a $50 prize for the first person to discover the private key to my digital signature.
  To do this, all you need to do is to find one of the two prime factors of my encoding modulus. My public exponent is 65537 and my encoding modulus is the following small number:

  23321680547237225443713525721224584148665866330023524154360369181729452176249 96103385626498674629326288055713436516501479310848653803934728265021726372294 29442459642645282778448888896437772603901327650651185497794346270444876304210 35208423337776242804927764218195414173345386511219439787886101490508289061116 32736609730760383212056016899799620199509047160706797135421083890287484948962 47636160505623651803170397909798289974867510937900033170267321525569657216937 03360898621986838871091236507919939781960611168013517457493751969193634213340 178321446168127058612602730007308362491141503782643973149471343893309705271619

  To assist you in this task, I am prepared to tell you that, to 6 significant figures, the smaller prime factor is 1.44702×10^{308}. (That just leaves 304 digits for you to work out.) The prize must be claimed before the digital signature expires on 22 February 2008.

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