Advanced s
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This document contains 10 math problems with solutions. The problems cover topics like probability, combinatorics, geometry, and complex analysis. The solutions provide detailed step-by-step workings to arrive at the final answers.
Advanced s
- 1. SMT 2009 Advanced Topics Test and Solutions February 28, 2009 1. Let a, b, c, and d be the numbers that show when four fair dice, numbered 1 through 6 are rolled. What is the probability that |(a − 1)(b − 2)(c − 3)(d − 6)| = 1? Answer: 1 324 The conditions implies that |a − 1| = |b − 2| = |c − 3| = |d − 6| = 1. a can equal 2, b can equal 1 or 3, c can equal 2 or 4, and d can equal 5. So the probability is 1 6 ∗ 2 6 ∗ 2 6 ∗ 1 6 = 1 324 . 2. Find all possibilities for the second-to-last digit of a number whose square is of the form 1 2 3 4 5 6 7 8 9 0 (each is a digit). Answer: 3, 7 Zero is the only digit with square ending in 0. The square of a number ending in zero will therefore end in two zeros. Next digit of the number therefore needs a square ending in 9, so it is 3 or 7. 3. Ten gears are lined up in a single file and meshed against each other such that the ith gear from the left has 5i + 2 teeth. Gear i = 1 (counting from the left) is rotated 21 times. How many revolutions does gear 10 make? Answer: 147 52 The number of teeth meshed does not vary. Thus, if n is the number of revolutions that gear 10 make, then (5(1) + 21)(21) = (5(10) + 2)n ⇒ n = 7×21 52 = 147 52 . 4. In the game Pokeymawn, players pick a team of 6 different Pokeymawn creatures. There are 25 distinct Pokeymawn creatures, and each one belongs to exactly one of four categories: 7 Pokeymawn are plant- type, 6 Pokeymawn are bug-type, 4 Pokeymawn are rock-type, and 8 Pokeymawn are bovine-type. However, some Pokeymawn do not get along with each other when placed on the same team: bug-type Pokeymawn will eat plant-type Pokeymawn, plant-type Pokeymawn will eat rock-type Pokeymawn, and bovine-type Pokeymawn will eat anything except other Bovines. How many ways are there to form a team of 6 different Pokeymawn such that none of the Pokeymawn on the team want to eat any of the other Pokeymawn? Answer: 245 If we make our team all the same type, then there are 7 6 + 6 6 + 4 6 + 8 6 = 7 + 1 + 0 + 28 = 36 ways to do this. If we make our team partially bug and partially rock type, there are 6 2 4 4 + 6 3 4 3 + 6 4 4 2 + 6 5 4 1 = 15 ∗ 1 + 20 ∗ 4 + 15 ∗ 6 + 6 ∗ 4 = 15 + 80 + 90 + 24 = 209 ways. Any other combination of types will not work. This gives a total of 245 ways. 5. Four cards are drawn from a standard deck (52 cards) with suits indistinguishable (for example, A♠ is the same as A♣). How many distinct hands can one obtain? Answer: 1820, or 13 1 + 3 13 2 + 3 13 3 + 13 4 We proceed by casework. Case 1 All cards have the same face value. There are 13 1 ways to choose the face values. Case 2 Some cards have face value A; some have face value B. There are 13 2 ways to choose A and B. One can have the combinations ABBB, AABB, AAAB, so there are 3 13 2 distinct ways for this case. Case 3 Some cards have value A, some B, and some C. There are 13 3 ways to choose the A, B, C. One can have the combinations ABCC, ABBC, and AABC. There are 3 13 2 distinct ways for this case. Case 4 The cards are distinct: ABCD. There are 13 4 ways to do this. Since these cases are mutually exclusive, we have 13 1 + 3 13 2 + 3 13 3 + 13 4 = 1820 distinct hands. 6. Find all complex numbers z such that z5 = 16¯z, where if z = a + bi, then ¯z = a − bi. Answer: 0, ±2, 1 ± i √ 3, −1 ± i √ 3
