Math Contest Round Two Solutions Chapter 2

Chapter 2 9. Solution by CwH. Let 𝑓 𝑥 = 2𝑥 ! + 28𝑥 + 91. We are told that there are two real solutions a and b to the equation 𝑓𝜊𝑓𝜊𝑓 𝑥 = 1, and we are asked to find |𝑎 − 𝑏|. It is too complicated to write out the equation 𝑓𝜊𝑓𝜊𝑓 𝑥 = 1. So I decided to follow a different approach: First find whatever values 𝑥! such that 𝑓(𝑥! ) = 1. Then find all possible values 𝑦! so that 𝑓(𝑦! ) = 𝑥! . Then find all the possible 𝑧! so that 𝑓(𝑧! ) = 𝑦! . Then any of the 𝑧! will be a solution for 𝑓𝜊𝑓𝜊𝑓 𝑥 = 1. Now for 𝑓 𝑥 = 1 or 2𝑥 ! + 28𝑥 + 91 = 1, or 2𝑥 ! + 28𝑥 + 90 = 0, or or 𝑥 ! + 14𝑥 + 45 = 0. Using the quadratic formula, we can easily find that 𝑥 = −7 ± 2, or 𝑥 = −5 or 𝑥 = −9. Now for the 𝑦! ’s, we need to solve 2𝑥 ! + 28𝑥 + 91 = −5, or 2𝑥 ! + 28𝑥 + 96 = 0. Note that we don’t have to worry about or 2𝑥 ! + 28𝑥 + 91 = −9, since it is not difficult to see that 𝑓 𝑥 = 1 or 2𝑥 ! + 28𝑥 + 91 is a parabola with its absolute minimum at 𝑥 = −7 and 𝑓(−7) = −7. Thus, 2𝑥 ! + 28𝑥 + 91 = −9 would not have any real solution. Now for the equation 2𝑥 ! + 28𝑥 + 91 = −5, or 2𝑥 ! + 28𝑥 + 96 = 0. By the quadratic formula, we have solution 𝑥 = −6 or 𝑥 = −8. Again, since the absolute minimum for 𝑓 𝑥 is −7. We don’t have to worry about 𝑥 = −8. Use the value 𝑥 = −6 and solve 2𝑥 ! + 28𝑥 + 91 = −6, or 2𝑥 ! + 28𝑥 + 97 = 0. Using quadratic formula again, we have 𝑥 = −7 ± answer is (E). ! ! . These are the two real solutions a and b, and 𝑎 − 𝑏 = 2 . The 9. Second solution by MF. First complete the square on the given function: 𝑓 𝑥 = 2 𝑥 ! + 14𝑥 + + 91 1 Recall the pattern for a perfect square, you take half of the middle coefficient, 14, and square that, giving 49. So the blank in equation (1) on the right side needs to be filled in with 49. But since the outer common factor is 2, you are actually adding 2 x 49 = 98 to the right side, so you have to compensate by subtracting 98 outside, yielding the result: 𝑓 𝑥 = 2 𝑥 ! + 14𝑥 + 49 + 91 − 98 2 Now write the enclosed quadratic on the right side as a square, with the result: 𝑓 𝑥 =2 𝑥+7 ! −7 (3) Now compute the composition: 𝑓 𝑓 𝑥 =2 𝑓+7 ! −7 (4) Notice that in order to compute this, we substitute f for x in equation (3). But now notice that from equation (3): 𝑓+7=2 𝑥+7 ! (5) Substitute equation (5) into equation (4), eliminating f with the result: 𝑓 𝑓 𝑥 =8 𝑥+7 ! −7 (6) Now compose this function with f one more time by substituting f for x in equation (6): 𝑓 𝑓 𝑓 𝑥 =8 𝑓+7 ! −7 (7) Now use equation (5) again to eliminate f from the right side of equation (7); 𝑓 𝑓 𝑓 𝑥 =8 2 𝑥+7 ! ! −7 (8) and after some algebra, this becomes: 𝑓 𝑓 𝑓 𝑥 = 128 𝑥 + 7 ! −7 (9) Now set the left side of equation (9) equal to one, as specified in the problem: 1 = 128 𝑥 + 7 ! −7 (10) This can be re-arranged as: 1 = 𝑥+7 16 ! (11) Take the square root of both sides of equation (11) with the result: 1 =± 𝑥+7 4 ! (12) The problem asks for real roots, so we drop the minus sign, since this gives an even power equal to a negative number, which is complex. Take the square root of equation (12) with the result: 1 = 𝑥+7 2 ! (13) where again the minus sign has been dropped to eliminate the complex roots. Now take one final square root, and this time we keep both signs because both such roots are real. So equation (13) becomes: ± 1 We see that the two roots are: ! 2 𝑥! = −7 + ! and 𝑥! = −7 − cancel and we get the result: =𝑥+7 ! ! (14) . If we take the difference 𝑥! − 𝑥! , the seven’s will 𝑥! − 𝑥! = 2 2 = 2 (15) and the answer is (E). 