MA 103-005

Topics in Contemporary Mathematics

Mathematics

Hints and Answers

Chapter 1

Exercise 52: (a) C (b) B (c) C (d) Condorcet criterion is violated: C is a Condorcet candidate but B wins under Borda Count. IIA criterion is violated: B wins original election, but when D drops out, C wins.

Exercise 58: If X is the first choice of a majority of the voters, then X can beat every other candidate in a one-on-one election. Therefore X gets the most points in the method of pairwise comparisons.

Chapter 2

Exercise 58: (a) Before merger: P_1 3/5, P_2 1/5, P_3 1/5. After merger: P_1 1/2, P* 1/2.
(b) Before merger: P_1 1/2, P_2 1/2, P_3 0. After merger: P_1 1/2, P* 1/2.
(c) Before merger: P_1 1/3, P_2 1/3, P_3 1/3. After merger: P_1 1/2, P* 1/2.

Exercise 64: Mayor 5/13, others 2/13 each.

Chapter 3

Exercise 10: (a) 1, 3, 5.
(b) Cut 1: either. Cut 3: bottom. Cut 5: right.

Exercise 12: Many answers.

Exercise 16: (a) DiPalma s_1, Childs s_2, Choate s_3, Chou s_4.
(b) Chou must get s_4. Then Choate must get s_3. Then Childs must get s_2. DiPalma is left with s_1.

Exercise 18: The three choosers only selected two slices among them. Give one of the unselected slices (for example, s_4) to the divider. The three choosers divide the remainder of the cake using the Lone-Divider Method. (One of them will be the divider.)

Exercise 24: (a) Ch 1: s_3, s_4. Ch 2: s_1, s_2. Ch 3: s_2, s_4.
(b) Many answers.

Chapter 4

Exercise 4: (a) The shifts.
(b) 12. The standard divisor represents the average number of patients per shift.
(c) A: 72.583; B: 85.750; C: 50.833; D: 15.833.

Exercise 10: A: 72; B: 86; C: 51; D: 16.

Exercise 18: A: 73; B: 86; C: 51; D: 15. Any divisor between 11.876 and 11.931 will work.

Exercise 30: A: 72; B: 86; C: 51; D: 16. Any divisor between 12.014 and 12.035 will work.

Exercise 56a: Say there are originally N seats. Jefferson's Method produces a divisor that gives a total of N seats when every's state's quota is rounded down. Now we make the number of seats N+1. Our previous divisor only gives a total of N seats, which is now too few. So the divisor must be made even smaller. This will make all the quotas bigger, so no state will lose a seat.

Chapter 5

Exercise 60: Yes, there are only two vertices of odd degree.

Chapter 6

Exercise 8: (a) ABCFEDA, ADFEBCA, and ABEDFCA, along with their mirror-image circuits.
(b) DABCFED, DFEBCAD, and DFCABED, along with their mirror-image circuits.
(c) You always get the same circuits, just starting a different vertex.

Chapter 7

Exercise 48: (a) There are 11 different spanning trees.
(b) 29

Chapter 8

Exercise 14: (a) E (b) A

Exercise 58: Red team 20 hours, green team 19 hours. The red team is better, but the green team wins.

Chapter 12

Exercise 2(a) 4 x 5^n
Exercise 4(a) X + (4/9)X
(b) X + (4/9)X + (5/9)(4/9)X
(c) X + (4/9)X + (5/9)(4/9)X + (5/9)^2(4/9)X
(d) X + (4/9)X + (5/9)(4/9)X + (5/9)^2(4/9)X + (5/9)^3(4/9)X
Exercise 6(b) 3 x (4/3)^n
(c) infinite
Exercise 8(a) A - (1/3)A
(b) A - (1/3)A - (4/9)(1/3)A
(c) A - (1/3)A - (4/9)(1/3)A - (4/9)^2(1/3)A
Exercise 10(b) 4 x 7^n
Exercise 12 At every step the area is X.
Exercise 18 At step N there are 1 + 8 + 8^2 + ... + 8^(N-1) holes.
Exercise 20(a) At step N the fraction of the area that remains is (8/9)^N
(b) (8/9)A
Exercise 22 At step N the length of the boundary is 2^N x P
Exercise 24(a) 3 x 2^N
(b) sqrt(3)/4 or about 0.433
(c) (2/3)^N x sqrt(3)/4

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