answer：Okay, so I have this problem here about a geometric sequence. Let me first recall what a geometric sequence is. It's a sequence where each term after the first is found by multiplying the previous term by a constant called the common ratio, usually denoted as 'r'. So, if I denote the first term as 'a', then the sequence goes like a, ar, ar², ar³, and so on.The problem says that a₆ = 3, and I need to find the value of the product a₃ a₄ a₅ a₆ a₇ a₈ a₉. Hmm, that's seven terms multiplied together. Let me write down what each of these terms is in terms of 'a' and 'r'.First, let's express each term:- a₃ is the third term, which is ar².- a₄ is the fourth term, which is ar³.- a₅ is the fifth term, which is ar⁴.- a₆ is the sixth term, which is ar⁵, and that's given as 3.- a₇ is the seventh term, which is ar⁶.- a₈ is the eighth term, which is ar⁷.- a₉ is the ninth term, which is ar⁸.So, the product we need is:a₃ * a₄ * a₅ * a₆ * a₇ * a₈ * a₉ = (ar²)(ar³)(ar⁴)(ar⁵)(ar⁶)(ar⁷)(ar⁸)Let me compute this product step by step. First, I can multiply all the 'a's together and then all the 'r's together.There are seven 'a's, so multiplying them gives a⁷.Now, for the 'r's: the exponents are 2, 3, 4, 5, 6, 7, 8. Let me add those exponents together.2 + 3 + 4 + 5 + 6 + 7 + 8.Let me compute that:2 + 3 = 55 + 4 = 99 + 5 = 1414 + 6 = 2020 + 7 = 2727 + 8 = 35So, the exponent on 'r' is 35. Therefore, the product is a⁷ * r³⁵.But wait, I know that a₆ = ar⁵ = 3. So, ar⁵ = 3. Maybe I can express 'a' in terms of 'r' or vice versa to simplify this product.Let me solve for 'a' from a₆ = 3:ar⁵ = 3 => a = 3 / r⁵So, if I substitute this into the product expression:a⁷ * r³⁵ = (3 / r⁵)⁷ * r³⁵Let me compute (3 / r⁵)⁷ first:(3)⁷ / (r⁵)⁷ = 3⁷ / r³⁵So, the product becomes:(3⁷ / r³⁵) * r³⁵Ah, the r³⁵ in the numerator and denominator will cancel each other out. So, we're left with 3⁷.What's 3⁷? Let me compute that:3¹ = 33² = 93³ = 273⁴ = 813⁵ = 2433⁶ = 7293⁷ = 2187So, the product is 2187.Wait, let me double-check my steps to make sure I didn't make a mistake.1. Expressed each term in terms of 'a' and 'r' correctly.2. Multiplied all the terms together: a⁷ * r^(2+3+4+5+6+7+8) = a⁷ * r³⁵. That seems right.3. Noted that a₆ = ar⁵ = 3, so a = 3 / r⁵.4. Substituted into the product: (3 / r⁵)^7 * r³⁵ = 3⁷ / r³⁵ * r³⁵ = 3⁷. That seems correct.5. Calculated 3⁷ as 2187. Let me verify that:3^1 = 33^2 = 93^3 = 273^4 = 813^5 = 2433^6 = 7293^7 = 2187. Yes, that's correct.So, I think my answer is correct. The product a₃ a₄ a₅ a₆ a₇ a₈ a₉ is 2187.**Final Answer**The value of (a_3 a_4 a_5 a_6 a_7 a_8 a_9) is boxed{2187}.

