1. (4 pts) (x4)^2.

2. (15 pts,3,3,3,6) a) 3, b) remove 3 from 3rd pile, b) could use pile 2 or 3 or 4- remove 1, d) Gr = 2, remove 4 from pile 4.

3. (12 pts) a (-,oo), b(g-,1), c(h-,1), d(a+,1), e(j-,1), f(k-,1), g(c+,1) h(d +,1), i(b+ or f+,1), j(d+,1), k(e+,1), z(i+,1). One new solution using b-i is b-i,

4. (12 pts) (1/6)[(x1)^9 + 2(x3)^3 + 3x1(x2)^4], xi = (b^i + w^i);

5. (8 pts) supplies L 1, T 0, V 3: supplies of wins less than demand of games to be played, impossible:

6. (10 pts: 2,2,2,4): a) x11=20,x21=10,x22=20,x23=10,x33=30, b) u1=10, u2=8, u3=10, v1=14,v2=15,v3=13, c) increase x12, New solution obtained by increasing x12 and x21 by 20, reduce x11 and x22 by 20.

1.(4 pts) : (x sub 8).

2. (12 pts) a (-,oo), b(g-,1), c(i-,1), d(j-,1), e(k-,1), f(a+,1), g(d+,1) h(b+ or c+,1), i(e+,1), j(f+,1), k(f+,1), z(h+,1). One new solution using b-h is b-h, c-i, d-g, e-k, f-j.

3. (15 pts) a) initial digital sum is 011 = 3; b) remove 3 from pile 3; c) remo ve 1 from pile 2 or 3 or 4; d) g(0)=0, g(1)=1, g(2)=0, g(3)=1, g(4)=2, g(5)=3, g(6)=2, g(7) = 0, sum is 2, remove 4 from pile 4.

4. (12 pts) (1/4)[(b+w)^11 + 2(b+w)(b^2+w^2)^5 + (b+w)^3(b^2 + w^2)^4].

5. (8 pts) Possible. Number of available wins (L 2, T 2, V 4). Solution has L beat T twice, T beat L twice, and V beat L and T twice each.

6. (10 pts) Initial solution: x11=20, x21=10, x22=20, x23=10, x33=30. Prices u1=$10, u2=$8, u3=$10,v1=$14, v2=$15, v3=$13. Add x12: increase x12, x21 by 20, decrease x11 and x22 by 20.

1.(4 pts) : (x sub 3)^3.

2. (12 pts) a (-,oo), b(h-,1), c(g-,1), d(i-,1), e(ki-,1), f(a+,1), g(b+,1) h(f+,1), i(f+,1), j(c+ or e+,1), k(d+,1), z(j+,1). One new solution using c-j is b-g, c-j, d-i, e-k, f-h; other is b-h, c-g, d-k, e-j, f-i.

3. (15 pts) a) initial digital sum is 011 = 3; b) remove 1 from pile 3; c) remove 2 from pile 1 or 3 or 4; d) g(0)=0, g(1)=1, g(2)=0, g(3)=1, g(4)=2, g(5)=3, g(6)=2, sum is 1, remove 1 from pile 1.

4. (12 pts) (1/4)[(b+w)^8 +(b^2+w^2)^4 + 2(b+w)^2(b^2 + w^2)^3].

5. (8 pts) Possible. Number of available wins (L 2, T 2, V 3) equals number of games to be played. Many possible solutions.

NOTE that there is an error in the table because it says the Bears have 8 more total games but only 6 games against the Lions, Tigers and Vampires.

6. (10 pts) Initial solution: x11=20, x12=10, x22=20, x23=20, x33=20. Prices u1=$10, u2=$8, u3=$9, v1=$15, v2=$14, v3=$13. Select either x13 or x32. For x31, increase x31, x12, and x23 by 20. and reduce x11, x22, x33 by 20.

1. x sub 10

2. a) digital sum of piles is 011=3, b) remove 3 from pile 3, c) take 1 from pile 3, d) g(0)=0, g(1)=1, g(2)=0, g(3)=1, g(4)=2, g(5)=3, g(6)=2, g(7)=0, sum is 10, remove 4 from pile 1.

3. a (-,oo), b(g-,1), c(h-,1), d(i-,1), e(j-,1), f(a+,1), g(c+) h(f+,1), i(f+,1), j(d+,1), k(b+ or e+,1), z(k+,1). One new solution b-k, c-g, d-i, e-j, f-h)

4.(1/8)[(b+w)^12 + 2(b^4+w^4)^3 + 3(b^2+w^2)^6 + 2(b+w)^2(b^2+w^2)^5].

5. Not possible,L & T play 4 games but they can collectively only win 3 games.

6. Initial solution: x11=30, x21=30, x22=20, x23=10, x33=20. Prices u1=$10, u2=$9, u3=$12, v1=$15, v2=14, v3=16. Select x13. Increase x13 and x21 by 10, reduce x11 and x23 by 10.