💡Okay, so I have this problem where the first four terms of an arithmetic sequence are given as ( x + y ), ( x - y ), ( xy ), and ( frac{x}{y} ). I need to find the fifth term. Hmm, arithmetic sequence means that the difference between consecutive terms is constant. So, the difference between the first and second term should be the same as between the second and third, and so on.Let me write down the terms:1. ( x + y )2. ( x - y )3. ( xy )4. ( frac{x}{y} )Since it's an arithmetic sequence, the difference between each term is the same. Let's denote the common difference as ( d ). So, the difference between the first and second term is:( (x - y) - (x + y) = x - y - x - y = -2y )So, ( d = -2y ).Now, the difference between the second and third term should also be ( d ). Let's compute that:( xy - (x - y) = xy - x + y )This should equal ( d = -2y ). So,( xy - x + y = -2y )Let me rearrange this equation:( xy - x + y + 2y = 0 )( xy - x + 3y = 0 )( x(y - 1) + 3y = 0 )Hmm, that's one equation. Let's see if I can get another equation from the next difference. The difference between the third and fourth term should also be ( d = -2y ):( frac{x}{y} - xy = -2y )Let me write that down:( frac{x}{y} - xy = -2y )Multiply both sides by ( y ) to eliminate the denominator:( x - x y^2 = -2y^2 )Bring all terms to one side:( x - x y^2 + 2y^2 = 0 )( x(1 - y^2) + 2y^2 = 0 )So now I have two equations:1. ( x(y - 1) + 3y = 0 )2. ( x(1 - y^2) + 2y^2 = 0 )I need to solve these two equations to find ( x ) and ( y ). Let me try to express ( x ) from the first equation and substitute into the second.From equation 1:( x(y - 1) = -3y )( x = frac{-3y}{y - 1} )Simplify the denominator:( x = frac{-3y}{y - 1} = frac{3y}{1 - y} )Okay, so ( x = frac{3y}{1 - y} ). Let's substitute this into equation 2.Equation 2:( x(1 - y^2) + 2y^2 = 0 )Substitute ( x ):( frac{3y}{1 - y} (1 - y^2) + 2y^2 = 0 )Notice that ( 1 - y^2 = (1 - y)(1 + y) ), so:( frac{3y}{1 - y} times (1 - y)(1 + y) + 2y^2 = 0 )The ( (1 - y) ) terms cancel out:( 3y(1 + y) + 2y^2 = 0 )Expand ( 3y(1 + y) ):( 3y + 3y^2 + 2y^2 = 0 )( 3y + 5y^2 = 0 )Factor out ( y ):( y(3 + 5y) = 0 )So, either ( y = 0 ) or ( 3 + 5y = 0 ). But ( y = 0 ) would make the fourth term ( frac{x}{y} ) undefined, which isn't possible. Therefore,( 3 + 5y = 0 )( 5y = -3 )( y = -frac{3}{5} )Now, substitute ( y = -frac{3}{5} ) back into the expression for ( x ):( x = frac{3y}{1 - y} )( x = frac{3(-frac{3}{5})}{1 - (-frac{3}{5})} )( x = frac{-frac{9}{5}}{1 + frac{3}{5}} )( x = frac{-frac{9}{5}}{frac{8}{5}} )( x = -frac{9}{5} times frac{5}{8} )( x = -frac{9}{8} )So, ( x = -frac{9}{8} ) and ( y = -frac{3}{5} ).Now, let's verify the terms with these values to ensure consistency.1. First term: ( x + y = -frac{9}{8} + (-frac{3}{5}) = -frac{9}{8} - frac{3}{5} )Convert to common denominator:( -frac{45}{40} - frac{24}{40} = -frac{69}{40} )2. Second term: ( x - y = -frac{9}{8} - (-frac{3}{5}) = -frac{9}{8} + frac{3}{5} )Convert to common denominator:( -frac{45}{40} + frac{24}{40} = -frac{21}{40} )3. Third term: ( xy = (-frac{9}{8})(-frac{3}{5}) = frac{27}{40} )4. Fourth term: ( frac{x}{y} = frac{-frac{9}{8}}{-frac{3}{5}} = frac{9}{8} times frac{5}{3} = frac{45}{24} = frac{15}{8} )Wait a minute, let me check the fourth term again. ( frac{x}{y} = frac{-9/8}{-3/5} = frac{9/8}{3/5} = frac{9}{8} times frac{5}{3} = frac{45}{24} = frac{15}{8} ). Yes, that's correct.Now, let's check the differences:Difference between first and second term: ( -frac{21}{40} - (-frac{69}{40}) = frac{48}{40} = frac{12}{10} = frac{6}{5} ). Wait, but earlier, we had ( d = -2y = -2(-frac{3}{5}) = frac{6}{5} ). So that's consistent.Difference between second and third term: ( frac{27}{40} - (-frac{21}{40}) = frac{48}{40} = frac{6}{5} ). Good.Difference between third and fourth term: ( frac{15}{8} - frac{27}{40} ). Let's compute this:Convert to common denominator 40:( frac{15}{8} = frac{75}{40} )So, ( frac{75}{40} - frac{27}{40} = frac{48}{40} = frac{6}{5} ). Perfect, consistent.So, the common difference ( d = frac{6}{5} ).Now, to find the fifth term, we just add the common difference to the fourth term:Fifth term = ( frac{15}{8} + frac{6}{5} )Convert to common denominator 40:( frac{15}{8} = frac{75}{40} )( frac{6}{5} = frac{48}{40} )So, ( frac{75}{40} + frac{48}{40} = frac{123}{40} )Therefore, the fifth term is ( frac{123}{40} ).**Final Answer**The fifth term is boxed{dfrac{123}{40}}.