answer：Okay, so I have this problem here about stratified sampling. Let me try to understand it step by step. First, the problem says there are three types of households in a region: farmer households, worker households, and intellectual households. The numbers given are 1500 farmer households, 401 worker households, and 99 intellectual households. So, in total, how many households are there? Let me add them up: 1500 + 401 + 99. Hmm, 1500 plus 401 is 1901, and then adding 99 gives 2000. So, there are 2000 households in total. Now, the problem mentions using stratified sampling to draw a sample of size n from all the households. I remember that stratified sampling is a method where the population is divided into subgroups, or strata, and then samples are taken from each stratum proportionally. So, each stratum is represented in the sample according to its size in the population. It's given that 75 farmer households were drawn in the sample. So, I need to find the total sample size n. Let me recall the formula for stratified sampling. If we have different strata with sizes N1, N2, ..., Nk, and we want a total sample size n, then the number of samples from each stratum is n1 = (N1 / N) * n, where N is the total population. In this case, the farmer households are one stratum with N1 = 1500, worker households N2 = 401, and intellectual households N3 = 99. The total population N is 2000. We know that the number of farmer households sampled, n1, is 75. So, using the formula:n1 = (N1 / N) * nPlugging in the numbers:75 = (1500 / 2000) * nLet me compute 1500 divided by 2000. That simplifies to 3/4 or 0.75. So,75 = 0.75 * nTo find n, I can divide both sides by 0.75:n = 75 / 0.75Calculating that, 75 divided by 0.75 is 100. So, n should be 100.Wait, let me double-check. If the total sample size is 100, and 75 are farmers, then the remaining 25 should be from the other strata. Let me see if that makes sense proportionally.The proportion of farmer households is 1500/2000 = 0.75, so 75/100 = 0.75, which matches. For the worker households, the proportion is 401/2000 ≈ 0.2005, so the number of worker households in the sample should be approximately 0.2005 * 100 ≈ 20.05, which we can round to 20. Similarly, for intellectual households, the proportion is 99/2000 ≈ 0.0495, so the number in the sample should be approximately 0.0495 * 100 ≈ 4.95, which we can round to 5. Adding those up: 75 + 20 + 5 = 100. That seems consistent. So, it looks like 100 is the correct total sample size. I think that makes sense. So, n is 100.**Final Answer**The value of ( n ) is boxed{100}.

answer：First, I need to determine how to arrange the four basil plants and four tomato plants in a row with the condition that all tomato plants must be next to each other.Since all tomato plants must be together, I can treat them as a single unit. This means I have the four basil plants plus one unit of tomato plants, making a total of five units to arrange.The number of ways to arrange these five units is 5 factorial, which is 5! = 120.However, within the tomato unit, the four tomato plants can be arranged among themselves in 4! = 24 ways.To find the total number of possible arrangements, I multiply the number of ways to arrange the units by the number of ways to arrange the tomato plants within their unit: 120 × 24 = 2880.Therefore, there are 2880 different ways April can arrange the plants in a row with all tomato plants next to each other.

answer：Alright, so I've got this problem about an 8x8 chessboard with two identical pieces on a1 and c3. Petya and Vasya are taking turns moving these pieces either upwards or to the right by any number of cells, and the goal is to be the one to move a piece to h8. Petya goes first. I need to figure out which player can always win, no matter how the opponent plays.Hmm, okay. Let me start by understanding the movement. Each piece can only move up or to the right. So, from a1, a piece can move right along the first rank or up along the a-file. Similarly, from c3, it can move right along the third rank or up along the c-file. The pieces can't jump over each other, so if one piece is blocking the path, the other can't move past it.The target is h8, which is the top-right corner. So, to get there, a piece needs to move right 7 times and up 7 times, but in any order. But since each move can be any number of cells, it's more about the distance each piece has to cover.Let