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Classic Brain Teasers and Riddles

(Click on the question to see the answer.)

 16. What comes next in the following sequence?

 17 At a math teacher conference, 100 teachers lined up in a hall right in front of a row of 100 locker cabinets. The first teacher ran down the hall opening all 100 lockers. The second teacher then proceeded to run down the hall closing every 2nd locker. The third teacher ran down the hall doing the opposite to every 3rd locker (meaning the teacher closed the third locker because it was currently open, and then opened the 6th locker because it was closed). The 4th teacher did the opposite to every 4th locker (meaning if a locker was closed the teacher opened it and if a locker was open the teacher closed it). If this process repeats all the way down to the 100th teacher. Which lockers would be open at the end of the contest?
 ANSWER Each possible factor touches a locker (e.g. the 10th locker will be touched by the 1st, 2nd, 5th, and 10th teachers). So, most of the lockers will be closed since factors come in pairs. If an even number of teachers open, close, open close, and so on the locker will end up closed. It would only be lockers that were touched an odd number of times that will be open at the end of this sequence of events and that will only happen with the perfect square numbers from 1 to 100 (e.g. the 25th locker will be touched by the 1st, 5th, and 25th teacher which would leave it open). The following lockers will be open: 1st, 4th, 9th, 16th, 25th, 36th, 49th, 64th, 81st and 100th lockers.

 18 Three boxes are labeled “Oranges”, “Apples”, and “Apples & Oranges”. All boxes have been labeled incorrectly. One box actually contains only oranges, another contains only apples, and the last contains both apples and oranges. If you cannot see inside any box but you are allowed to pick exactly one fruit out of a box and see what it is (without peaking or feeling around when you pick the one fruit out) , how can you determine the how to rearrange the labels so that they are correct?
 ANSWER Since we know the boxes are labeled incorrectly, we should grab one piece of fruit from the box labeled "Apples & Oranges". Two outcomes are possible. (Outcome 1): We pull out an Orange from the box labeled "Apples & Oranges" then, we know that box should be labeled "Oranges". The box labeled "Apples" can't be just apples because it labeled incorrectly and it can't be just oranges because we just determined which box must be oranges only. So, the box labeled "Apples" should be labeled "Apples & Oranges". This only leaves the box labeled "Oranges" which should then be labeled "Apples & Oranges". ....Similarly.... (Outcome 2): We pull out an Apple from the box labeled "Apples & Oranges" then, we know that box should be labeled "Apples". The box labeled "Oranges" can't be just oranges because it labeled incorrectly and it can't be just apples because we just determined which box must be apples only. So, the box labeled "Oranges" should be labeled "Apples & Oranges". This only leaves the box labeled "Apples" which should then be labeled "Apples & Oranges"

 19. You have 6 glasses. 3 are filled with water and 3 empty as shown below. Touching only one glass. How can you arrange it so that the filled and empty glasses alternate? (full, empty, full, empty, etc.)
 ANSWER Pick up the second class with water and pour the water in to the 5th glass. Then, put the empty second glass back.

 20 What is the most change that you can have without being able to make change for a dollar? (using only U.S. coins Half dollars (50¢), Quarters(25¢), Dimes (10¢), Nickels (5¢), and Pennies (1¢) (HINT: It is more than a dollar.)
 ANSWER 3 Quarters, 4 Dimes, and 4 Pennies for a total of \$1.19. With these coins it is impossible to give someone precisely \$1.00.

 21. Three tribes live on a remote island. Each tribe has two campsites. Each tribe cannot stand any of the other tribes. Each tribe wants to create a path or walk way that connects their two campsites but they do not want any of their paths to cross over another tribes path. How might they create their paths so that no 2 paths intersect? (It is possible.)

 22 Determine the Expanded form of the following: (a – x)(b – x)(c – x)(d – x)……(z –x) = (You may assume that every letter in the alphabet is used.)
 ANSWER The answer would be 0. If the pattern is consistent one of the binomial factors is (x - x) which is 0 and the product of anything and 0 is 0.

 23 There are 3 light switches on a panel. You wish to determine which light switch turns on an attic light which you can not see at all from the panel. You have no idea what any of the switches do. How can you determine which switch turns on the attic light by adjusting the switches and then only making one trip up to the attic to determine which is the correct switch (also, we are assuming the light bulb has not burned out)?
 ANSWER First, turn on the first light switch and wait about 5 minutes. Then, turn off the first switch and turn on the second switch. This will leave 3 possible outcomes. Outcome #1:The light is off and it is hot to the touch. This would suggest it is light switch #1. Outcome #2:The light is on. This would suggest it is light switch #2. Outcome#3:The light is off and cool to the touch. This would suggest it is light switch #3. You could argue that we must also assume it is a incandescent light bulb that gets hot.

 24 How many pieces of pizza can a pizza be portioned in to using only 3 straight (linear) cuts? ...and with 4 straight cuts? (The portions don't have to be equal)
 ANSWER 7 pieces with 3 cuts ANSWER 11 pieces with 4 cuts

 25 Two babies are born to the same mother on the same exact day of the same year. How is possible that the two babies would not be described as twins?
 ANSWER The two children could be 2 members of triplets. If the mom had 3 babies, they wouldn't be considered twins.