💡Okay, so I have this problem where I need to find the sine of angle GAD in a cube labeled ABCDEFGH. Hmm, cubes can be a bit tricky, but let me try to visualize it. I remember that in a cube, all edges are equal, and all angles are right angles. First, I need to figure out where points G, A, and D are located in the cube. Let me try to sketch a cube in my mind. Let's say point A is at the origin (0,0,0). Then, moving along the x-axis, y-axis, and z-axis, the other points can be labeled accordingly. So, if A is (0,0,0), then B would be (1,0,0), D would be (0,1,0), and E would be (0,0,1). Wait, no, that might not be correct. Let me think again. In a cube, each vertex can be represented by coordinates where each coordinate is either 0 or 1. So, if A is (0,0,0), then moving along the x-axis, we reach B (1,0,0). Moving along the y-axis from A, we reach D (0,1,0). Moving along the z-axis from A, we reach E (0,0,1). Now, point G is a bit further. Since G is diagonally opposite to C, which is (1,1,0), so G would be (1,1,1). Wait, no. Let me clarify. In a cube, each vertex is connected to three others. So, if A is (0,0,0), then the vertices connected to A are B (1,0,0), D (0,1,0), and E (0,0,1). Then, moving from B, we can reach C (1,1,0), and from D, we can reach C as well. From E, moving along the x-axis, we reach F (1,0,1), and along the y-axis, we reach H (0,1,1). Then, G would be connected to C, F, and H. So, G is (1,1,1). So, points:- A: (0,0,0)- B: (1,0,0)- C: (1,1,0)- D: (0,1,0)- E: (0,0,1)- F: (1,0,1)- G: (1,1,1)- H: (0,1,1)Alright, so angle GAD is the angle at point A between points G and D. So, it's angle GAD, meaning vertex at A, with segments AG and AD forming the angle.To find the sine of this angle, I can use vector analysis. If I can find the vectors AG and AD, then I can use the cross product formula to find the sine of the angle between them.First, let's find the vectors:Vector AG goes from A to G. Since A is (0,0,0) and G is (1,1,1), vector AG is (1,1,1).Vector AD goes from A to D. A is (0,0,0) and D is (0,1,0), so vector AD is (0,1,0).Now, the formula for the sine of the angle between two vectors is:sin(theta) = |AG × AD| / (|AG| |AD|)First, let's compute the cross product AG × AD.AG = (1,1,1)AD = (0,1,0)The cross product is calculated as:|i   j   k||1    1    1||0    1    0|= i*(1*0 - 1*1) - j*(1*0 - 1*0) + k*(1*1 - 1*0)= i*(0 - 1) - j*(0 - 0) + k*(1 - 0)= -i - 0j + k= (-1, 0, 1)Now, the magnitude of this cross product is sqrt((-1)^2 + 0^2 + 1^2) = sqrt(1 + 0 + 1) = sqrt(2)Next, let's find the magnitudes of AG and AD.|AG| = sqrt(1^2 + 1^2 + 1^2) = sqrt(3)|AD| = sqrt(0^2 + 1^2 + 0^2) = sqrt(1) = 1So, sin(theta) = |AG × AD| / (|AG| |AD|) = sqrt(2) / (sqrt(3) * 1) = sqrt(2)/sqrt(3)To rationalize the denominator, multiply numerator and denominator by sqrt(3):sqrt(2)/sqrt(3) = (sqrt(2)*sqrt(3))/(sqrt(3)*sqrt(3)) = sqrt(6)/3Wait, but earlier I thought the answer was sqrt(3)/3. Hmm, maybe I made a mistake.Wait, let's double-check the cross product:AG × AD = (-1, 0, 1). Its magnitude is sqrt(1 + 0 + 1) = sqrt(2). That's correct.|AG| is sqrt(3), |AD| is 1. So sin(theta) = sqrt(2)/sqrt(3) = sqrt(6)/3.But in the initial problem, the user had an answer of sqrt(3)/3. So, where did I go wrong?Wait, maybe I misassigned the points. Let me check the cube labeling again.In some cube labelings, the order might be different. Maybe G is not (1,1,1). Let me confirm.In a standard cube labeling, starting from A at (0,0,0), moving along x to B, y to D, and z to E. Then, C is (1,1,0), F is (1,0,1), G is (1,1,1), and H is (0,1,1). So, G is indeed (1,1,1).Wait, but in the initial problem, the user had a different approach, considering triangle AGD as a right triangle, but I think that might be incorrect because AGD is not a right triangle. Let's see.If we consider triangle AGD, points A(0,0,0), G(1,1,1), D(0,1,0). Let's compute the lengths:AG: distance from A to G = sqrt((1)^2 + (1)^2 + (1)^2) = sqrt(3)AD: distance from A to D = sqrt((0)^2 + (1)^2 + (0)^2) = 1GD: distance from G to D = sqrt((1-0)^2 + (1-1)^2 + (1-0)^2) = sqrt(1 + 0 + 1) = sqrt(2)So, triangle AGD has sides of length 1, sqrt(2), sqrt(3). That doesn't form a right triangle because 1^2 + (sqrt(2))^2 = 1 + 2 = 3, which equals (sqrt(3))^2. So, actually, triangle AGD is a right triangle with right angle at D.Wait, that contradicts my earlier thought. If AGD is a right triangle with right angle at D, then angle GAD is not the right angle. So, in that case, we can use the definition of sine as opposite over hypotenuse.In triangle AGD, right-angled at D, angle at A is angle GAD. The side opposite to angle GAD is GD, which is sqrt(2), and the hypotenuse is AG, which is sqrt(3). So, sin(angle GAD) = opposite/hypotenuse = sqrt(2)/sqrt(3) = sqrt(6)/3.But earlier, using vectors, I got the same result. So, why did the initial problem mention a different answer? Maybe the initial problem had a different labeling of the cube.Alternatively, perhaps the angle is being considered differently. Let me double-check.Wait, in the initial problem, the user considered triangle AGD as a right triangle with right angle at G, but that's not correct because the right angle is at D, not at G.So, perhaps the initial problem had a mistake in identifying the right angle. Therefore, the correct approach is to recognize that triangle AGD is right-angled at D, and thus sin(angle GAD) = GD/AG = sqrt(2)/sqrt(3) = sqrt(6)/3.Alternatively, using vectors, we found the same result.Wait, but in the initial problem, the user had an answer of sqrt(3)/3, which is approximately 0.577, while sqrt(6)/3 is approximately 0.816. These are different values, so there must be a misunderstanding.Let me try another approach. Maybe using coordinates, find the vectors and compute the sine.Vectors:AG = G - A = (1,1,1) - (0,0,0) = (1,1,1)AD = D - A = (0,1,0) - (0,0,0) = (0,1,0)The angle between AG and AD is angle GAD.We can use the dot product to find the cosine of the angle, and then use the identity sin^2(theta) + cos^2(theta) = 1 to find sine.First, compute the dot product:AG · AD = (1)(0) + (1)(1) + (1)(0) = 0 + 1 + 0 = 1The magnitudes:|AG| = sqrt(1^2 + 1^2 + 1^2) = sqrt(3)|AD| = sqrt(0^2 + 1^2 + 0^2) = 1So, cos(theta) = (AG · AD) / (|AG| |AD|) = 1 / (sqrt(3)*1) = 1/sqrt(3)Therefore, sin(theta) = sqrt(1 - (1/sqrt(3))^2) = sqrt(1 - 1/3) = sqrt(2/3) = sqrt(6)/3So, that confirms the earlier result. Therefore, the correct value is sqrt(6)/3, not sqrt(3)/3.Wait, but in the initial problem, the user had an answer of sqrt(3)/3. So, perhaps there was a mistake in their reasoning. They considered triangle AGD as a right triangle with right angle at G, but that's incorrect. The right angle is at D, making the opposite side GD = sqrt(2), and hypotenuse AG = sqrt(3), leading to sin(theta) = sqrt(2)/sqrt(3) = sqrt(6)/3.Therefore, the correct answer should be sqrt(6)/3.But to be thorough, let me consider another approach. Maybe using coordinates to find the angle.We can find the coordinates of points A, G, and D:A: (0,0,0)G: (1,1,1)D: (0,1,0)We can compute the vectors AG and AD as before:AG = (1,1,1)AD = (0,1,0)The angle between them can be found using the dot product formula:cos(theta) = (AG · AD) / (|AG| |AD|) = 1 / sqrt(3)Therefore, sin(theta) = sqrt(1 - (1/3)) = sqrt(2/3) = sqrt(6)/3Yes, that's consistent.Alternatively, using the cross product method:|AG × AD| = sqrt(2)|AG| |AD| = sqrt(3)*1 = sqrt(3)sin(theta) = sqrt(2)/sqrt(3) = sqrt(6)/3Same result.Therefore, I think the initial answer of sqrt(3)/3 was incorrect, and the correct answer is sqrt(6)/3.But wait, let me check if the angle is indeed at A. Yes, angle GAD is at A, between points G, A, D.So, in triangle AGD, with right angle at D, angle at A is angle GAD, opposite side GD = sqrt(2), hypotenuse AG = sqrt(3), so sin(theta) = sqrt(2)/sqrt(3) = sqrt(6)/3.Yes, that seems correct.Therefore, the correct answer is sqrt(6)/3.

💡To determine the range of values for ( x ) that make the expression ( sqrt{2 + x} ) meaningful within the real number range, we need to ensure that the expression inside the square root is non-negative. This is because the square root of a negative number is not defined in the set of real numbers.1. **Start with the condition for the expression under the square root to be non-negative:** [ 2 + x geq 0 ]2. **Isolate ( x ) to find the range of values it can take:** [ x geq -2 ]Thus, for the expression ( sqrt{2 + x} ) to be meaningful within the real number range, ( x ) must be greater than or equal to (-2).

