Tip the Balance

Source: Gazette of the Australian Mathematical Society

Problem: A balance scale sits on the teacher’s table, currently tipped to the right. There is a set of weights on the scales, and on each weight is the name of at least one student. The teacher chooses a set of students to enter the classroom. One by one, each chosen student will move each weight carrying his or her name to the opposite side. Prove that the teacher can always choose a set of students to tip the scales to the left.

Update (June 17, 2010)
Solution: Posted by me in comments!

Comments

  1. Pratik Poddar11:28 PM, June 17, 2010

    We will use the Averaging argument.

    Consider all subsets of the set of students including the empty set and the full set. Each weight
    will be on the left side of the balance half of the time. So, the total weight on the left for all these subsets is the same as total weight on the right side of the balance. Since the empty set results in a tip on the right, some other set must tip it on the left.

    Hence, by averaging argument, there exists a set of students which the teacher can call to make
    sure that the balance tips to the left.

    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