Hard | Discrete Maths |

An optimist and a pessimist are examining a sequence of real numbers. The optimist remarks, ‘Oh jolly! The sum of any eight consecutive terms is positive!’ But the pessimist interjects, ‘Not so fast, the sum of any five consecutive terms is negative.’ Can they both be right? Atmost how large can this sequence be?

Hint

One easy proof is to construct rows, starting from ai, and satisfying one of the sum properties - optimist or pessimist. The column sum can follow the other property. Try making a contradiction.

Answer

11

Solution

Suppose such a sequence with terms {ai} has length 12

a1+a2+...+a5<0

a2+a3+...+a6<0

..

a8+a9+...+a12<0

Note that rows sums are negative and column sums are positive. This is contradiction to the total sum! So max length of sequence is 11. This sequence can be constructed but we wont cover the instruction. Example: 1 at positions 1,3,4,6,8,9,11 and -1.6 at positions 2,5,7,10.

a1+a2+...+a5<0

a2+a3+...+a6<0

..

a8+a9+...+a12<0

Note that rows sums are negative and column sums are positive. This is contradiction to the total sum! So max length of sequence is 11. This sequence can be constructed but we wont cover the instruction. Example: 1 at positions 1,3,4,6,8,9,11 and -1.6 at positions 2,5,7,10.

Source: Arthur Engel

Enable Like and Comment Hard | Discrete Maths |

We want to construct a structure made as follows: imagine that two long cylindrical pillars each with radius 1 intersect at right angles and their centers also intersect. What is the volume of this intersection?

Answer

16/3

Solution

If you cut the intersection by a horizontal plane at distance z from center, the cut will be a square with side-length 2*sqrt( 1-z^2). Integrate to get volume 16/3.

Another way is to imagine the largest possible sphere inscribed at the center of intersection. The sphere should have a radius of 1. At each cut perpendicular to the z-axis, the circle from the sphere is inscribed in the square from the intersection as well, So Area of cut-circle = (Pi/4)*Area of cut-square. This is true for all z, hence Volume of sphere = (Pi/4)*Volume of Intersection, this also gives 16/3

Another way is to imagine the largest possible sphere inscribed at the center of intersection. The sphere should have a radius of 1. At each cut perpendicular to the z-axis, the circle from the sphere is inscribed in the square from the intersection as well, So Area of cut-circle = (Pi/4)*Area of cut-square. This is true for all z, hence Volume of sphere = (Pi/4)*Volume of Intersection, this also gives 16/3

Hard | Probability |

You are given an urn with 100 balls (50 black and 50 white). You pick balls from urn one by one without replacements until all the balls are out. A black followed by a white or a white followed by a black is "a colour change". Calculate the expected number of colour changes if the balls are being picked randomly from the urn.

Hint

Linearity of expectation

Solution

There are 99 positions. Let X_i be a random variable taking value 1 if i_th position has a colour change and zero otherwise.

We have to find expected value of E[X_1 + X_2 + ... + X_99]

Since all X_i are equivalent, the answer is 99*E[X_i]

E[X_i] = ((50/100)*(50/99)+(50/100)*(50/99)) = 50/99

So, Answer is 50.

We have to find expected value of E[X_1 + X_2 + ... + X_99]

Since all X_i are equivalent, the answer is 99*E[X_i]

E[X_i] = ((50/100)*(50/99)+(50/100)*(50/99)) = 50/99

So, Answer is 50.

Source: Placement test

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Color Switches Weird Sequences Intersecting Pillars Consecutive sums Scaling a Square Difficulty Level

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