💡Okay, so I need to find the equation of a hyperbola that's centered at the origin, with its axes along the coordinate axes. The hyperbola has an eccentricity of 2 and a real axis of length 4. Hmm, let me think about how to approach this.First, I remember that the standard form of a hyperbola centered at the origin can be either (frac{x^2}{a^2} - frac{y^2}{b^2} = 1) or (frac{y^2}{a^2} - frac{x^2}{b^2} = 1), depending on whether it opens horizontally or vertically. Since the problem mentions the real axis, I think that refers to the transverse axis, which is the one that passes through the vertices.The length of the real axis is given as 4. I recall that the length of the transverse axis is (2a), so if (2a = 4), then (a = 2). That seems straightforward.Next, the eccentricity (e) is given as 2. I remember that for hyperbolas, eccentricity is defined as (e = frac{c}{a}), where (c) is the distance from the center to each focus. So, if (e = 2) and (a = 2), then (c = e times a = 2 times 2 = 4). So, (c = 4).Now, I need to find (b), the length of the semi-minor axis. I recall that for hyperbolas, there's a relationship between (a), (b), and (c): (c^2 = a^2 + b^2). Let me plug in the values I have:(c^2 = a^2 + b^2)(4^2 = 2^2 + b^2)(16 = 4 + b^2)Subtracting 4 from both sides:(b^2 = 12)So, (b = sqrt{12}), which simplifies to (2sqrt{3}).Now, I need to write the equation of the hyperbola. Since the problem doesn't specify whether it opens horizontally or vertically, I think I need to consider both possibilities. If it opens horizontally, the equation would be:(frac{x^2}{a^2} - frac{y^2}{b^2} = 1)Plugging in (a = 2) and (b = 2sqrt{3}):(frac{x^2}{4} - frac{y^2}{12} = 1)If it opens vertically, the equation would be:(frac{y^2}{a^2} - frac{x^2}{b^2} = 1)Plugging in the same values:(frac{y^2}{4} - frac{x^2}{12} = 1)Wait, but the problem mentions the real axis is of length 4. I think the real axis is the transverse axis, which is the one that determines the direction the hyperbola opens. Since the real axis is along the coordinate axes, it could be either x or y. But the problem doesn't specify, so maybe both equations are possible?But actually, in standard terminology, the real axis is usually the transverse axis, and it's associated with the term under which the positive term is. So if the real axis is along the x-axis, the equation is (frac{x^2}{4} - frac{y^2}{12} = 1). If it's along the y-axis, it's (frac{y^2}{4} - frac{x^2}{12} = 1).But the problem says "axes along the coordinate axes," which might mean that both axes are along the coordinate axes, but I think that's always the case for hyperbolas centered at the origin. So, maybe it's just specifying that the hyperbola is aligned with the coordinate axes, meaning it's not rotated.So, considering all that, I think both equations are possible, depending on the orientation. But since the problem doesn't specify, perhaps I should present both?Wait, but in the initial problem statement, it just says "axes along the coordinate axes," which is a bit ambiguous. It could mean that both the transverse and conjugate axes are along the coordinate axes, which is standard, but it doesn't specify which one is which. So, perhaps I need to assume a standard orientation, like horizontal.But to be thorough, maybe I should present both possibilities.Alternatively, maybe the real axis being of length 4 implies that the transverse axis is along the x-axis, making the equation (frac{x^2}{4} - frac{y^2}{12} = 1).I think that's the more likely case, as usually, unless specified otherwise, the real axis is considered along the x-axis.So, to sum up, the equation of the hyperbola is (frac{x^2}{4} - frac{y^2}{12} = 1).But just to double-check, let me verify the eccentricity. We have (a = 2), (c = 4), so (e = c/a = 4/2 = 2), which matches the given eccentricity. And (b^2 = c^2 - a^2 = 16 - 4 = 12), so that's correct.Therefore, I'm confident that the equation is (frac{x^2}{4} - frac{y^2}{12} = 1).

💡Okay, so I have this equation here: 2y equals negative 3x plus 6. I need to find the slope of the line represented by this equation. Hmm, I remember that the slope of a line is a measure of its steepness, right? It tells me how much y changes for every unit change in x. But wait, how do I actually find the slope from an equation like this? I think it has something to do with rearranging the equation into a specific form. Oh yeah, I think it's called the slope-intercept form. What's that look like again? I believe it's y equals mx plus b, where m is the slope and b is the y-intercept. So, if I can rewrite the given equation in that form, I can easily identify the slope.Alright, let's start with the given equation: 2y equals negative 3x plus 6. To get y by itself, I need to divide every term by 2. So, dividing both sides by 2, I get y equals negative 3 over 2 x plus 3. Wait, let me make sure I did that correctly. Starting with 2y equals -3x plus 6. If I divide both sides by 2, that would be (2y)/2 equals (-3x)/2 plus 6/2. Simplifying that, I get y equals (-3/2)x plus 3. Okay, that looks right. So now the equation is in the slope-intercept form, y equals mx plus b, where m is the slope.Looking at my equation, y equals (-3/2)x plus 3, the coefficient of x is -3/2. That must be the slope, right? So, the slope of the line is -3/2. Wait a second, let me double-check to make sure I didn't make any mistakes. Starting from the original equation: 2y equals -3x plus 6. Dividing both sides by 2 gives y equals (-3/2)x plus 3. Yep, that seems correct. The coefficient of x is indeed -3/2, so the slope is -3/2.I think that's it. I don't see any errors in my steps. I converted the equation into slope-intercept form, identified the coefficient of x as the slope, and confirmed my calculations. So, the slope of the line is negative three halves.

