Puzzles

Column 1

100

How many times a day do the minute and hour hands of a clock overlap?

Did you think the answer was 24 times? Well if you did, it’s time you think again. Let’s do some math. In T hours, the minute hand completes T laps. In the same amount of time, the hour hand completes T/12 laps. The first time the minute and hour hands overlap, the minute hand would have completed 1 lap more than the hour hand. So we have T = T/12 + 1. This implies that the first overlap happens after T = 12/11 hours (~1:05 am). Similarly, the second time they overlap, the minute hand would have completed two more laps than the hour hand. So for N overlaps, we have T = T/12 + N. Since we have 24 hours in a day, we can solve the above equation for N 24 = 24/12 + N 24 = 2 + N N = 22 Thus, the hands of a clock overlap 22 times a day. Thus the hands of the clock overlap at 12:00, ~1:05, ~2:10, ~3:15, ~4:20, ~5:25, ~6:30, ~7:35, ~8:40, ~9:45, ~10:50. Note that there is no ~11:55. This becomes 12:00.

200

Three ants are sitting at the three corners of an equilateral triangle. Each ant starts randomly picks a direction and starts to move along the edge of the triangle. What is the probability that none of the ants collide?

So let’s think this through. The ants can only avoid a collision if they all decide to move in the same direction (either clockwise or anti-clockwise). If the ants do not pick the same direction, there will definitely be a collision. Each ant has the option to either move clockwise or anti-clockwise. There is a one in two chance that an ant decides to pick a particular direction. Using simple probability calculations, we can determine the probability of no collision. P(No collision) = P(All ants go in a clockwise direction) + P( All ants go in an anti-clockwise direction) = 0.5 * 0.5 * 0.5 + 0.5 * 0.5 * 0.5 = 0.25

300

A certain town comprises of 100 married couples. Everyone in the town lives by the following rule: If a husband cheats on his wife, the husband is executed as soon as his wife finds out about him. All the women in the town only gossip about the husbands of other women. No woman ever tells another woman if her husband is cheating on her. So every woman in the town knows about all the cheating husbands in the town except her own. It can also be assumed that a husband remains silent about his infidelity. One day, the mayor of the town announces to the whole town that there is at least 1 cheating husband in the town. What do you think happens?

Say there was only 1 cheating husband in the town. There will be 99 women who know exactly who the cheater is. The 1 remaining woman, who is being cheated on, would have assumed there are no cheaters. But now that the mayor has confirmed that there is at least one cheater, she realizes that her own husband must be cheating on her. So her husband gets executed on the day of the announcement. Now let’s assume there are 2 cheaters in the town. There will be 98 women in the town who know who the 2 cheaters are. The 2 wives, who are being cheated on, would think that there is only 1 cheater in the town. Since neither of these 2 women know that their husbands are cheaters, they both do not report their husbands in on the day of the announcement. The next day, when the 2 women see that no husband was executed, they realize that there could only be one explanation – both their husbands are cheaters. Thus, on the second day, 2 husbands are executed. Through induction, it can be proved that when this logic is applied to n cheating husbands, they all die on the n th day after the mayor’s announcement.

Column 2

100

A train leaves City X for City Y at 15 mph. At the very same time, a train leaves City Y for City X at 20 mph on the same track. At the same moment, a bird leaves the City X train station and flies towards the City Y train station at 25 mph. When the bird reaches the train from City Y, it immediately reverses direction. It then continues to fly at the same speed towards the train from City X, when it reverses its direction again, and so forth. The bird continues to do this until the trains collide. How far would the bird have traveled in the meantime?

Knowing that the bird is the faster than both the trains, you would only imagine that theoretically, the bird could fly an infinite number of times between the trains before they collide. This is because you know that no matter how close the trains get, the bird will always complete its trip before the crash happens. At the time of the crash, the bird would probably get squashed between the trains! I bet sometime in school, you learnt how to sum up an infinite series. But do we have to do that? The concept of relative speed (rings a bell?) can work handy here. Let’s assume that the distance between City X and City Y is d miles. The trains are approaching each other at a relative speed of (20 + 15) = 35 mph. The sum of the distances covered by the trains when they collide is d (i.e. the distance between the cities). Since distance/speed gives us time, we know that the trains collide d/35 hours after they start. Since the speed of the bird is constant at 25 mph, we know that the bird would have covered 25 * (d/35) miles = 5d/7 miles before the trains collide.