9. Third solution by GM. Find |a – b| if a and b are the two real solutions to (f ◦ f ◦ f)(x) = 1 for f(x) = 2x2 + 28x + 91. Solution: To find the real solutions to (f ◦ f ◦ f)(x) = 1, I will simplify: f(f(f(x))) − 1 = 0 using a transformation: x = y − 7 and apply it to simplify: f(f(f(y − 7))) − 1 = 0. Observe that: f(y − 7) = 2(y − 7)2 + 28(y − 7) + 91 = 2y2 − 7. Thus the original composition equation now becomes: f(f(2y2 − 7)) − 1 = 0. Next observe that: f(2y2 − 7) = 2(2y2 − 7)2 + 28(2y2 − 7) + 91 = 8y4 − 7. So that the previous composition equation now becomes: f(8y4 − 7) − 1 = 0. Finally observe that: f(8y4 − 7) = 2(8y4 − 7)2 + 28(8y4 − 7) + 91 = 128y8 − 7. The previous composition equation now becomes: 128y8 − 7 − 1 = 128y8 − 8 = 0. In factored form, it becomes: 8(2y2 − 1)(2y2 + 1)(4y4 + 1) = 0. Only the first variable factor has real roots: y = ±(√2)/2. Since x = y − 7 with y = ±(√2)/2, then x = –7 ± (√2)/2. If we set a = –7 + (√2)/2 and b = –7 − (√2)/2, then |a − b| = √2. The solution is: (E), √2. Incidentally: (f ◦ f ◦ f)(x) − 1 in factored form is: 8(2x2 + 26x + 85)(2x2 + 28x + 97)(2x2 + 28x + 99)(2x2 + 30x + 113). The bold factor is the one with real roots. 10. Solution by MF. We examine three cases and look for contradictions: Case 1: Suppose that X is a knight. X says “I am a knight”, which is true and consistent with X being a knight. Y says “X is telling the truth”, which is also true, so Y must be the spy, who is allowed to tell the truth (we cannot have two knights). Finally, Z says “I am a spy” which must be false since Y is the spy and X is a knight. This is consistent with Z being a knave. No contradiction was arrived at, so the answer is (D). Is this the only possible answer? Let us look at two more cases: Case 2: Suppose X is a knave. X says “I am a knight”, which is false and consistent with X being a knave. Y says “X is telling the truth”, which is false, so Y must be the spy, who is allowed to lie. The means Z must be the knight. But Z says “I am a spy”, which is false, contradicting Z being a knight. So X cannot be a knave. Case 3: Suppose X is the spy. X says “I am a knight”, which is false and consistent with X being a spy. Y says “X is telling the truth”, which is false, so Y must be the knave, and again Z must be a knight. But Z says “I am a spy”, which is false, contradicting Z being a knight. So answer (D) is the only possible one. 11. Solution by CwH. Solve the equation 𝑒 !! sin 𝑥 − 𝑒 !! cos 𝑥 = 0. Note that 𝑒 !! is never zero, the given equation is equivalent sin 𝑥 − cos 𝑥 = 0 or sin 𝑥 = cos 𝑥. Considering sin 𝑥 𝑎nd cos 𝑥 as the 𝑦 and 𝑥 coordinates of a point on the unit circle, the equation sin 𝑥 = cos 𝑥 means that the point is on the 45° line. Thus, 𝑥 = 𝜋/4 (when both sin 𝑥 and cos 𝑥 are positive) or 𝑥 = 5𝜋/4 (when both sin 𝑥 and cos 𝑥 are negative). ! !! !! Thus, 𝑀 = + = and csc 𝑀 = −1. The answer is (A). ! ! ! 12. Note to calculator users by MF. If you do this problem by plotting the given polar expression on your calculator, you need to put your calculator into polar mode and make sure you are using radians, not degrees. If you are unsure how to do this with the graphing calculator you are using, there are YouTube videos available. For example, if you have a TI-84 Plus calculator, you can visit the site: https://www.youtube.com/watch?v=PZwiiZQhM0c This is a video that explains step by step how to make the polar plot using this calculator. For any graphing calculator, such as a TI-89, or Casio, etc., you can enter the search phrase: “Polar graphing with (your calculator name) “ and YouTube videos will come up. However, it is important to realize that a calculator is not necessary to do this problem. To see this, study the next two solutions below: ! 12. Solution by CwH. Determine the graph of 𝑟 = . We could cite the standard !!!.! !"# ! conic equation in polar form. However, most students may not be familiar with that. Let us transform it back into the rectangular coordinates. Clearing the denominator, we have ! 