💡Alright, so I've got this problem here about surveys and people's answers. It's divided into two parts, both dealing with 20 questions, but part a) has two possible answers per question, and part b) has 12 possible answers per question. The main question is whether it's necessarily true that there will be two people whose answers don't match on any question, given that for any 10 questions and any combination of answers, there's at least one person who provided exactly those answers.Okay, let's start with part a). So, each question has two possible answers, say "Yes" or "No." There are 20 questions, so the total number of possible answer combinations is 2^20, which is a pretty large number. The problem states that for any 10 questions and any combination of answers to those 10 questions, there exists a person who provided exactly those answers. That means the survey results are very comprehensive in covering all possible subsets of 10 questions.Now, the question is whether there must be two people whose answers don't match on any question. In other words, is there necessarily a pair of people where, for every single question, one answered "Yes" and the other answered "No"? That would mean their answer vectors are exact opposites.Hmm, to approach this, maybe I can think about it in terms of binary vectors. Each person's set of answers can be represented as a 20-dimensional binary vector. The condition given is that for any 10-dimensional subspace, all possible combinations are covered. So, the set of vectors is such that every possible 10-dimensional projection is surjective.Now, does this imply that there are two vectors that are complements of each other? That is, for every bit, one is 0 and the other is 1. In coding theory, this might relate to error-correcting codes or something like that. Maybe I can use the pigeonhole principle or some combinatorial argument.Wait, if every possible combination of 10 answers is covered, then the number of people must be at least 2^10, which is 1024. But 2^20 is much larger, so actually, the number of people could be up to 2^20, but the condition is that every 10-question combination is covered. So, the minimal number of people needed to satisfy this condition is 2^10, but the actual number could be larger.But does having this coverage necessarily lead to having two people who are complements? I'm not sure. Maybe I can think about it probabilistically. If you have a large enough set of vectors, the chance that two are complements increases. But is it guaranteed?Alternatively, maybe I can construct an example where this doesn't happen. Suppose I have a set of vectors where none are complements of each other. Is that possible while still covering all 10-question combinations? If I can construct such a set, then the answer would be no, it's not necessarily true.Let me try to think of such a construction. Suppose I fix one question, say the first one, and only include vectors where the first answer is "Yes." Then, for any 10 questions, as long as they include the first question, the combination must have "Yes" in that position. But wait, the problem states that for any 10 questions and any combination, there's a person with exactly those answers. So, if I fix the first answer to "Yes," then for any 10 questions that include the first question, the combination where the first answer is "No" wouldn't be covered. Therefore, fixing one answer doesn't work.Hmm, maybe a different approach. What if I partition the 20 questions into two groups of 10 and ensure that within each group, all combinations are covered. But then, across the two groups, maybe the answers don't necessarily have to be complementary.Wait, but the condition is for any 10 questions, not just specific partitions. So, it's more stringent. It has to cover all possible 10-question subsets, not just one fixed subset.Maybe I can think about the concept of covering codes or something similar. A covering code ensures that every possible word is within a certain distance from a codeword. But I'm not sure if that's directly applicable here.Alternatively, perhaps I can use the probabilistic method. If I assume that no two people are complements, then what does that imply about the number of people? If no two people are complements, then the set of people's answers is such that for every vector, its complement is not in the set. So, the maximum size of such a set is 2^19, since for every vector, you exclude its complement.But in our case, the number of people is at least 2^10, which is much smaller than 2^19. So, it's possible to have a set of vectors where no two are complements and still cover all 10-question combinations.Wait, but does such a set actually exist? I mean, just because the maximum size is 2^19 doesn't necessarily mean that a set of size 2^10 can avoid having complements. Maybe it's possible, maybe not.Alternatively, maybe I can use the concept of orthogonality. If two vectors are complements, their dot product is 20, which is even. But I'm not sure if that helps.Wait, another idea. If I consider the set of all vectors, and I want to ensure that for any 10 questions, all combinations are covered. So, the set must intersect every 10-dimensional subspace. Now, if I have a set that doesn't contain any complementary pairs, does that interfere with covering all subspaces?I'm not sure. Maybe I need to think about specific properties. For example, if I have a vector and its complement, they both cover different subspaces. But if I exclude one, does that leave some subspace uncovered?Actually, no, because the condition is that for any 10 questions, all combinations are covered. So, even if I don't have a complementary vector, as long as all combinations are covered, it's fine.Wait, but if I have a vector, say, all "Yes" answers, then its complement, all "No" answers, would also be a vector. If I don't include the all "No" vector, then for the 10 questions, the combination of all "No" answers wouldn't be covered. But the problem states that for any 10 questions and any combination, there exists a person with exactly those answers. So, if I don't include the all "No" vector, then for the 10 questions, the all "No" combination wouldn't be covered, which violates the condition.Wait, that's a good point. If I don't include the all "No" vector, then for any 10 questions, the all "No" combination wouldn't be covered. Therefore, the all "No" vector must be included. Similarly, the all "Yes" vector must be included because for any 10 questions, the all "Yes" combination must be covered.So, both the all "Yes" and all "No" vectors are included. Now, are these two vectors complements of each other? Yes, because one is all "Yes" and the other is all "No." So, in this case, there are two people whose answers don't match on any question.Wait, but does this mean that in any such set, the all "Yes" and all "No" vectors must be included? Because if they weren't, then for the 10 questions, the all "Yes" or all "No" combination wouldn't be covered.Yes, that seems to be the case. Therefore, in any such set, the all "Yes" and all "No" vectors must be present. Hence, there must be two people whose answers don't match on any question.But wait, is that necessarily true? Suppose I have a set where instead of including the all "Yes" vector, I include a vector that's all "Yes" except for one question. Then, for any 10 questions that don't include that one question, the all "Yes" combination would still be covered. But for the 10 questions that do include that one question, the combination where all are "Yes" except for that one would be covered, but the all "Yes" combination wouldn't be covered.Therefore, to cover the all "Yes" combination for any 10 questions, you must have a vector that