me think about the positions of the two pieces. The first piece is on a1, which is the bottom-left corner. The second is on c3, which is a bit further along. So, the piece on c3 is closer to h8 in both the horizontal and vertical directions. From a1, the piece has to move 7 squares right and 7 squares up. From c3, it's 5 squares right and 5 squares up.Wait, so the piece on c3 is actually closer to h8. That might give it an advantage. But since both players can choose which piece to move on their turn, maybe the starting position isn't the only factor.I remember something about impartial games and the concept of Nimbers or Grundy numbers. Maybe this can be modeled as a game where each piece is a heap, and moving a piece is like reducing the heap size. But in this case, each piece has two dimensions: horizontal and vertical distance to h8.Let me try to model each piece's position as coordinates. Let's say the board is a grid with coordinates (x, y), where x is the column (a=1, b=2, ..., h=8) and y is the row (1=1, 2=2, ..., 8=8). So, a1 is (1,1), c3 is (3,3), and h8 is (8,8).Each move either increases x or y by some amount, but you can't move past h8 or beyond. So, for each piece, the distance to h8 is (8 - x, 8 - y). For the piece on a1, that's (7,7). For the piece on c3, that's (5,5).Wait, so each piece can be thought of as having two independent distances: right and up. But since each move affects only one of them, maybe this is a two-dimensional game. But I'm not sure how to handle that.Alternatively, maybe I can think of each piece as a separate game, and the overall game is the combination of these two. So, each piece is a pile in a Nim game, but with two dimensions. Hmm, that might complicate things.Alternatively, perhaps I can convert the two-dimensional distances into a single number. For example, the number of moves required if you could only move one square at a time. But since you can move any number of squares, it's more like each direction is a heap where you can take any number from the heap.Wait, that might make sense. For each piece, the horizontal distance is a heap, and the vertical distance is another heap. So, each piece is a pair of heaps. Then, the game is the combination of two such pairs.But I'm not sure how that would translate into a winning strategy. Maybe I need to think in terms of the Sprague-Grundy theorem, which says that any impartial game under the normal play condition is equivalent to a Nimber.So, perhaps each piece's position can be assigned a Grundy number, and the XOR of these numbers will determine the winning position.Let me try to compute the Grundy numbers for each piece.For a single piece, the Grundy number would be the XOR of the horizontal and vertical distances. Wait, is that right? Because moving in one direction doesn't affect the other.Wait, actually, for a single piece, the game is similar to a two-dimensional Nim game, where you can decrease either the x or y coordinate. The Grundy number for such a position (a, b) is a XOR b.So, for the piece on a1, which is (1,1), the distance to h8 is (7,7). So, the Grundy number would be 7 XOR 7, which is 0.For the piece on c3, which is (3,3), the distance is (5,5). So, the Grundy number is 5 XOR 5, which is also 0.Wait, so both pieces have a Grundy number of 0? That would mean that the overall Grundy number is 0 XOR 0, which is 0. So, the starting position is a losing position for the first player.But that can't be right because the first player can make a move. Maybe I'm misunderstanding how to compute the Grundy numbers for the combined game.Alternatively, perhaps each piece is a separate game, and the overall game is the XOR of their Grundy numbers.But if each piece has a Grundy number of 0, then the overall game is 0, which would mean it's a losing position for the first player.But that contradicts my intuition because the first player can move either piece towards h8.Wait, maybe I'm miscalculating the Grundy numbers. Let me think again.For a single piece, the Grundy number is indeed the XOR of the horizontal and vertical distances. So, if a piece is at (x, y), the Grundy number is (8 - x) XOR (8 - y).So, for a1, which is (1,1), it's 7 XOR 7 = 0.For c3, which is (3,3), it's 5 XOR 5 = 0.So, both pieces have Grundy number 0. Therefore, the total Grundy number is 0 XOR 0 = 0.In combinatorial game theory, a position with a Grundy number of 0 is a losing position for