𝑟 + 𝑟 cos 𝜃 = 2. ! Recall that 𝑟 = 𝑥 ! + 𝑦 ! and 𝑟 𝑐𝑜𝑠 𝜃 = 𝑥. The above equation becomes 𝑥 ! + 𝑦 ! = ! ! ! !" ! ! − + 2. Square both sides and simplify to get 𝑥 ! + 𝑥 + 𝑦 ! = , or (𝑥 + )! + 𝑦 ! = !" ! ! ! ! ! ! ! . Thus, it is an ellipse. The answer is (E). 12. Second solution by GM. Consider the following function in polar coordinates: r = 2/(1 + 0.5·cos θ). Which of the following best describes the graph of this function? A. Line B. Two Lines C. Hyperbola D. Parabola E. Ellipse Solution: Since 1 ≥ cos θ ≥ –1, then 1.5 ≥ 1 + 0.5·cos θ ≥ 0.5 Thus 4/3 ≤ 2/(1 + 0.5·cos θ) ≤ 4. Therefore 4/3 ≤ r ≤ 4. This means that the graph of r = 2/(1 + 0.5·cos θ) is bounded within or on a circle of radius 4. The graphs of a “Line”, “Two Lines”, “Hyperbola” and “Parabola” are unbounded, while the graph of an “Ellipse” is bounded. Therefore the graph is that of an ellipse. The solution is: (E), Ellipse. 13. Solution by SV. As stated in the problem, the rolling ends with two successive “1”s or until a “1” and then a “2” are rolled in succession. So, the second-to-last roll will always be a “1”. The last roll will either be a “1” or “2”. What is the probability that the last roll is a “1” (or that the rolling ends with two successive “1”s)? Let A and B be two events. The conditional probability, denoted 𝑃 𝐵|𝐴 refers to the probability of event B, given that event A has happened. If A and B were independent events then we would have 𝑃 𝐵|𝐴 = 𝑃 𝐵 . In other words, for independent events, the probability of event B happening is not changed by whether or not A happened. However, if events A and B depend on each other, then the overall sample space must be restricted to the cases where event A happened. This gives rise to the following definition of conditional probability: 𝑃 𝐵|𝐴 = 𝑃 𝐴∩𝐵 𝑃 𝐴 1 where 𝑃 𝐴 ∩ 𝐵 is the probability that both events A and B happened. For this problem let A be the event that the last roll of the die was either a “1” or a “2”. Since these two events are mutually exclusive, then this probability is the sum of the individual probabilities of getting a “1” or a “2”. So we have: 𝑃 𝐴 = 0.5 + 0.2 (2) Let B be the event that the last roll of the die is a “1”. Then we see that the intersection 𝐴 ∩ 𝐵 is the event: “The last roll was either a “1” or a “2” AND the last roll was a “1”. We can see that this probability is the same as the probability of getting a “1”: 𝑃 𝐴 ∩ 𝐵 = 0.5 (3) We can now substitute equations (2) and (3) into (1), with the result: 𝑃 𝐵|𝐴 = 5 7 (4) This conditional probability gives us the desired probability that the game will end with two consecutive “1” s and the answer is (E). 13. Second solution by MF. There are two ways for this game to end after N rolls. In both cases, the (N – 1) roll must be a “1”, and then the Nth roll is either a “1” or a “2”. Let us assume that there have been (N – 1) rolls and the (N – 1) roll was a “1”. If the (N – 1) roll is either “something else” or “2”, keep rolling until a “1” appears. Then re-label this roll as (N – 1). If the Nth roll is “something else”, keep rolling until a “1” appears and again relabel this roll as (N – 1). So we can always arrange a situation where the (N – 1) roll is a “1”. The game can end on the Nth roll, and betting enthusiasts would place 5 to 2 odds on the successive “1” outcome for the Nth roll over the 1 followed by 2 strategy. The reason for these odds is that “1” has a probability of 50%, while “2” has probability 20%. Regardless of how many rolls the game ultimately takes, it must end with the next to last move being a “1” followed by either a “1” or a “2” if it is to end. These are the only two ways the game can end. So let 𝑃!! be the probability that the game ends with two successive “1”s and 𝑃!" the probability that the game ends with “1” followed by “2”. Then we have the two equations: 𝑃!! + 𝑃!" = 1 1 and because of the 5 to 2 odds, we have 2 𝑃!" = 𝑃!! 5 Eliminate 𝑃!" between equations (1) and (2) with the result: 2 𝑃!! = 5 7 (3) and the answer is (E). ! 