Hard | Discrete Maths |

Several gas stations on a circular trek have between them just enough gas for one car to make a complete round trip. Prove that if you start at the right station with an empty tank you shall be able to make it all the way around.

Hint

Induction

Solution

Take N=2 gas stations. Assume their capacities are x1 and x2 (in terms of distance). Now, x1+x2 = 1. Assume y1 and y2 be clockwise distances from 1 to 2 and 2 to 1 (clockwise), y1+y2 = 1. Hence we cant have x1<y1 and x2<y2 both. Thus, let x1 >= y1. In this case we start with station 1 and complete the journey. Assume the path is possible for N=K stations, we want to prove for N = K+1 stations, using same argument as above, xi>=yi for some i, and hence reaching xi with zero fuel ensures reaching x(i+1). Thus we merge xi and x(i+1), and the problem is reduced to N=K stations, which we assumed!

Another method is to imagine having a big tank with enough gas for a round trip and enough room for going through the motions of emptying every gas station on your way. Start at any station and mind to record the amount of gasoline on reaching gas stations on your way around. At the end of the trip, when you pull into the station of departure with the original amount of gas, check your list. The station marked with the least number is the one where you want to start on an empty tank.

Another method is to imagine having a big tank with enough gas for a round trip and enough room for going through the motions of emptying every gas station on your way. Start at any station and mind to record the amount of gasoline on reaching gas stations on your way around. At the end of the trip, when you pull into the station of departure with the original amount of gas, check your list. The station marked with the least number is the one where you want to start on an empty tank.

Source: P. Winkler

Enable Like and Comment Hard | Discrete Maths |

Can you pack 53 bricks of dimensions 1x1x4 into a 6x6x6 box? The faces of the bricks are parallel to the faces of the box

Hint

See its cousin called Domino Covering in which two opposite corners of a checkerboard are removed.

Answer

No!

Solution

Divide the volume of 6x6x6 box into 2x2x2 mini cubes. Imagine each of these mini cubes is either fully red or fully blue such that it forms a 3D checkerboard pattern on the box. This will restrict 14-13 restriction on cube colors, say 14 blue and 13 red. Now, putting bricks into box, parallel to faces, each brick will be half blue and half red, so 52 bricks fill all the red cubes and there is no way to place a 53rd brick.

Source: Xinfeng Zhou

Enable Like and Comment Medium | Discrete Maths |

There is a crazy clock in Alice's Dream, it has two hands initially pointing at 12. The minute hand moves clockwise, making 5 rounds (with varying speeds) and comes back to 12. In the same time, the hour hand goes anti-clock wise, finishing 4 rounds and returns to 12. How many times did the two cross each other ? (Cross means meet & pass through, hence ignore start & end)

Answer

8

Solution

In the reference frame of minute hand, hour hand moves exactly (5+4) = 9 rounds anti-clockwise with varying speeds (by adding total angular distance covered). 'Cross' occurs just in between two consecutive rounds. Thus hour hand crosses minute hand exactly 9-1=8 times. Same answer in ref. plane of hour-hand.

Source: Self

Enable Like and Comment Medium | Discrete Maths |

A group of students are sitting in a circle with the teacher in the center. They all have an even number of candies (not necessarily equal). When the teacher blows a whistle, each student passes half his candies to the student on his left. Then the students who have an odd number of candies obtain an extra candy from the teacher. Show that after a finite number of whistles, all students have the same number of candies.

Hint

Look at minimum & maximum count of candies.

Solution

1. The maximum number of candies held by a single student can never increase.

2. The minimum number of candies held by a single student always strictly increases, unless the student to his right also has the minimum number of candies, in which case the length of the longest consecutive segment of students who have minimum number of candies strictly decreases. Thus eventually the minimum has to strictly increase.

3. Since the minimum has to strictly increase in a finite number of steps and cannot go beyond the maximum, all the numbers must eventually be equal in atmost n(max-min) steps.

2. The minimum number of candies held by a single student always strictly increases, unless the student to his right also has the minimum number of candies, in which case the length of the longest consecutive segment of students who have minimum number of candies strictly decreases. Thus eventually the minimum has to strictly increase.

3. Since the minimum has to strictly increase in a finite number of steps and cannot go beyond the maximum, all the numbers must eventually be equal in atmost n(max-min) steps.

Source: puzzletweeter

Enable Like and Comment Medium | Discrete Maths |

A rectangular table has 100 coins with unit radius, placed on it such that none of the coins overlap, and it is impossible to place any more coins on the table without causing an overlap. Using this specific configuration, find a special configuration of 400 coins which covers the table with overlaps.

Covering means for every point on table there is a coin above it.

Covering means for every point on table there is a coin above it.

Hint

Create coins of radius 2 from the center of all coins. Notice that these coins fill up entire table, they are just bigger than what we are given.

Solution

Consider just one of these coins, with center P. It follows that the center Q of any other coin cannot lie within the coin of radius 2 with center P because it must be at least 2 units away. Thus, we construct all of these coins of radius 2, concurrent with each of the coins of radius 1. If the set of coins of radius 2 did not cover the rectangle entirely, then we could place a coin of radius 1 in this region, contradiction. Thus, the set of coins of radius 2 entirely covers the rectangle.

We now have 100 coins of radius 2 that entirely covers the rectangle. Scale this by a factor of 1/2 in both planar dimensions. Now we have 100 coins of radius 1 that entirely covers a rectangle that is a quadrant of the original rectangle. By placing four of these sets together, we get 400 coins of radius 1 that entirely covers the original rectangle.

We now have 100 coins of radius 2 that entirely covers the rectangle. Scale this by a factor of 1/2 in both planar dimensions. Now we have 100 coins of radius 1 that entirely covers a rectangle that is a quadrant of the original rectangle. By placing four of these sets together, we get 400 coins of radius 1 that entirely covers the original rectangle.

Source: CSEblog

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