The snub dodecahedron, one of the Archimedean solids, has eighty faces which are triangles, and twelve pentagonal faces as well. In the view above, the pentagons are rendered invisible, allowing the interior to be viewed as the solid rotates.

The eighty triangles are of two types: the sixty yellow ones share an edge with a pentagon, and the twenty blue ones do not. If the blue triangles are also hidden, the “transparency” of this solid becomes even greater, as seen below.

Both of these images were created using *Stella 4d*, software you may try for free at this website.

Yep! We all had faced the same question for years when we were in schools/colleges. I had friends who were telling that they haven’t studied anything for the exam. They are simply coming and attending it, whatever they get in this exam is their own ability blah blah blah. I wondered how is this possible? Attending an exam without actually studying? Years passed stepped in to college and had new friends. Then once I tried the same sentence **“I haven’t studied anything for the exam. I’m simply coming and attending it, whatever I get in this exam is my own ability” **Lol in most of the cases this is how it works!

Technically one cannot top the exam without actually studying! They might be studying smarter but that doesn’t mean they are not at all studying for the exam! Every living being is different, everyone has their own abilities and capacities which is called individual differences. If all are same then how could you differ yourself from others? How could you differ yourself in the crowd?

here are some best ways to get rid of these feelings,

**1.Accept yourself: **

Tell yourself that you are unique. Believe in yourself. You have to be your first priority not others and their achievements. Without hard work nothing better can happen. Work hard, trust yourself and wait for the result.

**2. Stop Comparing:**

Individual differences differentiate one another. There people who are extremely bright, who are average and below average along with everyone has their own hidden abilities. Scoring 90+ is not the true value of education, applying the essence of education to real life is. Compare yourself with your previous result not with others!

**3. Do your best:**

If you spend your time thinking about others such as your classmates/friends, when will you think about yourself to improve you? You might not get enough time to do what you need to! Quit comparing and start working harder than.

**4. Proper study schedule:**

Though examination marks can never evaluate human being, but it is very important to get necessary grades and knowledge to lead a standard life. Don’t underestimate yourself, don’t get discouraged, don’t compare yourself to others, don’t give-up, just change your study habits. Applying proper study habits or effective learning can surely improve your academic grades. For more click here – Proper study schedule

Dont get fooled by others who make you feel things look easier. Without actually studying academics one cannot get 90+/100 marks as academics are new concepts which has to be learnt by all. So stop thinking alternative way to get answer to your question **“How to get 90+/100 as my classmates without studying?”**

*“Closures and Cavities in the Human Connectome”*

by Ann Sizemore, Chad Giusti, Richard F. Betzel, and Danielle S. Bassett.

]]>**(i) (16)**^{-0.75 }**x ****(64)**^{4/3 }

(16)^{-0.75 }x (64)^{4/3 }[16 = 2^{4} ]

= (24)^{-0.75 }x (24)^{4/3 }[64 = 2^{6} ]

= 2^{4} x ^{-3/4 }x 2^{6} x ^{4/3 }

= 2^{-3 }x 2^{8}

= 2^{5}

**= 32**

**(ii) (0.25)**^{0.5}**x ****(100)**^{-1/2}

(0.25)^{0.5} x (100)^{-1/2}

= (0.25)^{1/2} x (^{1}/_{100} )^{1/2}

= (0.25)^{1/2} x (^{1}/_{10}^{2} )^{1/2}

= (0.5) x ( ^{1}/_{10} )

= ( ^{5}/_{10} ) x ( ^{1}/_{10})

= ^{5}/_{100}

**= **^{𝟏}/_{𝟐𝟎}

**(iii) (6.25)**^{0.5}**x ****10**^{2}**x ****(100)**^{-1/2}**x ****(0.01)**^{-1}

= ( ^{625}/_{100} )^{1/2} x 10^{2} x (^{1}/_{10})^{1/2} x ( ^{1}/_{100} )^{-1}

= ( ^{25}/_{10} )^{1/2} x 10^{2} x (^{1}/_{10} )^{1/2} x (^{1}/_{100})^{-1}

= ( 25^{2}/_{10} )^{1/2} x 10^{2} x (^{1}/_{10}^{2} )^{1/2} x (100)^{1}

= ( ^{25}/_{10} ) x 100 x 1 10 x 100

**= 2500 **

**(iv) (3**^{-1/2}**x ****2**^{-1/3}**) ÷ (3**^{-3/4}**x ****2**^{-5/6}**) **

= 3^{−1/2}× 2^{−1/3} 3^{−3/4}× 2^{−5/6}

= 3^{-1/2+3/4} x 2^{-1/3+5/6 }

= 3^{1/4 }x 2^{3/6 }

= 3^{1/4 }x 2^{1/2 }

= 3^{1/4 }x 2^{1/4 }

=(3 × 2^{2})^{1/4}

= (3 × 4)^{1/4}

**= **𝟏𝟐^{1/4}

**Find the value of the expression.**

**[ 3 ^{1/3 }{5^{-1/2 }x 3^{-1/3 }x (225^{2})^{1/3}}^{1/2}]^{6}**

Solution:

= [ 3^{1/3 }{5^{-1/2}^{x}^{-1/2 }x 3^{-1/3}^{x}^{1/2 }x (225^{2})^{2/3}^{x}^{1/2}]^{6}

= [ 3^{1/3 }{5^{-1/4 }x 3^{1/6 }x (225^{2})^{-2/6}]^{6}

= [ 3^{1/3 }^{x }^{6 }{5^{1/4 }^{x }^{6 }x 3^{1/6 }^{x }^{6 }x (225^{2})^{-2/6 }^{x }^{6}]

= [ 32 x 5^{1/4 }^{x }^{6 }x 3^{1/6 }^{x }^{6 }x 225^{2-2/6 }^{x }^{6}]

= [32 x 5^{3/2} x 3^{1} x 225^{-2}]

= [32 x 5^{2/3} x 3^{1 }x 15^{-4}]

= [32 x 5^{-2/3 }x 31 x 3^{-4} x 5^{-4}]

= [3^{2 + 1 – 4 }x 5^{3/2} – 4]

= ^{1}/_{3}^{1} x ^{1}/_{5}^{5/2}

= ^{1}/_{3}^{1} x ^{1}/_{5}^{5/2}

= ^{1}/_{3}^{1} x ^{1}/_{√5}^{5}

= ^{1}/_{3} x ^{1}/_{√3125}

= ^{1}/_{3√3125}

**Simplify:**

**[{(3**^{5/2 }**x ****5**^{3/4}**) ÷ 2**^{-5/4}**} ÷ {16 / (5**^{2}**x ****2**^{1/4}**x ****3**^{1/2}**)}]**^{1/5}

Solution:

= [{(3^{5/2} × 5^{3/4}) ÷ 2^{−5/4}} ÷ {16÷ (5^{2}× 2^{1/4}× 3^{1/2})}] ^{1/5 }

= [{(3^{5/2} × 5^{3/4})/(_{2}^{−54}) ÷ ^{16}/(_{5}^{2}× _{2}^{1/4}× _{3}^{1/2})}] ^{1/5}

= [(3^{5/2} × 5^{3/4})/(_{2}^{−54}) ÷ (5^{2}× 2^{1/4}× 3^{1/2})/_{2}^{4}]^{1/5}

= [(3^{5/2} × 5^{3/4} × 5^{2/1} × 2^{1/4} × 3^{1/2})/(_{2}^{−5/4} × _{2}^{4 })]^{1/5}

= [2^{1/4} × 3^{5/2 + 1/2} × 5^{3/4 + 2/1}/_{2}^{−5/4 + 4/1}]^{1/5}

= 2^{1/4} × 3^{5/2 + 1/2} × 5^{11/4}/_{2}^{11/4}]^{1/5}

= [2^{1/4 }^{– }^{4/11} × 3^{3} × 5^{11/4 }] ^{1/5}

= [2^{10/4 × 1/5} × 3^{3 }^{× 1/5} × 5^{11/4 }^{+ }^{1/5}]

= 2^{−}^{1/2} × 3^{1/5} × 5^{11/20}

]]>

Further, let us consider the mixed strategies for Clinton, and for Trump. That is, Clinton predominantly attacks Trump with probability , and Trump predominantly attacks Clinton with probability .

