Question: There are 3 light bulbs in a hidden room and 3 switches outside the room that correspond to those light bulbs. You do not know which switch affects which bulb and you cannot see inside of the room. You are allowed to go inside of the room only one time. How do you find out which switch corresponds to which bulb?
Answer: Turn on two switches and wait for a while. Then turn off one switch and go inside the room. The bulb that is still on corresponds to the switch that is still on. Touch the remaining bulbs. The hotter bulb is the switch that you turned off, and the remaining bulb is the switch that you never turned on.
Showing posts with label puzzle. Show all posts
Showing posts with label puzzle. Show all posts
Wednesday, July 23, 2008
Monday, June 23, 2008
9 Minutes
Question: You are given two hourglasses. One measures 4 minutes and one measures 7 minutes. How would you measure exactly 9 minutes?
Answer: we want to measure the 9 minutes right from the start.
Answer: we want to measure the 9 minutes right from the start.
| Step | Time | 4 minute timer | 7 minute timer |
|---|---|---|---|
| 1 | 0 min | Start | Start |
| 2 | 4 mins | Flip | 3 minutes left |
| 3 | 7 mins | 1 minute left | Flipt |
| 4 | 8 mins | Stop | Flip (1 minute left) |
| 5 | 9 mins | Stop | Stop |
Sunday, March 23, 2008
12 Identical Balls Problem
Question: You are give 12 identical looking balls. One of them is fake (could be heavier or lighter) than the rest of the 11 (all the others weight exactly the same). You a provided with a simple mechanical balance and you are restricted to only 3 uses. Find the fake ball.
Answer: For convenience sake, let’s name the balls 1-12. This question is quite tricky. If you want to warm up your skills, you can try this easier 8 ball problem first.
Ok, let’s get on with it.
First we weigh {1,2,3,4} on the left and {5,6,7,8} on the right. There are three scenarios which can arise from this.
If they balance, then we know 9, 10, 11 or 12 is fake. Weigh {8, 9} and {10, 11} (Note: 8 is surely not fake)
If they balance, we know 12 is the fake one. Just weigh it with any other ball and figure out if it is lighter or heavier.
If {8, 9} is heavier, then either 9 is heavy or 10 is light or 11 is light. Weigh {10} and {11}. If they balance, 9 is fake (heavier). If they don’t balance then whichever one is lighter is fake (lighter).
If {8, 9} is lighter, then either 9 is light or 10 is heavy or 11 is heavy. Weigh {10} and {11}. If they balance, 9 is fake (lighter). If they don’t balance then whichever one is heavier is fake (heavier).
Phwww, that was confusing enough but we are not done yet.
If {1,2,3,4} is heavier, we know either one of {1,2,3,4} heavier or one of {5,6,7,8} is lighter but it is guarantees that {9,10,11,12} are not fake. This is where it gets really tricky, watch carefully. Weigh {1,2,5} and {3,6,9} (Note: 9 is surely not fake).
If they balance, then either 4 is heavy or 7 is light or 8 is light. Following the last step from the previous case, we weigh {7} and {8}. If they balance, 4 is fake(heavier). If they don’t balance then whichever one is lighter is fake (lighter).
If {1,2,5} is heavier, then either 1 is heavy or 2 is heavy or 6 is light. Weigh {1} and {2}. If they balance, 6 is fake (lighter). If they don’t balance then whichever one is heavier is fake (heavier).
If {3,6,9} is heavier, then either 3 is heavy or 5 is light. Weigh {5} and {9}. They won’t balance. If {5} is lighter, 5 is fake (lighter). If they balance, 3 is fake (heavier).
If {5,6,7,8} is heavier, it is the same situation as if {1,2,3,4} was heavier. Just perform the same steps using 5,6,7 and 8. Unless maybe you are too lazy to try and reprocess the steps, then you continue reading the solution. Weigh {5,6,1} and {7,2,9} (Note: 9 is surely not fake).
If they balance, then either 8 is heavy or 3 is light or 4 is light. Following the last step from the previous case, we weigh {3} and {4}. If they balance, 8 is fake(heavier). If they don’t balance then whichever one is lighter is fake (lighter).
If {5,6,1} is heavier, then either 5 is heavy or 6 is heavy or 2 is light. Weigh {5} and {6}. If they balance, 2 is fake (lighter). If they don’t balance then whichever one is heavier is fake (heavier).
If {7,2,9} is heavier, then either 7 is heavy or 1 is light. Weigh {1} and {9}. If they balance, 7 is fake (heavier). If they don’t balance then 1 is fake (lighter).
Answer: For convenience sake, let’s name the balls 1-12. This question is quite tricky. If you want to warm up your skills, you can try this easier 8 ball problem first.
Ok, let’s get on with it.
First we weigh {1,2,3,4} on the left and {5,6,7,8} on the right. There are three scenarios which can arise from this.
If they balance, then we know 9, 10, 11 or 12 is fake. Weigh {8, 9} and {10, 11} (Note: 8 is surely not fake)
If they balance, we know 12 is the fake one. Just weigh it with any other ball and figure out if it is lighter or heavier.
If {8, 9} is heavier, then either 9 is heavy or 10 is light or 11 is light. Weigh {10} and {11}. If they balance, 9 is fake (heavier). If they don’t balance then whichever one is lighter is fake (lighter).
If {8, 9} is lighter, then either 9 is light or 10 is heavy or 11 is heavy. Weigh {10} and {11}. If they balance, 9 is fake (lighter). If they don’t balance then whichever one is heavier is fake (heavier).
Phwww, that was confusing enough but we are not done yet.
