Midterm Review: Math 210G-07, Spring 2009 - Prof. Joseph D. Lakey, Exams of Mathematics

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Midterm Review,

Math 210G‐07, Spring 2009

1) Explain why the following pictures

prove the Pythagorean theorem

• SOLUTION: For the picture on the right, the area of the big square is (a+b) 2 =a 2 + 2ab + b 2 while the area of its parts amount to c 2 +4x(axb/2) =2ab+c 2 . SubtracUng 2ab from both sides gives the Pythagorean idenUty.

3) The following Escher picture is an

example of

• Hyperbolic geometry
• Euclidean geometry
• SOLUTION: this is an example of hyperbolic geometry, as was discussed earlier in class.

4) Propose a soluUon to the following

problem:

A king decides to give 100 of his prisoners a test. If they pass, they can go free. Otherwise, the king will execute all of them. The test goes as follows: the prisoners stand in a line, all facing forward. The king puts either a black or a white hat on each prisoner. The prisoners can only see the colors of the hats in front of them. Then, in any order they want, each one guesses the color of the hat on their head. Other than that, the prisoners can not speak. To pass, no more than 1 of them may guess incorrectly. If they can make their strategy before hand, how can they be assured that they will survive? SOLUTION: The last person in line has 99 people in front. So either the black or white hats in front of him are even in number but not both. The last person in line will say “white” if there are an even number of white hats, and “black” if an even number of black hats. If #100 says “black” and #99 sees an even number of black hats, then his hat must be white. Otherwise his hat must be black. The reasoning is similar for #98 on down to the front.

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

• The outcomes of rolling a pair of dice are as follows

6) Bayes rule states that

• Find P(A|B) is P(B|A)=1/2, P(A)=1/3 and P(B)=2/3.
• SOLUTION: the right hand side is
• P(B|A)xP(A)/P(B)=(1/2 x 1/3 ) / 2/3 = ¼ so
• P(A|B) = ¼.

8) Fill in the next row of Pascal’s

triangle (the soluUon is highlighted in red)

1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1

9) Poker hands

• Compute the number of possible ways of choosing 5 cards from a deck of 52 cards.
• Compute the number of possible ways of gemng 4 of a kind in a five card poker hand. Explain your result and its probability of happening.
• SOLUTION: The number of ways of choosing 5 from 52 is 52! / (5!)(52‐5!) = 52! / (5!)(47!)
• =52x51x50x49x48x47! / (5!)(47!)
• =52x51x50x49x48 / (5x4x3x2)=10x12x17x26x
• =2,598,
• The number of ways of choosing 4 of a given rank is 48, the number of ranks is 13 so the number of ways of gemng 4 of a kind is 48x13=

11) Use the Euclidean algorithm to

find the following:

• GCD(69,96)

• GCD(35,53)

• GCD(123,321)

• SOLUTION: By the Euclidean algorithm, GCD(69,96)=GCD(69, 96‐69)=GCD(69,27)=GCD( 27,69‐54)=GCD(15,27)=GCD(15,27‐15)=G CD(15,12)=GCD(12,15‐12)=GCD(12,3)=3.
• Similarly, GCD(35,53)=1 and
• GCD(123,321)=

12) Which of the following numbers

are prime numbers?

• Here are some divisibility tests
• SOLUTION: 11,101 are prime, 1001=71113; 10001=73*
• 13,31,113,311 are prime, 1331=1111
• 41 is prime,141=347,1441=11