Let us first deal with the general case of arbitrary payoffs, thus, generating the following payoff matrix:

That is, if Clinton attacks Trump and Trump attacks Clinton, the payoff to Clinton is , while the payoff to Trump is . If Clinton attacks Trump, and Trump ignores and discusses policy positions instead, the payoff to Clinton is , while the payoff to trump is . If Clinton discusses policy positions while Trump attacks, the payoff to Clinton is , while the payoff to Trump is , and if both candidates discuss policy positions instead of attacking each other, the payoff to them both will be and respectively.

With this information in hand, we can calculate the payoff to Clinton as:

while the payoff to Trump is:

With these payoff functions, we can compute each candidate’s **best response** to the other candidate by solving the following equations:

where indicates the best response strategy to a fixed strategy for the other player.

Solving these equations, we obtain the following:

If

then,

Clinton’s best response is to choose .

If

then,

Clinton’s best response is to choose .

If

then,

Clinton’s best response is to choose .

While for Trump, the best responses are computed as follows:

If

Trump’s best response is to choose .

If

Trump’s best response is to choose .

If

Trump’s best response is to choose .

**To demonstrate this, let us work out an example. Assume (for this example) that the payoffs for each candidate are to sway independent voters / voters that have not made up their minds. Further, let us assume that these voters are more interested in policy positions, and will take attacks negatively. Obviously, this is not necessarily true, and we have solved the general case above. We are just using the following payoff matrix for demonstration purposes:**

Using the above equations, we see that if , Clinton’s best response is to choose . While, if , Trump’s best response is to choose . That is, no matter what Trump’s strategy is, it is always Clinton’s best response to discuss policy positions. No matter what Clinton’s strategy is, it is always Trump’s best response to discuss policy positions as well. The two candidates’ payoff functions take the following form:

What this shows for example is that there is a Nash equilibrium of:

.

The expected payoffs for each candidate are evidently

.

**Let us work out an another example. This time, assume that if Clinton attacks Trump, she receives a payoff of , while if Trump attacks Clinton, he receives a payoff of . While, if Clinton discusses policy, while being attacked by Trump, she receives a payoff of , while Trump receives a payoff of . On the other hand, if Trump discusses policy while being attacked by Clinton, he receives a payoff , while Clinton receives a payoff of . If Clinton discusses policy, while Trump discusses policy, she receives a payoff of , while Trump receives a payoff of . The payoff matrix is evidently:**

In this case, if , then Clinton’s best response is to choose . While, if , then Trump’s best response is to choose . The Nash equilibrium is evidently

.

The expected payoffs for each candidate are evidently

.

In this example, even though it is the optimal strategy for each candidate to play a mixed strategy of 50% attack, 50% discuss policy, Clinton is expected to benefit, while Trump is expected to lose.

The interested reader is invited to experiment with different scenarios using the general results derived above.

]]>**Section: Quantitative Aptitude – Arithmetic Ability**

**Choose the correct option.**

**Q.1 ) A man is three (3) years older than his wife and four (4) times as old as his son. If the son attains an age of fifteen (15) years after three (3) years, what is the present age of the mother?**

**A: 60 years B: 51 years C: 48 years D: 45 years E: 65 years**

**Q.2 ) If Thursday was the day after the day before yesterday five days ago, what is the least number of days ago when sunday was three days before the day after tomorrow?**

**A: Two days ago B: Three days ago C: Four days ago D: Five days ago E: None of these**

**Q.3 ) A train started from station ‘A’ and proceeded towards station ‘B’ at a speed of 48 Km/hr. Forty-five minutes later another train started from station ‘B’ and proceeded towards station ‘A’ at 50 Km/hr. If the distance between the two stations is 232 Km, at what distance from station ‘A’ will the trains meet?**

**A: 132 Km B: 144 Km C: 108 Km D: 160 Km E: None of these**

**Q.4) If a number is five time as great as another number which is four less than forty, then the number is:**

**A: 220 B: 180 C: 144 D: 200 E: 196**

**Q.5 ) How many such 3’s are there in the following number sequence which are immediately Preceded by an odd number and immediately followed by an even number? 5 3 8 9 4 3 7 2 3 8 1 3 8 4 2 3 5 7 3 4 2 3 6**

**A: One B: Two C: Three D: Four E: More than four**

**Q.6 ) The sum of two consecutive number is 87. Which is the larger number?**

**A: 42 B: 43 C: 44 D: 45 E: 46**

**Q.7 ) If ratio of profit of A and B is 4:5 and they together invested Rs.90,000 then money invested by B is**

**A: Rs.48,000 B: Rs.45,000 C: Rs.30,000 D: Rs.50,000 **

**Q.8 ) If 10 litres of an oil of Rs.50 per litres be mixed with 5 litres of another oil of Rs.66 per litre then what is the rate of mixed oil per litre?**

**A: Rs.49.17 B: Rs.51.03 C: Rs.54.17 D: Rs.55.33 **

**Q.9 ) A runs twice as fast as B and gives B a start of 50m.How long should the racecourse be so that A and B might reach in the same time?**

**A: 75 m. B: 80 m. C: 150 m. D: 100 m. **

**Q.10) Sum of ages of father and his son is 60 years while difference is 30 years. what is father’s age?**

**A: 60 years B: 45 years C: 50 years D: 40 years **

**Q.11) The average of all odd numbers upto 100 is :**

**A: 49 B: 49.5 C: 50 D: 51 **

**Q.12 ) The largest 5-digit number exactly divisible by 91 is**

**A: 99921 B: 99918 C: 99981 D: 99971 E: None of these**

**Q.13 ) My age becomes half that of my brother`s if we simply add 2 years to his present age. If I am 25 years old today, my brother will be**

**A: 46 B: 48 C: 44 D: 36 E: 38**

**Q.14) The ratio between the present ages of A and B is 5 : 3 respectively. The ratio between A`s age 4 years ago and B`s age 4 years hence is 1 : 1. What is the ratio between A`s age 4 years hence and B`s age 4 years ago ?**

**A: 1:3 B: 2:1 C: 3:1 D: 4:1 E: None of these**

**Q.15) Find the sub-triplicate ratio of [8x]^3 :64y^3**

**A: x : 2y B: x^2 : 4y^2 C: 4y^2 : x^2 D: 2y : x **

** Section: Verbal Reasoning – General**

**In each of the following series determine the order of the letters. Then from the given options select the one which will complete the given series.**

**Q.1) DUE ,CVV,BWU,AXT,?**

**A: BYS B: CZV C: XYS D: ZYS **

**Which letter(s) in each of the following series is wrong or is misfit in the series ?**

**Q.2) Z T P K H F**

**A: Z B: P C: T D: F **

**Instructions for Question No. 3 :**

**In the following questions select the right option which indicates the correct code for the word or letter given in the question.**

**Q.3) If PHILOSOPHY is coded as HPLISOPOYH, ORNAMENTAL will be coded as :**

**A: ROANEMNTLA B: ONRAMNEALT C: ROANEMTNLA D: ROANEMNATL **

**Instructions for Question No. 4 to 5 :**

**In the following questions study the coded patterns and then select the right option from the given alternatives.**

**Q.4) In a certain code language,(1)’lo ni hie pun’stands for’he is drinking coke’;(2)’hol ful gui pun’stands for’she is eating food’;and(3)’ne ful ni lo’stands for ‘drinking coke and food’.Which of the following words is the code for ‘he’?**

**A: hie B: lo C: pun D: ni **

**Q.5) In a certain code language,’col tip mot’means’singing is appreciable’,’mot baj min’means ‘dancing is good’and’tip nop baj’means ‘singing and dancing’.Which of the following means ‘Good’in that code language?**

**A: mot B: min C: baj D: cannot be determined **

**Instructions for Question No. 6 to 7 :**

**In each of the following questions keenly study the relationship mentioned between the persons, and then from the given options select the right relationship as the answer.**

**Q.6) Pointing to Suman,Amit said,”He is my sister’s only brother’s son”.How is Suman related to Amit?**

**A: Grandson B: Son C: Nephew D: Cannot be determined E: None of these**

**Q.7 ) B is maternal uncle of A. C is maternal grandfather of B. D is grandson of C. How is D related to A?**

**A: Maternal uncle B: Cousin brother C: Nephew D: Maternal grandfather E: None of these**

**Instructions for Question No. 8 to 9 :**

**Choose the correct option.**

**Q.8 ) In a telephone directory,which of the following names will appear in the middle?**

**A: Randhir B: Randesh C: Rama D: Ramesh E: Renmurthi**

**Q.9 ) Which one would be a meaningful order of the following?**

**Snake 2. Grass 3. Eagle 4. Frog 5. Insect**

**A: 3 2 1 4 5 B: 5 2 1 4 3 C: 2 5 4 1 3 D: 2 4 5 3 1 E: 2 4 5 1 3**

**Instructions for Question No. 10 :**

**Find out the one word among the options which cannot be formed by using the letters of the word as given in each question.**

