Math Contest Solutions

Math Contest Solutions 1. The first discount is 0.04 d, and the second discount is 0.10 (2d). Adding these together gives 0.24 d. But we have to express this as a percentage of 3d, which was the total amount paid. So we have 0.24 d = 0.08 (3d) and the overall discount is 8%, which is answer (D). 2. Let x be the price of a gallon of gas, and y the price of a quart of oil. Also let p be the price that Jo pays, which we want to find. Since Jo bought 2 gallons of gas and 2 quarts of oil, we have: p = 2x + 2y. But we are also given that x = y + 1.20, and y = 0.6 x. Eliminating y between these two equations, we have: x = 0.6 x + 1.20, which gives x = 3.00. Then y = 3.00 – 1.20 = 1.80. So we get: p = 2 (3.00) + 2 (1.80) = 9.60, and the answer is (C). 3. In the figure above, if we add together angles BEF, DCE, and ABC, we must get 360 because if we do the following rotations: 1. Segment AE into segment FE (x – 15) 2. Segment FC into segment DC (x + 10). 3. Segment DB into segment AE, we have done one complete clockwise revolution. 1 So we have: (x – 15) + (x + 10) + x = 360, or 3 x – 5 = 360, which gives x = 121.7. This is in the range [120, 123), and the answer is (C). 4. Let x = number of dimes, y = number of nickels, and z = number of pennies. We are given the conditions: x = y, and z = 0. Now Ed spends 15 cents. Without pennies available, this can be done in two ways. Case 1: Ed uses three nickels. So now we have x = y + 3, and we are also told that x = 2y. Eliminating x between these two equations, we have: 2y = y + 3, which yields y = 3 after three nickels have been spent. So the original number of nickels was 6. Case 2: Ed uses one dime and one nickel. So we still have x = y, but now we also have x = 2y. The only solution of this is x = y = 0. Since Ed spent one dime and one nickel, in this case, he originally had 1 dime and 1 nickel. Since one is not listed as one of the possible answers, we conclude that the desired answer is (D), six nickels. 5. Here, I first give a solution provided by Gregory Melblom, which does not require logarithms. This is followed by Marc’s longer solution, in which logarithms are used. First solution by Gregory: Note that the number AA has an “A” in the ten’s place and also an “A” in the ones place, so the value of the number can be expressed as: AA = 10A + A = 11 A. It follows that AA + A = 12 A. So the equation given in the problem becomes: 12 A = B AB. Since A is nonzero, we may divide both sides by A with the result: 12 = B A(B-1). We see that B must a factor of 12, since A(B-1) is an integer. We cannot have A = 1, because this would give B = 12, and B must be a one-digit integer. With A > 1, we see that B cannot be 4 or greater, because 12 = 22 x 3 cannot have any integer factors that are cubed or to any higher power than squared. This leaves choices 1, 2, or 3 for B. B = 1 in the equation 12 = B A(B-1) does not work, because it gives 12 = A0, which does not have integer solutions. 2 B = 2 allows A to be 6 in the equation, but 6 is not listed among the possible answers. B = 3 gives A = 2, which is listed, and the answer is (B). Here is Marc’s longer solution using logarithms: There are more cases to check here, so the solution is less efficient than Gregory’s solution above. As above, we obtain the equation: 12 A = B AB. We take the natural log of both sides, using the properties of logarithms, and obtain: Ln 12 + Ln A = Ln B + B Ln A, which can be written as: Ln 12 – Ln B = (B – 1) Ln A. We write this as: 12 𝐿𝑛 # ' = 𝐿𝑛 𝐴!"# 𝐵 #$ = 𝐴!"# . The right side of this equation ! must be an integer, so the left side must be too. So B has to be a factor of 12, which gives the possibilities: B = 1, 2, 3, 4, 6, and 12. Take exponential of both sides and get: If B = 1, the equation becomes 12 = A0, which has no solution for A. If B = 2, we can satisfy the equation with A = 6. If B = 3, we can satisfy the equation with A = 2. If B = 4, the equation becomes 3 = A3, which has no integer solution for A. If B = 6, the equation becomes 2 = A5, which also has no integer solution for A. Finally, if B = 12, we can satisfy the equation with A = 1. In the statement of the problem, A and B are listed as single non-zero digits, so we have to disallow B = 12. Also, A = 6 is not listed as one of the answers, so the remaining answer we can choose is A = 2, which is (B). 6. Let us assume that each person at the table can see everyone except themselves. One person makes the statement: “I see at least one person who sees only knaves.”. Let us first assume that this is a true statement, so that the person making the statement is a knight. But he says that another person in the group sees only knaves, 3 and if that is true, then the person making the statement is a knave, contradicting the assumption that the speaker is a knight. So the speaker must be a knave, and there is at least one knight in the group, and this person cannot be the speaker. However, if that person were the only knight in the group, then she/he would see only knaves, and then the speaker’s statement would be true (at least one person sees only knaves) and the speaker would be a knight, contradicting our earlier finding that the speaker is a knave. So there must be at least two knights. If there is a minimum of two knights in the group, then no one will see only knaves, and the speaker’s statement will be false, as it must be. So the minimum number of knights is two, and the answer is (C). 