14. Solution by CwH. Find the sum of all complex zeros of 𝑓 𝑥 = ! ! ! ! ! ! !! !! ! ! . This can be simplified to 𝑥 ! + + = 0. There is a theorem that connects the roots (zeros) and factors of a polynomial of arbitrary degree n. The theorem states: Let P(x) be a polynomial of arbitrary degree. Then the element a is a zero or root of P(x), P(a) = 0, if and only if (x – a) is a factor of P(x). Note that if R and S are propositions, the phrase R if and only if S means two things: R implies S and S implies R. One also says that R and S are equivalent. So to prove this theorem, we have to prove two things: 1. First assume that (x – a) is a factor of P(x). This means that (x – a) divides P(x) with zero remainder, and we can write: P(x) = (x – a) q(x), where q(x) is another polynomial of degree one less than P(x). If we insert x = a into this equation, we get: P(a) = (a – a) q(a) = 0, and a is a zero of P(x). 2. Now assume that x = a is a zero of P(x). We can perform a long division of P(x) by the linear expression (x – a) with the result: P(x) = (x – a) q(x) + r(x), where the remainder, r(x), must be one degree less than the degree of the divisor (x – a), so it is a constant. Now substitute x = a into the expression for P(x) with the result: 0 = r(a), and since the remainder is a constant, we have r(x) = 0, and (x – a) is a factor of P(x). The set of complex numbers is complete, meaning that any polynomial of degree n can be factored down completely into a list of linear factors, so if P(x) is given by: 𝑃 𝑥 = 𝐴𝑥 ! + 𝐵𝑥 !!! + ⋯ + 𝐶𝑥 + 𝐷 Then P(x) can be factored into the list: 𝑃 𝑥 = 𝐴 𝑥 − 𝑎! 𝑥 − 𝑎! ⋯ 𝑥 − 𝑎! where by the root and factor theorem above, the numbers 𝑎! are the n roots of P(x), which do not all have to be distinct. If this list of factors is multiplied back together, and the leading coefficient A = 1, then the coefficient of the linear term will be minus the sum of all the roots. For example, using a quadratic, as in this problem, we see that: 𝑥 − 𝑎 𝑥 − 𝑏 = 𝑥 ! − 𝑎 + 𝑏 𝑥 + 𝑎𝑏 and the coefficient of the linear term x is – (a + b). Recall that in this problem we have the equation: 𝑥! + 𝑥 1 + = 0. 2 4 By the root and factor theorem, the sum of all zeros of this equation is the negative of the ! coefficient of the 𝑥 term, namely, − . The answer is then (B). ! 14. Second solution by MF. In order to have a zero of this function, the numerator must be zero, while the denominator is nonzero. Recall the difference of cubes factorization: 𝑎! − 𝑏 ! = 𝑎 − 𝑏 𝑎! + 𝑎𝑏 + 𝑏 ! 1 ! With 𝑎 = 𝑥 and 𝑏 = , equation (1) becomes: ! 𝑥! − 1 2 ! = 𝑥− 1 2 𝑥! + 𝑥 1 + 2 4 (2) ! The factor of 𝑥 − cancels with the denominator, so we set the quadratic factor on the ! right side of equation (2) equal to zero and solve it using the quadratic formula, with the result: 1 𝑖 𝑥=− ± 3 (3) 4 4 We see that the sum of these two roots is 2 − ! ! ! = − , and the answer is (B). ! For students studying the polar form of complex numbers, let us solve the problem this way. The equation to be solved is: 𝑥! = 1 8 (4) ! We put in polar form: ! 1 1 ! = 𝑒 8 8 !!" (5) where 𝑛 = 1,2,3, ⋯. Substitute equation (5) into the right side of equation (4) and take the cube root of both sides, yielding 1 !!" 𝑥 = 𝑒! ! 