AMS 303 SECOND TEST REVIEW Solutions to Past Tests


Fall 2016 AMS 303 Second Test
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(c+,1), k(f+,1), z(h+,1). One new solution using b-h is b-h, c-i, d-k, 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.

Spring 2016 AMS 303 Second Test
1.(4 pts) : (x sub 3)^3.
2. (12 pts) a (-,oo), b(h-,1), c(g-,1), d(i-,1), e(i-,1), f(a+,1), g(b+,1) h(f+,1), i(f+,1), j(c+ or e+,1), k(d+,1), z(g+,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 x31. For x13, increase x13 and x22 by 10, reduce x12 and x23 by 10. For x31, increase x31, x12, x23 by 20, reduce x11, x22, x33 by 20.


Spring 2015 AMS 303 Second Test
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.