💡First, I need to determine the total volume of the cube-shaped tank. Since each side of the cube measures 12 inches, the volume is calculated as:[ 12 times 12 times 12 = 12^3 = 1728 text{ cubic inches} ]Next, I calculate the volume of water in the tank when it is one-third full:[ frac{1}{3} times 1728 = 576 text{ cubic inches} ]Then, I find the volume of one spherical marble. The formula for the volume of a sphere is ( frac{4}{3}pi r^3 ), where the radius ( r ) is half of the diameter (0.5 inches):[ frac{4}{3} pi (0.5)^3 = frac{4}{3} pi left(frac{1}{8}right) = frac{pi}{6} text{ cubic inches per marble} ]With fifteen marbles, the total volume occupied by all the marbles is:[ 15 times frac{pi}{6} = frac{15pi}{6} text{ cubic inches} ]Adding the volume of water and marbles gives:[ 576 + frac{15pi}{6} text{ cubic inches} ]Finally, to find the unoccupied volume, I subtract the total occupied volume from the tank's total volume:[ 1728 - left( 576 + frac{15pi}{6} right) = 1152 - frac{15pi}{6} approx 1152 - 7.854 = 1144.146 text{ cubic inches} ]Therefore, the volume unoccupied by water or marbles in the tank is approximately 1144.146 cubic inches.

💡Okay, so I have this triangle ABC, and I need to solve three parts based on the given conditions. Let me start by understanding what's given.First, the sides opposite angles A, B, and C are a, b, and c respectively. The first condition is:(2a - c) cos B = b cos CAnd the second condition is:sin² A = sin² B + sin² C - λ sin B sin C, where λ is a real number.Alright, so part 1 is to find the measure of angle B. Hmm, okay. Let me think about how to approach this.I remember that in triangles, the Law of Cosines relates the sides and angles. Maybe I can use that here. Also, the Law of Sines relates the sides and sines of the angles. Maybe that can help too.Looking at the first equation: (2a - c) cos B = b cos C.I can try to express everything in terms of sines using the Law of Sines. The Law of Sines says that a/sin A = b/sin B = c/sin C = 2R, where R is the radius of the circumscribed circle. So, a = 2R sin A, b = 2R sin B, c = 2R sin C.Let me substitute these into the first equation:(2a - c) cos B = b cos CSubstituting a, b, c:[2*(2R sin A) - (2R sin C)] cos B = (2R sin B) cos CSimplify:[4R sin A - 2R sin C] cos B = 2R sin B cos CFactor out 2R:2R [2 sin A - sin C] cos B = 2R sin B cos CDivide both sides by 2R (since R ≠ 0):[2 sin A - sin C] cos B = sin B cos COkay, so now I have:2 sin A cos B - sin C cos B = sin B cos CHmm, maybe I can rearrange this equation:2 sin A cos B = sin C cos B + sin B cos CWait, the right-hand side looks familiar. Isn't that the sine of (B + C)? Because sin(B + C) = sin B cos C + cos B sin C.Yes! So, sin(B + C) = sin C cos B + sin B cos C.So, the equation becomes:2 sin A cos B = sin(B + C)But in a triangle, A + B + C = π, so B + C = π - A.Therefore, sin(B + C) = sin(π - A) = sin A.So, substituting back:2 sin A cos B = sin AAssuming sin A ≠ 0 (which it can't be in a triangle, since A is between 0 and π), we can divide both sides by sin A:2 cos B = 1So, cos B = 1/2Therefore, angle B is π/3 radians, or 60 degrees.Alright, that seems straightforward. So, part 1 is solved: angle B is 60 degrees.Moving on to part 2: If λ = √3, determine the shape of triangle ABC.So, we have the second