7. By the Pythagorean theorem, we have the condition: x2 + y2 + z2 = 172. We need integer solutions to this equation, which is known as a Diophantine equation, after the mathematician Diophantus, who lived in the 2nd century AD and made some progress in solving such equations. Sometimes, we can save a bit of work by using modular arithmetic, in which each term in the equation is replaced by the remainder it leaves when divided by some chosen divisor. In this case, I decided to use divisor four. The possible remainders that can occur when a number is divided by four are 0, 1, 2, and 3. Consider the first term x. We can write the following equation: x = 4q + r, where q is the quotient, and r is the remainder. If x is squared, what possible remainders can occur? To answer this, write x2 = (4q + r)2 = 16q2 + 8qr + r2. The possible values of r2 are: 0, 1, 4, 9, all of which are either divisible by four with no remainder, or they leave remainder one when divided by four. So when an integer is squared, the only possible remainders that can occur are zero or one. We can re-write the Pythagorean equation as: rx + ry + rz = 1, where rx is the remainder left when x2 is divided by four, and similarly for ry and rz. The right side of the equation is one because 172 = 289 leaves remainder one when divided by four. We see that if x2 leaves remainder one when divided by four, then y2 and z2 must be multiples of four (leave remainder zero). This leaves the following possibilities for y and z: 2, 4, 6, 8, 10, 12, 14, and 16, all the even numbers less than 17. This must be expected, since any even number already has a factor of two, so if it is squared, it must then have a factor of four. Suppose x2 is assumed to leave remainder one. Then x could take on all the odd numbers less than 17, since the evens leave remainder zero when squared. I proceeded as follows: 4 Let x = 1: Then the original equation becomes: y2 = 288 – z2. Immediate success! I found the triple (x, y, z) = (1, 12, 12). So we search for the second solution. Just a reminder of what we are doing. Note that 12 + 122 + 122 = 172. Let x = 3: We have y2 = 280 – z2, and I found no perfect squares when the allowed even values of z were substituted into this equation. Let x = 5: Then the equation becomes: y2 = 264 – z2, and again I found no perfect squares. Let x = 7: The equation is: y2 = 240- z2, and again, no perfect squares on the left. Let x = 9: The equation is: y2 = 208 – z2, and I found the second triple: (x, y, z) = (9, 12, 8). I checked the remaining possibilities x = 11, 13, and 15, and found no further solutions. Note that the volume of the rectangular solid is V = xyz, and we want the ratio of %&#$&' volumes, so we take: = 6, and the answer is (E). #&#$&#$ 8. Here I give a shorter, more direct solution by Chungwu Ho, followed by Marc’s longer solution: Chungwu’s solution: Since 527, 622, and 698 give the same nonzero remainder when divided by the integer N, then the difference of any pair of these three integers must be a multiple of N. This follows because the common remainder will cancel when the difference between any pair is taken. We see that 622 – 527 = 95 = 19 x 5, and 698 – 622 = 76 = 19 x 4, and 698 – 527 = 171 = 19 x 9. The only common factor is 19, so we conclude that N = 19. Similarly, the integer M must divide the pairwise differences: 881 – 736 = 145 = 29 x 5, 997 – 886 = 116 = 29 x 4, and 997 – 736 = 261 = 29 x 9. The only common factor is 29, so we have M = 29. So M + N = 29 + 19 = 48, and the answer is (B). Marc’s solution to problem 8: By the quotient remainder theorem, there exist three integers, q1, q2, and q3, known as quotients, such that the following three equations hold: 5 527 = q1 N + r, 622 = q2 + r, and 698 = q3 + r. Now we use the given condition that the nonzero remainder is the same in all three equations and we eliminate the remainder: 527 – q1 N = 622 – q2 N, 527 – q1 N = 698 – q3 N, and 622 – q2 N = 698 – q3 N. After rearranging terms, we arrive at the following three equations: (q2 – q1) N = 95, (q3 – q1) N = 171, and (q3 – q2) N = 76. Since each of the differences of two quotients on the left sides are integers, we see that N must simultaneously be a factor of 95, 171, and 76 on the right. Consider the first equation, containing 95 = 5 x 19. Since 5 and 19 are both prime numbers, the possible values that N could take on are: N = 1, 5, 19, and 95. From the second equation, we see that 171 = 3 x 3 x 19, and so the possible values of N from this equation are: N = 1, 3, 9, 19, 57, and 171. Notice how we are including all possible combinations of the prime factors of 171. For example, 57 = 3 x 19. From the third equation, we see that 76 = 2 x 2 x 19, and the possible values of N are: N = 1, 2, 4, 19, 38, and 76. Only two numbers appear in each of the three lists of possibilities for N. These numbers are 1 and 19. However, since the number one leaves remainder r = 0 when divided into any number, we exclude that possibility, since r was given as nonzero. So we arrive at the conclusion that N = 19. Now we do the same procedure for M. We have: 736 = v1 M + s, 881 = v2 M + s, and 997 = v3 M + s, where v1, v2, and v3 are the integer quotients, and again the remainder s is given as nonzero. As we did above, we eliminate the remainder s between these equations, and arrive at the following three equations: (v2 – v1) M = 145, (v3 – v1) M = 261, and (v3 – v2) M = 116. In the first equation, 145 = 5 x 29, so M = 1, 5, 29, or 145. In the second equation, 261 = 3 x 3 x 29, so M = 1, 3, 9, 29, 87, or 261. Finally, in the third equation, 116 = 2 x 2 x 29, so M = 1, 2, 4, 29, 58, or 116. 