If {1,2,3,4} is heavier, we know either one of {1,2,3,4} heavier or one of {5,6,7,8} is lighter but it is guarantees that {9,10,11,12} are not fake. This is where it gets really tricky, watch carefully. Weigh {1,2,5} and {3,6,9} (Note: 9 is surely not fake).
If they balance, then either 4 is heavy or 7 is light or 8 is light. Following the last step from the previous case, we weigh {7} and {8}. If they balance, 4 is fake(heavier). If they don’t balance then whichever one is lighter is fake (lighter).
If {1,2,5} is heavier, then either 1 is heavy or 2 is heavy or 6 is light. Weigh {1} and {2}. If they balance, 6 is fake (lighter). If they don’t balance then whichever one is heavier is fake (heavier).
If {3,6,9} is heavier, then either 3 is heavy or 5 is light. Weigh {5} and {9}. They won’t balance. If {5} is lighter, 5 is fake (lighter). If they balance, 3 is fake (heavier).
If {5,6,7,8} is heavier, it is the same situation as if {1,2,3,4} was heavier. Just perform the same steps using 5,6,7 and 8. Unless maybe you are too lazy to try and reprocess the steps, then you continue reading the solution. Weigh {5,6,1} and {7,2,9} (Note: 9 is surely not fake).
If they balance, then either 8 is heavy or 3 is light or 4 is light. Following the last step from the previous case, we weigh {3} and {4}. If they balance, 8 is fake(heavier). If they don’t balance then whichever one is lighter is fake (lighter).
If {5,6,1} is heavier, then either 5 is heavy or 6 is heavy or 2 is light. Weigh {5} and {6}. If they balance, 2 is fake (lighter). If they don’t balance then whichever one is heavier is fake (heavier).
If {7,2,9} is heavier, then either 7 is heavy or 1 is light. Weigh {1} and {9}. If they balance, 7 is fake (heavier). If they don’t balance then 1 is fake (lighter).
Saturday, February 23, 2008
8 Identical Balls Problem
Question: You are given 8 identical looking balls. One of them is heavier than the rest of the 7 (all the others weigh exactly the same). You a provided with a simple mechanical balance and you are restricted to only 2 uses. Find the heavier ball.
Answer: For convenience sake, let’s name the balls 1-8. First we weigh {1,2,3} on the left and {4,5,6} on the right. There are three scenarios which can arise from this.
If the left side is heavier, then we know that one of 1, 2 or 3 is the heavier ball. Weigh {1} on the left and {2} on the right. By doing this, we will know if 1 or 2 is heavier. If they balance out, then 3 is the heavier one.
If the right side is heavier, then we know that either 4, 5 or 6 is the heavier ball. Weigh {4} on the left and {5} on the right. By doing this we will know if 4 or 5 is heavier. If they balance out, then 6 is the heavier one.
If {1,2,3} and {4,5,6} balance out, then we know either 7 or 8 is the heavier one. Weigh both of them to find out which one is heavier.
Answer: For convenience sake, let’s name the balls 1-8. First we weigh {1,2,3} on the left and {4,5,6} on the right. There are three scenarios which can arise from this.
If the left side is heavier, then we know that one of 1, 2 or 3 is the heavier ball. Weigh {1} on the left and {2} on the right. By doing this, we will know if 1 or 2 is heavier. If they balance out, then 3 is the heavier one.
If the right side is heavier, then we know that either 4, 5 or 6 is the heavier ball. Weigh {4} on the left and {5} on the right. By doing this we will know if 4 or 5 is heavier. If they balance out, then 6 is the heavier one.
If {1,2,3} and {4,5,6} balance out, then we know either 7 or 8 is the heavier one. Weigh both of them to find out which one is heavier.
Wednesday, January 23, 2008
A Box of Defective Balls
Question: You have 10 boxes of balls (each ball weighing exactly10 gm) with one box with defective balls (each one of the defective balls weigh 9 gm). You are given an electronic weighing machine and only one chance at it. How will find out which box has the defective balls?
Answer: For convenience sake, let’s name the boxes from 1 to 10. In order to solve this problem, you have to leverage the fact that you know exactly what each good ball is supposed to weigh and what each defective ball is supposed to weigh. Many of us instinctively will take one ball out of each box and try to find a way to make it work but the trick to take different number of balls from each box.
The number of balls you pick from each bag is equal to the box number. For example, pick 1 ball from box 1, 2 balls from box 2 and so on. In total you will have 55 balls. If all of the boxes have good balls, then the total weight of these balls would be 550gm.
If box 1 has defective balls, then the total weight should be 1gm less than expected (only one ball weighing 9 gm). If box 2 has defective balls, then the total weight should be 2gm less than expected (two balls weighing 9 gm). So once you weigh the set of chosen balls, find out the difference between the total weight and the expected weight. That number represents the box number which contains the defective balls.
Answer: For convenience sake, let’s name the boxes from 1 to 10. In order to solve this problem, you have to leverage the fact that you know exactly what each good ball is supposed to weigh and what each defective ball is supposed to weigh. Many of us instinctively will take one ball out of each box and try to find a way to make it work but the trick to take different number of balls from each box.
The number of balls you pick from each bag is equal to the box number. For example, pick 1 ball from box 1, 2 balls from box 2 and so on. In total you will have 55 balls. If all of the boxes have good balls, then the total weight of these balls would be 550gm.
If box 1 has defective balls, then the total weight should be 1gm less than expected (only one ball weighing 9 gm). If box 2 has defective balls, then the total weight should be 2gm less than expected (two balls weighing 9 gm). So once you weigh the set of chosen balls, find out the difference between the total weight and the expected weight. That number represents the box number which contains the defective balls.
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