condition:sin² A = sin² B + sin² C - λ sin B sin CGiven that λ = √3, let's substitute that in:sin² A = sin² B + sin² C - √3 sin B sin CHmm, I need to figure out what this implies about the triangle.I remember that in a triangle, using the Law of Cosines, we have:a² = b² + c² - 2bc cos ASimilarly, sin² A can be related to the sides using the Law of Sines. Let me recall that sin² A = (a²)/(4R²), and similarly for sin² B and sin² C.But maybe instead of going that route, I can use the identity for sin² A in terms of cos 2A, but that might complicate things.Alternatively, perhaps I can use the fact that in a triangle, angles add up to π, so A = π - B - C.Given that we already found angle B is π/3, so A = π - π/3 - C = (2π/3) - C.So, maybe I can express sin² A in terms of C.Alternatively, let me think about the given equation:sin² A = sin² B + sin² C - √3 sin B sin CSince we know angle B is π/3, sin B = sin(π/3) = √3/2.So, let's substitute sin B = √3/2 into the equation:sin² A = ( (√3/2)² ) + sin² C - √3*(√3/2)*sin CSimplify:sin² A = (3/4) + sin² C - (3/2) sin CSo, sin² A = sin² C - (3/2) sin C + 3/4Hmm, let me rearrange this:sin² A - sin² C + (3/2) sin C - 3/4 = 0I wonder if I can express sin² A - sin² C in terms of something else.Recall that sin² A - sin² C = sin(A - C) sin(A + C)Since A + C = π - B = π - π/3 = 2π/3.So, sin(A + C) = sin(2π/3) = √3/2.Therefore, sin² A - sin² C = sin(A - C) * (√3/2)So, substituting back:sin(A - C) * (√3/2) + (3/2) sin C - 3/4 = 0Hmm, not sure if that helps directly. Maybe another approach.Alternatively, let's consider that in triangle ABC, with angle B = π/3, sides are related by the Law of Cosines:b² = a² + c² - 2ac cos BBut we also have the given equation in terms of sines. Maybe I can relate the two.Wait, the given equation is:sin² A = sin² B + sin² C - √3 sin B sin CLet me express this in terms of sides using the Law of Sines.We know that a = 2R sin A, b = 2R sin B, c = 2R sin C.So, sin A = a/(2R), sin B = b/(2R), sin C = c/(2R)Therefore, sin² A = a²/(4R²), sin² B = b²/(4R²), sin² C = c²/(4R²)Substituting into the equation:a²/(4R²) = b²/(4R²) + c²/(4R²) - √3*(b/(2R))*(c/(2R))Multiply both sides by 4R²:a² = b² + c² - √3 bcSo, we have:a² = b² + c² - √3 bcHmm, comparing this to the Law of Cosines:a² = b² + c² - 2bc cos ASo, equating the two expressions for a²:b² + c² - 2bc cos A = b² + c² - √3 bcSubtract b² + c² from both sides:-2bc cos A = -√3 bcDivide both sides by -bc (assuming bc ≠ 0, which they aren't in a triangle):2 cos A = √3Therefore, cos A = √3 / 2So, angle A is π/6, or 30 degrees.Since angle B is π/3, angle C is π - π/6 - π/3 = π - π/2 = π/2, or 90 degrees.So, angle C is 90 degrees. Therefore, triangle ABC is a right-angled triangle at C.So, the shape of triangle ABC is a right triangle.Alright, that makes sense. So, part 2 is solved: it's a right triangle.Now, part 3: If triangle ABC is an obtuse triangle, determine the range of values for the real number λ.Hmm, so triangle ABC is obtuse. That means one of the angles is greater than 90 degrees. Since we already know angle B is π/3, which is 60 degrees, the obtuse angle must be either A or C.So, either angle A > π/2 or angle C > π/2.Given that angle B is fixed at π/3, let's see.We have the second condition:sin² A = sin² B + sin² C - λ sin B sin CAgain, we can