6 Again, the common value M = 1 is rejected because it gives remainder s = 0, contrary to the given condition. The only other common number is 29, so M = 29. The problem asks for the sum N + M = 19 + 29 = 48, and the answer is (B). 9. Here I give a shorter more direct solution by Gregory Melblom, followed by Marc’s longer solution. In my longer solution, I introduce some information about linear algebra for students who are studying that subject. Gregory’s solution: Using the concept of place holding, we can write the three given conditions as: 100 a + 10 b + c = 29 (a + b + c), 100 b + 10 c + a = 68 (a + b + c), and 100 c + 10 a + b = 14 (a + b + c). We add 7 times the first equation, 19 times the second equation, and 28 times the third equation together, with the result: 999 (a + 2b + 3c) = 1887 (a + b + c) Note that 999 and 1887 have 111 as a common factor, so we divide this factor out, yielding: 9 (a + 2b + 3c) = 17 (a + b + c). Since 17 is not a factor of 9, it must be a factor of (a + 2b + 3c). The only choice among the answers given with 17 as a factor is answer (C). It is possible that a + 2b + 3c could be 34, which would correspond to a + b + c = 18 in the last equation above. But note we cannot have a + 2b + 3c = 51 = 17 x 3 or higher because this would require a + b + c = 27 or greater. We cannot have a = b = c = 9, because this would require simultaneously: The number abc = 29 (a + b + c) and also the number abc = 68 (a + b + c). Here is Marc’s longer solution, which uses techniques of linear algebra: As above, obtain the three equations: 100 a + 10 b + c = 29 (a + b + c) 100 b + 10 c + a = 68 (a + b + c), 7 100 c + 10 a + b = 14 (a + b + c). We rearrange terms in each of these equations to get: 71 a – 19 b – 28 c = 0, -67 a + 32 b – 58 c = 0, -4 a – 13 b + 86 c = 0. This is known as a homogeneous system of equations, since the right sides are all zero. By inspection, it is clear that we have a trivial solution (a, b, c) = (0, 0, 0). However, we need a non-trivial (non-zero) solution in which each of a, b, and c is a positive, single-digit number. Using the techniques of linear algebra, we can write the system of equations as a matrix equation. 𝑎 71 −19 −28 0 ! −67 32 −58 - .𝑏 2 = !0𝑐 −4 −13 86 0 In the usual way of solving systems of equations taught in elementary algebra classes, we add multiples of one equation to the others in order to cancel out terms and eventually isolate one variable. In linear algebra, this is equivalent to performing row operations on a matrix, known as the augmented matrix, in which an extra column is added to the right side of the matrix of coefficients above. In this case, the extra column will have all zero’s. Row operations consist of adding a multiple of one row to another row of the matrix. We can also interchange two rows. Here is the augmented matrix for the system of equations above: 71 −19 −28 0 !−67 32 −58 0−4 −13 86 0 There is a theorem in linear algebra, which says that a homogeneous system of equations has a non-trivial solution if and only if it contains a free variable. Such a variable is one whose value is not determined. Instead, we arbitrarily choose a value for the free variable, and then the values of the other two variables are determined. We will see shortly that the variable c will be free, and we can choose values of c to calculate the values of a and b. The restriction that a, b, and c all be positive, singledigit numbers will allow only one suitable solution for this problem. To get the solution, we perform row operations on the augmented matrix above. As mentioned, this is equivalent to adding multiples of one equation to another in the 8 system. Our goal is to put the matrix into a form in which the solution can be read off immediately. Such a form is known as reduced row echelon form of the matrix. You can reach this site by doing a Google search on “linear algebra toolkit”. So I submitted the above augmented matrix into the site and obtained the following matrix in the desired form: 1 0 −2 0 !0 1 −6 00 0 0 0 The interpretation of this matrix in terms of equations is as follows: First row: a – 2 c = 0. Second row: b – 6 c = 0, and the bottom row of all zeros represents 0 = 0, indicating that variable c is free. We have eight possible choices for c (1 – 9), so we start with c = 1. In this case we get a = 2, and b = 6. Note that the choice of c = 2 or higher would result in b > 9, which is not allowed, so we have found our desired solution for a, b, and c. The problem asks us to calculate the quantity: a + 2 b + 3 c = 2 + 2 (6) + 3 (1) = 17, and the answer is (C). This is a good practice problem for students taking linear algebra. 