- 2. SMT 2009 Advanced Topics Test and Solutions February 28, 2009 Clearly 0 is a solution. Now we assume z = 0. We have |z5 | = |16¯z|. By DeMoivre’s Theorem, |z5 | = |z|5 . The left hand side becomes |z5 | = 16|¯z| = 16|z|. Equating the two sides, 16|z| = |z|5 ⇒ |z|4 = 16 ⇒ |z| = 2. Multiplying both sides of the given equation by z, z6 = 16|z|2 = 64. Let z = r(cos θ + i sin θ). Then r6 (cos(6θ) + i sin(6θ)) = 64. Thus, r = 2 and 6θ = 360k, for k = 0, 1, 2, 3, 4, 5. So our other solutions are 2, 2cis(60◦ ), 2cis(120◦ ), −2, 2cis(240◦ ), 2cis(300◦ ), which are equal to ±2, 1 ± i √ 3, −1 ± i √ 3. 7. Evaluate 1+ √ 3i 2 Answer: e π 6 i , or ± √ 3+i 2 Let x = 1+ √ 3i 2 . Then x2 = 1+ √ 3i 2 . Converting to polar form,1+ √ 3i 2 = (e π 3 i ) 1 2 = e π 6 i = √ 3+i 2 8. Frank alternates between flipping a weighted coin that has a 2 3 chance of landing heads and a 1 3 chance of landing tails and another weighted coin that has a 1 4 chance of landing heads and a 3 4 chance of landing tails. The first coin tossed is the “2/3 − 1/3” weighted coin. What is the probability that he sees two heads in a row before he sees two tails in a row? Answer: 13 33 If the first toss comes up heads (2/3 probability), Frank has a 1/4 chance of getting another heads, a (3/4) ∗ (1/3) = 1/4 chance of getting two successive tails, and a (3/4) ∗ (2/3) = 1/2 chance of getting tails-heads and winding up back at his current position of tossing the “1/4−3/4” coin with the previous toss being a heads. Expressing the probabilities as geometric series (or just the weighted probability of the two nonrepeating options), he has a 1/2 chance of getting HH first and a 1/2 chance of getting TT first. If instead, the first toss comes up tails (1/3 probability), he has a 3/4 chance of getting another tails, a (1/4) ∗ (2/3) = 2/12 chance of getting two successive heads, and a (1/4) ∗ (1/3) = 1/12 chance of getting heads-tails and winding up back at my current state. Expressing the probabilities as a geometric series, he has a 2/11 chance of getting HH first and a 9/11 chance of getting TT first. The probability of getting HH before TT is (2/3) ∗ (1/2) + (1/3) ∗ (2/11) = 13/33. 9. The triangular numbers Tn = 1, 3, 6, 10, . . . are defined by T1 = 1 and Tn+1 = Tn +(n+1). The square numbers Sn = 1, 4, 9, 16, . . . are defined by S1 = 1 and Sn+1 = Tn+1 + Tn. The pentagonal numbers Pn = 1, 5, 12, 22, . . . are defined by P1 = 1 and Pn+1 = Sn+1 + Tn. What is the 20th pentagonal number P20? Answer: 590 Expanding out the recurrence relations, we confirm that the triangular numbers are Tn = 1 + 2 + 3 + · · · + n = n(n+1) 2 and the square numbers are Sn = n2 . A general formula for the pentagonal numbers is therefore Pn = n2 + n(n − 1)/2 = n(3n − 1)/2. Substituting n = 20 gives P20 = 20(60 − 1)/2 = 590. 10. Evaluate eiπ/3 + 2e2iπ/3 + 2e3iπ/3 + 2e4iπ/3 + e5iπ/3 + 9e6iπ/3 . Answer: 6 eiπ/3 + e2iπ/3 + e3iπ/3 + e4iπ/3 + e5iπ/3 + e6iπ/3 sum to 0 because the terms are sixth roots of unity (i.e. they satisfy z6 − 1 = 0, which is a 6th degree polynomial whose 5th degree coefficient is 0). Likewise, e2iπ/3 + e4iπ/3 + e6iπ/3 sum to zero because the terms are cubic roots of unity. e3iπ/3 + e6iπ/3 sum to 0 because they are square roots of unity. Subtracting these sums from the original expression, we are left with only 6e6iπ/3 , which is 6(cos(2π) + i sin(2π)) = 6.