2 (6) where the three distinct values of x correspond to 𝑛 = 0,1,2. If you put 𝑛 = 3 or higher into equation (6), the three roots just keep cycling through the same three values. Again, ! we throw out the root for 𝑛 = 0, or 𝑥 = because this value also makes the denominator ! of the function equal to zero. The two remaining complex roots can be expanded as: 𝑥 = cos 2𝜋 2𝜋 + 𝑖 sin 3 3 (7) 𝑥 = cos 4𝜋 4𝜋 + 𝑖 sin 3 3 (8) and for 𝑛 = 1 and 𝑛 = 2 respectively. Using the unit circle or some other method, you can ! verify that the sum of the two values in equation (7) and (8) is again equal to − , and the ! answer is (B). 15. Solution by CwH. How many ordered lists of (a, b, c, d, e, f) of nonnegative integers are there whose sum is 12? Since the integers can be zero, we need to consider the cases when some of them are zero. a) Only one of the six integers is nonzero. Since the sum is 12, this nonzero integer is 12. It can be any one of these six, and thus, there are 6 such possibilities. b) Two of the six integers are nonzero. There are six possible combinations of these two nonzero integers: (11, 1), (10, 2), (9, 3), (8, 4), (7, 5), and (6, 6). For the first five combinations, we have to decide which two of the six symbols, a, b, c, d, e, f are to be non-zeros (there are 𝐶(6,2) choices), and for the two non-zero symbols which one will be for the larger number (there are 2 choices). For the combination (6, 6) there are only 𝐶(6,2) possible choices. Thus, there are 5×C 6,2 ×2 + 𝐶 6,2 = 5×15×2 + 15 = 165 many listings. c) Three of the six integers are nonzero. The combinations are (10, 1,1), (9,2,1), (8,3,1), (8,2,2), (7,4,1), (7,3,2), (6,5,1), (6,4,2), (6,3,3), (5,5,2), (5,4,3), (4,4,4). Among these, there is one triple, (4,4,4), four consist pairs, and the remaining consist of three distinct elements. For the triple, we must decide which three of the six symbols are to be the triple, and there are 𝐶 6, 3 such choices. For the four combinations consisting a pair, we must decide which three of the symbols are to be nonzero, and for the three nonzero numbers which two are to be the pair. Since there are 4 such combinations, the possible listings with a pair are 4×𝐶 6, 3 ×C(3, 2). For the remaining 7 combinations of three distinct elements, there are 7×C(6,3)×3!. We need the 3! In the product because after choosing the three symbols for the nonzero elements, there are 3! many ways to fit the three nonzero elements in the three positions. Thus, the total number of listings In this case is C 6,3 + 4×C 6,3 ×C 3,2 + 7×𝐶 6,3 ×3! = 20 + 4×20×3 + 7×20×6 = 1,100. d) Four of the six integers are nonzero. The combinations are (9,1,1,1,), (8,2,1,1), (7,3,1,1), (7,2,2,1), (6,4,1,1), (6,3,2,1), (6,2,2,2), (5,5,1,1), (5,4,2,1), (5,3,3,1), (5,3,2,2), (4,4,3,1), (4,4,2,2), (4,3,3,2), (3,3,3,3). There are 𝐶(6,4) listings for the four equal numbers (3,3,3,3); Listings for a triple plus an element: 2×𝐶(6,4)×𝐶(4,3) (There are two of them.) Listings for two distinct pairs: 2×𝐶(6,4)×𝐶(4,2). listings for a pair plus two distinct elements; 8×𝐶(6,4)×𝐶(4,2)×2!. Listings for 4 distinct elements: 2×𝐶(6,4)×4! Thus, the total number is this case is 𝐶 6,4 + 2×𝐶 6,4 ×𝐶 4,3 + 2×𝐶 6,4 ×𝐶 4,2 + 8×𝐶 6,4 ×𝐶 4,2 ×2! + 2×𝐶 6,4 ×4! = 15 + 120 + 180 + 1,440 + 720 = 2,475 e) Five of the integers are nonzero. The combinations are (8,1,1,1,1), (7,2,1,1,1), (6,3,1,1,1), (6,2,2,1,1), (5,4,1,1,1), (5,3,2,1,1), (5,2,2,2,1), (4,4,2,1,1), (4,3,3,1,1), (4,3,2,2,1), (4,2,2,2,2), (3,3,3,2,1), (3,3,2,2,2). The listings for 4 of a kind: 2×𝐶 6,5 ×C 5,4 (There are two of them). Listings for a triple and a pair: 𝐶 6,5 ×C 5,3 . Listings for a triple and two distinct elements: 5×𝐶 6,5 ×C 5,3 ×C 2,1 . Listings for two pairs plus a single element: 3×C 6,5 ×C 5,2 ×C 3,2 . Listings for a pair and three distinct elements: 2×C 6,5 ×C 5,3 ×3!. Listings for five distinct elements: 0 The total number is 2×𝐶 6,5 ×C 5,4 + 𝐶 6,5 ×C 5,3 + 5×𝐶 6,5 ×C 5,3 ×C 2,1 + 3×C 6,5 × C 5,2 ×C 3,2 + 2×C 6,5 ×C 5,3 ×3!. = 60 + 60 + 600 + 