10. Solution by Chungwu Ho: The part of DTYC which is not congruent must be a side. Otherwise, the two triangles would have all equal sides, which would force them to be congruent triangles. This is known as the SSS rule for congruence in elementary geometry. I do not include the proof here. Since the three angles are equal for the two triangles, they must be similar, by the AAA rule for similar triangles (proof also not included here). The given triangle SML has sides of length SM = 8, ML = 12, and LS = 18. We look at the ratios of sides, and we find that ML/SM = 3/2, and LS/ML = 3/2. Let the sides of DTYC have length s, m, and b, from smallest to largest. Since this is similar to the given triangle, we must have m/s = 3/2, and b/m = 3/2, and two of these sides must be equal to two of the given sides. We look at the possibilities: 9 s = 8, m = 12: Since b/m = 3/2, this would require b = 18 and the resulting triangle would be congruent to the given one. But recall we are given that the two triangles are not congruent, so we throw out this case. s = 8, m = 18: This does not satisfy m/s = 3/2, so we throw it out. s = 8, b = 12: Since m/s = 3/2, and b/m = 3/2, we have b/s = 9/4, so b cannot be 12. s = 8, b = 18: This would give m = 12, and again, the two triangles would be congruent. So this is not allowed. s = 12, m = 18: This satisfies m/s = 3/2, and if we choose b = 3/2 m = 27, we satisfy all of the desired ratios. So the long side of DTYC could be 27. The only other possible case would be: s = 12, b = 18, but this again would contradict b/s = 9/4, so it is not allowed. So the long side is 27, and the answer is (C). 11. Solution courtesy of Chungwu Ho. The given expression can be written as: ab – (a + b) = (a – 1) b –a. Since 𝑎 ≤ 𝑏 and both of them are positive integers, the smallest possible value of a is 2. With this choice of a, the given expression becomes: (a – 1) b –a = b – 2. It is clear that every positive integer N can be represented in this way, using b = 2, 3, 4, etc. So every integer now has one possible representation in the desired form. The next choice of a is a = 3, and the given expression becomes: (a – 1) b – a = 2b – 3. The definition of an odd integer is one which can be written in the form 2n + 1 for some integer n. But 2b – 3 = (2b – 4) +1, and by closure of the integers under multiplication and subtraction, (2b – 4) is an integer if b is. So (2b – 3) represents positive odd integers for b = 3, 4, etc. Now we see that every positive odd integer, starting with 3, has two representations in the desired form. Will some positive odd integer be the first one to receive a third such representation? Read on! The next choice of a is a = 4, and in this case the expression becomes: (a – 1) b – a = 3b – 4, which can be either even or odd, depending on the choice of b. Since 𝑎 ≤ 𝑏, the smallest choice of b is b = 4, so 3b – 4 = 8, an even integer, which is 10 the first to even to have two representations in the desired form. Recall that every odd integer starting with three already has two such representations. We go to the next possible value of b, which is b = 5, and 3b – 4 = 11, an odd. So we see that the number 11 is the first positive integer to receive three representations in the desired form, and this is the answer to the question. Note that by contest rules, this question is marked either blank, if the student left it blank, or correct. It is never marked wrong. 12. Gregory Melblom kindly supplied three excellent solutions to this problem. These are followed by a longer solution by Marc. First solution: By the definition of division with remainder, we may write the following equation: x2000 – 2x15 + 2 = q(x) (x2 – 1) + r(x), where q(x) is the quotient, and r(x) is the remainder when the original polynomial is divided by (x2 – 1). If we put x2 = 1 into this equation, the result is: (x2)1000 – 2x (x2)7 + 2 = r(x), where I used the laws of exponents to write x2000 = (x2)1000 and x15 = x(x7)2. Since x2 = 1 and 1 to any power is still 1, the equation reduces to: -2x + 3 = r(x), and the answer is (A). This solution is remarkably creative and anyone, whether they are a student or the math contest moderator, is well advised to study it carefully! Second Solution: Notice that the exponents of x2000 and x14 are both even. Therefore, I claim that (x2000 – 1) and (x14 – 1) both have (x2 – 1) as a factor. To see this, note the following factorization: x14 – 1 = (x2 – 1) (x12 + x10 + x8 + x6 + x4 + x2 + 1). In fact, for any even exponent, we have the factorization: 𝑥 !" − 1 = (𝑥 ! − 1)9𝑥 !("$%) + 𝑥 !("$!) + ⋯ + 𝑥 ' + 𝑥 ! + 1< Factoring polynomials can be a very creative process! The usefulness of this factoring result here is that if you have a number or expression that leaves a certain remainder when divided by some divisor, you can add or subtract multiples of the divisor without changing the remainder. For example, we see that 5 leaves remainder 2 when divided by 3. If we add a multiple of three, say 12, to 5, we get 17, which also leaves remainder two since 17 = 5 x 3 + 2. Since this result is plausible, I will not give a formal proof of it here. So in the problem at hand, we can add or subtract any multiple of (x2 – 1) to the given polynomial, without changing the remainder left when the expression is divided by (x2 – 1). 