**Q.10) COMMENTATOR**

**A: COMMON B: MOMENT C: COSMOS D: TART**

**Instructions for Question No. 11 to 12 :**

**In the following questions select the number(s) from the given options for completing the given series.**

**Q.11) 3,5,9,15,23,?,45**

**A: 37 B: 35 C: 31 D: 33 **

**Q.12) 0,2,10,30,68,?**

**A: 125 B: 120 C: 175 D: 130 **

**Instructions for Question No. 13 :**

**In the following questions, select the number from the given options which follows the same relationship as shared between the first two numbers.**

**Q.) 13 92 : 69 :: 46 :?**

**A: 57 B: 23 C: 31 D: 19 **

**Instructions for Question No. 14 to 15 :**

**In the questions given below establish the relationship between two words. Then from the given option select the one which has the same relationship as of the given two words.**

**Q.14 ) SYMPATHY : EQUITY**

**A: Common : Exclusive B: Certain : Precise C: Snub : Admire D: Beloved : Paltry E: Unity : Strife**

**Q.15) JUPITER : PLANET**

**A: Earth : Orbit B: Saturn : Sun C: Cosmos : Star D: Moon : Satellite E: Milky way : Pole**

**Instructions for Question No. 16 to 20 :**

**In each of the following questions four words are alike in some manner. Spot the odd one out.**

**Q.16) **

**A: Van B: Aeroplane C: Helicopter D: Transport E: Scooter**

**Q.17) **

**A: Horse B: Cow C: Donkey D: Pig E: Turkey**

**Q.18) **

**A: Picture B: Poster C: Photograph D: Scenery E: Painting**

**Q.19) **

**A: Kite B: Vulture C: Hawk D: Bat E: Duck**

**Q.20) **

**A: Tool B: Rule C: Cool D: Pool E: Fool**

** **

**Instructions for Question No. 21 :**

**Directions : Choose the correct answer :**

**Q.21) BLOOD is related to ARTERY in the same way as WATER is related to**

**A: Vein B: Pipe C: Aorta D: Straw **

**Instructions for Question No. 22 :**

**Directions : Find out the wrong number in the series**

**Q.22) 1,2,8,33,148,760,4626**

**A: 2 B: 8 C: 33 D: 148 E: 760**

**Instructions for Question No. 23 :**

**Directions : Insert the missing number**

**Q.23) 10,5,13,10,16,20,19,(……)**

**A: 22 B: 40 C: 38 D: 23 **

**Instructions for Question No. 24 :**

**Directions : In each of the following questions, there is some relationship between the two terms to the left of : : and the same relationship holds between the two terms to its right. Also, in each question, one term either to the right of : : of to the left of it is missing. This term is given as one of the alternatives given below each question. Find out this term.**

**Q.24 ) KORT is related to PJWO in the same way as FINR is related to …….?……….**

**A: KCSM B: KDSM C: JSMR D: JCRN **

**Instructions for Question No. 25 :**

**Directions: In each questions below, two pairs of numbers are given but one number in the second pair is missing. Identify the relationship between the two numbers in the first pair and find the missing number in the second pair such that the numbers in second pair also follow the same relationship.**

**Q.25) 25: 125 :: 64 : ?**

**A: 521 B: 512 C: 513 D: 520 **

**Instructions for Question No. 26:**

**Directions: Each of the following questions has a pair of CAPITALIZED words followed by four pairs of words. Choose the pair of words which best expresses the relationship similar to that expressed in the capitalized pair.**

**Q.26) SPEAR : DART**

**A: knife : sword B: door : window C: mountain : molehill D: cannon : gun **

**Instructions for Question No. 27 :**

**Directions : In each of these questions, only one of the four alternatives given under each question satisfies the same relationship as is found between two terms on the left side of the sign (::). Find the correct alternative to fill in the right side of the sign (::).**

**Q.27) Spasm : Pain :: ____________________ : ____________________**

**A: Touch : Feel B: Fast : Quick C: Flash : Light D: Water : Wet**

**Instructions for Question No. 28 to 29 :**

**Directions : In each of the following questions, five words have been given out of which four are alike in some manner, while the fifth one is different. Choose out the odd one.**

**Q.28) Choose the odd one out**

**A: Flute B: Guitar C: Sitar D: Violin E: Veena**

**Q.29) Choose the odd one out**

**A: Sahara B: Thar C: Gobi D: Sunderban E: Kalahari**

**Instructions for Question No. 30 to 31 :**

**Directions : In each of the following questions, four words have been given, out of which three are alike in some manner and the fourth one is different. Choose out the odd one.**

**Q.30) Choose the odd one out**

**A: Platform B: Dock C: Bus-stand D: Park**

**Instructions for Question No. 31 to 30 :**

**Directions : Choose the odd numeral pair/group in each of the following questions:**

**Q.31 ) **

**A: 48:134 B: 40:110 C: 18:48 D: 30:80 **

**Instructions for Question No. 32 to 31 :**

**Directions : In each of the following questions, certain pairs of words are given, out of which the words in all pairs except one, bear a certain common relationship. Choose the pair in which the words are differently related.**

**Q.32 **

**A: Mercury : Sun B: Moon : Earth C: Star : Galaxy D: Wheel : Axle **

**Instructions for Question No. 33 to 36 :**

**Directions: In each of the following questions, five words have been given out of which four are alike in some manner, while the fifth one is different. Choose out the odd one.**

**Q.33) Choose the odd one out**

**A: Feathers B: Tentacles C: Scales D: Pseudopodia E: Flagella**

**Q.34) Choose the odd one out**

**A: Kiwi B: Eagle C: Emu D: Penguin E: Ostrich**

**Q.35) Choose the odd one out**

**A: Mustard B: Rapeseed C: Sesame D: Cashewnut E: Groundnut**

**Q.36) Choose the odd one out**

**A: Indigo B: Orange C: Yellow D: Pink E: Green**

**Instructions for Question No. 37 to 38 :**

**Directions : In each of the following questions, four words have been given, out of which three are alike in some manner and the fourth one is different. Choose out the odd one.**

**Q.37) Choose the odd one out**

**A: Kleptomania B: Schizophrenia C: Agoraphobia D: Alzheimer’s disease**

**Instructions for Question No. 38 :**

**Directions : Read the information carefully and answer the questions:**

**M,N,O,P,Q,R,T and V are eight sports persons. Three of them play cricket, two of them play hockey, one play badminton and two play volleyball. There are three lady members among them. No lady either plays hockey or badminton and at least one lady plays the remaining two games. None of the games is played by only ladies. P plays badminton Q`s sister V plays cricket. M plays volleyball with T. N plays cricket. O and T are sisters.**

**Q.38 ) Which of the following persons play cricket ?**

**A: NRV B: NQV C: MPV D: Data inadequate E: None of these**

**Instructions for Question No. 39 :**

**Directions : Read the following information to answer the given questions:**

**A bag contains coins of four different denominations, viz. 1 rupee, 50-paise, 25-paise and 10-paise. There are as many 50-paise coins as the value of 25-paise coins in rupees. The value of 1-rupee coins is 5 times the value of 50-paise coins. The ratio of the number of 10-paise coins to that of 1-rupee coins is 4 : 3, while the total number of coins in the bag is 325.**

**Q.39 How many 10-paise coins are there?**

**A: 25 B: 50 C: 75 D: 100 **

**Instructions for Question No. 40 :**

**Directions : study the following information carefully and answer the question.**

**B, M, T, R, K, H and D are travelling in a train compartment with III-tier sleeper berth. Each of them is of a different profession of engineer,doctor,architect,pharmacist,lawyer,journalist and pathologist. They occupied two lower berths,three middle berths and two upper berths. B,the engineer is not on the upper berth. The architect is the only other person who occupies the same type of berth so that B,M and H are not on the middle berth and their professions are pathologist and lawyer respectively. T is a pharmacist. D is neither a journalist nor an architect. K occupies same type of berths as that of the doctor.**

