answer：Alright, so I've got this chessboard problem to solve. It's a 1983 by 1984 chessboard, which is pretty large, but the specifics of the dimensions might be important. Each white square is filled with either a 1 or a -1. The condition given is that for every black square, the product of the numbers in all the adjacent white squares is equal to 1. I need to prove that all the numbers in the white squares are 1.First, let me visualize the chessboard. It's a standard chessboard pattern, alternating black and white squares. Since it's 1983 by 1984, it's almost square but slightly longer in one dimension. The key here is that each black square has adjacent white squares, and the product of those adjacent white squares is 1. So, if I think about a single black square, it has up to four adjacent white squares: above, below, left, and right. The product of the numbers in these four white squares must be 1. Since each white square can only be 1 or -1, the product being 1 means that there must be an even number of -1s among those adjacent white squares. Because multiplying an even number of -1s gives 1, and an odd number would give -1.Now, if I consider the entire chessboard, every black square enforces this condition on its adjacent white squares. So, if I can show that this condition propagates through the entire board, forcing all white squares to be 1, then I've proven the statement.Let me think about how to approach this. Maybe I can use induction or some kind of propagation argument. If I can show that assuming one white square is -1 leads to a contradiction, then all white squares must be 1.Alternatively, perhaps I can model this as a system of equations. Each black square gives an equation that the product of its adjacent white squares is 1. Since all these equations must be satisfied simultaneously, maybe the only solution is that all white squares are 1.But with such a large board, setting up equations for each black square might be too cumbersome. Maybe there's a pattern or a property of the chessboard that I can exploit.Let me consider the checkerboard pattern. Each white square is adjacent to black squares, and each black square is adjacent to white squares. So, the conditions imposed by the black squares affect the white squares, and vice versa.If I start from one corner, say the top-left corner, which is a white square. Let's assume it's 1. Then, the black squares adjacent to it will have their product conditions satisfied based on their other adjacent white squares. But if I assume that one white square is -1, how does that affect the neighboring black squares?Suppose I have a white square with -1. Then, all the black squares adjacent to it will have their product condition altered. Each of those black squares will now require an odd number of -1s among their adjacent white squares to maintain the product as 1. But this could create a chain reaction, where each subsequent black square's condition forces another white square to be -1, potentially leading to a contradiction.Wait, but how do I formalize this? Maybe I can use the idea of parity. Each black square's condition imposes a parity constraint on its adjacent white squares. If I can show that these constraints are only consistent if all white squares are 1, then that would solve the problem.Another thought: since the chessboard is connected, meaning there's a path from any white square to any other white square through adjacent squares, the constraints imposed by the black squares would propagate throughout the entire board. So, if one white square is -1, it would force a whole chain of other white squares to be -1, which might not be possible due to the board's dimensions or the initial conditions.But I'm not sure if that's the right way to go. Maybe I need to think about the dual graph of the chessboard, where white squares are nodes and black squares are edges connecting them. Then, the condition becomes that for each edge (black square), the product of the two nodes (white squares) it connects is 1. But that might not capture all the adjacencies, since each black square can connect up to four white squares.Hmm, perhaps I'm overcomplicating it. Let me try a different approach. Let's consider that each black square's condition is a constraint on its adjacent white squares. If I can show that these constraints force all white squares to be 1, then I'm done.Suppose, for contradiction, that there exists at least one white square with -1. Then, consider the black squares adjacent to this white square. Each of these black squares must have an even number of -1s among their adjacent white squares. Since one of them is already -1, the others must be arranged such that the total number of -1s is even.But this creates a kind of dependency chain. Each black square adjacent to a -1 white square must have another -1 white square to satisfy the even count. This could lead to a domino effect where more white squares are forced to be -1, potentially covering the entire board.However, the board has an odd number of rows (1983) and an even number of columns (1984). This might create a problem because the domino effect could lead to a contradiction when trying to cover the entire board with an odd number of rows.Wait, let me think about that again. If I start with a -1 in a white square, the adjacent black squares require another -1 in their other adjacent white squares. This would propagate the -1s across the board. But because the number of rows is odd, this propagation might lead to a situation where a black square requires an odd number of -1s, which contradicts the initial condition.Alternatively, maybe the key is that the chessboard has an even number of squares in one dimension and odd in the other, which could lead to an