11 So the original polynomial given in the problem and the following expression must leave the same remainder: (x2000 – 2x15 + 2) – (x2000 – 1) + 2x (x14 – 1). But this last expression is just (-2x + 3), so this is the remainder itself, since it is of lower degree than (x2 – 1). The answer is (A). Third solution: Since (x2 – 1) divides itself with remainder zero, we see that x2 must leave remainder 1 when divided by (x2 – 1). We can write this as an equation: 𝑥 ! ≡ 1, where is it understood here that the divisor is (x2 – 1). This is called a congruence relation in the subject of modular arithmetic, which is taught in discrete mathematics. We can take both sides of the congruence to the 1000th power and obtain: 𝑥 !((( ≡ 1. Similarly, we have 𝑥 %' ≡ 1. We can multiply both sides of the last congruence by -2x and get: −2𝑥 %) ≡ −2𝑥. If we add the three congruences together, we get the result that the original polynomial given in the problem is congruent to (-2x + 3), which is the remainder left when the polynomial is divided by (x2 – 1). So the answer is (A). Here is Marc’ longer solution. I include it here because it is a method that some students may attempt. To do this problem, I set up a long division of the given polynomial by (x2 – 1). Since the entire division calculation would be very lengthy, I started it and looked for patterns. The result is show in the figure below: 12 From the calculation, we can see that the remainder is -2x + 3, and the answer is (A). 13. Using the properties of logs, the given equation can be written as: 𝑙𝑛(𝑘𝑥) = 𝑙𝑛(𝑥 + 1)$ . We take exponential of both sides and get: 𝑘𝑥 and get the quadratic equation: = (𝑥 + 1)$ . We multiply this out 𝑥 $ + (2 − 𝑘)𝑥 + 1 = 0. 13 By quadratic formula, the solutions are: 𝑥 = ()"$)±,($"))! "- . In order to have $ exactly one real solution, we need the expression under the radical to be zero: (2 − 𝑘)$ = 4 This equation has solutions k = 0 and k = 4. But we have to throw out k = 0 because we have ln(kx) in the original equation, and ln(0) is not a real number. So the only result is k = 4, and the answer is (A). 14. Solution courtesy of Chungwu Ho: If we add the two given equations together, we obtain the result: x (1 – y) + z (y – 1) = 2. Let n = y – 1, and the last equation above becomes: n (z – x) = 2. Note that this equation gives the number 2 as the product of two integer factors. Since 2 is the smallest prime number, the integer factors can only be ±1 and ±2. The gives four possibilities: n = -1, z – x = -2, so we have y = 0. We put y = 0 back into the original set of equations and obtain the pair: x = 107, z = 105, so we have one solution in all integers. n = 1, z – x = 2, so we have y = 2. Putting y = 0 into the original equations gives the pair: 2x + z = 105, and x + 2z = 107. We multiply the second equation by (-2) to get -2x -4z = -214, and then add this to the first equation in the pair to eliminate x, with the result: -3z = -109. But this yields a non-integer value for z, since 109 is not a multiple of 3. So we have to throw out this solution. n = -2, z – x = -1, so we have y = -1. We put y = -1 into the original equations, and get the resulting pair: -x + z = 105, and x – z = 107. If we add these two equations together, we get the impossible equation: 0 = 212 ! Well, I guess we better throw out that solution. One case remains. n = 2, z – x = 1, so we have y = 3. 14 With y = 3, the original equations now become: 3x + z = 105, and x + 3z = 107. We multiply the second equation by -3 to get: -3x – 9z = -321, and add it to the first equation to eliminate x, with the result: -8z = -216. This time, fortune favors us, since 216 is indeed a multiple of eight, and we have z = 27 and x = 26. So the total number of solutions was two, and the answer is (C). 15. This problem can be done purely geometrically, and I am sure that would be elegant, but I wanted to make use of a great breakthrough by Rene Descartes, who lived between 1596 and 1650. That breakthrough is the connection between geometry and algebra. In this method, I place the given triangle onto a set of coordinate axes, as shown in the figure below. I did use one property of 30-60-90 triangles that we learned in middle school, namely that the short side is one-half of the hypotenuse (the longest side). Since the hypotenuse is given as 2√6, we see that the short side along the positive x-axis, CB, must be √6. The longer side, y = AC, is obtained by the Pythagorean theorem: 6 + y2 = 24, which has solution 𝑦 = 3√2. Since the segment CD bisects a 90 degree angle, it makes an angle of 45 degrees with respect to the positive x-axis. Since point C is at the origin, we can immediately write the equation of the line containing CD as: y = x, which has an inclination of 45 degrees. The line containing segment AB contains point A = (0, 3√2), and point B = (√6, 0), so using techniques from elementary algebra, we write the equation of this line as: 𝑦 = −√3 𝑥 + 3√2. Then point D is the intersection of this line with the previous line y = x. This gives the equation: 𝑥 = −√3 𝑥 + 3√2. .