**Q.40) What is D`s profession ?**

**A: Doctor B: Engineer C: Lawyer D: Pharmacist E: Data inadequate**

** Section: Verbal Reasoning – Paragraph**

**Paragraph : 1**

**(i) A,B,C,D,E,F and G are sitting on a wall and all of them are facing east. (ii) C is on the immediate right of D. (iii) B is at an extreme end and has E as his neighbour. (iv) G is between E and F. (v) D is sitting third from the south end.**

**Please refer paragraph 1 for the next question**

**Q.1 Which of the following pairs of people are sitting at the extreme ends?**

**A: AE B: AB C: FB D: CB E: Cannot be determined**

**Please refer paragraph 1 for the next question**

**Q.2 Immediately between which of the following pairs of people is D sitting?**

**A: CE B: AC C: CF D: AF E: None of these**

**Please refer paragraph 1 for the next question**

**Q.3 Name the person who should change places with C such that he gets the third place from the north end?**

**A: G B: F C: E D: Can’t be determined E: None of these**

**Please refer paragraph 1 for the next question**

**Q.4 Who is sitting to the right of E?**

**A: F B: D C: C D: A E: None of these**

**Please refer paragraph 1 for the next question**

**Q.5 Which of the following pairs of people are sitting next to F?**

**A: EG B: DC C: ED D: GD E: None of these**

** **

** **

** **

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**(i) A = {1, 2, 3, 4, 8, 9}, B = {1, 2, 3, 5} **

**(ii) A = {1, 2, 3, 4, 5}, B = {4, 5, 7, 9} **

**(iii) A = {1,2,3}, B = {4, 5, 6} **

**(iv) A = {1, 2, 3, ,4 ,5}, B = {1, 3, 5} **

**(v) A = {a, b, c, d}, B = {b, d, e, f} **

Solution:

(i) A U B = {1, 2, 3, 4, 5, 8, 9}

(ii) A U B = {1, 2, 3, 4, 5, 7, 9}

(iii) A U B = {1, 2, 3, 4, 5, 6}

(iv) A U B = {1, 2, 3, 4, 5}

(v) A U B = {a, b, c, d, e, f}

**Find the intersection of A and B, and respect it by Venn diagram:**

**(i) A = {a, c, d, e}, B = {b, d, e, f} **

**(ii) A = {1, 2, 4, 5}, B = {2, 5, 7, 9} **

**(iii) A = {1, 3, 5, 7}, B = {2, 5, 7, 10, 12} **

**(iv) A = {1, 2, 3}, B = {5, 4, 7} **

**(v) A = {a, b, c}, B = {1, 2, 9}**

Solution:

(i) A ∩ B = {d, e}

(ii) A ∩ B = {2, 5}

(iii) A ∩ B = {5, 7}

(iv) A ∩ B = { }

**Find A**∩**B and A**∩**B when:**

**(i) A is the set of all prime numbers and B is the set of all composite natural numbers: **

**(ii) A is the set of all positive real numbers and B is the set of all negative real numbers: **

**(iii) A = N and B = Z: **

**(iv) A = {x /x **∩ **Z and x is divisible by 6} and **

**B = {x / x **∩ **Z and x is divisible by 15} **

**(v) A is the set of all points in the plane with integer coordinate and B is the set of all points with rational coordinates **

Solution:

(i) A = {2, 3, 5, 7….}

B = {1, 4, 6….}

A U B = {1, 2, 3, 4} = N

A ∩ B = { }

(ii) A = R+

B = R+

A U B = R – {10}

i.e. A U B ={set of non zero real numbers}

A U B = { }

(iii) A = N B = N

A U B = Z

A ∩ B = N

(iv) A U B = {x / x € Z and x is divisible by 6 and 15} and

A ∩ B = {x / x € Z and x is divisible by 30}

[LCM of 6 and 15 = 30]

(v) A U B = the set of all points with rational co-ordinates = B.

A ∩ B = the set of all points with rational co-ordinates = A.

**Give examples to show that**

**(i) A **U **A = A and A **∩ **A = A **

**(ii) If A **⸦ **B, then A **U **B = B and A **∩ **B =A. can you prove these statements formally? **

Solution:

(i) If A = {2 4 6 8}

Then A U A = {2, 4, 6, 8……} = A

A ∩ A = {2, 4, 6, 8…….} = A

Hence A U A = A and A ∩ A = A

(ii) A = {1, 3, 5, 7, 9……}

B = {1, 2, 3, 4, 5……}

We see that A ⸦ B

A U B = {1 2 3 4…..} = B

A ∩ B = B

A ∩ B = {1, 3, 5……} = A

A ∩ B = A

**What is A**U Φ**and A**∩ Φ**for a set A?**

**Solution:**

** **AUΦ = A ; A ∩Φ = Φ

]]>

The next step is to ask whether these singularities rob general relativity of its predictive power. The ‘cosmic censorship hypothesis’, proposed by Penrose in 1969, claims they do not.

In this final post I’ll talk about cosmic censorship, and conclude with some big questions… and a place where you can get all these posts in a single file.

To say what we want to rule out, we must first think about what behaviors we consider acceptable. Consider first a black hole formed by the collapse of a star. According to general relativity, matter can fall into this black hole and ‘hit the singularity’ in a finite amount of proper time, but nothing can come out of the singularity.

The time-reversed version of a black hole, called a ‘white hole’, is often considered more disturbing. White holes have never been seen, but they are mathematically valid solutions of Einstein’s equation. In a white hole, matter can come *out* of the singularity, but nothing can fall *in*. Naively, this seems to imply that the future is unpredictable given knowledge of the past. Of course, the same logic applied to black holes would say the past is unpredictable given knowledge of the future.

If white holes are disturbing, perhaps the Big Bang should be more so. In the usual solutions of general relativity describing the Big Bang, *all matter in the universe* comes out of a singularity! More precisely, if one follows any timelike geodesic back into the past, it becomes undefined after a finite amount of proper time. Naively, this may seem a massive violation of predictability: in this scenario, the whole universe ‘sprang out of nothing’ about 14 billion years ago.

However, in all three examples so far—astrophysical black holes, their time-reversed versions and the Big Bang—spacetime is globally hyperbolic. I explained what this means last time. In simple terms, it means we can specify initial data at one moment in time and use the laws of physics to predict the future (and past) throughout all of spacetime. How is this compatible with the naive intuition that a singularity causes a failure of predictability?

For any globally hyperbolic spacetime one can find a smoothly varying family of Cauchy surfaces ($t \in \mathbb{R}$) such that each point of lies on exactly one of these surfaces. This amounts to a way of chopping spacetime into ‘slices of space’ for various choices of the ‘time’ parameter For an astrophysical black hole, the singularity is in the future of all these surfaces. That is, an incomplete timelike or null geodesic must go through all these surfaces before it becomes undefined. Similarly, for a white hole or the Big Bang, the singularity is in the past of all these surfaces. In either case, the singularity cannot interfere with our predictions of what occurs in spacetime.

A more challenging example is posed by the Kerr–Newman solution of Einstein’s equation coupled to the vacuum Maxwell equations. When

this solution describes a rotating charged black hole with mass charge and angular momentum in units where However, an electron violates this inequality. In 1968, Brandon Carter pointed out that if the electron were described by the Kerr–Newman solution, it would have a gyromagnetic ratio of much closer to the true answer than a classical spinning sphere of charge, which gives But since

this solution gives a spacetime that is not globally hyperbolic: it has closed timelike curves! It also contains a ‘naked singularity’. Roughly speaking, this is a singularity that can be seen by arbitrarily faraway observers in a spacetime whose geometry asymptotically approaches that of Minkowski spacetime. The existence of a naked singularity implies a failure of global hyperbolicity.

The cosmic censorship hypothesis comes in a number of forms. The original version due to Penrose is now called ‘weak cosmic censorship’. It asserts that in a spacetime whose geometry asymptotically approaches that of Minkowski spacetime, gravitational collapse cannot produce a naked singularity.

In 1991, Preskill and Thorne made a bet against Hawking in which they claimed that weak cosmic censorship was false. Hawking conceded this bet in 1997 when a counterexample was found. This features finely-tuned infalling matter poised right on the brink of forming a black hole. It *almost* creates a region from which light cannot escape—but not quite. Instead, it creates a naked singularity!