the player about to move. So, since Petya is the first player, and the starting position has a Grundy number of 0, Petya is in a losing position if both players play optimally.Wait, but that seems counterintuitive because Petya can make a move. Maybe I'm missing something.Alternatively, perhaps the Grundy number for each piece is not just the XOR of the distances, but something else.Wait, let me think about the game as a two-dimensional game where each move affects one coordinate. So, each piece is a separate game, and the overall game is the disjunctive sum of these two games.In that case, the Grundy number of the overall game is the XOR of the Grundy numbers of each piece.But if each piece's Grundy number is 0, then the overall is 0, meaning it's a losing position for the first player.But let me test this with smaller examples.Suppose we have a 2x2 board, with pieces on a1 and c3 (but in 2x2, c3 doesn't exist). Maybe a 3x3 board.Wait, maybe I should consider a simpler case. Let's say we have two pieces on a1 and a2 on a 2x2 board. The target is b2.Wait, maybe that's too small. Alternatively, think of each piece as a separate game where you can move right or up, and the Grundy number is the XOR of the right and up distances.So, for a piece at (x, y), the Grundy number is (8 - x) XOR (8 - y). So, for a1, it's 7 XOR 7 = 0. For c3, it's 5 XOR 5 = 0.So, the total Grundy number is 0 XOR 0 = 0, which is a losing position.Therefore, the first player cannot force a win, meaning Vasya can always win.Wait, but that seems to contradict the idea that the first player can make a move. Maybe the issue is that the two pieces are identical, so moving one is the same as moving the other.Wait, in standard combinatorial game theory, when you have multiple identical games, the overall Grundy number is the XOR of each individual game's Grundy number. So, if both games have Grundy number 0, the total is 0.But in this case, the two pieces are on different positions, so they are different games. So, their Grundy numbers are both 0, so the total is 0.Therefore, the first player cannot make a winning move, because any move they make will change the Grundy number of one of the pieces, making the total Grundy number non-zero, allowing the second player to respond accordingly.Wait, let me think about that. If the starting position is 0, then any move will change one of the pieces' Grundy numbers, say from 0 to some non-zero number, so the total becomes non-zero. Then, the second player can mirror the move or adjust accordingly to bring the total back to 0.But in this case, since the two pieces are on different positions, the second player might not be able to mirror the move exactly, but can adjust one of the other pieces to balance the Grundy numbers.Alternatively, perhaps the second player can always respond in such a way to maintain the total Grundy number as 0.Wait, let me test this with an example.Suppose Petya moves the piece on a1. Let's say he moves it right to, say, a8. So, the piece is now on a8, which is (1,8). The distance to h8 is (7,0). So, the Grundy number is 7 XOR 0 = 7.Now, the other piece is still on c3, which has a Grundy number of 0. So, the total Grundy number is 7 XOR 0 = 7.Now, Vasya can move the other piece. He can move it in such a way to make its Grundy number 7, so that the total becomes 7 XOR 7 = 0.How can he do that? The piece on c3 has a distance of (5,5). To get a Grundy number of 7, he needs to move it such that (8 - x) XOR (8 - y) = 7.Let me compute what x and y would need to be.Let’s denote dx = 8 - x, dy = 8 - y. So, dx XOR dy = 7.We need to find dx and dy such that dx XOR dy = 7, and the move is either increasing x or y.Currently, dx = 5, dy = 5.If Vasya moves the piece on c3 to the right, he can increase x, which decreases dx. Similarly, moving up increases y, which decreases dy.Wait, so he can choose to either decrease dx or dy.He wants to change either dx or dy such that the new dx XOR dy = 7.Let me see. Currently, dx = 5, dy = 5. So, 5 XOR 5 = 0.He needs to change either dx or dy to make the XOR 7.Let’s see, if he decreases dx by some amount, say k, so new dx = 5 - k. Then, new dy remains 5.So, (5 - k) XOR 5 = 7.We need to find k such that (5 - k) XOR 5 = 7.Let me compute 5 in binary: 101.7 is 111.So, (5 - k) XOR 5 = 7.Let me denote a = 5 - k.So, a XOR 5 = 7.Therefore, a = 7 XOR 5.Compute 7 XOR 5:7: 1115: 101XOR: 010, which is 2.So, a = 