express this in terms of sides using the Law of Sines.As before, a² = b² + c² - λ bcBut from the Law of Cosines, a² = b² + c² - 2bc cos ATherefore, equating the two:b² + c² - 2bc cos A = b² + c² - λ bcSimplify:-2bc cos A = -λ bcDivide both sides by -bc:2 cos A = λSo, λ = 2 cos ATherefore, λ is twice the cosine of angle A.Now, since triangle ABC is obtuse, either angle A or angle C is obtuse.Case 1: angle A > π/2Then, cos A < 0, since cosine is negative in the second quadrant (angles between π/2 and π).So, in this case, λ = 2 cos A < 0Case 2: angle C > π/2Similarly, angle C > π/2 implies cos C < 0.But angle C = π - A - B = π - A - π/3 = (2π/3) - ASo, if angle C > π/2, then (2π/3 - A) > π/2Which implies:2π/3 - A > π/2Subtract 2π/3:-A > π/2 - 2π/3 = (3π/6 - 4π/6) = -π/6Multiply both sides by -1 (inequality sign flips):A < π/6So, if angle C is obtuse, then angle A < π/6.In this case, angle A is acute, so cos A > 0.But we have λ = 2 cos A.So, in this case, λ > 0.But we need to find the range of λ when the triangle is obtuse, regardless of which angle is obtuse.So, combining both cases:If angle A is obtuse, λ < 0If angle C is obtuse, λ > 0But we need to find the exact range.Wait, let's think about the possible values of angle A.Since angle B is π/3, the sum of angles A and C is 2π/3.If angle A is obtuse, then angle A > π/2, so angle C = 2π/3 - A < 2π/3 - π/2 = π/6So, angle C < π/6Similarly, if angle C is obtuse, angle C > π/2, so angle A = 2π/3 - C < 2π/3 - π/2 = π/6So, in both cases, one angle is greater than π/2, and the other is less than π/6.Now, let's find the range of λ.We have λ = 2 cos ACase 1: angle A > π/2 (obtuse)Then, angle A is between π/2 and π - π/3 - 0 (since angle C must be positive). Wait, actually, angle A can be at most less than π - π/3 = 2π/3.Wait, no, angle A can be up to just less than π, but in reality, since angle C must be positive, angle A must be less than π - π/3 = 2π/3.Wait, no, if angle A is obtuse, it's greater than π/2 but less than π - π/3 = 2π/3.Wait, actually, if angle A is obtuse, then angle C = 2π/3 - A must be positive, so:2π/3 - A > 0 => A < 2π/3Therefore, angle A is between π/2 and 2π/3.So, angle A ∈ (π/2, 2π/3)Therefore, cos A is between cos(2π/3) and cos(π/2)cos(2π/3) = -1/2, cos(π/2) = 0So, cos A ∈ (-1/2, 0)Therefore, λ = 2 cos A ∈ (-1, 0)Case 2: angle C > π/2 (obtuse)Then, angle C is between π/2 and π - π/3 = 2π/3So, angle C ∈ (π/2, 2π/3)Then, angle A = 2π/3 - C ∈ (2π/3 - 2π/3, 2π/3 - π/2) = (0, π/6)So, angle A ∈ (0, π/6)Therefore, cos A ∈ (cos(π/6), cos(0)) = (√3/2, 1)Thus, λ = 2 cos A ∈ (√3, 2)So, combining both cases, λ can be in (-1, 0) or (√3, 2)Therefore, the range of λ is (-1, 0) ∪ (√3, 2)So, that's the range when the triangle is obtuse.Let me double-check.If angle A is obtuse, λ = 2 cos A, and cos A is between -1/2 and 0, so λ is between -1 and 0.If angle C is obtuse, then angle A is less than π/6, so cos A is greater than √3/2, so λ = 2 cos A is greater than √3, but less than 2, since cos A < 1.Yes, that seems correct.So, the range of λ is (-1, 0) union (√3, 2)Alright, that makes sense.**Final Answer**1. The measure of angle ( B ) is (boxed{dfrac{pi}{3}}).2. The triangle ( triangle ABC ) is a (boxed{text{right triangle}}).3. The range of values for ( lambda ) is (boxed{(-1, 0) cup (sqrt{3}, 2)}).