Given the delicate nature of this construction, Hawking did not give up. Instead he made a second bet, which says that weak cosmic censorshop holds ‘generically’ — that is, for an open dense set of initial conditions.

In 1999, Christodoulou proved that for spherically symmetric solutions of Einstein’s equation coupled to a massless scalar field, weak cosmic censorship holds generically. While spherical symmetry is a very restrictive assumption, this result is a good example of how, with plenty of work, we can make progress in rigorously settling the questions raised by general relativity.

Indeed, Christodoulou has been a leader in this area. For example, the vacuum Einstein equations have solutions describing gravitational waves, much as the vacuum Maxwell equations have solutions describing electromagnetic waves. However, gravitational waves can actually form black holes when they collide. This raises the question of the stability of Minkowski spacetime. Must sufficiently small perturbations of the Minkowski metric go away in the form of gravitational radiation, or can tiny wrinkles in the fabric of spacetime somehow amplify themselves and cause trouble—perhaps even a singularity? In 1993, together with Klainerman, Christodoulou proved that Minkowski spacetime is indeed stable. Their proof fills a 514-page book.

In 2008, Christodoulou completed an even longer rigorous study of the formation of black holes. This can be seen as a vastly more detailed look at questions which Penrose’s original singularity theorem addressed in a general, preliminary way. Nonetheless, there is much left to be done to understand the behavior of singularities in general relativity.

In this series of posts, we’ve seen that in every major theory of physics, challenging mathematical questions arise from the assumption that spacetime is a continuum. The continuum threatens us with infinities! Do these infinities threaten our ability to extract predictions from these theories—or even our ability to formulate these theories in a precise way?

We can answer these questions, but only with hard work. Is this a sign that we are somehow on the wrong track? Is the continuum as we understand it only an approximation to some deeper model of spacetime? Only time will tell. Nature is providing us with plenty of clues, but it will take patience to read them correctly.

To delve deeper into singularities and cosmic censorship, try this delightful book, which is free online:

• John Earman, *Bangs, Crunches, Whimpers and Shrieks: Singularities and Acausalities in Relativistic Spacetimes*, Oxford U. Press, Oxford, 1993.

To read this whole series of posts in one place, with lots more references and links, see:

• John Baez, Struggles with the continuum.

]]>The *residue *of an integer n modulo an integer d > 1 is the remainder r left when n is divided by d. That is, if n = dq + r for integers q and r with 0 __<__ r < d, we write for the residue of n modulo d. Show that the residue modulo 7 of a (large) integer n can be found by separating the integer into 3-digit blocks .(Note that b(s) may have 1, 2, or 3 digits, but every other block must have exactly three digits.) Then the residue modulo 7 of n is the same as the residue modulo 7 of . For example,

.

Explain why this works and show that the same trick works for residues modulo 13.

Visit our main site for more breaking news http://wearechange.org/

]]>Using data science / machine learning methodologies, it basically showed that the most important factors in characterizing a team’s playoff eligibility are the opponent field goal percentage and the opponent points per game. This seems to suggest that defensive factors as opposed to offensive factors are the most important characteristics shared among NBA playoff teams. It was also shown that championship teams must be able to have very strong defensive characteristics, in particular, strong perimeter defense characteristics in combination with an effective half-court offense that generates high-percentage two-point shots. A key part of this offensive strategy must also be the ability to draw fouls.

Some people have commented that despite this, teams who frequently attempt three point shots still can be considered to have an efficient offense as doing so leads to better rebounding, floor spacing, and higher percentage shots. We show below that this is not true. Looking at the last 16 years of all NBA teams (using the same data we used in the paper), we performed a correlation analysis of an individual NBA team’s 3-point attempts per game and other relevant variables, and discovered:

One sees that there is very little correlation between a team’s 3-point attempts per game and 2-point percentage, free throws, free throw attempts, and offensive rebounds. In fact, at best, there is a somewhat “medium” anti-correlation between 3-point attempts per game and a team’s 2-point attempts per game.

As you all know, I have been very busy with putting final touches to my manuscript on geometry and number theory. I am planning to call it Elements. The publisher says that he does not anticipate much demand for this series, except maybe some orders from libraries. I am, of course, not bothered by it and feel that only time can tell the true impact of my work. Printing press has not been invented yet, so publishing book is not that easy. Someones needs to copy each page and this is a laborious task. Anyways, I want to take my mind off from geometry and start thinking about numbers (you have to wait till my Book IX comes out, but I wanted to give my followers a sneak-peek).

I was in my office today (which by the way is the court of Ptolemy I) and my colleagues were asking if I had anything interesting on prime numbers. There were all sorts of questions: is there a way to generate all prime numbers, what is the nth prime, how many prime numbers are less than given a number, what is Amazon Prime and so on? These are all very deep questions and I am not sure I can answer all of these at this point.

Since prime numbers (II, III, V, VII,XI,XIII and so on) are building blocks, it does make sense to ask how many of them are there. I started asking myself if I can start with any finite list of prime numbers and then show that there exists a prime number not on the list. If this is the case, then there must be infinitely many primes . To be honest, I suspected there are an infinite number of primes and this suspicion led me to this methodology. I think, in general, this is a good line of attack for problems of this sort. Given any list of some numbers with certain property, it is always good to ask oneself, if we can generate another number with that property from this list.

Start with any finite list of prime numbers. Multiply all the prime numbers in the list and add I (the number one and not Euclid) to the product. Let us call this number P. Now, P is not divisible by any of the prime numbers on the list as it leaves a reminder of I when divided by any prime on the list. So, either P should be prime or should have another prime number not on the list as a factor. Either way, we exhibited another prime number not on the list. Since, the above logic applies to any finite list of primes, we can see that there must be an infinite number of primes.

*For the curious mind, I should mention that we never wrote the dates as BC back in our time. This is just for your convenience.

PS: It has been brought to my notice that there are other ways to prove the above assertion (see Euclid’s theorem). However, it seems that my proof has endured owing to its beauty. When Dr.Watson asked Sherlock what is so great about my proof, Holmes replied that the elegance of the proof lies in the fact it is ‘elementary, my dear Watson’.

]]>

His dawn to dusk warbling has also been… **(CONTINUE)**

**Factorial is an important topic in various competitive exams based quantitative aptitude.Factorial topic finds its application in Permutation and Combinations , Probability , Number theory ,etc…**

**Factorial number or n! is the product of n consecutive natural numbers starting from 1 to n.**

**The notation of factorial was introduced by Christian Kramp in 1808.**

** For example : 7 ! =7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040 **

**Here are some factorial values of some values.It is good if you can remember (because they come handy in fast calculation) .Try to keep in mind these values similar to multiplication table you did in your schools.**

**1! = 1**

**2! = 2**

**3! = 6**

**4!= 24**

**5!=120**

**6!= 720**

**7! = 5040**

**8! = 40320**

**Some of the commonly asked questions based on the concept of factorials and number of trailing zeros in the various exams :**

**1.Highest power of a number in a factorial or in a product**

** 2. Number of zeros in the end of a factorial or a product**

** 3.Number of factors of any factorial**

** 4. Highest power of a number in a factorial or in a product.**

**” Highest power of a prime number in a factorial or in a product “**

**Let us first understand what do we mean by ” Highest Power ” ?**

**Consider a number N= p^2 k^ 1.**

**Here the highest power of “p” in N will be 2 and the highest power of “k” in N will be 1.**

**To find the highest Power of factor in the given factorial can be obtained by continuously dividing the number by whose highest power of factor is to be found & we will be approximating the quotients obtained after each division and adding up ! This result after the adding all quotients , we will be obtaining the highest power of a number in the given factorial.**

