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Logical Reasoning Basics – How to solve coin-picking / matchstick-related problems?



COIN PICKING

In this post, we will figure out how to tackle Logical Reasoning Problems dependent on coins and matchsticks picking puzzles. To see how precisely these sorts of puzzles resemble, we should begin the post with a basic model. The technique to understand the model will give better knowledge in order to how to move toward these puzzles.

Two savvy players An and B are playing a coin game in which they can get 1, 2, 3 or 4 coins. They have 78 coins and the player who picks the last coin will lose the game. An and B play on the other hand and A plays the main move. What number of coins should A pick from the outset so his success is free of number of coins B picks in his first move?


Method to solve:

To win the game A must leave only 1 coins for B to pick up for B’s last move. If A leaves 2 coins then B will pick up 1 coin forcing A to pick up the last coin. Similar condition is when 3, 4 or 5 coins are left for B to pick after A’s move. Hence, A can leave only 1 coin for B to pick which will make A as the winner. For A’s last move, he should be left with 2, 3, 4 or 5 coins which means his second last move should ensure that only 6 coins are left.



Start analyzing this structure from the bottom. For the B’s last turn only 1 coins is left which means A must have picked up either 1 or 2 or 3 or 4 coins in A’s last turn. This is only possible if A has left only 6 coins for B to pick up in A’s second last turn.

Going with B’s second last turn, he has an option of picking from 6 coins. If he picks 4 coins, A will be left with 2 coins and so A can pick up just 1 coin forcing B to pick up the last one. If B picks up 1 coin, A will be left with 5 coins and so A can pick up 4 coins which again forces B to pick up the last coin.

To get even better understanding, let us make A and B play the game. The number of coins each picks is a number among 1,2,3 or 4. Every time A will try to control the right column of the above table. Since they have 78 coins initially and from the column we see that A wants to leave 76 coins after his first turn, A will pick up 2 coins. Note that number of coins picked by B is a random among 1,2,3 and 4.






Here we can see that A is trying to maintain the controlling factor column shown previously.

A’s controlling factor can be generalized as: (MIN + MAX)*k + MIN

where k is a natural number

In our example: min =1, max =4 so the controlling factor is 5k + 1.

In order to find the number of coins A should pick up, the controlling factor is to be the largest number possible but less than the total number of coins. In our example this number is 76.

( 5k + 1) + Numer of coins to pick in first turn         = Total coins

Numer of coins to pick in first turn                          = Total coins – (5k + 1)

Number of coins to pick in first turn                        = (Total coins – 1) -5k

The number of coins to pick in first turn, from the above equation is the remainder (when total coins -1) is divided by 5, i.e.





The controlling factor is the key to win. This case illustrates the situation when the person who picks the last coin will lose the game. If we change the minimum coins to pick from 1 to 2 and maximum remain as 4, then A should pick 4 coins in his first turn.

This example illustrates the situation when the one who picks the last coin loses the game. What if it’s the other way around? SOMEONE WHO PICKS THE LAST COIN WINS THE GAME. Let us take that example as well.

QUESTION: Two smart players A and B are playing a coin game in which they can pick up 1, 2, 3 or 4 coins. They have 78 coins and the player who picks the last coin will win the game. A and B play alternately and A plays the first move. How many coins should A pick at first so his win is independent of a number of coins B picks in his first move?

Method: In this case A would want to pick the last coin (he can even pick upto last 4 coins). He wants to leave 0 coins after his last turn. Hence, his control factors will start at 0 from the bottom of the table.

If we go through the following table, we can see that the method to solve remains similar to the previous approach, only the control factor changes.



n this case, the controlling factor changes as: (MIN + MAX)*K




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