💡Okay, so I have this function f(x) = sin x (cos x - sqrt(3) sin x). There are three parts to this problem. Let me try to tackle each part step by step.Starting with part (I): Find the smallest positive period of the function f(x).Hmm, the function f(x) is given as sin x multiplied by (cos x - sqrt(3) sin x). That looks a bit complicated, but maybe I can simplify it first. Let me expand the expression:f(x) = sin x * cos x - sqrt(3) sin^2 x.Alright, so f(x) is the sum of two terms: sin x cos x and -sqrt(3) sin^2 x. I remember that both sin x and cos x have a period of 2π, but when they are multiplied or combined, the period can change. Also, sin^2 x has a period of π because of the square.Wait, so sin x cos x is actually equal to (1/2) sin 2x, right? Because sin 2x = 2 sin x cos x, so sin x cos x is half of that. Similarly, sin^2 x can be written using the double-angle identity: sin^2 x = (1 - cos 2x)/2.Let me rewrite f(x) using these identities:f(x) = (1/2) sin 2x - sqrt(3) * (1 - cos 2x)/2.Simplifying that:f(x) = (1/2) sin 2x - (sqrt(3)/2) + (sqrt(3)/2) cos 2x.So, f(x) = (1/2) sin 2x + (sqrt(3)/2) cos 2x - sqrt(3)/2.Hmm, this looks like a combination of sine and cosine functions with the same argument 2x. I remember that A sin θ + B cos θ can be written as C sin(θ + φ), where C is the amplitude and φ is the phase shift. Let me try that.Let me denote A = 1/2 and B = sqrt(3)/2. Then, the amplitude C is sqrt(A^2 + B^2) = sqrt((1/2)^2 + (sqrt(3)/2)^2) = sqrt(1/4 + 3/4) = sqrt(1) = 1.Okay, so the amplitude is 1. Now, the phase shift φ can be found using tan φ = B/A. So tan φ = (sqrt(3)/2)/(1/2) = sqrt(3). Therefore, φ = π/3.So, the expression (1/2) sin 2x + (sqrt(3)/2) cos 2x can be written as sin(2x + π/3). Therefore, f(x) = sin(2x + π/3) - sqrt(3)/2.Now, to find the period of f(x). The function sin(2x + π/3) has a period of 2π divided by the coefficient of x, which is 2. So, the period is π. The constant term -sqrt(3)/2 doesn't affect the period. Therefore, the smallest positive period of f(x) is π.Alright, that seems solid. Let me move on to part (II).Part (II): The graph of y = sin 2x is translated to the left by a units (0 < a < π/2) and downwards by b units to obtain the graph of y = f(x). Find the value of ab.Hmm, okay. So, starting from y = sin 2x, we translate it left by a units and down by b units to get f(x). Let me recall how translations affect the function.Translating left by a units means replacing x with x + a. So, y = sin[2(x + a)] = sin(2x + 2a). Then, translating down by b units means subtracting b from the entire function: y = sin(2x + 2a) - b.But from part (I), we have f(x) = sin(2x + π/3) - sqrt(3)/2. Therefore, comparing the two expressions:sin(2x + 2a) - b = sin(2x + π/3) - sqrt(3)/2.So, for these two functions to be equal for all x, their arguments inside the sine function must be equal (up to a multiple of 2π, but since we're dealing with the smallest positive a, we can ignore that), and their constants must be equal.Therefore:2a = π/3 + 2π k, where k is an integer. But