√$ = y. Then we see that the length √.0# of segment CD is given by the Pythagorean theorem as: We solve this for x and get the result: 𝑥 𝐶𝐷! = = 18 18 36 + = ! ! ! 9√3 + 1< 9√3 + 1< 9√3 + 1< We take the square root and obtain: 𝐶𝐷 1 . But of course, one must not #0√. tolerate an irrational denominator, as demanded in the elementary algebra courses, so we do extra work to rationalize it: 𝐶𝐷 = 1 = #"√. 8 9 = 3√3 − 3, #0√. #"√. 15 and the answer is (A). 16. We are given the condition: sin(x)cos(x) = 4sin(x) + 4cos(x). Recall the trigonometric identity: sin(2x) = 2sin(x)cos(x). So we can re-write the given condition as: sin(2x) = 8sin(x) + 8cos(x). We square both sides of this last equation, with the result: sin2(2x) = 64sin2(x) + 128sin(x)cos(x) + 64cos2(x). Using the trigonometric identity sin2(x) + cos2(x) = 1, we see that this simplifies to: sin2(2x) = 64 + 64sin(2x), where I replaced 128 sin(x)cos(x) = 64 sin(2x). We make the substitution u = sin(2x), and obtain a quadratic equation: u2 – 64u – 64 = 0, which by quadratic formula has two solutions which simplify to: 32 ± 8√17. We choose the plus sign, since that is the form desired in the problem, and we find that 32 + 8 + 17 = 57, and the answer is (E). 16 17. The three circles are shown in the figure below. We see that segment AP = AO - 10 = 15. Using the similar triangles, we form the ratio: 23 42 = #5 $6 = 7" 48 , which gives the result y1 = 6. Then we observe that QP = QE = radius of the second largest circle. Also note that AQ = AP – QP = AP – QE. So we set up another ratio: 7" 48 = 1 #6 = 9: 49 = 9: 48"9: . With AP = 15, we can solve for QE, with the result QE = 30/7. For algebra students, the last equality above is easily solved by cross-multiplying. So the equation: 6 𝑄𝐸 = 15 15 − 𝑄𝐸 is equivalent to: 6 (15 – QE) = 15 QE. Then we continue with the observation that: AQ = AP – QE = 15 – 30/7 = 75/7. Again, for algebra students, the equation: 15 – 30/7 is equivalent to again. #6 # − .5 ; = (#6)(;)".5 ; = ;6 ; , where I cross-multiplied This technique works for addition or subtraction of any two fractions. The resulting denominator is always equal to the product of the two individual denominators. However, such a product is not always the least common denominator. For some reason I do not understand, many algebra instructors wish to emphasize obtaining the least common denominator before attempting to add or subtract two fractions. Mathematically, this is not necessary, since the result can always be reduced to lowest terms later if desired. It is just an additional cognitive task imposed on the students for no good reason as far as I can see. An emeritus mathematician at UC Berkeley, Dr. Hung-Hsi Wu, has written extensively about this peculiar practice in elementary algebra. He has written a book: “Understanding Numbers in Elementary School Mathematics”, (2011), AMS, Providence RI. Anyway, back to the problem at hand. Now we observe that AR = AQ – RQ = AQ – QE = 75/7 – 30/7 = 45/7. 17 Setting up another ratio, we have: 𝑦! 𝑦% 6 = = 𝐴𝑅 𝐴𝑃 15 Since AR = 45/7, we can solve for y2 with the result y2 = 18/7. Now we observe that AS = AR – SR, and SR = SG = radius of the smallest circle. We can set up another ratio: 𝑦! 6 𝑆𝐺 = = 𝐴𝑅 15 𝐴𝑅 − 𝑆𝐺 With AR = 45/7, we can solve for SG as above by cross-multiplying, with the result SG = 90/49. Now we have the ratio: <= 4< = 23 42 = #5 . $6 With SG = 90/49, we solve for AS with the result AS = 225/49. We take the square root of this and get the result 15/7, and the answer is (B). 18 18. Solution courtesy of Gregory Melblom, followed by Marc’s solution. We are given M = a + b, and N = ab. Let a = 1 and b = 0, which gives M = 1 and N = 0. Then the given equation: a4 + b4 = PM4 – QM2 N + RN2 reduces to 1 = P. Now let a = 1 and b = -1, so that M = 0 and N = -1. In this case, the given equation reduces to 2 = R. Finally, let a = 1 and b = 1, so that M = 2 and N = 1. Then the given equation with P = 1 and R = 2 becomes: 2 = 24 – Q (22) + 2, which yields Q = 4. Thus P + Q + R = 1 + 4 + 2 = 7, and the answer is (D). Here is Marc’s solution: Let us use the definitions of M and N and write out the desired equation as: a4 + b4 = P (a4 + 4a3b + 6a2b2 + 4ab3 + b4) – Q (a2 + 2ab + b2) ab + Ra2b2 We see immediately that P = 1, because the coefficient of a4 on the left side is one, and the first term on the right is the only one that contains fourth powers. So set us put P = 1 in the above equation and then cancel common fourth power terms: 0 = 4a3b + 6a2b2 + 4ab3 – Qa3b – 2Qa2b2 – Qab3 + Ra2b2 We collect similar terms, with the result: 0 = (4 – Q) a3b + (4 – Q) ab3 + (6 – 2Q + R) a2b2 Since the left side is zero, each coefficient on the right side must be zero, since a and b are assumed to be non-zero. This gives us 4 – Q = 0, or Q = 4, and 6 – 2Q + R = 0, which gives R = 2, if Q = 4. Thus, we have P + Q + R = 1 + 4 + 2 = 7, and the answer is (D). 