**Lets look at some examples..**

**Q. 1 ) Find the highest power of 5 in 30!**

**Solution: 30/5= 6 ; 6/5=1;**

**Adding the quotients, its 6+1=7**

**So highest power of 5 in 30! = 7**

**Q. 2 ) Find the highest power of 2 in 50 !**

**Solution: 50/2= 25 ; 25/2=12; 12/2 =6 ; 6/2 = 3 ; 3/2 = 1**

**Adding the quotients, its 25+12+6+3 +1= 47**

**So highest power of 2 in 50! = 47**

**Q.3) ****What is the highest power of 5 such that the number divided 125! ?**

**Solution : ****[125/5] + [125/25] + [125/125] = 25+5+1 = 31**

** Highest power of 5 in 125! = 31**

** Hence 5^31 divided 125!**

**Q.4) ****Find the highest power of 3 in 100!**

**Solution :**

** = [100/3] + [100/3 ^{2}] + [100/3^{3}] + [100/3^{4}] + [100/3^{5}] + …**

** = 33 + 11 + 3 + 1 + 0 (from here on greatest integer function evaluates to zero)**

** = 48.**

**Q.5) Find the highest power of 2 in 100!**

**Solution :- = ****[100/2]+[100/2^2]+[100/2^3]+[100/2^4]+[100/2^5]+[100/2^6]**

** = 50+25+12+6+3+1**

** =97.**

**” Highest number of a composite number in the given factorial value”**

**Finding the highest power of composite number in the given factorial value is as follows.**

** Step 1 : Factorize the composite number into prime factorization ( or canonical form) .**

** Step 2 : Find the highest power of all the prime numbers in that factorial using the previous method.**

** Step 3 : Take the least power.**

**Q.1 ) What is the highest power of 10 in 100!**

**Solution: Lets first factorize 10 as 5 x 2.**

** Step 1 :Highest power of 5 in 100! =24**

** Step 2 : Highest power of 2 in 100! =97 ( as already obtained ) .**

** Step 3 : Therefore, the answer will be 24, because to get a 10, you need a pair of 2 and 5, and only 24 such pairs are available. ****So take the lesser value i.e. 24 is the answer.**

**Q.2 ) What is the highest power of 12 in 100!**

**Solution: Lets first factorize 12 as 2^2 x 3.**

** Step 1 :Highest power of 2 in 100! = 97 ( for method look at above examples)**

**So , the highest power of 4 ( i.e 2^2 ) is 97 /2 = 48**

** Step 2 : Highest power of 3 in 100! = 48 ( as already obtained ) .**

** Step 3 : Therefore, the answer will be 48, because to get a 10, you need a pair of 2 and 5, and only 48 such pairs are available. ****So take the lesser value i.e. 48 is the answer.**

**Q.3 ) What is the highest power of 12 in 30!**

**Solution: Lets first factorize 12 as 3 x 2^2 .**

** Step 1 :Highest power of 3 in 30! **

**(30/3) + ( 30/9) + ( 30/27) +……**

** = 14**

** Step 2 : Highest power of 2 in 30! = ( as already obtained ) .**

= **[30/2] + [30/4] + [30/8] + [30/16]**

**= 26**

** Therefore , the highest power of ( 2^2) i.e 4 is 13**

** Step 3 : Therefore, the answer will be 13, because to get a 10, you need a pair of 2 and 5, and only 13 such pairs are available. ****So take the lesser value i.e. 13 is the answer.**

**Q.4) What is the highest power of 72 in 100!**

**Solution: Lets first factorize 72 as 3^2 x 2^3 .**

** Step 1 :Highest power of 3 in 100! **

**(100/3) + ( 100/9) + ( 100/27) +……**

** = 48**

** But , here the power of 3 is 2 . Therefore highest power of 3^2 (i.e 9) is 48/2 = 24.**

** Step 2 : Highest power of 2 in 30! = ( as already obtained ) .**

= **[100/2] + [100/4] + [100/8] + [100/16]+…..**

**=97**

** Therefore , the highest power of ( 2^3) i.e 8 is 97/3 = 32.**

** Step 3 : Therefore, the answer will be 24, because to get a 10, you need a pair of 2 and 5, and only 24 such pairs are available. ****So take the lesser value i.e. 24 is the answer.**

** ” Number of zeros in the end of a factorial value or a product”**

**Find the number of trailing zeros ( zeros at the end ) of the factorial value is simple as discussed earlier. We can say it as special case of the above problems.**

** Number of zeros in the end depends on the number of 10s( i.e. effectively, on the number of 5s).**

** In this method , we need to find the highest power of 5 in the factorial value by successive divison.**

**Q.1) Find the number of zeros in 30! .**

** Solution : [30/5] + [30/25]**

** = 6+1 = 7**

**Q.2) Find the number of zeros in 100! .**

** Solution : [100/5] + [100/25] + … **

** = 20 + 4 + 0 + … = 24.**

**Q.3) Find the number of zeros in 250! .**

** Solution : ****[ 250 / 5 ] + [ 250 / 5^2 ] + [ 250 / 5^3 ]**

** = 50 + 10 + 2**

** = 62.**

**Q.4) Find the number of zeros in 15! .**

** Solution : [15/5] + [15/25] + …**

** = 3 + … **

** =3**

**Above questions were related to the base 10 system , i,e they need 5 and 2 to get 10.**

** Now , what if it is other than base 10 . Like base 7 , base 13 ,etc….**

**Lets take base N as general one and learn the concept**

**In base N system, number of zeroes in the end is nothing but the highest power of N in that product.**

**Q.1) Find the highest power of 7 in 100!.**

**Solution: ****[100/7] + [100/49] + … = 14 + 2 = 16.**

** There will be 16 zeros at the end of 100! in base 7**

**Q.2) ****Find the trailing zeros in 13! as in base 10 .**

**Solution: ****[13/5] + [13/25] + … = 2 +.. = 2**

** There will be 2 zeros at the end of 13! in base 10**

**Q.3) ****Find the number of zeros in the end of 15! as in the base 12.**

**Solution: ** **Highest power of 2 in 15!**

**= 15/2=7**

**7/2=3**

**3/2=1**

** Total =11.**

**Highest power of 2 ^{2} (i.e 4) = 11/2=5**

**Highest power of 3 in 15!= 5**

**Thus, the highest power of 12 in 15! = 5**

**Some more examples :-**

**Q.1) Find the total no of divisors of 17!**

**Solution :- ****Prime factors are 2,3,5,7,11,13,17.**

**Highest power of 2 =[17/2] + [ 17/2^2] + [ 17/2^3] + [17/2^4]**

**8+4+2+1 = 17.**

**Highest power of 3 = [17/3] + [17/3^2]**

**5+1 = 6.**

**Highest power of 5 = [17/5] + [17/5^2] = 3.**

**Highest power of 7 = [17/7] + [17/7^2] = 2.**

**Highest power of 11 = [17/11] = 1.**

**Highest power of 13 = [17/13] = 1.**

**Highest power of 17 = [17/17] = 1.**

**17! = 2^17 x 3^6 x 5^3 x 7^2 x 11 x 13 x 17**

**Factors = (17+1)(6+1)(3+1)(2+1)(1+1)(1+1)(1+1)**

**= 18 x 7 x 4 x 3 x 2 x 2 x 2**

**= 126 x 96**

**= 12096**

**Q.2) Find the least number which has highest power of 6 as 23.**

**Solution : **

** 23 = ( n/6) +(n/ 6^2) +…**

** 23 = (n/6) +( n/36) **

** By trail and error or by solving the above equation , we get 121 as the answer.**

**Wilson’s theorem**

**For every prime number p, (p-1)! + 1 is divisible by p**

Consider for example: -p =7

According to Wilson’s theorem, 7 is prime only if (6!) + 1 is divisible by 7 .

some useful results derived from Wilson’s theorem

**Remainder [ (p-1)!/p] = p – 1 ( where p is a prime number )**

Remainder [ 6!/7 ] = 7; Remainder [100!/101] = 100 and so on…

**Remainder [ (p-2)!/p] = 1 ( where p is a prime number )**

Remainder [ 5!/7 ] = 1; Remainder [99!/101] = 1

**“To find the non zero digit from right** “

**If we have to find first non zero digit from right in any number n!**

**First of all divide n by 5 and represent in the form shown below and follow next steps.. **

** n = 5 x p + k.**

** Rightmost Digit in ‘n! ‘ is given as 2^p x p! x k! .**

** **

**Now , let’s look into examples to understand the concept.**

**Q.1) Find the rightmost digit in 6!**

** Solution :**

** 6 = 5 x 1 +1 **

** Here , n = 6 , p = 1 , k =1**

** Right most digit = 2^p x p! x k !**

** = 2^1 x (1!) x( 1!) **

** =2**

**Q.2) Find the rightmost digit in 11!**

** Solution :**

** 11 = 5 x 2 +1 **

** Here , n = 11, p = 2 , k =1**

** Right most digit = 2^p x p! x k !**

** = 2^2 x (2!) x( 1!) **

** =8**

**Q.3) Find the rightmost digit in 9 !**

** Solution :**

** 9 = 5 x 1 +4 **

** Here , n = 9 , p = 1 , k =4**

** Right most digit = 2^p x p! x k !**

** = 2^1 x (1!) x( 4!) **

** =8**

**Q.4) Find the rightmost digit in 24 !**

** Solution :**

** 24= 5 x 4 +4 **

** Here , n = 24 , p = 4, k =4**

** Right most digit = 2^p x p! x k !**

** = 2^4 x (4!) x( 4!) **

** = 16 x 24 x24**

= **xxx6 ( Check on Wolfram Alpha)**

** ( Don’t waste your time in finding the whole answer, you will not be rewarded for the value of it).**

]]>

Solution:

**Definition: **A set is a collection of well defined objects. The objects of a set are called elements or members of the set.