since 0 < a < π/2, let's solve for a:2a = π/3 => a = π/6.Similarly, the constants must satisfy:-b = -sqrt(3)/2 => b = sqrt(3)/2.Therefore, a = π/6 and b = sqrt(3)/2. So, ab = (π/6)(sqrt(3)/2) = (π sqrt(3))/12.Wait, is that correct? Let me double-check.Yes, because when translating left by a, the function becomes sin[2(x + a)] = sin(2x + 2a). Comparing to f(x) which is sin(2x + π/3) - sqrt(3)/2, so 2a must equal π/3, so a = π/6. Then, the vertical translation is subtracting b, so b must equal sqrt(3)/2. Multiplying them gives (π/6)(sqrt(3)/2) = π sqrt(3)/12.Okay, that seems right.Now, part (III): Find the range of the function f(x) on [0, π/2].Alright, so f(x) = sin(2x + π/3) - sqrt(3)/2. We need to find the range when x is between 0 and π/2.First, let's find the range of the argument inside the sine function. When x is in [0, π/2], then 2x is in [0, π]. Adding π/3, the argument 2x + π/3 is in [π/3, 4π/3].So, the sine function sin(θ) where θ is in [π/3, 4π/3]. Let me recall the behavior of sine in this interval.At θ = π/3, sin(π/3) = sqrt(3)/2.As θ increases to π/2, sin(θ) increases to 1.Then, as θ increases beyond π/2 to π, sin(θ) decreases back to 0.At θ = π, sin(π) = 0.Then, as θ increases to 4π/3, which is π + π/3, sin(4π/3) = -sqrt(3)/2.So, the maximum value of sin(θ) in [π/3, 4π/3] is 1, and the minimum is -sqrt(3)/2.Therefore, sin(2x + π/3) ranges from -sqrt(3)/2 to 1.Therefore, f(x) = sin(2x + π/3) - sqrt(3)/2 ranges from (-sqrt(3)/2 - sqrt(3)/2) to (1 - sqrt(3)/2).Simplifying:Lower bound: -sqrt(3)/2 - sqrt(3)/2 = -sqrt(3).Upper bound: 1 - sqrt(3)/2.Therefore, the range of f(x) on [0, π/2] is [-sqrt(3), 1 - sqrt(3)/2].Wait, let me confirm that.Yes, because when sin(2x + π/3) is at its maximum 1, f(x) is 1 - sqrt(3)/2. When sin(2x + π/3) is at its minimum -sqrt(3)/2, f(x) is -sqrt(3)/2 - sqrt(3)/2 = -sqrt(3). So, the range is from -sqrt(3) to 1 - sqrt(3)/2.Therefore, the range is [-sqrt(3), 1 - sqrt(3)/2].I think that's correct.So, summarizing:(I) The smallest positive period is π.(II) The product ab is (π sqrt(3))/12.(III) The range of f(x) on [0, π/2] is [-sqrt(3), 1 - sqrt(3)/2].**Final Answer**(I) The smallest positive period is boxed{pi}.(II) The value of ( ab ) is boxed{dfrac{sqrt{3}}{12} pi}.(III) The range of the function ( f(x) ) on ([0, dfrac{pi}{2}]) is boxed{[-sqrt{3}, 1 - dfrac{sqrt{3}}{2}]}.

💡First, I need to determine the amount of flour required for one batch of cookies. Let's denote this amount as ( F ) cups.For one batch, the recipe calls for ( F ) cups of flour and 1.5 cups of sugar.When making 8 batches, the total amount of flour needed would be ( 8F ) cups, and the total amount of sugar needed would be ( 8 times 1.5 = 12 ) cups.The combined total of flour and sugar for 8 batches is given as 44 cups. This leads to the equation:[8F + 12 = 44]To solve for ( F ), I'll subtract 12 from both sides of the equation:[8F = 32]Finally, I'll divide both sides by 8 to find the value of ( F ):[F = frac{32}{8} = 4]Therefore, 4 cups of flour are needed for one batch of cookies.