200

You’ve got someone working for you for seven days and a gold bar to pay them. You must pay the worker for their work at the end of every day. If you are only allowed to make two breaks in the gold bar, how do you pay your worker? (Assuming equal amount of work is done during each day thus requiring equal amount of pay for each day)

The trick is not to try and how to cut in such a way to make 7 equal pieces but rather to make transactions with the worker. Make two cuts on the gold bar such that you have the following sizes of bars. 1/7, 2/7 and 4/7. For convenience sake, I would just refer to the bars as 1, 2 and 4. At the end of Day 1: Give Bar 1 (You- 2 and 4, Worker- 1) At the end of Day 2: Give Bar 2, Take back Bar 1 (You- 1 and 4, Worker- 2) At the end of Day 3: Give Bar 1 (You- 4, Worker- 1 and 2) At the end of Day 4: Give Bar 4, Take back Bar 1 and Bar 2 (You- 1 and 2, Worker- 4) At the end of Day 5: Give Bar 1 (You- 2, Worker- 1 and 4) At the end of Day 6: Give Bar 2, Take back Bar 1 (You- 1, Worker- 2 and 4) At the end of Day 7: Give Bar 1 (You- Empty, Worker- 1, 2 and 4) That should take care of everything.

300

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?

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.

Column 3

100

If you had an infinite supply of water and a 5 quart and 3 quart pails, how would you measure exactly 4 quarts? and What is the least number of steps you need?

This question is very simple actually. Since we can’t hold 4 quarts in the 3 quart pail, we have to look to filling up the 5 quart pail with exactly 4 quarts. Lets count the steps as we move along 1. Fill 3 quart pail ( 5p – 0, 3p – 3) 2. Transfer to 5 quart pail (5p – 3, 3p – 0) 3. Fill 3 quart pail ( 5p – 3, 3p – 3) 4. Transfer to 5 quart pail (5p – 5, 3p – 1) 5. Empty 5 quart pail (5p – 0, 3p – 1) 6. Transfer to 5 quart pail (5p – 1, 3p – 0) 7. Fill 3 quart pail ( 5p – 1, 3p – 3) 8. Transfer to 5 quart pail (5p – 4, 3p – 0) We are done!!!

200

Four people need to cross a rickety bridge at night. Unfortunately, they have only one torch and the bridge is too dangerous to cross without one. The bridge is only strong enough to support two people at a time. Not all people take the same time to cross the bridge. Times for each person: 1 min, 2 mins, 7 mins and 10 mins. What is the shortest time needed for all four of them to cross the bridge? Let’s brainstorm a little further. To reduce the amount of time, we should find a way for 10 and 7 to go together. If they cross together, then we need one of them to come back to get the others. That would not be ideal. How do we get around that? Maybe we can have 1 waiting on the other side to bring the torch back. Ahaa, we are getting closer. The fastest way to get 1 across and be back is to use 2 to usher 1 across. So let’s put all this together. 1 and 2 go cross 2 comes back 7 and 10 go across 1 comes back 1 and 2 go across (done) Total time = 2 + 2 + 10 + 1 + 2 = 17 mins

The initial solution most people will think of is to use the fastest person as an usher to guide everyone across. How long would that take? 10 + 1 + 7 + 1 + 2 = 21 mins. Is that it? No. That would make this question too simple even as a warm up question. Let’s brainstorm a little further. To reduce the amount of time, we should find a way for 10 and 7 to go together. If they cross together, then we need one of them to come back to get the others. That would not be ideal. How do we get around that? Maybe we can have 1 waiting on the other side to bring the torch back. Ahaa, we are getting closer. The fastest way to get 1 across and be back is to use 2 to usher 1 across. So let’s put all this together. 1 and 2 go cross 2 comes back 7 and 10 go across 1 comes back 1 and 2 go across (done) Total time = 2 + 2 + 10 + 1 + 2 = 17 mins

300

How many points are there on the globe where, by walking one mile south, then one mile east and then one mile north, you would reach the place where you started?

The trivial answer to this question is one point, namely, the North Pole. But if you think that answer should suffice, you might want to think again! :) Let’s think this through methodically. If we consider the southern hemisphere, there is a ring near the South Pole that has a circumference of one mile. So what if we were standing at any point one mile north of this ring? If we walked one mile south, we would be on the ring. Then one mile east would bring us back to same point on the ring (since it’s circumference is one mile). One mile north from that point would bring us back to the point were we started from. If we count, there would be an infinite number of points north of this one mile ring. So what’s our running total of possible points? We have 1 + infinite points. But we’re not done yet! Consider a ring that is half a mile in circumference near the South Pole. Walking a mile along this ring would cause us to circle twice, but still bring us to back to the point we started from. As a result, starting from a point that is one mile north of a half mile ring would also be valid. Similarly, for any positive integer n, there is a circle with radius r = 1 / (2 * pi * n) centered at the South Pole. Walking one mile along these rings would cause us to circle n times and return to the same point as we started. There are infinite possible values for n. Furthermore, there are infinite ways of determining a starting point that is one mile north of these n rings, thus giving us (infinity * infinity) possible points that satisfy the required condition. So the real answer to this question is 1 + infinity * infinity = infinite possible points!