**Examples: **

- a) The collection set of all prime numbers between 100 and 200.
- b) The collection of all planets in the universe.
- c) The collection of all fair people in the city

Here (a) and (b) are examples of sets but (c) is not one cannot define fair.

**Which of the following collection are sets?**

**(a) All the students of your school. **

**(b) Members of Indian parliament. **

**(c) The colures of rainbow. **

**(d) The people of Karnataka having green ration card. **

**(e) Good teachers in a school **

**(f) Honest persons of your village. **

Solution:

(a), (b) and (c) are sets.

(d), (e) and (f) are not sets.

**Represent the following sets in roster method:**

**(a) Set of all alphabet in English language. **

**(b) Set of all odd positive integers less than 25. **

**(c) The set of all odd integers. **

**(d) The set of all rational numbers divisible by 5. **

**(e) The set of all colors in the Indian flag. **

**(f) The set of letters in the word ELEPHANT. **

**Solution:**

** **(a) A = {a, b, c……..x, y, z}

(b) Z = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23}

(c) P = {±1, ±3, ±5….}

(d) R = {5, 10, 15……}

(e) Y = {saffron, white, green}

(f) S = {E, L, P, H, A, N, T}

**Represent the following sets by using their standard notations.**

**(a) Set of natural numbers **

**(b) Set of integers **

**(c) Set of positive integers **

**(d) Set of rational numbers **

**(e) Set of real numbers **

Solution:

(a) N = {1, 2, 3……}

(b) Z = {0, ±1, ±2, ±3…….}

(c) Z + = {1, 2, 3…….}

(d) Q = {p/q, p, q z and q ≠ 0}

(e) R = (Q U Z}

**Write the following sets in set builder form:**

**(a) {1, 4, 9, 16, 25, 36} **

**(b) {2, 3, 5, 7, 11, 13, 17, 19, 23……} **

**(c) {4, 8, 12, 16, 20, 24……} **

**(d) {1, 4, 7, 10, 13, 16……} **

Solution:

(a) A = {x: /x=k^{2} for some k€N, 1 ≤ k ≤6}

(b) P = {x/x is a prime number}

(c) X = {x/x is a multiple of 4}

(d) Z = {x/x=3n – 2 when n = 1, 2, 3…..}

**State whether the set is finite or infinite:**

**(a) The set of all prime numbers. **

**(b) The set of all sand grains on this earth. **

**(c) The set of points on a line. **

**(d) The set of all school in this world. **

Solution:

(a) infinite set

(b) finite set

(c) Infinite set

(d) finite set

**Check whether the sets A and B are disjoint**

**(a) A is the set of all even positive integers. B is the set of all prime numbers. **

**(b) A = {3, 6, 9, 12, 15……} **

**B = {19, 24, 29, 34, 39……..} **

**(c) A is the set of all perfect squares; B is the set of all negative integers. **

**(d) A = {1, 2, 3} and B = {4, 5,{1, 2, 3}} **

**(e) A is the set of all hydrogen atoms in this universe; B is the set of all water molecules on earth. **

Solution:

(a) A and B are not disjoint sets since A ∩ B = {2}

(b) A and B are not disjoint since A ∩ B = {24….}

(c) A and B are disjoint.

(d) A ∩ B = {1, 2, 3}. Hence they are not disjoint.

(e) A and B is disjoint.

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The first stellation of the icosahedron can be stellated again, and again, and so on. The “final stellation” of the icosahedron is the one right before the stellation-series “wraps around,” back to where it started:

This final stellation of the icosahedron would serve pretty well as a “polyhedral porcupine,” but I was seeking something even better, so I turned my attention to polyhedral compounds. This is the compound of five icosahedra:

The program I use to manipulate these solids is called *Stella 4d: Polyhedron Navigator* (free trial download available here). My next move, using Stella, was to create the final stellation of this five-icosahedron compound . . . and, when I saw it, I knew I had found my “polyhedral porcupine.”

**AB = 4.8cm. BC = 4.4cm. CD = 7cm. DA = 3.4cm. BD = 6.2cm.**

Steps of construction

- Draw AB = 4.8cm.
- With A as centre and 3.4cm radius draw an arc.
- With B as centre and 6.2cm as radius drawn an arc to intersect the first at D.
- Join AD and BD.

- With B as centre and 4.4cm as radius drawn an arc.
- With D as centre and 7cm as radius draw an arc to intersect the above at C.
- Join DC and BC.

ABCD is the required quadrilateral

**AB = 6.3cm. CD = 6.5cm. BC = 3.8cm. AC = 4.5cm. AD = 3.6cm.**

**Solution:**

Steps of construction:

- Draw AB = 6.3cm
- With A as centre and 4.5cm radius draw an arc.
- With B as centre and 3.8cm as radius drawn another arc to intersect the above at C.
- Join AC and BC.
- With C as centre and 6.5cm as radius drawn an arc.
- With A as centre and 3.6cm as radius draw another arc to intersect the above at D.
- Join AD and CD.

ABCD is the required quadrilateral.

**AB = CD = 5.2cm. BC = AD = 3.2cm. BD = 7.2cm.**

**Solution:**

Steps of construction:

- Draw AB = 5.2cm
- With A as centre and 3.2cm radius draw an arc.
- With B as centre and 7.2cm as radius drawn another arc to intersect the above at D.
- Join AD and BD.
- With B as centre and 3.2cm as radius drawn an arc.
- With D as centre and 5.2cm as radius draw another arc to intersect the above at C.
- Join BC and DC.

ABCD is the required quadrilateral

**AC = 5.8cm. BD = 5cm. AB = 3.8cm. BC = 5.5cm. CD = 4.5cm**

**Solution:**

Steps of construction:

- Draw AB = 3.8cm
- With A as centre and 5.8cm radius draw an arc.
- With B as centre and 5.5cm as radius drawn another arc to intersect the above at C.
- Join AC and BC.
- With B as centre and 5cm as radius drawn an arc.
- With C as centre and 4.5cm as radius draw another arc to intersect the above at D.
- Join CD, BD and AD.

ABCD is the required quadrilateral.

**AC = 5.8cm. BD = 5.6cm. AB = 3.5cm. AD = 4.9cm. CD = 4cm.**

**Solution:**

Steps of construction:

- Draw AB = 3.5cm
- With A as centre and 4.9cm radius draw an arc.
- With B as centre and 5.6cm as radius drawn another arc to intersect the above at D.
- Join AD and BD.
- With A as centre 5.8cm as radius draw an arc.
- With D as centre and 4cm as radius draw another arc to intersect the above at C.
- Join DC, AC and BC.

ABCD is the required quadrilateral.

**AC= 6cm. BD = 5.8cm. AB = 3.4cm. AD = 5.2cm. BC = 4.2cm.**

**Solution:**

Steps of construction:

- Draw AB = 3.4cm
- With A as centre and 6cm radius draw an arc.
- With B as centre and 4.2cm as radius drawn another arc to intersect the above at C.
- Join AC and BC.
- With A as centre 5.2cm as radius draw an arc.
- With B as centre and 5.8cm as radius draw another arc to intersect the above at D.
- Join AD, BD and CD.

ABCD is the required quadrilateral

]]>

I decided to implement a random move computer AI as I was working on tic-tac-toe in Swift this week. Not only could I use the AI within my suite of acceptance tests, but I could also watch a random move AI play against a first available move AI to see who was dominant.

Of course, if you are going to implement a random move AI, it helps to be able to generate random numbers. A quick Google search yielded that Swift could call two methods from the standard library to do just that: `arc4random()`

and `arc4random_uniform(n)`

. The first of these functions generates a random number between 0 and RAND_MAX (a variable set in cstdlib). The second function generates a random number between 0 and n-1 where n is some value provided by the user.