💡Alright, so I've got this problem here about Petya and Vasya with the candies. Let me try to wrap my head around it. There are 28 candies on a table, and Petya thinks some of them are tasty. Vasya can ask Petya about any set of candies and find out how many are tasty in that set. The goal is for Vasya to figure out which specific candies are tasty, and the challenge is to do it in 21 moves for part (a) and 20 moves for part (b).Okay, so first, I need to understand what a "move" is. It seems like a move is when Vasya points to a set of candies and asks Petya how many are tasty. So, each move gives Vasya some information about the number of tasty candies in a specific subset. The key here is that Vasya needs to design these subsets strategically so that after a certain number of moves, he can deduce exactly which candies are tasty.Let me think about part (a) first, which asks for a solution in 21 moves. Since there are 28 candies, and each move gives some information, I need to figure out how to cover all 28 candies efficiently. Maybe I can divide the candies into groups and ask about each group separately. If I divide them into smaller groups, say groups of 4, then I can ask about each group and narrow down the tasty candies within each group.So, if I have 7 groups of 4 candies each, and for each group, I ask about different subsets, maybe I can figure out which ones are tasty. For example, for each group of 4, I could ask about three different subsets: {a, c, d}, {b, c, d}, and {a, b, c}. Then, based on the answers, I can determine which candies are tasty.Wait, let me think through that. If I ask about {a, c, d}, {b, c, d}, and {a, b, c}, and compare the answers, maybe I can figure out the status of each candy. For instance, if the answers for the first two sets are different, that might tell me something about a and b. If they're the same, maybe a and b are both tasty or both not tasty. Then, using the third answer, I can figure out c, and then go back to the first question to figure out d.Hmm, that seems plausible. So, for each group of 4, it takes 3 questions to figure out which ones are tasty. Since there are 7 groups, that would be 3 times 7, which is 21 moves. That matches the requirement for part (a). So, that seems like a solid strategy.Now, moving on to part (b), which asks for a solution in 20 moves. That's one fewer move than part (a). So, I need to find a way to be more efficient. Maybe there's a way to overlap some of the information or reuse some of the questions across different groups.I remember that in information theory, each yes/no question gives one bit of information. But in this case, each question gives more information because it's not just a yes/no answer but a count. So, maybe I can leverage that to get more information per question.Alternatively, maybe there's a way to structure the questions so that each question provides information about multiple groups or candies. For example, instead of asking about each group separately, I could design questions that cover multiple groups at once, thereby reducing the total number of questions needed.Wait, another thought: if I can find a way to determine the status of multiple candies with a single question, that could help reduce the number of moves. Maybe by using a binary search approach or something similar, I can eliminate large portions of candies at once.But I need to ensure that I can still uniquely identify each tasty candy. So, it's a balance between getting enough information to distinguish each candy while minimizing the number of questions.Maybe I can use a method similar to binary representation. If I assign each candy a unique binary identifier and then ask questions that correspond to each bit position, I can determine the status of each candy based on the answers. For 28 candies, I would need log2(28) ≈ 5 bits, so 5 questions might be enough. But that seems too optimistic because each question only gives a count, not a binary answer.Wait, perhaps I can use a system where each candy is included in a specific combination of sets, and based on the counts, I can solve for which candies are tasty. This is similar to how error-correcting codes work, where each bit is covered by multiple parity checks.So, if I design the sets such that each candy is in a unique combination of sets, I can set up a system of equations where the counts correspond to the sum of tasty candies in each set. Then, solving this system would give me the exact set of tasty candies.But setting up such a system might require more questions than 20, especially since I need to ensure that the system is solvable. Maybe there's a more efficient way to design the sets.Alternatively, I could use a divide and conquer strategy. Split the candies into two groups, ask about each group, and then recursively apply the same strategy to the groups that contain tasty candies. But I'm not sure if that would necessarily get me down to 20 moves.Wait, going back to the initial strategy for part (a), where I used 3 questions per group of 4. If I can find a way to overlap some of these questions or reuse some information across groups, maybe I can save a move.For example, if I have overlapping groups, a single question could provide information about multiple groups. But I need to ensure that the information is still sufficient to determine the status of each candy without ambiguity.Another idea: maybe I can use the answers from previous questions to inform the design of