inconsistency if we try to assign -1s.Another idea: consider the entire product of all white squares. Each black square's condition contributes a factor of 1 to the overall product. But the overall product can also be expressed in terms of the white squares. Maybe this can lead to a contradiction if not all white squares are 1.Wait, let's formalize that. Let me denote each white square as ( w_{i,j} ) where ( i ) and ( j ) are the row and column indices. Each black square at position ( (i,j) ) has adjacent white squares, say ( w_{i-1,j} ), ( w_{i+1,j} ), ( w_{i,j-1} ), ( w_{i,j+1} ) (depending on the exact position). The product of these four white squares is 1.If I multiply all these conditions together for every black square, I get the product of all white squares raised to some power. But I need to figure out how many times each white square is included in the product.Each white square is adjacent to four black squares (except for those on the edges). So, each white square's value is multiplied four times in the overall product. Therefore, the product of all black square conditions is equal to the product of all white squares raised to the fourth power.But since each black square's product is 1, the overall product is 1. Therefore, the product of all white squares raised to the fourth power is 1. Since 1 raised to any power is still 1, this doesn't give us new information.Hmm, maybe I need a different approach. Let's consider the dual of the chessboard. If I color the chessboard in the standard alternating black and white pattern, then each white square is surrounded by black squares and vice versa.If I assign variables to the white squares, the conditions from the black squares give me equations that these variables must satisfy. Specifically, for each black square, the product of its four adjacent white squares is 1.This forms a system of equations where each equation is a product of four variables equal to 1. Solving this system would give me the values of all white squares. But with such a large system, it's impractical to solve directly.However, maybe I can find that the only solution is all variables equal to 1. To do this, I can assume that there's a white square with -1 and show that this leads to a contradiction.Suppose there's a white square ( w_{i,j} = -1 ). Then, all black squares adjacent to ( w_{i,j} ) must have an odd number of -1s among their adjacent white squares to maintain the product as 1. This means that each of these black squares must have another -1 in their adjacent white squares.This creates a chain reaction where each black square adjacent to a -1 white square forces another white square to be -1. This propagation continues across the board. However, because the chessboard has an odd number of rows, this propagation might lead to a contradiction.Imagine starting from the top-left corner, which is a white square. If I set it to -1, the black squares adjacent to it require another -1 in their other adjacent white squares. This would force the white squares to the right and below to be -1. Continuing this process, I would end up propagating -1s across the entire board.But since the number of rows is odd, when I reach the bottom of the board, the last row would have an odd number of -1s, which might not satisfy the conditions of the black squares in the last row. This could lead to a contradiction because the product condition might not hold for those black squares.Alternatively, maybe the key is that the chessboard has an even number of columns, which allows for a consistent propagation, but the odd number of rows disrupts this consistency.Wait, perhaps I need to think about the parity of the number of -1s. Each black square requires an even number of -1s among its adjacent white squares. If I have an odd number of rows, the total number of -1s might end up being odd, which contradicts the even requirement.But I'm not sure if that's the exact contradiction. Maybe I need to consider the overall parity of the entire board.Another approach: consider the chessboard as a bipartite graph, where white squares are one partition and black squares are the other. Each black square is connected to its adjacent white squares. The condition is that for each black square, the product of its neighbors is 1.In graph theory terms, this is similar to a constraint satisfaction problem where each node (black square) enforces a constraint on its neighbors (white squares). The solution to this problem is an assignment of values to the white squares that satisfies all constraints.If I can show that the only solution is all white squares being 1, then I'm done. To do this, I can assume that there's a white square with -1 and show that this leads to an inconsistency.Suppose there's a white square ( w ) with -1. Then, all black squares adjacent to ( w ) must have an odd number of -1s among their neighbors. This forces another white square adjacent to each of these black squares to be -1. This creates a chain reaction where -1s propagate across the board.However, because the chessboard is finite and has specific dimensions, this propagation might lead to a contradiction. Specifically, if the number of rows is odd, the propagation might end up requiring a white square to be both 1 and -1, which is impossible.Alternatively, maybe the key is that the chessboard has an even number of columns, which allows for a consistent propagation, but the odd number of rows disrupts this consistency.Wait, let me think about the specific dimensions: 1983 rows and 1984 columns. 