IBM Ponder This Puzzle

IBM Launches one puzzle every month in its "Ponder This" section. I just came to know about it day before yesterday and sent my solution for the active challenge.

Check my name at http://domino.research.ibm.com/Comm/wwwr_ponder.nsf/challenges/October2009.html

The Puzzle is not mathematical though :P

Complete the following sequence:
1, 13, 2, 1, 5, 3, 2, 1, 4, 4, 7, 3, 1, 5, 3, 5, ?, ?

Hint: The solution is related to a quote about mathematicians.
You can do it only if you know it. Thinking does not really help!! :)

PandeyJi (Nikhil Pandey) said that he was able to solve it. :)

Solution: IBM guys will post it on their link in 2 days anyways! :)

Comments

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