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 :)

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

ReplyDeleteDude..

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

:)

so what about

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

will this do?

nice solution Piyush..

ReplyDeleteThe 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!!

Here's mine :) :

ReplyDeletelg(2)/.2

( base 2 logarithm )

Jonathan Busby

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 ;)

ReplyDeletesec(arctan(2))

ReplyDeleteor

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

A small mistake..

ReplyDeletesec(arctan(2))^2

or

square(sec(arctan(sqrt(2*2))))

cos(2-2)=1

ReplyDeleteso, 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