1983 is odd, and 1984 is even. This might play a role in the contradiction.If I start propagating -1s from one corner, the even number of columns allows the propagation to continue without immediate contradiction, but the odd number of rows might cause a problem when reaching the end.Alternatively, maybe the key is that the product of all white squares must be 1, but if there's an odd number of -1s, the product would be -1, leading to a contradiction.Wait, earlier I thought about the product of all white squares. If I consider the product of all black square conditions, which is 1, and relate it to the product of all white squares, I might get some information.Each white square is adjacent to four black squares, except for those on the edges. So, each white square is included in four black square conditions. Therefore, the product of all black square conditions is equal to the product of all white squares raised to the fourth power.Since the product of all black square conditions is 1, we have:( left( prod_{text{white squares}} w_{i,j} right)^4 = 1 )This implies that the product of all white squares is either 1 or -1, because ( ( pm 1 )^4 = 1 ).But if the product of all white squares is 1, then it's possible that all are 1, or there's an even number of -1s. However, if the product is -1, that would mean an odd number of -1s, which might be impossible due to the board's dimensions.Wait, but the product of all white squares is either 1 or -1, but the condition from the black squares only tells us that the fourth power is 1, which doesn't distinguish between 1 and -1.So, maybe this approach isn't sufficient. I need another way to show that all white squares must be 1.Perhaps I can use the fact that the chessboard is connected and that the conditions force a consistent assignment. If I start assigning 1s and -1s based on the conditions, I might find that the only consistent assignment is all 1s.Let me try to formalize this. Suppose I pick a white square and assign it a value, say 1. Then, based on the adjacent black squares, I can determine the values of the neighboring white squares. If I propagate this assignment across the entire board, I might find that all white squares must be 1.Alternatively, if I assume a white square is -1, the propagation might lead to a contradiction, such as requiring a white square to be both 1 and -1.Given the chessboard's dimensions, with an odd number of rows and even number of columns, the propagation might not close consistently, leading to a contradiction.Wait, let's think about the parity of the number of -1s. Each black square requires an even number of -1s among its adjacent white squares. If I have an odd number of rows, the total number of -1s might end up being odd, which contradicts the even requirement.But I'm not sure if that's the exact contradiction. Maybe I need to consider the overall parity of the entire board.Another idea: consider the chessboard as a grid graph, where each white square is a vertex, and edges connect adjacent white squares. The conditions from the black squares impose constraints on the vertices.Each black square corresponds to a hyperedge connecting four white squares, requiring their product to be 1. This forms a system of hyperequations.In such a system, the only solution might be all variables equal to 1, especially if the system is overdetermined and the constraints are tight.Alternatively, maybe I can use linear algebra over the field of two elements, but since we're dealing with products (multiplicative) rather than sums (additive), it's a bit different.Wait, if I take the logarithm of the products, turning them into sums, I could potentially model this as a linear system over the field of two elements. But since we're dealing with 1 and -1, which are multiplicative inverses, this might complicate things.Alternatively, I can think of the problem in terms of linear algebra modulo 2, where 1 corresponds to 0 and -1 corresponds to 1. Then, the product condition becomes a sum condition modulo 2.Each black square's condition is that the sum of its adjacent white squares (mod 2) is 0, since an even number of -1s corresponds to an even number of 1s in the modulo 2 system.So, if I set up a system where each black square gives an equation that the sum of its adjacent white squares is 0 mod 2, then solving this system would give me the values of the white squares.If this system has only the trivial solution where all white squares are 0 (which corresponds to 1 in the original problem), then all white squares must be 1.But setting up such a large system is impractical, but maybe I can argue that the system is such that only the trivial solution exists.Given that the chessboard is connected and the system of equations is overdetermined, the only solution might be the trivial one.Alternatively, maybe I can use the fact that the chessboard has no cycles of odd length, which would imply that the system is bipartite and the only solution is all variables equal.Wait, a chessboard is a bipartite graph, so it doesn't have any odd-length cycles. This might help in showing that the only solution is all variables equal to 1.In a bipartite graph, any cycle has even length, which might imply that the system of equations has only the trivial solution.But I'm not entirely sure how to formalize this. Maybe I need to think about the connected components. Since the chessboard is connected, the only solution is the trivial one.Alternatively, perhaps I can use induction on the number of rows or columns. Start with a smaller chessboard and show that the condition forces all white squares to be 1, then extend it to the larger board.Let me try that. Suppose I have a 1x1 chessboard. It's just a single white square. There are no black squares, so the condition is trivially satisfied. But this doesn't help much.What about a 2x2 chessboard? It has two white squares and two black squares. Each black square is adjacent to two white squares. The product of the two white squares