19. Solution courtesy of Gregory Melblom: The prime factorization of 2016 is: 2016 = 25 x 32 x 7. We see that 2016 is not a perfect cube. In order to be a perfect cube, each exponent would need to be a multiple of three. So we cannot write 2016 as the product of three equal factors. However, by examining possible groupings of the prime factors, we see that there are six distinct ways to write 2016 as a product of three factors with two of the factors equal. I will write these triplets of numbers as follows: (2016, 1, 1), (504, 2, 2), (224, 3, 3), (126, 4, 4), (56 (6, 6), and (14, 12, 12). 19 A permutation of any of these triplets is a rearrangement of the numbers. For example, one permutation of (126, 4, 4) is (4, 126, 4). We are instructed not to count permutations of a triplet among our total. If we allowed different permutations to count, there would be three permutations for each of the triplets above. For example, we could write the triplet (504, 2, 2) as: (504, 2, 2), (2, 504, 2), or (2, 2, 504). We will use this result below. Now we will count the total number of three-factor arrangements, including all permutations, which we will then need to subtract out. Note that for any triplet of factors, there are six permutations of that triplet: (p, q, r), (p, r, q), (q, p, r), (q, r, p), (r, q, p), and (r, p, q). How do we count the total number of permutations? Let us first consider all possible three-factor arrangements of the number 25. We write these as (2a, 2b, 2c), where a + b + c = 5. We find that there are 21 distinct permutations with these restrictions. Next, we find the possible three-factor arrangements of 32: There are only six of these: (32, 1, 1), (1, 32, 1), (1, 1, 32), (3, 3, 1), (3, 1, 3), and (1, 3, 3). Finally, we find three arrangements involving 7: (7, 1, 1), (1, 7, 1), and (1, 1, 7). If we multiply permutations together by multiplying corresponding components, we get permutations, whose product of components gives 2016. For example, we take the product (22, 23, 1) x (1, 3, 3) x (1, 7, 1) = (4, 168, 3). Since we have 21, 6, and 3 possible three-factor arrangements of each prime factor, the total number of permutations is 21 x 6 x 3 = 378. This includes all permutations in which the three factors are distinct, as well as those for which two of the factors are equal. But we want only the number of distinct three-factor triplets without regard to order of the factors, so we have to subtract off permutations. To do this, we first subtract all three-factor triplets, where exactly two factors are equal, together with their permutations. Recall that there are six possible threefactor arrangements with exactly two of the factors equal, and each of these arrangements has three permutations (re-ordering of the factors). So there are 6 x 3 = 18 of these altogether. After subtraction we have 378 – 18 = 360 three-factor arrangements, in which the three factors are all distinct. But recall that each three-factor arrangement with distinct factors has six permutations, so we have to divide 360 by 6 to get 60 three-factor arrangements with distinct factors, but no extra permutations included. But recall that we subtracted out all three-factor arrangements with exactly two factors equal. There are six of these, not counting permutations. So we add back 20 these six arrangements to get a total of 66 possible three-factor arrangements, in which the product of the three factors is 2016, and different ordering of factors does not count. So the correct answer is 66. However, 66 was not listed among the possible answers to this question. By contest rules, I was instructed to mark each student answer to this question correct, even if they left it blank. 20. To compute an average, we can use probabilities. For example, consider a common die, which is a small cube with the number 1 – 6 on its six faces, one number per face. When we toss the die, one face will be facing upwards. We see that there is an equal probability of 1/6 that any given face will face upwards when the die is tossed. How do we compute the average number which will occur on the upward facing face after a large number of tosses? We compute a weighted average as follows: For each number that can occur, (1 – 6), we multiply it by its corresponding probability of occurring (this is the weighting factor). Then we do a summation as shown below: Average number = (1/6) 1 + (1/6) 2 + (1/6) 3 + (1/6) 4 + (1/6) 5 + (1/6) 6 = 7/2 = 3.5. Note that the average number appearing does not have to equal any of the individual numbers. This method of computing averages generalizes in an obvious way if the various probabilities are not equal (for example, if the die is “loaded” to favor one face over the others). In the problem at hand, we have a series of time intervals, and at the end of each successive intervals, there is a probability that the element “1” will be removed from the set S. We wish to compute a weighted average using these probabilities in order to get the average time of survival for the element “1”. This average will appear as an infinite sum of terms, since we have an infinite number of possible time intervals. Let us see how to compute the first few terms of this summation, which is referred to as an infinite series. The first time interval is one minute, and at the end of this interval, the set S has two elements, 1 and 2. So there is a probability of ½ that element “1” will be removed. Thus, the first term of our infinite series will be ½. Note that the probability of element “1” surviving the first time interval is 1 – ½ = ½. In order for element “1” to reach the second time interval and then be removed, it must survive the first interval and then be removed at the end of the second interval. 