 26 Your parents have six sons including you and each son has one sister. How many people are in the family?
 ANSWER "9 people" 1 Father, 1 Mother, 6 sons, 1 daughter would meet the requirements above.

 27 Using only two 2’s and any mathematical operators, create an expression equal to 5.

 28 Given a simple balance and 12 identical looking coins. If one of the coins is counterfeit and has a weight that is slightly different from the rest, how can you determine which coin is counterfeit if you are only permitted to use the simple balance 3 times. You should also be able to determine if the counterfeit coin is heavier or lighter.
 ANSWER First, consider labeling the coins 1 - 12. Then using the scale for our first time, weigh coins 1 - 4 against 5 - 8. Three outcomes are possible; (Outcome #1) the group 1 - 4 could be heavier, (Outcome #2) the group 1-4 could be lighter, or (Outcome #3) the first two groups balance. Let's look at Outcome #3 first. (Outcome #3) would suggest that the counterfeit coin is in the coin group 8 - 12. Since we know coins 1 - 8 are authentic coins we will use them as a standard. For our second use of the simple scale, we could weigh coins 1 - 3 against coins 9 - 11. Again, we are presented 3 outcomes at this step; (Outcome 3A) coins 9-11 are heavier, (Outcome 3B) coins 9 - 11 are lighter, (Outcome 3C) coins 9 - 11 balance. (Outcome 3A): We know the counterfeit coin is heavier. Weigh coin 9 against coin 10 and which ever coin is heavier is the counterfeit. If coins 9 and 10 balance, then coin 11 is counterfeit and heavier. (Outcome 3B): We know the counterfeit coin is lighter. Weigh coin 9 against coin 10 and which ever coin is lighter is the counterfeit. If coins 9 and 10 balance, then coin 11 is counterfeit and lighter. (Outcome 3C): Weigh coin 1 against coin 12. Coin 12 must be the counterfeit and this will tell us if it is heavier or lighter. (Outcome #2) In this outcome we know coins 9-12 must be authentic and we will use coins in this group as a standard. We also know if the counterfeit coin is in the group 1-4 it is lighter and if it is in 5 - 8 is heavier. For our second use of the scale in this option, let's weigh coins 1,2, 5,6 against coins 8,9,10,11. On one side we have 2 potential light counterfeits with 2 potential heavy counterfeits and on the other side 3 authentic coins with one potential heavy. This weighing will present 3 options; (Outcome 2A) the coins 1,2,5,6 are heavier, (Outcome 2B) the coins 1,2,5,6 are lighter, (Outcome 2C) the coins on both side balance. (Outcome 2A): We know that the only way coins 1,2,5,6 are heavier are if 5 or 6 is a heavier counterfeit coin. So, for our final scale use we could weigh coin 9 against coin 5. If the scale is unbalanced, coin 5 is counterfeit and heavier. If it balances, coin 6 is counterfeit and heavier. (Outcome 2B): We know that the only way coins 1,2,5,6 can be lighter is if coins 1, 2 are lighter or coin 8 on the other side is heavier. So, for our final scale use we could weigh coins 1,8 against coins 9,10. If coins 1,8 are heavier then coin 8 is counterfeit and heavier. If coins 1,8 are lighter then coin 1 is counterfeit and lighter. If they balance then coin 2 is counterfeit and lighter. (Outcome 2C): We know that the only way coins 1,2,5,6 can balance with coins 8,9,10,11 is if one of the coins 3,4,7 is counterfeit. Next, we should weigh coins 3,7 against coins 9,10. If coins 3,7 are heavier then coin 7 is counterfeit and heavier. If coin 3,7 are lighter then coin 3 is counterfeit and lighter. If coins 3,7 balance then coin 4 is counterfeit and lighter. (Outcome #1) In this outcome, you can apply the exact same strategy as we did in outcome #2 to determine which coin is counterfeit.

 29 One apartment is directly above a second apartment. The resident living downstairs calls his neighbor living above him and states, “If one of you is willing to come downstairs, we’ll have the same number of people in both apartments.” The upstairs resident responds, “We are all too tired to move. Why don’t one of you come up here? Then we’ll have twice as many people up here as you’ve got down there.” How many people are in each apartment?
 ANSWER Let U = number of people upstairs and D = number of people down stairs. Then, based on the statements: Equation 1: U - 1 = D + 1 and Equation 2:U + 1 = 2(D - 1) Solve the system. Based on Equation 1 we know, U = D +2. Then, using substitution we could rewrite Equation 2 as (D+2) +1 = 2D - 2. Then, solving that equation would give us D = 5. Then, plug that in to U = D + 2 = 5 + 2 = 7. So, there are 5 people downstairs and 7 people upstairs.

 30 If two days after the day before tomorrow is Sunday, what day is today?
 ANSWER "Friday" "The day before tomorrow" = "today". So, we could rewrite the sentence to make it a little easier. If two days after today is Sunday, what is today? Then, today must be Friday.

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