subsequent questions, thereby making each question more informative based on what I've already learned.But I need to make sure that this adaptive questioning doesn't end up requiring more questions overall. It has to be structured in a way that guarantees completion in 20 moves regardless of Petya's answers.Hmm, this is tricky. I think the key is to find a systematic way to cover all candies with as few questions as possible, ensuring that each candy's status can be uniquely determined from the answers.Perhaps using a combination of group questions and individual questions. For example, first ask about large groups to narrow down the possible candidates, and then ask about smaller subsets to pinpoint the exact tasty candies.But I need to calculate how many questions that would take and ensure it's within 20.Wait, maybe I can use a binary search approach. Start by dividing the candies into two groups of 14 and ask about each group. Based on the counts, I can determine which groups contain tasty candies and focus on those. Then, recursively apply the same process to smaller groups.But binary search typically requires log2(n) steps, which for 28 would be around 5 steps. However, each step involves asking about multiple groups, so the total number of questions might be higher.I think I need to find a balance between the number of groups and the number of questions per group. Maybe a hybrid approach where I use some initial questions to narrow down the possibilities and then use more detailed questions to identify the exact candies.Alternatively, perhaps there's a known algorithm or method for this type of problem that I can apply. It seems similar to group testing, where the goal is to identify defective items (in this case, tasty candies) with as few tests as possible.In group testing, there are strategies like binary search, hierarchical testing, and combinatorial testing. Combinatorial group testing might be particularly useful here because it allows for more efficient identification by testing combinations of items.In combinatorial group testing, the idea is to design tests (questions) such that each item is included in a unique combination of tests. Then, based on the results of the tests, you can determine which items are defective (tasty).For 28 candies, if I can design 20 tests where each candy is included in a unique subset, I might be able to uniquely identify the tasty ones. However, designing such a set of subsets is non-trivial and requires careful planning to ensure that the system of equations is solvable.Alternatively, I could use a method where each candy is included in a specific number of tests, and the counts from the tests can be used to solve for the individual statuses. This might involve setting up a system of linear equations where each equation corresponds to a test, and the variables correspond to the candies.But solving such a system would require that the number of tests is at least equal to the number of variables (candies), which in this case is 28. However, we're limited to 20 tests, so this approach might not work directly.Wait, maybe I can use a probabilistic method or some form of compressed sensing, where I can recover the exact set of tasty candies with fewer measurements (questions) than the total number of candies. But I'm not sure if that's applicable here since we need a deterministic method that guarantees finding all tasty candies.Another thought: if I can find a way to represent the candies in a lower-dimensional space, I might be able to reduce the number of questions needed. For example, using binary representations or other encoding schemes to map the candies into a smaller set of questions.But again, I'm not sure how to apply that directly to this problem. It seems like I'm going in circles here.Let me try to summarize what I've thought so far:1. For part (a), dividing the candies into groups of 4 and asking 3 questions per group seems to work, requiring 21 moves in total.2. For part (b), I need to find a way to reduce the number of moves by 1, to 20. This likely involves finding a more efficient questioning strategy, possibly by overlapping questions or using a different grouping method.3. Potential strategies include combinatorial group testing, binary search, or designing a system of questions that provides maximum information per question.4. However, designing such a system is complex and requires ensuring that the information from each question can be used to uniquely determine the status of each candy.Maybe I need to look for a specific method or algorithm that's known for this type of problem. I recall that in information theory, the minimum number of questions needed to identify a specific item is related to the logarithm of the number of possibilities. But in this case, we're dealing with multiple items (the tasty candies), so it's more complex.Wait, perhaps I can think of it in terms of binary representations. If I assign each candy a unique binary code, and then design questions that correspond to each bit position, I can determine which candies are tasty by looking at the parity of the counts.For example, if I assign each candy a 5-bit binary number (since 2^5 = 32 > 28), then I can ask 5 questions, each corresponding to a bit position. For each question, I include all candies that have a '1' in that bit position. The count from each question would give me the number of tasty candies with a '1' in that bit position. By combining these counts, I can