I wondered they would create two functions for random numbers. Maybe the second function was a convenience function. After all, couldn’t you just use the modulus function to generate random numbers between zero and some maximum value? A closer inspection of the documentation yielded this quote:

arc4random_uniform() will return a uniformly distributed random number less than upper_bound.

arc4random_uniform() is recommended over constructions like “arc4random() % upper_bound” as it avoids “modulo bias” when the upper bound is not a power of two.

Apparently, there is a problem using the modulus function on a random number generator. Let’s explore why.

It’s easiest to explain what modulo bias is with an example. I want to generate a random number between 0 and 8 (i.e. the nine indices of the nine cells on my tic-tac-toe board). Just to keep the numbers simple, let’s pretend that `arc4rand`

is a perfect random number generator that will return a random number between 0 and RAND_MAX = 19; each number in the range (0 to 19) has an equal probability of being picked:

P(number) = 1/20 = 5%

Since I want to generate a random number between 0 and 8, I will simply take the result of `arc4rand`

modulo 9.

Let’s go through the results:

If `arc4rand`

returns a 0, 9, or 18, I will receive a 0 as my final output.

If `arc4rand`

returns a 1, 10, or 19, I will receive a 1 as my final output.

If `arc4rand`

returns 2 or 11, I will receive a 2 as my final output.

Wait a minute…

The probability of getting a zero is P(0) = 3/20 = 15%. The probability of getting a two is P(2) = 2/20 = 10%. That is a really big difference between those probabilities! In the graph below, I have calculated the difference between the maximum and minimum probabilities that are returned using the modulus method of random number generation. The source code for the Matlab script can be found here.

The difference between the maximum and minimum probabilities due to modulo bias is inversely proportional to RAND_MAX. Even when RAND_MAX is 100, there is a noticeable difference between the maximum and minimum probabilities. Furthermore, we see that when RAND_MAX is a multiple of eight, the error drops down to zero. The periodicity of the drops depends on the number with which you are taking the modulus (e.g. in our case we are using nine cells, so n = 9 and n – 1 = 9 – 1 = 8).

“Wait a minute,” I hear you say. “Using a RAND_MAX of 100 is ridiculous. No modern computer uses a number so small.”

I cannot deny that. The function `arc4tan`

uses a RAND_MAX value of 2^32 – 1 = 4,294,967,295. That means that the difference between the probabilities is on the order of 10^-10 (a really tiny number). However, things change drastically when you need to compute *a lot* of random numbers.

Suppose my tic-tac-toe game explodes in popularity, and I decide to calculate random moves as a service. Of course, since people love my tic-tac-toe game so much, the traffic to my servers reaches the same levels as Google searches.

This site reports that Google hosts 100 billion searches per month or 1.2 trillion per year. Let’s use these numbers to make some rough estimates.

In the ideal case, each cell on a tic-tac-toe board would have a probability of P(cell) = 1/9 of being selected. That means that over the course of a year, the expected number of times a cell is selected is:

(1.2 trillion requests)(1/9) ≈ 133,333,333,333 times (i.e. 133 billion and change).

However, if I were to use the modulus method of calculating random numbers from `arc4rand`

, the probability of selecting one of the first four numbers is given by the ceiling function:

ceil(2^32 / 9) / 2^32 = 0.11111111124046

The probability of selecting one of the final five numbers is given by the floor function:

floor(2^32 / 9) / 2^32 = 0.1111111110076308

The first four cells are selected 133,333,334,886 times each while the last five numbers are selected 133,333,332,092 each. That is a difference of over 1000 from the expected value of 133,333,333,333!

Long story short: for my little game of tic-tac-toe, modulo bias really makes no difference. However, if lives hang in the balance, modulo bias might be unacceptable.

Modulo bias can occur when you use a modulus function to calculate a random number from a range of values. When RAND_MAX is small, the bias can be very pronounced and can heavily skew your results. However, when RAND_MAX is very large, the bias tends towards zero. This might be fine for most situations (i.e. a game of tic-tac-toe), but it might become a security risk in systems that depend on perfectly uniform random distributions.

I can think of a few solutions to the problem of modulo bias:

- Implement an algorithm that eliminates the modulo bias from the calculation.
- Use a RAND_MAX such that RAND_MAX % n == n – 1
- Use a really big RAND_MAX and accept an approximate solution.

Until next time!

]]>*Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). *

The question:

Timmy and Chet head back to the Glass House in Denver while Sarah, Barry, and Sam continue to ski. Cynthia got confused and ended up in Moab … somehow. Anyways, Timmy and Chet have a cook off. Timmy wakes up early and makes 5 slices of bacon and then Chet gets up and starts cooking right along side him. Chet can produce 7 slices of bacon per hour while Timmy can only make 5. The cook off ends three hours after Chet gets up. Who cooked more bacon?

The answer:

To answer this question we have to make equations for Timmy and Chet’s cooking. Let us set the variable t equal to the time in hours. We will set the variable b equal to the amount of bacon produced after t hours. So our equation for Timmy becomes:

b = 5t + 5 (the plus 5 accounts for the 5 bacon slices Timmy cooked before Chet got up)

And the equation for Chet becomes:

b = 7t

Plugging in 3 for t (since the competition ends 3 hours after Chet wakes up) we get:

b = 5(3) + 5 = 15 + 5 = 20 for Timmy

And

b = 7(3) = 21 for Chet. So Chet wins the cook off by 1 slice of bacon!

If you liked this excerpt, then please be sure to check out, “The Book of Timmy – Time to Burn Off the Bacon,” coming Sunday Sept. 25th!

© Copyright 2016 I will solve that! All rights reserved. http://www.iwillsolvethat.com

]]>In one of those posts I argue that Creativity must be a quality of Everything. Another quality is (Self) Consciousness. Those qualities are inescapable because otherwise (the) Everything wouldn’t be relevant.

Creativity in itself does not make sense. These things only make sense as part of a Trinity. Where Trinity is like a ‘coin’. A coin is made of stuff and has 2 sides. Creativity is the stuff, and the 2 sides are Creator and Creation.

So what about π ?

I quote from Wikipedia:

The number π is a mathematical constant, the ratio of a circle’s circumference to its diameter, commonly approximated as 3.14159. It has been represented by the Greek letter “π” since the mid-18th century, though it is also sometimes spelled out as “pi” (/paɪ/). Being an irrational number, π cannot be expressed exactly as a fraction (equivalently, its decimal representation never ends and never settles into a permanent repeating pattern). Still, fractions such as 22/7 and other rational numbers are commonly used to approximate π. The digits appear to be randomly distributed. In particular, the digit sequence of π is conjectured to satisfy a specific kind of statistical randomness, but to date no proof of this has been discovered. Also, π is a transcendental number – a number that is not the root of any non-zero polynomial having rational coefficients. This transcendence of π implies that it is impossible to solve the ancient challenge of squaring the circle with a compass and straightedge.

I wonder, where does π come from? Who/what decided on the constant of π ? I remember reading Carl Sagan’s novel ‘Contact‘ in the mid 80’s. In that novel, Sagan has a mind boggling passage in which it is discovered that π contains a pattern, a ‘hidden message from the Gods’. Back then all I could think was ‘Wow’, because it was such a mind blowing idea.

But, Sagan’s Contact, is fiction. Not real, mind blowing or not. Yet, the intriguing mathematical constant π kept cropping up in my musings once in a while. Where did the constant come from? A constant is fixed. So what/who fixed it and how? Well, your simple answer could be, ‘God fixed it’. But unfortunately I can not go for that answer. It is too simple and too lazy. It is like saying ‘It is not my problem, God will take care of it’. Sure, but somehow it doesn’t feel right. As said, this way out is simply too easy.

No way out? Yes there is, if you look carefully that is. Because although the definition and the ratio of π are fixed, the number itself, the infinite row of digits, is irrational. And irrational is simply another word for …. Creativity. No matter how many digits of the row we will calculate, the next digit we find (by calculating) is a creative process. Only by calculating we find the next digit. No calculating, no next digit! So we, once again stumble on this ‘Trinity’ namely:

Creator (Calculator) – Creativity – Creation (Next digit)

As π is a center piece of Mathematics, it shows that Mathematics is founded on Creativity

~ Max

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