IBM Ponder This January 2010 Puzzle


I solved IBM Ponder This October 2009 Challenge and IBM Ponder This November 2009 Challenge and now solved this very interesting puzzle at IBM Ponder This January 2010 Challenge

I am happy!! My name on IBM's research site three times in 4 months. :)

Problem: Present a computation whose result is 5, being a composition of commonly used mathematical functions and field operators (anything from simple addition to hyperbolic arc-tangent functions will do), but using only two constants, both of them 2.
It is too easy to do it using round, floor, or ceiling functions, so we do not allow them.
Source: IBM Ponder This January 2010 Challenge
Solution: IBM guys would post it on their site by February 2010 :) Three different solutions given by Piyush (Fourth year, IITM), Aniesh Chawla (Fourth year, IITD) and me (Fourth year, IITB) :). All posted in comments!!

Thanx to Anirudh Shekhawat urf Maoo (CSE, IITB & Google) for help in the last step :)

Comments

  1. Piyush9:49 PM, January 22, 2010

    I'm not sure if i've understood the question correctly but will 2^2 + (2/2) do?

    ReplyDelete
  2. Pratik Poddar11:32 PM, January 22, 2010

    Dude..
    You need to use only 2 2's. You have used 4 2's in your expression :P
    :)

    ReplyDelete
  3. Piyush10:18 AM, January 23, 2010

    so what about

    (cosh(arcsinh(2)))^2 =5 ?

    will this do?

    ReplyDelete
  4. Pratik Poddar10:02 PM, January 23, 2010

    nice solution Piyush..
    The solution I gave was this:
    secinv(-cosec(secinv(2)))/cosecinv(2)

    Aniesh Chawla (Fourth Year, IIT Delhi) gave this solution:
    (sqrt(antilog(2))/2

    :)

    Happy!!

    ReplyDelete
  5. Unknown8:43 PM, January 29, 2010

    Here's mine :) :

    lg(2)/.2

    ( base 2 logarithm )

    Jonathan Busby

    ReplyDelete
  6. Pratik Poddar9:52 PM, January 29, 2010

    I was told by someone that there is nothing such as .2.. its 0.2 and hence it should not work.. Don't know whether IBM guys are accepting this ;)

    ReplyDelete
  7. avinash3:02 AM, June 02, 2011

    sec(arctan(2))
    or
    sec(arctan(sqrt(2*2)))

    ReplyDelete
  8. avinash2:46 PM, June 02, 2011

    A small mistake..
    sec(arctan(2))^2
    or
    square(sec(arctan(sqrt(2*2))))

    ReplyDelete
  9. Kapil Dubey3:20 AM, August 10, 2012

    cos(2-2)=1
    so, the following approach works for generating any number
    sec(arctan(sec(arctan(sec(arctan(sec(arctan(cos(2-2)))))))))

    this is basically sec(arctan(x)) applied on cos(2-2) 4 times. So, it takes 1 -> 2 -> 3 -> 4 -> 5

    ReplyDelete
  10. Unknown10:21 PM, September 18, 2017

    What about 2^2 + logx(x)?
    Is it acceptable?( It's log base x)

    ReplyDelete

Post a Comment

Popular posts from this blog

Fraction Brainteaser

Source: Sent to me by Gaurav Sinha Problem: Siddhant writes a Maths test and correctly answers 5 out of 6 Arithmetic questions and 20 out of 28 Geometry questions.  In total, Siddhant scores 25 out of 34.  Vaibhav writes another Maths test and correctly answers 20 out of 25 Arithmetic questions and 6 out of 9 Geometry questions. in total, Vaibhav scores 26 out of 34. Note that a) Vaibhav scores more than Siddhant b) Siddhant score better than Vaibhav in both individual topics -  5/6 > 20/25 and 20/28 > 6/9 How is it possible? 
Read more

Buying Dimsums

Source: Alok Goyal (Stellaris VP, Ex-Helion VC) puzzle blog Problem: A fast food restaurant sells dimsums in boxes of 7 and 3. What’s the greatest number of dimsums a person cannot buy. Generalize it for p and q where p and q are relatively prime. I loved the puzzle. Hope you enjoy it too.
Read more

Consecutive Heads

Let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process; B keep tossing another fair coin, until he get 3 consecutive heads, define Y to be the number of the tosses for this process. 1) Calculate P{X>Y} 2) What's the expected value of X 3) What's the expected value of Y This is probably the hardest puzzle I have ever put on my blog. Hence, I will post its solution in the post directly rather than on comment. Solution: 1) (Solved by me finally after 13 months :)) Make a state diagram. Let the state be (a,b) where a is the number of consecutive heads streak "A" is on currently and b is the number of consecutive heads streak "B" is on currently. So, (0,3) (1,3) are final accepted states and (2,0) (2,1) (2,2) (2,3) are failure states. If you get tails, your contribution to the state reaches to "0" f(State) = P(X>Y | "State" configuration initially)
Read more