2.Therefore, 5 - k = 2 => k = 3.So, Vasya needs to decrease dx by 3, meaning he moves the piece from c3 to c6 (since moving up 3 squares from c3 would take it to c6). Wait, no. Wait, dx is the horizontal distance, so moving right decreases dx.Wait, dx is 8 - x, so moving right increases x, which decreases dx.So, to decrease dx by 3, he needs to move the piece 3 squares to the right. From c3, moving right 3 squares would take it to f3. So, the new position is f3, which is (6,3). So, dx = 8 - 6 = 2, dy = 8 - 3 = 5.So, the new Grundy number is 2 XOR 5 = 7.Therefore, after Vasya's move, the two pieces have Grundy numbers 7 and 7, so the total is 0.Wait, no. The first piece was moved to a8, which has a Grundy number of 7. The second piece was moved to f3, which has a Grundy number of 2 XOR 5 = 7. So, the total Grundy number is 7 XOR 7 = 0.So, Vasya can always respond to Petya's move by adjusting the other piece to balance the Grundy numbers, bringing the total back to 0.Similarly, if Petya had moved the other piece first, Vasya could have done the same.Therefore, Vasya can always mirror Petya's moves in such a way to maintain the total Grundy number as 0, eventually forcing Petya into a position where he cannot make a move without losing.Wait, but in this case, the target is h8, so the game ends when a piece reaches h8. So, the player who moves a piece to h8 wins.So, in the above example, if Petya moves the a1 piece to a8, which is one square away from h8, but not h8 yet. Then Vasya can move the other piece to f3, maintaining the balance.But then, on Petya's next turn, he could move the a8 piece up to h8 and win. Wait, that can't be right because a8 is on the a-file, so moving up from a8 would go off the board. Wait, no, a8 is already on the 8th rank, so you can't move up anymore. So, the only move from a8 is to move right, but it's already on the 8th file, so it can't move right either. So, actually, moving a piece to a8 doesn't help because it can't move further.Wait, that's a good point. So, moving a piece to a8 doesn't win the game because it can't reach h8 from there. Similarly, moving a piece to h1 wouldn't help because it can't move up anymore.So, in my earlier example, moving the a1 piece to a8 doesn't win the game, it just blocks that piece. So, Vasya can then focus on moving the other piece towards h8.Wait, so maybe my initial analysis was flawed because moving a piece to the edge doesn't necessarily help.Let me try a different approach.Let me think of each piece's distance to h8 in terms of the number of moves required if you could only move one square at a time. But since you can move any number of squares, it's more like each direction is a heap where you can take any number.Wait, so for each piece, the horizontal distance is a heap, and the vertical distance is another heap. So, each piece is a two-heap Nim game.But the overall game is the combination of two such games, so the Grundy number is the XOR of the Grundy numbers of each piece.But for a single piece, the Grundy number is the XOR of its horizontal and vertical distances.So, for a1, it's 7 XOR 7 = 0.For c3, it's 5 XOR 5 = 0.So, the total Grundy number is 0 XOR 0 = 0.Therefore, the starting position is a losing position for the first player.So, Vasya can always win.Wait, but let me test this with a specific example.Suppose Petya moves the a1 piece to, say, a3. So, from (1,1) to (1,3). Now, the distance is (7,5). So, the Grundy number is 7 XOR 5 = 2.Now, the other piece is still on c3, which has a Grundy number of 0.So, the total Grundy number is 2 XOR 0 = 2.Now, Vasya can move the other piece to balance it. He needs to make the total Grundy number 0.So, he can move the c3 piece such that its Grundy number becomes 2.From c3, which is (3,3), the distance is (5,5). To get a Grundy number of 2, he needs (8 - x) XOR (8 - y) = 2.So, let's denote dx = 8 - x, dy = 8 - y. We have dx XOR dy = 2.Currently, dx = 5, dy = 5.He can choose to move right or up.If he moves right, he decreases dx. Let's say he moves right k squares, so new dx = 5 - k.Then, (5 - k) XOR 5 = 2.Let me compute 5 - k such that (5 - k) XOR 5 = 2.Let me denote a = 5 - k.So, a XOR 5 = 2.Therefore, a = 2 XOR 5 = 7.But 5 - k = 7 implies k = -2, which is impossible because you can't move left.So, moving right won't work.Alternatively, he can move up, which decreases dy.So, new dy = 5 - k.Then, 5 XOR (5 - k) = 2.So, 5 XOR (5 - k) = 2.Let