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Confusing Ride on the Ferris Wheel

There are 10 two-seater cars attached to a Ferris wheel. The Ferris wheel turns so that one car rotates through the exit platform every minute
The wheel began operation at 10 in the morning and shut down 30 minutes later. 
What's the maximum number of people that could have taken a ride on the wheel in that time period?


Confusing Ride on the Ferris Wheel

Here is count of people enjoying the ride!

Peoples Enjoying Ferris Wheel Ride


What was the puzzle?

Let's simplify the situation by naming 10 cars as A, B, C, D, E, F, G, H, I, J.

Suppose the Car A is at the exit platform at 10:00 AM. Obviously it can be 'loaded' with 2 peoples say A1-A2.

At 10:01 AM, the Car B will be at the exit platform which can be 'loaded' with 2 more peoples say B1-B2.

So continuing this way, at 10:09 AM the car J will be loaded with peoples J1-J2.

At 10:10 AM, the Car A will be again at the exit platform and now A1-A2 can get off the board to allow 2 more new peoples A3-A4 to get on the board.

At 10:11 AM, the Car B will be at the exit platform where B3-B4 will replace B1-B2. 

Continuing this way, at 10:19 AM the J1-J2 in the Car J will be replaced by J3-J4.

So far, 10 x 4 = 40 different peoples would have enjoyed the ride. 

It's clear that every car takes 10 minutes to be at the exit platform after once it goes through it. That's how Car A is at the exit platform at 10:10 AM, 10:20 AM and it can be at 10:30 AM as well when the wheel is supposed to be shut down.

Now, since wheel needs to be shut down 10:30 AM, emptying all the cars must be started except Car A for the reason stated above.

Therefore, at 10:20, peoples A5-A6 replace A3-A4. Thereafter, every car should be emptied. So, far 40 + 2 = 42 different people have boarded on the cars of the wheel.

So, at 10:21 AM, the Car B is emptied, at 10:22 AM, the Car C should be emptied and so on.

At 10:29 AM, J3-J4 of Car J get out of the car and finally at 10:30 AM, A5-A6 get out of the Car A

Now, the ferris wheel can be shut down with no one stuck at any of cars.

To conclude, 42 different peoples can enjoy the ride in given time frame.



Develop an Unbeatable Strategy!

Consider a two player coin game where each player gets turn one by one. There is a row of even number of coins, and a player on his/her turn can pick a coin from any of the two corners of the row. The player that collects coins with more value wins the game. 

Develop a strategy for the player making the first turn, such he/she never looses the game.

Note : The strategy to pick maximum of two corners may not work. In the following example, first player looses the game when he/she uses strategy to pick maximum of two corners.

============================================================

Example :

  18 20 15 30 10 14


First Player picks 18, now row of coins is
  20 15 30 10 14


Second player picks 20, now row of coins is
  15 30 10 14


First Player picks 15, now row of coins is
  30 10 14


Second player picks 30, now row of coins is
  10 14


First Player picks 14, now row of coins is
  10 


Second player picks 10, game over.



Develop an Unbeatable Strategy!

The total value collected by second player is more (20 +
30 + 10) compared to first player (18 + 15 + 14).
So the second player wins.

===================================================================
 
This is the unbeatable strategy! 

The Unbeatable Strategy for a Coin Game!


Why strategy needed in the case?

Let's recall the example given in the question.
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Example

  18 20 15 30 10 14


First Player picks 18, now row of coins is
  20 15 30 10 14


Second player picks 20, now row of coins is
  15 30 10 14


First Player picks 15, now row of coins is
  30 10 14


Second player picks 30, now row of coins is
  10 14


First Player picks 14, now row of coins is
  10 


Second player picks 10, game over.

The total value collected by second player is more (20 +
30 + 10) compared to first player (18 + 15 + 14).
So the second player wins.

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Here, it's very clear that the player who chooses coins numbered at even positions wins & the one who chooses odd position loses.

So first player who is going to choose coin first need to be smart. All that he/she need to do is make sum of values of all coin at even position, sum of values of all coins at odd positions and compare them. 

If he finds the sum of values of coins at odd position greater then he should choose 1st coin (odd position) followed by 3rd,5th.....(odd positions) & force other to choose even coins

And if he finds the sum of values of coins at even positions greater then he should choose
last coin (which is at even position as number of coins are even).

For example, in above case, 

  18 20 15 30 10 14

first player calculates 

Sum of odd coins = 18 + 15 + 10 = 43
Sum of even coins = 20 + 30 + 14 = 64.  


Since sum of even coins is greater, he should choose 6th coin (which will be followed by 4th and 2nd) and force other player to choose 5th,3rd and 1st coin.

Even in case second player selects 1st coin after 14 at other corner is selected by first, still he can be forced to choose the coin at 3rd position (odd position) if first player selects 2nd coin.

The Unbeatable Strategy for a Coin Game!


In short, first player need to make sure that he collects all the coins that are at even positions or odd position whichever has greater sum!   

Note that the total number of coins in the case can't be odd as then distribution of coins among 2 will be unequal.

An Island Of Puzzles

There is an Island of puzzles where numbers 1 - 9 want to cross the river.

There is a single boat that can take numbers from one side to the other.

However, maximum 3 numbers can go at a time and of course, the boat cannot sail on its own so one number must come back after reaching to another side.


Also, the sum of numbers crossing at a time must be a square number.

You need to plan trips such that minimum trips are needed.


Digits On An Island Of Puzzles - Maths Puzzles

This should be that minimum number! 

Numbers On An Island Of Puzzles


What was the challenge?

We need only 7 trips to send all digits across the river.

1. Send 2, 5, 9 (sum is 16).

2. Bring back the 9.

3. Send 3,4, 9 (sum is 16).

4. Bring back the 9.

5. Now send 1,7,8 (sum is 16).

6. Bring back the 1.

7. And finally send 1,6,9
(sum is 16).

Taking Numbers On An Island Of Puzzles - Maths Puzzles

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