must be 1 for each black square.So, if I have white squares ( w_1 ) and ( w_2 ), then ( w_1 times w_2 = 1 ) for both black squares. This implies that ( w_1 = w_2 ). So, both white squares must be the same. They can both be 1 or both be -1.But in the original problem, the chessboard is larger, so maybe the constraints are tighter.Wait, in the 2x2 case, both white squares can be -1, and the product condition is still satisfied. So, in that case, the conclusion that all white squares are 1 doesn't hold. But in the original problem, the chessboard is much larger, so maybe the constraints are different.Perhaps the key is the specific dimensions of the chessboard, 1983x1984, which are co-prime or have certain properties that make the only solution all 1s.Alternatively, maybe the chessboard's dimensions being one odd and one even leads to a contradiction if we assume a -1 exists.Wait, let me think about the number of white squares. In a chessboard, the number of white squares is equal to the number of black squares if the total number of squares is even. But 1983x1984 is an even number, so the number of white squares is equal to the number of black squares.But I'm not sure if that helps directly.Another idea: consider the fact that the chessboard can be tiled with dominoes, each covering one black and one white square. But I'm not sure how that relates to the problem.Wait, maybe I can use the fact that the chessboard is a bipartite graph and consider the connected components. Since it's connected, the only solution is the trivial one.Alternatively, perhaps I can use the fact that the system of equations is such that each white square is involved in an even number of equations, leading to the only solution being all 1s.Wait, each white square is adjacent to four black squares, so it appears in four equations. Each equation is that the product of four white squares is 1. So, each white square is involved in four equations.But I'm not sure how to use that to show that all white squares must be 1.Wait, maybe I can consider the product of all equations. Each equation is the product of four white squares equals 1. If I multiply all these equations together, I get the product of all white squares raised to the fourth power equals 1.But as I thought earlier, this doesn't give me new information because ( ( pm 1 )^4 = 1 ).Hmm, maybe I need to consider the fact that the chessboard has an odd number of rows. If I fix the value of one white square, the values of the others are determined by the conditions. But with an odd number of rows, this might lead to a contradiction.Let me try to formalize this. Suppose I fix the value of the top-left white square to 1. Then, based on the adjacent black squares, I can determine the values of the white squares in the first row and first column.But since the number of rows is odd, when I reach the bottom row, the values might not align correctly, leading to a contradiction.Alternatively, if I fix the top-left white square to -1, the same propagation might lead to a contradiction when reaching the end of the board.Wait, maybe I can use the fact that the chessboard has an odd number of rows to show that the number of -1s must be even, but the propagation forces an odd number, leading to a contradiction.Alternatively, perhaps the key is that the chessboard's dimensions are such that the only way to satisfy all the conditions is to have all white squares equal to 1.I think I need to take a different approach. Let me consider the problem in terms of linear algebra over the field of two elements, where 1 corresponds to 0 and -1 corresponds to 1.Each black square's condition is that the sum of its adjacent white squares is 0 mod 2. So, we have a system of linear equations over GF(2).The variables are the white squares, and each equation corresponds to a black square, stating that the sum of its adjacent white squares is 0.If I can show that this system has only the trivial solution (all variables 0, which corresponds to all white squares being 1), then I'm done.But to do this, I need to analyze the rank of the system's matrix. If the rank is equal to the number of variables, then the only solution is the trivial one.However, with such a large system, it's impractical to compute the rank directly. But maybe I can use properties of the chessboard's structure.Since the chessboard is a bipartite graph, the system might have certain symmetries or properties that allow me to conclude that the only solution is trivial.Alternatively, maybe I can use the fact that the chessboard is connected and that the system is such that each variable is connected to others in a way that forces all variables to be 0.Wait, another idea: consider that the chessboard can be colored in such a way that the white squares form a checkerboard pattern themselves, but I'm not sure if that helps.Wait, perhaps I can use the fact that the chessboard has an even number of columns, which allows for a consistent assignment, but the odd number of rows disrupts this.Alternatively, maybe I can use the fact that the chessboard has an odd number of rows, which means that the number of white squares in each column is odd, leading to a contradiction if we assume a -1 exists.Wait, let me think about the number of white squares in each column. Since the chessboard has 1983 rows, which is odd, each column has 1983 squares, alternating between black and white. So, each column has either (1983 + 1)/2 = 992 white squares and 991 black squares, or vice versa, depending on whether the column starts with white or black.But since the chessboard has 1984 columns, which is even, half of them start with white and half with black. So, there are 992 columns starting with white and 992 starting with black.Therefore, the total number