me compute 5 XOR (5 - k) = 2.Let me denote b = 5 - k.So, 5 XOR b = 2.Therefore, b = 5 XOR 2 = 7.So, 5 - k = 7 => k = -2, again impossible.Wait, that can't be right. Maybe I made a mistake.Wait, let me think differently. Maybe he can move the piece to a position where dx XOR dy = 2.So, starting from (5,5), he needs to reach a position where dx XOR dy = 2.Let me list possible (dx, dy) pairs that XOR to 2.Possible pairs:(0,2), (2,0), (1,3), (3,1), (4,6), (6,4), (5,7), (7,5), etc.But since dx and dy must be non-negative integers less than or equal to 7.So, starting from (5,5), he can move right or up.If he moves right, dx decreases, dy remains 5.So, new dx = 5 - k, dy = 5.We need (5 - k) XOR 5 = 2.Let me compute (5 - k) XOR 5 = 2.Let me denote a = 5 - k.So, a XOR 5 = 2.Therefore, a = 2 XOR 5 = 7.So, 5 - k = 7 => k = -2, which is impossible.So, moving right won't work.If he moves up, dy decreases, dx remains 5.So, new dy = 5 - k.We need 5 XOR (5 - k) = 2.So, 5 XOR (5 - k) = 2.Let me denote b = 5 - k.So, 5 XOR b = 2.Therefore, b = 5 XOR 2 = 7.So, 5 - k = 7 => k = -2, again impossible.Hmm, so moving either right or up from c3 won't allow him to reach a position where the Grundy number is 2.Wait, that's a problem. Maybe my initial assumption that the Grundy number is the XOR of the distances is incorrect.Alternatively, perhaps the Grundy number for each piece is not just the XOR, but something else.Wait, maybe I need to think of each piece's position as a separate game and compute the Grundy number for each.For a single piece, the game is equivalent to a Nim heap where you can move in two directions. So, the Grundy number for a position (x, y) is indeed (8 - x) XOR (8 - y).But in that case, from (5,5), moving right or up won't allow you to reach a position where the Grundy number is 2, as shown above.Wait, maybe I need to consider that moving a piece can affect both distances, but in reality, moving right only affects the horizontal distance, and moving up only affects the vertical distance.So, perhaps the Grundy number for each piece is indeed the XOR of the two distances.But then, in the example above, Vasya cannot make a move to balance the Grundy numbers, which would mean that the initial assumption that the starting position is losing is incorrect.Wait, maybe I need to think differently. Maybe the game is not equivalent to Nim, but to another impartial game.Alternatively, perhaps the key is to realize that the two pieces are symmetric in some way, and the second player can mirror the first player's moves.But in this case, the pieces start at different positions, so mirroring isn't straightforward.Wait, let me think about the parity of the distances.From a1, the piece has to move 7 right and 7 up. From c3, it's 5 right and 5 up.So, both are odd numbers. So, the total number of moves required for each piece is 14 and 10, respectively.But since each move can cover multiple squares, the number of moves isn't directly relevant.Wait, perhaps the key is that both pieces have an even number of squares to move in each direction? No, 7 and 5 are both odd.Wait, maybe the key is that the two pieces are both an even number of moves away from h8 in terms of individual directions.Wait, no, 7 and 5 are odd.Alternatively, perhaps the key is that the two pieces are both on squares of the same color. a1 is a dark square, c3 is also a dark square. h8 is a light square. So, moving a piece from a dark square to a light square requires an odd number of moves, but since each move can be any number of squares, the color doesn't matter.Wait, maybe that's not relevant.Alternatively, perhaps the key is to realize that the two pieces are both on the same diagonal. a1 to h8 is a diagonal, and c3 is also on that diagonal. Wait, no, c3 is not on the a1-h8 diagonal. The a1-h8 diagonal is a1, b2, c3, d4, e5, f6, g7, h8. Oh, wait, c3 is on that diagonal!So, both pieces are on the main diagonal from a1 to h8. So, the piece on a1 is at (1,1), and the piece on c3 is at (3,3). So, they are both on the same diagonal.So, maybe the key is that the second player can mirror the first player's moves with respect to this diagonal.Wait, but how? Because the pieces are on the same diagonal, but moving one affects its position, and the other can be moved accordingly.Wait, for example, if Petya moves the a1 piece right to, say, d1, then Vasya can move the c3 piece up to c6, maintaining some symmetry.But