540 + 720 =1,980. f) All six are nonzero. The combinations are (7,1,1,1,1,1), (6,2,1,1,1,1), (5,3,1,1,1,1), (5,2,2,1,1,1), (4,4,1,1,1,1), (4,3,2,1,1,1), (4,2,2,2,1,1), (3,3,3,1,1,1), (3,3,2,2,1,1), (3,2,2,2,2,1), (2,2,2,2,2,2) For six of a kind: 1 For five of a kind plus a single element: C(6,5). For four of a kind plus a pair: C(6,4). For four of a kind plus two distinct elements: 3×𝐶 6,4 ×C 2,1 . For two triples: 𝐶 6,3 . For a triple, a pair, plus a single element: 2×𝐶 6,3 ×C 3,2 . For a triple, plus 3 distinct elements: 𝐶 6,3 ×3! For 3 distinct pairs: 𝐶 6,2 ×C 4,2 . For 2 pairs, plus two distinct elements, one pair with 4 distinct elements, or 6 distinct elements: 0. The total number is 1 + 𝐶 6,5 + 𝐶 6,4 + 3×𝐶 6,4 ×C 2,1 + 𝐶 6,3 + 2×𝐶 6,3 ×C 3,2 + 𝐶 6,3 ×3! + 𝐶 6,2 ×C 4,2 .= 1 + 6 + 15 + 90 + 20 + 120 + 120 + 90 = 462. The total for all the cases: 6 + 165 + 1,100 + 2,475 + 1,980 + 462 = 6,188. The answer is (D). 15. Second solution by GM. How many ordered lists (a, b, c, d, e, f) of nonnegative integers satisfy a + b + c + d + e + f = 12? Solution: Define the notation Σ[f(k), k = 0, n] to mean the sum of f(k) from k = 0 to n. The derivation is a little long, but the idea is fairly straightforward and simple. Given a nonnegative number g: Case 1: If 1 nonnegative number sums to g, then there is only 1 way to do it: g = g. Case 2: If 2 nonnegative numbers (a and b) sum to g, and the first number is “a” (0 ≤ a ≤ g), then the other number b = g − a, and there is only 1 such number b paired with “a” that works. Thus there are Σ[1, a = 0, g] = (g+1) many pairs (a, b) that sum to g. Case 3. If 3 nonnegative numbers (a, b, and c) sum to g, and the first number is “a” (0 ≤ a ≤ g), then the other 2 numbers b + c = g − a. Based upon Case 2 above, for a given value of “a” (0 ≤ a ≤ g), there are (g−a+1) many pairs (b, c) that sum to g − a. Thus there are Σ[g−a+1, a = 0, g] = (g+1)(g+2)/2 many triples (a, b, c) that sum to g. Case 4. If 4 nonnegative numbers (a, b, c, and d) sum to g, and the first number is “a” (0 ≤ a ≤ g), then the other 3 numbers b + c + d = g − a. Based upon Case 3 above, for a given value of “a” (0 ≤ a ≤ g), there are (g−a+1)(g−a+2)/2 many triples that sum to g − a. Thus there are Σ[(g−a+1)(g−a+2)/2, a = 0, g] = (g+1)(g+2)(g+3)/6 many quadruples (a, b, c, d) that sum to g. Case 5. If 5 nonnegative numbers (a, b, c, d, and e) sum to g, and the first number is “a” (0 ≤ a ≤ g), then the other 4 numbers b + c + d + e = g − a. Based upon Case 4 above, for a given value of “a” (0 ≤ a ≤ g), there are (g−a+1)(g−a+2)(g−a+3)/6 many quadruples that sum to g − a. Thus there are Σ[(g−a+1)(g−a+2)(g−a+3)/6, a = 0, g] = (g+1)(g+2)(g+3)(g+4)/24 many quintuples (a, b, c, d, e) that sum to g. Case 6. If 6 nonnegative numbers (a, b, c, d, e, and f) sum to g, and the first number is “a” (0 ≤ a ≤ g), then the other 5 numbers b + c + d + e + f = g − a. Based upon Case 5 above, for a given value of “a” (0 ≤ a ≤ g), there are (g−a+1)(g−a+2)(g−a+3)(g−a+4)/24 many quintuples that sum to g − a. Thus there are Σ[(g−a+1)(g−a+2)(g−a+3)(g−a+4)/24, a = 0, g] = (g+1)(g+2)(g+3)(g+4)(g+5)/120 many sextuples (a, b, c, d, e, f) that sum to g. This contest problem corresponds to Case 6 with g = 12. Thus, there are 13×14×15×16×17/120 = 6188 many sextuples of nonnegative The solution is: (D), 6188. 16. Solution by CwH. In parallelogram ABCD, BC is extended beyond C to point E. Points F and G are the points of intersection of AE with BD and CD, respectively. If FG = 12 and EG = 15, find AF. Let AF = x, BC = AD = p and p B q C p B E 15 15 F x p A F 12 x ! = !!! D Figure 2 CE = q. From Figure 1, we have !"!!" 12 p A D Figure 1 ! q C ! !" = ! !!!" or 𝑞 = !"