is all "Yes" on those 10 questions. But since this has to be true for any 10 questions, the only way to ensure that is to have a vector that is all "Yes" on all 20 questions. Similarly, you must have a vector that is all "No" on all 20 questions.Therefore, in any such set, the all "Yes" and all "No" vectors must be included, and these two vectors are complements of each other. Hence, there must be two people whose answers don't match on any question.Wait, but the problem says "for any 10 questions and any combination of answers to these questions, there exists a person who provided exactly those answers." So, it's possible that the all "Yes" vector isn't included, but for any 10 questions, there's someone who answered all "Yes" on those 10. But to cover all possible 10-question combinations, including all "Yes," you don't necessarily need the all "Yes" vector, because for any specific 10 questions, you can have a different vector that is all "Yes" on those 10.But wait, no, because if you have a different vector for each set of 10 questions, then you would need an enormous number of vectors. The minimal number of vectors needed to cover all 10-question combinations is actually quite large, but it's possible that none of them are the all "Yes" or all "No" vectors.Wait, but if you have a vector that is all "Yes" on a specific set of 10 questions, it doesn't necessarily mean that it's all "Yes" on all 20 questions. So, maybe you can cover all 10-question combinations without including the all "Yes" or all "No" vectors.But then, how would you cover the all "Yes" combination for a different set of 10 questions? You would need a different vector for each set of 10 questions, which is not feasible because the number of 10-question combinations is C(20,10), which is 184,756. So, you would need at least that many vectors, which is way more than 2^10.Wait, no, the condition is that for any 10 questions and any combination, there exists a person with exactly those answers. So, it's not just covering all possible 10-question subsets, but for each subset, covering all possible answer combinations on those 10 questions.So, for each subset of 10 questions, there must be 2^10 different vectors that restrict to each possible combination on those 10 questions. Therefore, the total number of vectors must be at least 2^10, but actually, it's more complicated because vectors can cover multiple subsets.But regardless, the key point is that to cover all possible combinations on any 10 questions, you need a set of vectors that is sufficiently rich. Now, does this richness necessarily include a pair of complementary vectors?I think it does, because if you have a vector, you can always find another vector that is its complement on some subset of questions. But I'm not entirely sure.Wait, another approach. Suppose we have a set of vectors where no two are complementary. Then, for any vector v, its complement ~v is not in the set. Now, consider the projection of these vectors onto any 10 questions. For any combination, there must be a vector in the set that matches it on those 10 questions.But if we exclude ~v, does that affect the coverage? Maybe not directly, because the coverage is about matching any combination on any 10 questions, not about having both a vector and its complement.But if we have a vector v, and we don't have ~v, does that mean that for some 10 questions, the combination ~v on those 10 isn't covered? Not necessarily, because another vector could cover that combination.Wait, but if v is all "Yes," then ~v is all "No." If we don't have ~v, then for any 10 questions, the all "No" combination isn't covered, which violates the condition. Therefore, to cover the all "No" combination on any 10 questions, we must have a vector that is all "No" on those 10 questions. But to cover it for any 10 questions, we must have a vector that is all "No" on all 20 questions.Similarly, to cover the all "Yes" combination on any 10 questions, we must have a vector that is all "Yes" on all 20 questions.Therefore, both the all "Yes" and all "No" vectors must be included in the set. And these two vectors are complements of each other. Hence, there must be two people whose answers don't match on any question.So, for part a), the answer is yes, it is necessarily true that there will be two people whose answers don't match on any question.Now, moving on to part b), where each question has 12 possible answers instead of 2. The problem is similar: given that for any 10 questions and any combination of answers, there exists a person who provided exactly those answers, is it necessarily true that there will be two people whose answers don't match on any question?This seems more complex because now each question has 12 possible answers, so the total number of possible answer combinations is 12^20, which is enormous. The condition is similar: for any 10 questions and any combination of answers, there's a person with exactly those answers.Now, the question is whether there must be two people whose answers don't match on any question. In other words, for every question, their answers are different.This seems related to the concept of Latin hypercubes or something like that, where you want diversity across all dimensions.But let's think about it step by step. First, in part a), we saw that the all "Yes" and all "No" vectors must be included, and they are complements. But in part b), with 12 answers per question, the concept of complementarity is less straightforward because there are more options.Wait, in part a), the complement of a vector was unique because there were only two possible answers. But with 12 answers, the complement isn't uniquely defined. So, the idea of having two people whose answers don't match on any question is more about having, for each question, different answers, not necessarily inverses.So, in part b), we need to determine if there must exist two people such that for every question, their answers are different. This is a stronger condition than in part a), where it was about being exact inverses.Given that, let's see. The condition is that for any 10 questions and any combination of answers, there's a person with exactly those answers. So, the set of answer vectors is such that every possible 10-dimensional projection is surjective.Now, does this imply that there are two vectors that are completely different on all 20 questions? That is, for every question, their answers are different.This seems related to the concept of orthogonal arrays or something similar in combinatorics. An orthogonal array ensures that in every subset of columns, all possible combinations appear a certain number of times.But in our case, it's a bit different because we're not requiring a certain number of appearances, just that every combination appears at least once.Now, to determine if two completely different vectors must exist, let's think about the pigeonhole principle. If we have a large enough set of vectors, the probability that two are completely different increases. But is it guaranteed?Alternatively, maybe we can construct a set where no two vectors are completely different, yet still satisfy the condition that every 10-question combination is covered.Wait, but if we have a set where no two vectors are completely different, that means for any two vectors, there is at least one question where they have the same answer. So, the set is such that any two vectors share at least one common answer on some question.Is it possible to have such a set that still covers all 10-question combinations?Hmm, I'm not sure. Let's think about it differently. Suppose we fix one question, say question 1, and assign a specific answer to it, say answer A. Then, for any 10 questions that include question 1, the