I'm not sure if that's the case.Alternatively, perhaps the key is that the two pieces are both on the same diagonal, and the second player can always respond in a way that maintains the balance.Wait, but I'm not sure.Alternatively, perhaps the key is that the two pieces are both on the same diagonal, and the second player can always move the other piece to maintain the same distance from h8.Wait, let me think about the distances.From a1, the distance is 7 right and 7 up.From c3, the distance is 5 right and 5 up.So, the piece on c3 is 2 squares ahead in both directions.So, if Petya moves the a1 piece, Vasya can move the c3 piece in a way that maintains the 2-square lead.Alternatively, if Petya moves the c3 piece, Vasya can move the a1 piece to maintain the 2-square lead.Wait, that might be a strategy.For example, if Petya moves the a1 piece right by k squares, Vasya can move the c3 piece right by k squares as well, maintaining the 2-square lead.Similarly, if Petya moves the a1 piece up by k squares, Vasya can move the c3 piece up by k squares.Alternatively, if Petya moves the c3 piece, Vasya can move the a1 piece accordingly.Wait, but since the pieces are on the same diagonal, moving one affects the other's relative position.Wait, let me think with an example.Suppose Petya moves the a1 piece right to d1. So, from (1,1) to (4,1). Now, the distance for this piece is (4,7). So, right distance is 4, up distance is 7.Now, Vasya can move the c3 piece right to f3. From (3,3) to (6,3). Now, the distance for this piece is (2,5).Wait, but now the a1 piece is at (4,1), which is 4 right and 7 up from h8.The c3 piece is at (6,3), which is 2 right and 5 up.So, the distances are not the same, but perhaps Vasya can adjust in a way to maintain some balance.Alternatively, maybe Vasya should move the c3 piece up instead.If Petya moves a1 to d1, Vasya can move c3 up to c6. So, from (3,3) to (3,6). Now, the distance for c3 is (5,2).So, a1 is at (4,1): (4,7).c3 is at (3,6): (5,2).Hmm, not sure if that helps.Alternatively, maybe Vasya should move the c3 piece in such a way that the distances are mirrored.Wait, maybe the key is that the two pieces are on the same diagonal, and the second player can always move the other piece to maintain the same relative position.But I'm not sure.Alternatively, perhaps the key is that the two pieces are both on the same diagonal, and the second player can always respond to a move on one piece by making a symmetrical move on the other piece, effectively maintaining control over the game.Wait, but in this case, since the pieces are on the same diagonal, moving one affects the other's relative position.Wait, maybe the key is that the second player can always move the other piece in such a way that the first player is forced to make the last move.Wait, but I'm not sure.Alternatively, perhaps the key is that the two pieces are both on the same diagonal, and the second player can always move the other piece to maintain the same distance from h8.Wait, let me think about the total distance each piece has to cover.The a1 piece has to cover 14 squares (7 right, 7 up).The c3 piece has to cover 10 squares (5 right, 5 up).So, the total distance is 24 squares.But since each move can cover multiple squares, the number of moves isn't directly relevant.Wait, perhaps the key is that the second player can always respond to the first player's move in such a way that the total distance decreases by an even number, forcing the first player to make the last move.But I'm not sure.Alternatively, perhaps the key is that the two pieces are both on the same diagonal, and the second player can always move the other piece to maintain the same distance from h8.Wait, I'm going in circles here.Let me try to think differently.Since both pieces are on the same diagonal, and the target is on the same diagonal, perhaps the second player can always mirror the first player's moves with respect to the diagonal.But how?Wait, for example, if Petya moves the a1 piece right by k squares, Vasya can move the c3 piece up by k squares, maintaining the balance.Alternatively, if Petya moves the a1 piece up by k squares, Vasya can move the c3 piece right by k squares.Wait, let's test this.Suppose Petya moves the a1 piece right to d1 (k=3). Then, Vasya moves the c3 piece up to c6 (k=3). Now, the a1 piece is at (4,1), and the c3 piece is at (3,6).Now, the a1 piece has to move 4 right and 7 up.The