Plan an Unbeatable Strategy

Two people play a game of NIM. There are 100 matches on a table, and the players take turns picking 1 to 5 sticks at a time. The person who takes the last stick wins the game. (Both players has to make sure that the winner would be picking only 1 stick at the end) 

Who has a winning strategy?

And what must be winning strategy in the person who takes the last stick looses?

This could be the winning strategy! 

Planned The Unbeatable Strategy!

What is the game?

The first person can plan an unbeatable winning strategy.

CASE 1 : The person picking last stick is winner.

All that he has to do is pick 4 sticks straightaway at the start leaving behind 96 stick. Then, he has to make sure that the count of remaining stick will be always divisible by 6 like 96, 90, 84, 78......6. 

So if the opponent takes away 2 sticks in his first turn, then first person has to take 6 - 2 = 4 sticks leaving behind 90 sticks there. That is if the opponent takes away X stick the first person need to pick 6 - X sticks.

Now, when there are 6 stick left, even if opponent takes away 5 sticks then 1 stick will be left for the first person.

And even if the opponent picks 4 sticks then first person will take 2 remaining sticks.

CASE 2 : The person picking last stick is looser. 

Now the first person need to take away 3 sticks in first turn leaving behind 97. Next, he has to make sure the count of remaining sticks reduced by 6 after each of his turn. That is, the count should be like 91,85,79,72......7.

So if the opponent takes away 4 sticks in his first turn, then first person has to take 6 - 4 = 2 sticks leaving behind 91 sticks there. That is if the opponent takes away X stick the first person need to pick 6 - X sticks.

When there are 7 sticks are left then even if the opponent takes away 5 sticks then first person can force him to pick the last stick by picking only 1 stick of remaining 2. 

And if the opponent takes away 4 sticks at this stage, the first person still can force him to pick last stick by picking 2 of remaining 3 sticks.

Conclusion : The first person always has a chance to plan a winning strategy.

Fill in the Empty Boxes

Is it possible to fill each box in with an arithmetic operation so that this becomes a true equation?

Did you too find it true? 

Correct Operators in Empty Boxes!

What wasn't looking possible?

Yes, it's possible. All you need to do is recall BODMAS (Brackets, Of, Division, Multiplication, Addition, Subtraction) rule in mathematics that we learned in school.

What is the Weight of the Empty Jar?

A full jar of honey weighs 750 grams, and the same jar two-thirds full weighs 550 grams.

What is the weight of the empty jar in grams?

Find the correct way to find the weight here!

Calculation of Weight of the Empty Jar

Collect the given data.

Let J be the weight of empty jar and H be the weight of honey when jar was full.

J + H = 750                                            ......(1)

And in second case,

J + (2/3)H = 550                                     .....(2) 

Subtracting (2) from (1),

(1/3)H = 200

Hence, H = 600.

Putting this value in (1),

J = 750 - 600 = 150.

Therefore, the weight of the empty jar is 150 grams.

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.

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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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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.

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.

"What's The Area of The Triangle?"

If I place a 6 cm × 6 cm square on a triangle, I can cover up to 60% of the triangle. If I place the triangle on the square, I can cover up to 2/3 of the square. What is the area, in cm², of the triangle?

(a) 22 4/5
(b) 24
(c) 36
(d) 40
(e) 60

Here is that mathematical calculation!

Calculation of Area of Triangle

What was the question?

The most important thing to note here is the area that square overlaps on the triangle is equal to the area that triangle overlaps the square with the maximum contact area. That is both the areas are supposed to be equal in amount. 


So if T is the area of the triangle & S is area of the square then,

0.6 T = 2/3  x S = 2/3 x (6x6) = 2/3 x 36 = 24

T = 40 Square cm.

Hence, answer is option (d) 40. 
 

The Dice Date Indicator!

How can you represent days of month using two 6 sided dice? You can write one number on each face of the dice from 0 to 9 and you have to represent days from 1 to 31, for example for 1, one dice should show 0 and another should show 1, similarly for 29 one dice should show 2 and another should show 9.

This is how it can be indicated!

Making of The Dice Date Indicator

What was the challenge?

Dice 1: 0 1 2 3 5 7

Dice 2: 0 1 2 4 6 8

The number 0 has to be present on both the dice. The '0' on first die needed for dates from 1 to 9 to show them as 01,02,03......and the '0' on second die will be used for the dates 10,20,30.

The number 1 and 2 are repeated on the dates 11 and 22 so those 2 numbers has to be there on both dice.

Now, we are left with total 6 positions but 7 numbers - 3 to 9.