! !!!" . From Figure 2, we have or 𝑥(𝑝 + 𝑞) = 27𝑝. Substitute q from the first equation into the second equation to get 𝑥 𝑝 + !"! !!!" = 27𝑝. Dividing both sides by p and multiply both sides by (𝑥 + 12), we have 𝑥 𝑥 + 12 𝑝 + 15𝑥𝑝 = 27𝑝(𝑥 + 12), or 𝑥 ! 𝑝 + 27𝑥𝑝 = 27𝑥𝑝 + 324𝑝. Simplifying this, we have 𝑥 ! = 324 or 𝑥 = 18. The answer is (B). 17. Solution by CwH. Find the greatest possible value of 𝑛𝑏 if 𝑓 𝑥 = 𝑏(𝑥 − 𝑚)! + 𝑛 and 𝑔 𝑥 = 𝑥 − 𝑚 intersect. Now if these two functions intersect, there must some 𝑥 such that 𝑏(𝑥 − 𝑚)! + 𝑛 = 𝑥 − 𝑚, or 𝑏(𝑥 − 𝑚)! − 𝑥 − 𝑚 + 𝑛 = 0 has real solutions. This is a quadratic equation in 𝑥 − 𝑚 . By the quadratic formula, we have: 𝑥−𝑚 = 1 ± 1 − 4𝑏𝑛 2𝑏 In order to have a real solution, we need 1 − 4𝑏𝑛 ≥ 0. The greatest possible value for ! 𝑛𝑏 then is . The answer is (A). ! 18. Solution by CwH. Consider a random sample of 100 members in the Math Club, we can reasonably expect to see 69 of them are in the Physics Club and 79 of them are in the Quiz Team. Let Z be those of the 100 who are in the Physics Club, and Q be those who are in the Quiz Team, and P be those who are in both the Physics Club and Quiz Team. The largest possible size for P is when all the members of Z are also in Q, E Z 69% P Q 79% and P contains 69 members. On the other hand, the smallest possible size of P can be found by the relation #(𝑍 ∪ 𝑄) = # 𝑍 + # 𝑄 − #(𝑃). This is because in the sum # 𝑍 + # 𝑄 , we have counted their common part, P, twice, and thus, we need to subtract #(𝑃) from this sum in order to get the correct count of those who belong to either Z or Q. But #(𝑍 ∪ 𝑄) ≤ 100 since there may be members of the Math Club who do not belong to either Z or Q. Thus, 100 ≥ # 𝑍 + # 𝑄 − #(𝑃), and thus, #(𝑃) ≥ # 𝑍 + # 𝑄 − 100 = 69 + 79 − 100 = 48. We then have 48 ≤ # 𝑃 ≤ 69, and hence, the percentage of members belonging to P is between 𝑀 = 48% and 𝑁 = 69%. Thus, 𝑀 + 𝑁 = 117%. The answer is (C). 19. Solution by CwH. #19. We will show that it is impossible to add pennies so that all the sides of the octagon will carry the same number of pennies. First, observe that if we carry out a sequence of steps of adding pennies, the final numbers of pennies on the sides of the octagon do not depend on the particular order in which we carry out these steps. For instance, suppose we want to add a penny to each of the pairs BC-CD, CD-DE, and CD-DE, the final results will all be the same regardless which of these three pairs are added first, which the second, and which the third. B 1 C 3 1 A D 0 2 H E 0 1 G 0 F Now suppose there is a procedure for adding pennies so that all the sides of the octagon will carry the same number, say m, of pennies. We will show this will lead to a contradiction. Since the order of execution for the procedure is not important, we can execute all those steps involving the side CD. Suppose among all these steps, k of them involve the pair BC-CD (and 𝑚 − (𝑘 + 3) are from the pair CD-DE). When all these steps are done, the side CD will carry 𝑚 pennies, BC will carry 𝑘 + 1 pennies, and DE will carry 𝑚 − (𝑘 + 1) pennies. Now carry out all the steps involving the side BC. Since we have already carried out all the steps involving CD, the steps involving BC must all come from the pair AB-BC. There will be 𝑚 − (𝑘 + 1) such steps, and consequently, AB will carry 𝑚 − 𝑘 many pennies. Now carry out all the steps involving the side AB. Since we have already carried out all the steps involving BC, the steps involving AB must all come from the pair AB-HA. There will be 𝑘 such steps to bring the number of pennies on AB to 𝑚, and consequently, HA will carry 𝑘 pennies. The pattern is now clear, there must be 𝑚 − 𝑘 steps in the procedure involving the pair HA-GH, and after we carry them out, GH will carry 𝑚 − 𝑘 pennies. There must be 𝑘 steps involving the