Medium | Discrete Maths |

13 Apples, 15 Bananas and 17 Cherries are put in the magic hat. When ever a collision of two different fruits occurs, they both get converted into the third type. For example 1 Apple and 1 Banana can collide to form 2 cherries. No other collision is holy. Can a sequence of such magical collisions lead all 45 fruits to give just one type?

Hint

This can't be done. Try to create a function of A,B,C which remains constant during a collision to get contradiction.

Solution

Create the invariant function f(A,B,C) = (0A+1B+2C)mod3, this function remains constant during a collision. But f(13,15,17) = 1 is not same as any of final states f(45,0,0)=f(0,45,0)=f(0,0,45)=0. Hence this can not be done.

A refreshment to this old trick was given by Aritro Pathak:

call the no of times apples are increased by 2 as A, bananas increased by 2 as B, and cherries increased by 2 as C. if we need 45 apples,total increments - decrements of apple = 32

but increments of apple = 2*A;

when ever 2 banana's are created, 1 Apple is lost, similarly for 2 cherries

decrements = B+C

thus we have 2A-B-C=32, -2B+C+A=15, -2C+B+A=17. Subtract the last two to get 2=3*(B-C) which is impossible. similar cases for when you want 45 of either cherries or bananas.

A refreshment to this old trick was given by Aritro Pathak:

call the no of times apples are increased by 2 as A, bananas increased by 2 as B, and cherries increased by 2 as C. if we need 45 apples,total increments - decrements of apple = 32

but increments of apple = 2*A;

when ever 2 banana's are created, 1 Apple is lost, similarly for 2 cherries

decrements = B+C

thus we have 2A-B-C=32, -2B+C+A=15, -2C+B+A=17. Subtract the last two to get 2=3*(B-C) which is impossible. similar cases for when you want 45 of either cherries or bananas.

Medium | Discrete Maths |

There are 51 ants sitting on top of a square table with side length of 1. If you have a square card with side 1/5, can you put your card at a position on the table to guarantee that the card encompasses at least 3 ants?

(updated: square card was originally disk of radius 1/7)

(updated: square card was originally disk of radius 1/7)

Hint

Pigeonhole principle

Solution

To guarantee that the card encompasses at least 3 ants, we can separate the square into 25 smaller areas (squares of side 1/5 each). Applying the generalized Pigeon Hole Principle, we can show that at least one of the areas must have at least 3 ants. The card is large enough to cover any of the 25 smaller areas. Done!

Medium | Discrete Maths |

On a 2D complex plane, all the integer-component points are coloured either White or Black. Is Possible to find a rectangle parallel to axis which has all corners of same color?

Solution

asd

Source: Top Quant Interview

Enable Like and Comment Medium | Discrete Maths |

In a circle are light bulbs numbered 1 through n, all initially on. At time t, you examine bulb number t, and if it’s on, you change the state of bulb t + 1 (modulo n); i.e., you turn it off if it’s on, and on if it’s off. If bulb t is off, you do nothing. Prove that if you continue around and around the ring in this manner, eventually all the bulbs will again be on.

Hint

It doesn't matter what the rules are, only thing that matters is that all bulbs should not go off!

Solution

Suppose every orientation of light bulbs & position of current pointer (t modulo n) forms a different state. Since all bulbs don’t go off, we must repeat a state after finite number of steps. Also notice that every state has a unique pre-state. Suppose the two states are at time T1 & T2. Notice that at time 0 all bulbs are on. Thus moving backwards from both states, we arrive at T2-T1, where all bulbs are on again.

Source: P. Winkler

Enable Like and Comment Medium | Discrete Maths |

There is a 6x8 rectangular chocolate bar made up of small 1x1 bits. We want to break it into the 48 bits. We can break one piece of chocolate horizontally or vertically, but cannot break two pieces together! What is the minimum number of breaks required?

Answer

47

Solution

For a chocolate of size mxn, we need mn - 1 steps. By breaking an existing piece horizontally or vertically, we merely increase the total number of pieces by one. Starting from 1 piece, we need mn - 1 steps to get to mn pieces.

Another way to reach the same conclusion is to focus on "bottom left corners of squares": Keep the chocolate rectangle in front of you and start drawing lines corresponding to cuts. Each cut "exposes" one new bottom left corner of some square. Initially, only one square's bottom left corner is exposed. In the end, all mn squares have their bottom left corners exposed.

Another way to reach the same conclusion is to focus on "bottom left corners of squares": Keep the chocolate rectangle in front of you and start drawing lines corresponding to cuts. Each cut "exposes" one new bottom left corner of some square. Initially, only one square's bottom left corner is exposed. In the end, all mn squares have their bottom left corners exposed.

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