However, 6 and 9 can be represented by single die if it is written on one of the side of the die. In normal position it will represent 6 & in inverted position it will show 9 or say vice versa.

For example, the dates 6 and 9 can be indicated as - 

The Domino on The Chessboard Challenge

There is an 8 by 8 chessboard in which two diagonally opposite corners have been cut off.You are given 31 dominos, and a single domino can cover exactly two squares. Can you use the 31 dominos to cover the entire board?

Simple Arrangement? Check out it's possibility!

Impossible Dominos' Arrangement on Chessboard

What was the challenge given?

Initial mathematical calculations might suggest that the task is pretty simple. If 2 square are cut off from 64 squares then 62 squares will be left which are enough for 31 dominos (each covering 2 squares).

But, that is not the case. Since, 2 diagonally opposite squares are removed, they has to be either black or white like shown below with shaded regions.

We need 1 black and 1 White square for placement of 1 domino on the chessboard.That is 31 Black and 31 White squares are needed to give cover for 31 dominos.

 In the above 2 cases, there are either 32 Black and 30 White or 30 Black and 32 White squares are available.

This makes the task of placing 31 dominos on the chessboard (whose 2 diagonally opposite squares are removed) impossible! 

The Ping Pong Puzzle

Three friends (A, B and C) are playing ping pong. They play the usual way: the winner stays on, and the loser waits his/her turn again. At the end of the day, they summarize the number of games that each of them played:

A played 10
B played 15
C played 17.

Who lost the second game? 

This person played & lost the second game! 

Participant of the Second Game!

How games were played?

A played 10, B played 15 and C played 17 games. So total number of presences are 10 + 15 + 17 = 42. Every 2 presences form a game. Hence, the number of games played are 42/2 = 21.

Let's take into consideration the minimum number of games that a player can play. For that, he need to loose every game that he has played. That is, if he has played first game then he must have out in second but replaced looser of second in third game. In short, he must have played odd numbered of games like 1,3,5,7,9,11,13,15,17,19,21.That's 11 games in total.

And if he had made debut in second game then he must had played even numbered games like 2,4,6,8,10,12,14,16,18,20. That's 10 games in total.

Since, in the case only A has played 10 games, he must have made debut in second game where he lost that game to make comeback in 4th game thereby replacing looser of third game.

Maximum Runs That Batsman Can Score?

In a one day international cricket match, considering no extras(no wides, no ‘no’ balls, etc.) and no overthrows.

What is the maximum number of runs that a batsman can score in an ideal case ?

Note: Here we assume ideal and little practical scenario. We assume that batsman can not run for more than 3 runs in a ball, as otherwise there is no limit, he can run infinite runs(theoretically) in a ball, as far as opposite team does not catch the ball.”

Could be tricky! Here is correct number! 

Calculation of Maximum Runs by Batsman

What was the question?

It's not as straight forward as it seems at first glance. That is one might think that the maximum score that one can score by hitting 6 on every ball of 50 overs is 50 x 6 x 6 = 1800. 

No doubt, 1800 can be the maximum team score but not the individual score.Since batsman rotates strike every over, here both batsmen share these 1800 runs as 900 to each.

However, if the batsman on strike runs 3 runs on the last ball of the over then he can hit 5 more sixes in next over as strike is rotated back to him in next over. He can continue in this way till 49th over. And in 50th over he can hit 6 sixes on 6 balls.

Maximum Individual Score = 49 x [(5x6)+3]  + 36 = 1617 + 36 = 1653.

In this case, the batman at the non-striker end scores 0 runs as he doesn't get strike on a single ball.

"Who is telling the truth?"

King Octopus has servants with six, seven, or eight legs. The servants with seven legs always lie, but the servants with either six or eight legs always say the truth.

One day, 4 servants met :

The blue one says: “Altogether we have 28 legs”;

the green one says: “Altogether we have 27 legs”;

the yellow one says: “Altogether we have 26 legs”;

the red one says: “Altogether we have 25 legs”.

 
What is the color of the servant that says the truth?

"I'm telling the truth and my color is..." 

"Listen to me; I'm telling the truth!"

What others making statements?

Since, each of 4 servants telling different numbers then only 1 of them must be telling the truth & other must be lying.

In that case, all 3 must be having 7 legs i.e. 21 legs while the one who is making true statement must have either 6 or 8 legs. Therefore, there must be 21 + 6 = 27 or 21 + 8 = 29 legs altogether.

Nobody is saying that they altogether have 29 legs but the green octopus is saying that they altogether have 27 legs. Hence, the green octopus must be telling the truth.