sides GH-FG, and after we carry them out, FG will carry 𝑘 pennies. Finally, there will be 𝑚 − 𝑘 steps involving the pair FG-EF, and after all these are done, EF will carry 𝑚 − 𝑘 pennies. Recall when we execute all the steps involving CD, we noted that the side DE carried 𝑚 − (𝑘 + 1) pennies, and now, the side EF carries 𝑚 − 𝑘 pennies. Thus, the number of pennies on DE and EF are not the same. However, the only steps left are for the pair DEEF, which will add the same number of pennies to both sides, and hence, can never make them equal. Consequently, it is impossible to make all the sides carrying the same number of pennies. The answer is (A). 20. Solutions by CwH. Given five distinct integers a, b, c, d, e and we are supposed to arrange them in the order of least to the greatest. We are told that e, d, c, b, a has at least 3 of the 5 values correctly placed. I was a little puzzled by what “correctly placed” meant. Finally, I decided that it meant that if the number whose value is the kth smallest is at the kth place, then it is correctly placed. Once this is clear, we see that if an arrangement has 4 out of 5 values correctly placed, then the 5th value must be correctly placed as well. Thus, the arrangement e, d, c, b, a either has 3 values correctly placed or all 5 values correctly placed. We are also told that the arrangement a, b, c, d, e has an odd number of values correctly placed. We may conclude that the values, b and d, are either both correctly placed, or both incorrectly placed. This is because a, b, c, d, e is obtained from e, d, c, b, a by interchanging the positions of b and d. If only one of them was in the correct position, then interchanging their positions would make both placements incorrect. Thus, the number of correctly placed values of a, b, c, d, e can only differ from that of e, d, c, b, a by one. But as we observed that both a, b, c, d, e and e, d, c, b, a have an odd number of values correctly placed, this cannot happen. Thus, the values, b and d, are either both correctly placed, or both incorrectly placed in e, d, c, b, a. I contend that if we learn the relative order of b, c, d, then we can arrange all five numbers in their correct order. We do this by considering the following cases: 1). Suppose the orders for b, c, d are given such that 𝑏 < 𝑑. Then b and d are not in their correct places in e, d, c, b, a. But this arrangement has at least three values in their correct places, and thus, e, c, and a are in their correct places. All we need is to interchange the positions of b and d to make all 5 values in their correct places. In the following we need only consider the cases for 𝑑 < 𝑏. Observe that when this happens 𝑑 and 𝑏 must both in their correct places. This is because that if one of them is not, then the other is also not in its correct place. As noted above since e, c, and a are already in their correct places and since 𝑑 < 𝑏, there is no other places for 𝑑 and 𝑏 to go. 2). If the order given is 𝑐 < 𝑑 < 𝑏. Since 𝑑 and 𝑏 are in their correct places, this forces c to be the smallest element, which makes a not in its correct place. But a, b, c, d, e has at least 3 values correctly placed, the value e must be the greatest element, and the correct arrangement in this case is c, b, a, d, e. 3). If the order is 𝑑 < 𝑐 < 𝑏. Then the correct arrangement would either be a, b, c, d, e or e, b, c, d, a. But we are told that the first is not a solution, thus, the correct solution is e, b, c, d, a. 4). If the order is 𝑑 < 𝑏 < 𝑐, then c is the largest element, and both c and a are misplaced in e, d, c, b, a. In this case, the other 3 values are in their correct place. All we need is to trade places for c and a to get the correct arrangement. Thus, the answer is (C).