Notice the following selfie. It is a proper selfie – taken with the front camera of an iPhone (evident by the fact I could not use the flash). No timer was used and there was no third person involved as well as no editing was done. Yet, you can see both the arms of both the subjects fully.
How was this done?
Microsoft CEO Satya Nadella reportedly said that there will be “new battles in stores” in an interview yesterday.
And my first reaction was “What? People are already fighting outside the stores for the new iPads?” 🙂
Nadella maintained a studious silence when asked in a follow up question regarding what kind of respectable distance was Microsoft planning to watch the battles from. 🙂
Now, that is “good karma” 🙂
Found this interesting problem. See if you can solve this…
Four tanks going in a line on a very narrow bridge encounter four friendly tanks in a line coming from the other side. Unfortunately, they see each other only when there is exactly one tank worth of distance between the two leading tanks from either side. And there is no space on the bridge to go around each other.
Now here is a problem – None of the tanks can reverse. However, a tank can climb over another tank as long as there is space for the tank to land on the bridge after climbing over a tank. A tank cannot climb over more than one tank at a time. (meaning it has to come down to the bridge after climbing one tank). Also no tank can take the weight of more one one tank on top of it. (meaning you cannot have three tanks on top of each other).
How can the two sets of four tanks sort out the problem and eventually proceed their own way?
It has been two weeks since I left home. India trip, a week of business meetings and I am finally on my flight back home… It has been a long time since I posted a puzzle on my flight back home on a Thursday evening… so here goes the next one…
[Remember the rules – do not post your answers as Comments in FB. Either message me on FB or write a comment in my blogsite www.rajibroy.com]
A hotel wing has 17 consecutive rooms numbered 1 thru 17. Each of the rooms are connected to the two adjoining room(s) thru keyed (internal) doors and to the corridor with one (external) keyed door. Of course the two end rooms have only one internal door. A fugitive is holed up in one of the rooms. He has one of the two master keys that can open up the (internal) doors to the adjoining room(s). However, for security purpose, the key, once operated, is inactive for 24 hours. And the key has to be used once every 24 hours – else it gets completely deactivated. In other words, the fugitive has to shift to the next room every day using the internal door – on any one side, unless he is in room 1 or 17, in which case he has only one choice.
There is a detective who suspects that the fugitive is holed up there. He has the other master key. This key also works in the same way except it works only on the external door of the rooms (that opens to the corridor). Thus, the detective can open the external door of any room. If he finds the fugitive there, he apprehends him. If not, he has to immediately shut the door and stay in the corridor (so that the fugitive cannot escape thru the corridor). The detective, can choose to open the same room on consecutive days, if he so desires.
So day after day, this is what happens – the detective opens a room, checks for the fugitive and closes the door if he does not find him. The fugitive, meanwhile, day after day shifts to the room on the left or the right – once a day (as mentioned, unless he is in one of the end rooms, in which case, he can go in only one way).
Now, the detective’s assignment runs out in 30 days. So, he has to apprehend the fugitive within that time period or else he is recalled and the fugitive escapes.
Can you come up with a strategy for the detective (what logic will he follow to open doors) that will guarantee that he will catch the fugitive before he is recalled?
One more time (this time with a very different job) I find myself on a Thursday evening flying from DC to Atlanta. Which means time for another puzzle.
This time it is one of those about measuring arbitrary time with those dratted hourglasses. If you are reading this on FB, you probably remember that you can ask clarifying questions in the comments section but send answers by messaging me.
You have two hourglasses – one empties in four minutes and the other empties in seven minutes. You have to bake a cake but it must be baked for exactly nine minutes. What is the minimum time it will take you after I say Go! for you to bake the cake? (You don’t have to start baking the cake immediately when you turn the hourglass(es) but once you start baking, it has to finish in exactly nine minutes).
To further clarify, if you said “I will flip the four minute hourglass first and then moment it is done, I will start baking and flip both hourglasses….. And say you figured out how to bake the cake in nine minutes now using those two hourglasses”, then total time taken is thirteen minutes. Get it?
Go!
On a coast to coast flight. Figured I have enough time to come up with a puzzle.
As always send a personal message for your answer – do not put it in a public place like Comments section in Facebook.
I had heard this problem long time back and was struck by the simplicity of its answer. I am adding a twist to that question here.
You and fourteen more of your team members were having coffee at your office cafetaria one day. Your manager walked up to you and said “I have an interesting problem for you guys. All of you will get an increment if you can tell me the correct answer. I am going to pick a random person ten minutes from now and ask – ‘how many of the team members earn higher than the average salary of the team’ ?”
Now the trick is obviously nobody wants to divulge their own salary to anybody else. And nobody wants anybody to know whether he or she is making higher or lower than the average.
You can assume nobody is earning at exactly the average number.
The team members look up to you to devise a quick, easy and elegant way so that everybody will get that increment.
How would you do it?
This is one of those “shunting” puzzles. As you can see in my somewhat clumsy diagram, there is a railway track that goes indefinitely to the right. On the left, it splits into a “Y”, both ends of which merge into a finite straight railway track. There is a coach “A” sitting on one arm of the Y and another coach “B” sitting on the other arm. Both the coaches are 5 yards each.
The distance between the junction point on coach A side and the end of the finite straight track is 5 yards. On the other side, near coach B, the distance between the junction point and the end of the straight finite track is 15 yards. (See pic). In between the junction points on the straight track sits an engine “E” which is itself 10 yards in length.
Like every engine, E can go forwards or backwards and can pull or push the coaches. Any coach and engine can be attached to each other.
Here is the puzzle: Can you interchange the position of coach A and coach B? You have to get engine E back to its original position when you are done.
Two students of maths are playing the following game – each takes turn to pick a number from the following set {-4, -3, -2, -1, 0, +1, +2, +3, +4}. Once a number is picked, it is not replaced. And they keep playing till they run out of numbers. And they start all over again.
Whoever can come up with three numbers that add up to 0 first wins the game. It does not have to be the only three numbers she has. For example, she might have picked -3 and then in next chance picked +2. Hoping to pick +1 next. But other player seeing that, would  have taken away the +1 in the next move. So let’s say the original person next picked 4. And then if she got a chance to pick -1 in the next move, she would have -3, +2, +4 and -1. Note that -3, +4 and -1 adds to 0. So she wins.
Now, here is the question: If you are the first player, is there a strategy for you to win? If you are the second player, is there a strategy for you to win?
If you are seeing this in FB, send me message and do not post answers in Comments section.