c3 piece has to move 5 right and 2 up.Hmm, not sure if that's balanced.Alternatively, maybe Vasya should move the c3 piece right by k squares when Petya moves the a1 piece up by k squares.Wait, let me try.If Petya moves a1 up to a4 (k=3), then Vasya can move c3 right to f3 (k=3). Now, a1 is at (1,4), and c3 is at (6,3).Now, a1 has to move 7 right and 4 up.c3 has to move 2 right and 5 up.Still not balanced.Hmm.Alternatively, maybe the key is that the second player can always move the other piece to a position where the distances are equal.Wait, but in the starting position, the distances are not equal. The a1 piece has to move 7 right and 7 up, while the c3 piece has to move 5 right and 5 up.So, the c3 piece is closer.Wait, maybe the key is that the second player can always move the c3 piece to match the a1 piece's distance.Wait, if Petya moves the a1 piece, Vasya can move the c3 piece to reduce its distance by the same amount.But since the c3 piece is already closer, that might not work.Alternatively, maybe the key is that the second player can always move the c3 piece to a position where its distance is equal to the a1 piece's distance.Wait, let's see.If Petya moves the a1 piece right by k, increasing its x-coordinate, thus reducing its right distance by k.Vasya can then move the c3 piece right by k, reducing its right distance by k as well.But since the c3 piece's right distance is already less, this might not balance.Wait, for example, if Petya moves a1 right by 2 to c1, reducing its right distance from 7 to 5.Then, Vasya can move c3 right by 2 to e3, reducing its right distance from 5 to 3.Now, a1 has right distance 5, c3 has right distance 3.Not balanced.Alternatively, maybe Vasya should move the c3 piece up by k when Petya moves the a1 piece right by k.Wait, let me try.If Petya moves a1 right by 2 to c1, Vasya moves c3 up by 2 to c5.Now, a1 is at (3,1), right distance 5, up distance 7.c3 is at (3,5), right distance 5, up distance 3.So, now, both pieces have right distance 5, but different up distances.Hmm.Alternatively, maybe the key is that the second player can always make the two pieces have equal distances in both directions.But that seems difficult because the starting positions are different.Wait, maybe the key is that the second player can always move the other piece to a position where the XOR of the distances is the same.But earlier, I saw that both pieces have a Grundy number of 0, so the total is 0, which is a losing position.Therefore, Vasya can always win.But I need to confirm this.Wait, let me think about the game as a whole.Each piece is a separate game, and the overall game is the disjunctive sum of these two games.Each piece's Grundy number is the XOR of its right and up distances.So, for a1, 7 XOR 7 = 0.For c3, 5 XOR 5 = 0.So, the total Grundy number is 0.Therefore, the starting position is a losing position for the first player.Therefore, Vasya can always win.So, the answer is Vasya.

answer：First, I'll tackle the first problem: (0.25 + (-9) + (-frac{1}{4}) - (+11)). I'll start by combining the positive numbers and then the negative numbers. Adding (0.25) and (-frac{1}{4}) gives zero. Then, I'll add (-9) and (-11) to get (-20).Next, for the second problem: (-15 + 5 + frac{1}{3} times (-6)). I'll follow the order of operations, performing the multiplication first. (frac{1}{3}) multiplied by (-6) equals (-2). Then, I'll add (-15), (5), and (-2) together, which results in (-12).Finally, I'll solve the third problem: ((-frac{3}{8} - frac{1}{6} + frac{3}{4}) times 24). I'll first find a common denominator for the fractions, which is 24. Converting each fraction, I get (-9), (-4), and (18). Adding these together gives (5). Multiplying (5) by (24) results in (120).

answer：To determine the constant term in the binomial expansion of ((x - frac{1}{x})^8), I start by recalling the binomial theorem, which states that ((a + b)^n = sum_{k=0}^{n} binom{n}{k} a^{n-k} b^k).In this case, (a = x), (b = -frac{1}{x}), and (n = 8). Applying the binomial theorem, the general term in the expansion is (binom{8}{k} x^{8 - k} left(-frac{1}{x}right)^k).Simplifying the term, I get (binom{8}{k} (-1)^k x^{8 - 2k}). To find the constant term, I need the exponent of (x) to be zero. Therefore, I set (8 - 2k = 0) and solve for (k), which gives (k = 4).Substituting (k = 4) into the general term, the constant term is (binom{8}{4} (-1)^4 x^{0} = 70 times 1 = 70).