combination must have answer A on question 1. But the problem states that for any 10 questions and any combination, there's a person with exactly those answers. So, if we fix question 1 to answer A, then for any 10 questions that include question 1, the combination where question 1 has answer B (or any other answer) wouldn't be covered. Therefore, we can't fix any question to a specific answer.So, similar to part a), we can't fix any answer because that would prevent covering all combinations for subsets including that question. Therefore, the set of vectors must be such that for any question, all possible answers are represented across the vectors.But does that imply that there are two vectors that are completely different? Not necessarily. It just means that for each question, all answers are covered, but it doesn't guarantee that two vectors differ on all questions.Wait, but if for each question, all answers are covered, then for any two vectors, they might share some answers on some questions, but it's not guaranteed that they share answers on all questions.But the question is whether there must exist two vectors that differ on all questions.This seems similar to the concept of a code with a certain minimum distance. In coding theory, the minimum distance is the smallest number of positions in which any two codewords differ. If we can ensure that the minimum distance is 20, then any two codewords differ on all positions.But in our case, we're not constructing a code with a certain minimum distance; we're just given that every 10-question combination is covered.I think the key here is to realize that with 12 possible answers per question, the number of possible answer combinations is so large that it's impossible to avoid having two vectors that differ on all questions.But I'm not sure. Maybe I can think about it in terms of the pigeonhole principle. If we have enough vectors, the number of possible pairs is so large that some pair must differ on all questions.But how many vectors do we have? The minimal number of vectors needed to cover all 10-question combinations is 12^10, which is already 61,917,364,224. That's a huge number. So, the set of vectors is enormous.Given that, the number of possible pairs is astronomical, and the number of possible ways two vectors can differ on all questions is also large, but I'm not sure if it's guaranteed.Wait, another approach. Suppose we have a set of vectors where no two differ on all questions. Then, for any two vectors, there is at least one question where they have the same answer. Now, can such a set cover all 10-question combinations?I think it's possible, but I'm not sure. Maybe if we have a set where every pair of vectors shares at least one common answer on some question, but still, for any 10 questions, all combinations are covered.But I'm not sure how to construct such a set or prove that it's impossible.Alternatively, maybe we can use the concept of projective planes or finite geometries, but I'm not familiar enough with that to apply it here.Wait, another idea. If we have a set of vectors where no two differ on all questions, then the set is what's called a "code with no two codewords at maximal distance." In coding theory, the maximal distance for a code of length n over an alphabet of size q is n(q-1). In our case, n=20 and q=12, so the maximal distance is 20*11=220.But I'm not sure if that's directly applicable.Alternatively, maybe we can think about it in terms of the Lovász local lemma or something like that, but I'm not sure.Wait, maybe a simpler approach. If we have a set of vectors where no two differ on all questions, then for any vector v, there is no vector w such that w differs from v on all questions. That means, for every vector v, the set of vectors that differ from v on all questions is empty.But the total number of vectors that differ from v on all questions is (12-1)^20 = 11^20, which is a huge number. So, if our set is large enough, it's likely that it contains at least one such vector.But in our case, the set is required to cover all 10-question combinations, which means it's already very large. So, it's likely that it contains two vectors that differ on all questions.But I'm not sure if it's necessarily true.Wait, another angle. Suppose we have a set S of vectors that covers all 10-question combinations. We want to know if S must contain two vectors that differ on all questions.Assume, for contradiction, that S does not contain any two vectors that differ on all questions. That means, for any two vectors in S, there is at least one question where they have the same answer.Now, consider the size of S. Since S must cover all 10-question combinations, the size of S must be at least 12^10, as each 10-question combination must be represented.But 12^10 is a huge number, and the total number of possible vectors is 12^20, which is much larger. So, S is a relatively small subset of the entire space.But if S is small, does that mean it's possible that no two vectors in S differ on all questions? Or does the size of S being 12^10 force it to contain two vectors that differ on all questions?I think it's the latter. Because 12^10 is still a large number, and the number of possible pairs is enormous, so it's likely that some pair must differ on all questions.But I'm not sure how to formalize this.Wait, maybe I can use the probabilistic method. Consider randomly selecting vectors until we cover all 10-question combinations. What's the probability that no two vectors differ on all questions?But this seems complicated.Alternatively, think about it in terms of the inclusion-exclusion principle. The number of vectors that differ from a given vector on all questions is 11^20. So, if we have N vectors, the expected number of pairs that differ on all questions is C(N,2) * (11/12)^20.We want to know if this expectation is greater than zero, which it is for large N. But in our case, N is 12^10, which is large enough that the expectation is positive.Therefore, it's likely that there exists at least one pair of vectors that differ on all questions.But this is just a heuristic argument. I'm not sure if it's rigorous.Alternatively, maybe I can use the concept of Ramsey theory, which deals with conditions under which order must appear. But I'm not sure how to apply it here.Wait, another idea. If we have a set S of vectors that covers all 10-question combinations, then for any question, all 12 answers must appear in S. Because if an answer didn't appear on a question, then for any 10 questions including that one, the combination where that answer is used wouldn't be covered.Therefore, for each question, all 12 answers must appear in S. So, S is such that for every question, the projection onto that question is surjective.Now, does this imply that there are two vectors in S that differ on all questions?Not necessarily. It just means that for each question, all answers are represented, but it doesn't guarantee that two vectors differ on all questions.Wait, but if for each question, all answers are represented, then for any two vectors, they could potentially differ on all questions, but it's not guaranteed.But the question is whether it's necessarily true that such a pair exists.I think it is necessarily true, but I'm not sure how to prove it.Wait, maybe I can use the concept of derangements. A derangement is a permutation where no element appears in its original position. In our case, we want two vectors where no answer is the same in any position.But derangements are for permutations, and here we have vectors with possible repeats.Alternatively, maybe I can think of it as a Latin hypercube, where each dimension has all possible values, and no two points share the same value in any dimension.But in our case, we don't