21 Now we must use an elementary rule for probabilities. If we wish for both of two independent events to occur (in this case, surviving the first time interval, but not reaching the third), we multiply the two individual probabilities together to get the overall probability. This is called the multiplication rule for simple probabilities. At the end of the second time interval, set S now have five elements, because one was removed at the end of the first time interval, and then four more were added. If element “1” is going to be removed by choosing at random, two of the five elements of S, we now have two possibilities: Either element “1” gets removed on the first pick, or it survives the first pick and gets removed on the second pick. Now we need another rule for simple probabilities. When an event (in this case, picking two elements from set S) can happen in several independent ways, we must add the individual probabilities for each way the event can occur. This is called the addition rule for simple probabilities. As stated above, we have two possible ways that the element “1” can be removed. If the element “1” is removed on the first pick, there is a 1/5 probability for that, since S has five elements and the elements are being randomly picked. In the other way this can happen, element “1” survives the first pick, with probability 4/5, but then gets removed with the second pick, with probability ¼ (note that one element of S has been removed, so there are only 4 left). By the multiplication rule, the overall probability for this is 4/5 x ¼ = 1/5 So by the addition and multiplication rules the overall probability that element “1” will be picked at the end of the second time interval is: 1/5 + 1/5 = 2/5. But recall that to reach the second time interval, the element “1” have to survive the first time interval, with probability ½, so the overall probability of reaching the end of the second time interval, but no further is: ½ x 2/5 by the multiplication rule. I am not going to cancel the factor of two for reasons that will become clear below. Note that the total time elapsed at the end of the second time interval is 1 + ½ = 3/2. It follows that the first two terms in our series to compute the average time interval is: ½ (1) + ½ 2/5 (3/2), where the elapsed time is in parentheses. Note that the probability of surviving the second time interval is 1 – 2/5 = 3/5. Now to reach the third time interval, but no further, element “1” must survive the first two time intervals, with probability ½ 3/5 by the multiplication rule, then get picked at the end of the third interval. What is this probability? Note that at the end of the third time interval, set S now has 11 elements, because two were removed from the five elements, leaving three, and then eight more were 22 added. Recall that we add twice as many elements at the end of each succeeding time interval. This time, we are going to remove four elements, and there are according four ways in which the element “1” could get removed: Get picked on the first choice, or survive the first but get picked on the second, or survive the first two and get picked on the third, or survive the first three and get picked on the last choice. Let’s work this out: There is a 1/11 chance of getting picked on the first choice and 10/11 chance of surviving it. There is a 10/11 x 9/10 x 1/9 = 1/11 chance of surviving the first two picks, and then getting picked on the third choice. Notice that each way is going to give a contribution of 1/11, because if element “1” survives the first three picks and then gets removed, the probability of that is: 10/11 x 9/10 x 8/9 x 1/8 = 1/11. Using the multiplication and addition rules, the overall chance of getting picked during the third time interval is: 4/11. But remember, element “1” has to survive the first two time intervals, so the overall probability of reaching the end of the third time interval but no further is: ½ 3/5 4/11. Also note that the total elapsed time is 1 + ½ + ¼ = 7/4. So the first three terms of our summation to get the average time are: ½ + ½ 2/5 3/2 + ½ 3/5 4/11 7/4. Note that we can do some cancellations here, and I will show these in the figure below on the next page. Also note that in the fourth time interval, we will have removed four elements from set S, leaving seven, and then added sixteen more, giving a total of 23 elements. I worked out the probabilities the same way as described above for the first three time intervals. With common factors canceled, the series which emerged from this process is shown below. Although there is a pattern in the series, I did not see a way to sum the series to get the exact answer for the average time. It was necessary to sum the first 14 terms in order to get a result for the sum accurate to two decimal places. We say that the series does not converge rapidly. The result for the average time, correct to two decimal places, is 1.36 minutes, and the answer id (D). 23 24