determine the binary codes of the tasty candies.But wait, this would require 5 questions, which is way fewer than 20. However, this method assumes that the counts can be used to uniquely determine the binary codes, which might not be the case if multiple candies have overlapping bit patterns.Actually, this method is similar to non-adaptive group testing, where all questions are designed upfront without depending on previous answers. However, non-adaptive group testing might not always guarantee unique identification unless the test design is perfect.Given that, maybe I can use a non-adaptive approach with a number of questions equal to the number of bits needed to represent the candies, which is 5. But since we need to identify not just one candy but potentially multiple, the number of questions might need to be higher.Alternatively, I can use a method where each question is designed to split the remaining candidates as evenly as possible, maximizing the information gained from each question. This is similar to a binary search but applied to multiple items.But I'm not sure how to translate that into a fixed number of questions like 20.Wait, maybe I can use a combination of the initial grouping strategy and some overlapping questions to save a move. For example, instead of asking 3 separate questions for each group of 4, maybe I can reuse some of the questions across adjacent groups, thereby reducing the total number of questions needed.But I need to ensure that reusing questions doesn't lead to ambiguity in determining the status of individual candies.Alternatively, perhaps I can use a different grouping strategy, like dividing the candies into larger groups and asking questions that cover multiple groups at once, thereby reducing the total number of questions.But I need to make sure that the information from these larger groups can still be used to pinpoint the exact tasty candies.Another idea: maybe I can use the answers from previous questions to inform the design of subsequent questions, making each question more targeted and informative. This adaptive approach might allow me to save moves by focusing on the most informative subsets.However, designing such an adaptive strategy that guarantees completion in 20 moves is non-trivial and would require careful planning to ensure that no matter how Petya answers, Vasya can always determine the tasty candies in 20 moves.Perhaps I can start by asking about the entire set of 28 candies. That would give me the total number of tasty candies, which might help in designing subsequent questions. Then, I can ask about subsets that split the candies into halves or other divisions, using the counts to narrow down the possibilities.But I need to calculate how many questions that would take and ensure it's within 20.Wait, if I use a binary search approach, starting with the whole set and then dividing it into halves, quarters, etc., I might be able to identify the tasty candies in log2(28) ≈ 5 questions. But that's only if I'm looking for one candy. Since there could be multiple tasty candies, this approach might not directly apply.Alternatively, I can use a method where each question is designed to eliminate as many possibilities as possible, regardless of the number of tasty candies. This would involve asking about subsets that are likely to contain a significant number of tasty candies, thereby reducing the search space quickly.But again, without knowing the distribution of tasty candies, it's hard to design such questions upfront.Maybe I need to accept that part (b) requires a more advanced strategy, possibly involving combinatorial mathematics or information theory, to achieve the 20-move guarantee.In summary, for part (a), the grouping strategy with 3 questions per group of 4 seems solid, requiring 21 moves. For part (b), I need to find a more efficient method, possibly by overlapping questions, using a different grouping strategy, or applying a known algorithm from group testing or information theory.I think the key takeaway is that part (a) is manageable with a straightforward grouping approach, while part (b) requires a more sophisticated strategy to reduce the number of moves by 1. It might involve reusing information from previous questions or designing questions that cover multiple groups or candies simultaneously.I need to research more on group testing algorithms or information-efficient questioning strategies to find a method that can guarantee identifying all tasty candies in 20 moves. Alternatively, I might need to think outside the box and come up with a unique questioning pattern that maximizes the information gained per question.Overall, part (a) seems achievable with the grouping method, but part (b) will require more thought and possibly a different approach to minimize the number of moves while still ensuring all tasty candies are identified.

💡First, I notice that each term in the expression ( ab^2 + 10ab + 25a ) contains the common factor ( a ). I can factor out this ( a ) to simplify the expression.After factoring out ( a ), the expression becomes ( a(b^2 + 10b + 25) ). Next, I observe that the quadratic expression inside the parentheses, ( b^2 + 10b + 25 ), is a perfect square trinomial. A perfect square trinomial has the form ( (x + y)^2 = x^2 + 2xy + y^2 ). In this case, ( x = b ) and ( y = 5 ), since ( (b + 5)^2 = b^2 + 10b + 25 ).Finally, I substitute the perfect square back into the factored expression to obtain the fully factorized form.