have a Latin hypercube; we just have a set of vectors that cover all 10-dimensional projections.Wait, but if we have a set of vectors that cover all 10-dimensional projections, does that imply that they form a Latin hypercube? No, because a Latin hypercube requires that in every line parallel to any axis, each symbol appears exactly once, which is a stronger condition.So, our set is less structured than a Latin hypercube.But maybe the property that every 10-dimensional projection is surjective implies some kind of diversity that ensures two vectors differ on all questions.I'm still not sure.Wait, another approach. Suppose we have a set S of vectors that covers all 10-question combinations. Let's consider the size of S. The minimal size of S is 12^10, as each 10-question combination must be covered.Now, the total number of possible vectors is 12^20. So, the size of S is 12^10, which is much smaller than 12^20.Now, the number of possible pairs of vectors in S is C(12^10, 2), which is roughly (12^10)^2 / 2.The number of possible pairs of vectors that differ on all questions is 12^20 * (11/12)^20, which is roughly 12^20 * e^{-20/12} ≈ 12^20 * e^{-1.666}.But I'm not sure if this helps.Wait, maybe I can think about it in terms of the birthday problem. If we have a certain number of vectors, what's the probability that two of them differ on all questions?But again, this is a heuristic argument, not a proof.Alternatively, maybe I can use the concept of hash functions. If we have a set of hash functions that cover all possible 10-dimensional projections, does that imply that there are two functions that are completely different?I'm not sure.Wait, another idea. Suppose we fix 10 questions and consider the set of vectors restricted to those 10 questions. Since all combinations are covered, we have all 12^10 possible combinations on those 10 questions.Now, if we look at the remaining 10 questions, each vector has some answers. If we can find two vectors that differ on all 10 questions in the first set and also differ on all 10 questions in the second set, then they differ on all 20 questions.But I'm not sure if this is guaranteed.Alternatively, maybe we can use the fact that in the first 10 questions, we have all possible combinations, so for any two vectors, they can differ on those 10 questions. But we need them to differ on the remaining 10 as well.This seems complicated.Wait, maybe I can use the concept of orthogonality again. If we have two vectors that are orthogonal, meaning they differ on all questions, then they form a kind of basis. But I'm not sure.Alternatively, maybe I can think about it in terms of graph theory. Consider each vector as a vertex, and connect two vertices if they differ on all questions. Then, the problem reduces to whether this graph contains at least one edge.But I don't know the properties of this graph.Wait, another angle. Suppose we have a set S of vectors that covers all 10-question combinations. Then, for any question, all 12 answers are present in S. So, for any question, we can partition S into 12 subsets based on the answer to that question.Now, if we have two vectors that differ on all questions, then for each question, they are in different subsets.But I'm not sure how to use this.Wait, maybe I can use the concept of the pigeonhole principle across multiple questions. If we have enough vectors, then for each question, the answers are spread out, and it's likely that two vectors differ on all questions.But again, this is not a proof.I think I'm stuck here. Maybe I need to look for a different approach or recall some theorem that might apply.Wait, I remember something called the "combinatorial nullstellensatz," but I'm not sure if it's applicable here.Alternatively, maybe I can think about it in terms of linear algebra. If we represent the vectors as elements of a vector space over a finite field, but since we have 12 answers, which is not a prime power, it's not straightforward.Wait, 12 is not a prime power, but maybe we can use a different structure.Alternatively, maybe I can think of each answer as a symbol from a set of size 12, and consider the set of vectors as words over this alphabet.But I'm not sure.Wait, another idea. Suppose we have a set S of vectors that covers all 10-question combinations. Then, for any 10 questions, S contains all possible 12^10 combinations on those questions.Now, consider the set S restricted to the first 10 questions. It contains all possible combinations. Similarly, restricted to the last 10 questions, it also contains all possible combinations.Now, if we can find two vectors that are different on the first 10 questions and also different on the last 10 questions, then they are different on all 20 questions.But how can we ensure that?Well, since the first 10 questions contain all combinations, there must be two vectors that are different on the first 10 questions. Similarly, for the last 10 questions.But does that imply that there are two vectors that are different on both?Not necessarily. It could be that the two vectors that are different on the first 10 questions are the same on the last 10, and vice versa.But given that the set S is large, maybe it's possible to find two vectors that differ on both.Wait, but how?Alternatively, maybe we can use the fact that the set S is a covering code with covering radius 10, but I'm not sure.I think I'm going in circles here. Maybe I need to accept that I don't know the answer and look for hints or related problems.Wait, I recall that in part a), the answer was yes because the all "Yes" and all "No" vectors had to be included. But in part b), with 12 answers, the concept of all "Yes" and all "No" doesn't directly apply because there are more answers.However, the key idea from part a) was that to cover all combinations, you had to include vectors that were all the same on certain questions, leading to the necessity of having complementary vectors.In part b), maybe a similar idea applies, but instead of complementary vectors, we need vectors that are completely different on all questions.But I'm not sure how to formalize this.Wait, another thought. If we have a set S that covers all 10-question combinations, then for any question, all 12 answers are present. So, for any question, we can find a vector with any specific answer.Now, suppose we want to construct two vectors that differ on all questions. We can start by choosing an answer for the first question, say answer 1. Then, for the second question, choose an answer different from the first vector, say answer 2. Continue this process for all 20 questions, choosing answers different from the first vector.But wait, we need to ensure that such a vector exists in S. Since S covers all 10-question combinations, it's possible that such a vector exists, but I'm not sure.Alternatively, maybe we can use the fact that S is so large that it must contain two vectors that are completely different.But I'm not sure.Wait, maybe I can think about it in terms of the pigeonhole principle. If we have 12^10 vectors, and each vector has 20 answers, then the number of possible pairs is huge. The number of possible ways two vectors can differ on all questions is also huge, but I'm not sure if it's guaranteed.I think I need to conclude that, similar to part a), the answer is yes, it is necessarily true that there will be two people whose answers don't match on any question.But I'm not entirely confident in this conclusion. I think the key idea is that the set S is so large that it must contain two vectors that differ on all questions, but I can't formalize it properly.So, tentatively, I'll say that for part b), the answer is also yes, it is necessarily true that there will be two people whose answers don't match on any question.