Batch 2 - Class 97 - Prisoner Problem

• (Moscow - 174) Wheels A,B,C and D are connected with belts as shown. If wheel A starts to rotate clockwise, can all the 4 wheels rotate? In which direction would each wheel rotate?

• Can the wheels rotate if all 4 belts are crossed?
• What is belt 1 and 3 are crossed?

• A group of prisoners sits in a circle, numbered from 1 to n. Going around the circle repeatedly, the warden alternately says “skip, cell, skip, cell,” and so on, until finally only one person remains. That person is released from the prison. For instance, with 4 prisoners, you skip prisoner 1, send prisoner 2 to the cell, skip prisoner 3, send prisoner 4 to the cell, skip prisoner 1, send prisoner 3 to the cell, and thus prisoner 1 is released.
• In fact, on this day prisoner 1 is released, but the total number of prisoners is not 4: it is between 40 and 100. How many prisoners are there?

  Instructor Notes: Encourage kids to start with small numbers and identify a pattern. Can they articulate the pattern? Can they explain why the pattern exists?One way to think about it is to see what happens when you double from one power of two to another. What happens in one complete pass? What does the remaining problem look like?Another way is to write place positions in binary. Which numbers get knocked out on first pass? Which numbers get knocked out in second pass? And so on... Which number will remain? (Instructor Notes: Be careful to start the numbering from zero)
• On another day, when again there are between 40 and 100 prisoners, the very last prisoner (number n) is released. What is the value of n? Is there only one possibility?

  Instructor Notes: Again, encourage kids to start small and identify the pattern for last prisoner 
• On yet another day, prisoner 7 is released. How many prisoners were there that day? Make sure to analyze all possibilities.

  Instructor Notes: Identification of prisoner 7 release may require extending the pattern further 
• Generalize: find a rule for which prisoner number you want to be, if you know how many prisoners will be sitting in the circle.

  Instructor Notes: See if kids see the entire pattern now. Are they able to articulate it? Are they able to explain why that pattern exists?Answer: Essentially, all powers of 2 save the first prisoner. Each successive number till next power of two then advances the "saved prisoner" number by 2. One way to think about this is that on a successive number, you knock out prisoner 2 (moving two steps), and then the problem is same as previous number problem but starting at number 3 instead of number 1.
• What if the rule is changed, and instead of the last prisoner being released, it is the second-to-last prisoner who is released? Now what should be your strategy?

  Instructor Notes: This will require kids to redraw the pattern. The successive movement of 2's will still hold. However, the positions where prisoner 1 is saved is 3.(power of 2) - the first place is 3 prisoners. After that, encourage kids to think about what happens in first pass when you double the number of prisoners. And so on.
• A new warden is hired whose favorite number is 7. Thus, instead of counting “skip, cell, skip, cell”, the new warden counts “1, 2, 3, 4, 5, 6, cell,” over and over again, until all but one prisoner has been sent back to the cells. The one remaining prisoner is released. For example, with four people, the warden counts 1, 2, 3, 4, on person number 1, 2, 3, 4. Then the warden continues counting 5, 6, cell, so person 3 is sent to the cell. Continuing, person 4 is sent to the cell next, and then person 1, so number 2 is the desired number.
• One day there are 25 prisoners. What seat is the desired one?

  Instructor Notes: This is a more complex pattern. Give kids time to work if out. One thing to observe is that the "saved prisoner" advances 7 steps in each successive steps, which is analogous to advancing in 2 steps in previous problem (and for the same reason). 
• If you know how many prisoners there are, find a quick way to decide which number will go free.

• (Dudeney - 76) A man bought an odd lot of wine in barrels and one barrel containing beer. These are shown in the illustration, marked with the number of gallons that each barrel contained. He sold a quantity of the wine to one man and twice the quantity to another, but kept the beer to himself. Which barrel contained the beer?

• Answer: 20 gallons. This can be solve by trial and error. Or by noticing that the quantity of wine sold is a multiple of three, where as the digital root of sum of all barrels is 2. Hence the beer barrel must have a digital root of 2 + 0,3,6 i.e. 2, 5, or 8. That is only true of 20 gallon barrel.