of white squares is 992 columns * 992 white squares + 992 columns * 991 white squares = 992*(992 + 991) = 992*1983.But I'm not sure if that helps directly.Wait, maybe I can consider the product of all white squares in a column. If I fix the values of the white squares in one column, the values in the next column are determined by the conditions from the black squares.But with an odd number of rows, this might lead to a contradiction when trying to assign values to the last column.Alternatively, maybe I can use the fact that the chessboard's dimensions are co-prime, which might imply that the only solution is all 1s.Wait, 1983 and 1984 are consecutive integers, so they are co-prime. This might be relevant in some way.Perhaps I can use the fact that the chessboard can be traversed in a way that covers all squares, and the conditions force all white squares to be 1.Wait, another idea: consider that the chessboard can be tiled with 2x1 dominoes, each covering one black and one white square. But I'm not sure how that helps with the problem.Alternatively, maybe I can use the fact that the chessboard is a planar graph and apply some planarity arguments, but I'm not sure.Wait, going back to the original idea of assuming a -1 exists and propagating the conditions. Let's try to formalize this.Suppose there's a white square ( w_{i,j} = -1 ). Then, all black squares adjacent to ( w_{i,j} ) must have an odd number of -1s among their adjacent white squares. This means that each of these black squares must have another -1 in their adjacent white squares.This creates a chain reaction where each black square adjacent to a -1 white square forces another white square to be -1. This propagation continues across the board.However, because the chessboard has an odd number of rows, when we reach the bottom of the board, the propagation might require a white square to be both 1 and -1, which is impossible.Alternatively, the propagation might end up requiring an odd number of -1s in a column, which contradicts the even requirement from the black squares.Wait, let me think about this more carefully. If I start with a -1 in the top-left corner, the black squares adjacent to it require another -1 in their other adjacent white squares. This forces the white squares to the right and below to be -1.Continuing this process, I would end up propagating -1s across the entire board. But since the number of rows is odd, when I reach the bottom row, the last white square in that row would have to be -1, which might conflict with the conditions of the black squares in the last row.Specifically, the black squares in the last row would have their adjacent white squares in the last row and the row above. If the last row has an odd number of -1s, the black squares in the last row would require an even number of -1s, leading to a contradiction.Therefore, the assumption that there's a -1 in a white square leads to a contradiction, implying that all white squares must be 1.So, putting it all together, the key steps are:1. Assume, for contradiction, that there's a white square with -1.2. This forces adjacent black squares to have another -1 among their adjacent white squares.3. This creates a chain reaction propagating -1s across the board.4. Due to the odd number of rows, this propagation leads to a contradiction in the last row.5. Therefore, no such -1 can exist, and all white squares must be 1.I think this makes sense. The odd number of rows is crucial because it causes the propagation to end in a contradiction, whereas if both dimensions were even, it might be possible to have a consistent assignment with some -1s.

answer：First, I observe that both fractions have the same denominator, ( a - 2 ). This allows me to combine them into a single fraction by subtracting the numerators.Next, I notice that the resulting numerator, ( a^2 - 4 ), is a difference of squares. I can factor this expression into ( (a + 2)(a - 2) ).Since the denominator is also ( a - 2 ), I can simplify the fraction by canceling out the common factor of ( a - 2 ), provided that ( a neq 2 ) to avoid division by zero.After simplifying, I am left with the expression ( a + 2 ), which is the final simplified form of the original expression.

answer：Okay, so I need to find the largest prime factor of the expression (12^3 + 15^4 - 6^5). Hmm, that sounds a bit complicated, but let me break it down step by step.First, I should probably calculate each term separately to see what number I'm dealing with. Let me start by computing each exponent:1. (12^3): I know that (12 times 12 = 144), and then (144 times 12). Let me do that multiplication. 144 times 10 is 1440, and 144 times 2 is 288, so adding those together gives 1440 + 288 = 1728. So, (12^3 = 1728).2. (15^4): This is a bit bigger. I can compute it step by step. First, (15^2 = 225). Then, (15^3 = 15 times 225). Let me calculate that: 225 times 10 is 2250, and 225 times 5 is 1125, so adding those gives 2250 + 1125 = 3375. So, (15^3 = 3375). Now, (15^4 = 15 times 3375). Let me do that multiplication. 3375 times 10 is 33750, and 3375 times 5 is 16875. Adding those together: 33750 + 16875 = 50625. So, (15^4 = 50625).3. (6^5): Let me compute this as well. (6^2 = 36), (6^3 = 36 times 6 = 216), (6^4 = 216 times 6 = 1296), and (6^5 = 1296 times 6). Let me calculate that: 1296 times 6. 1000 times 6 is 6000, 200 times 6 is 1200, 90 times 6 is 540, and 6 times 6 is 36. Adding those together: 6000 + 1200 = 7200, 7200 + 540 = 7740, and 7740 + 36 = 7776. So, (6^5 = 7776).Now, I have all the components:- (12^3 = 1728)- (15^4 = 50625)- (6^5 = 7776)So, the expression becomes (1728 + 50625 - 7776). Let me compute that step by step.First, add (1728 + 50625). Let me do that: 1728 + 50625. Adding 1728 to 50625. Let me break it down: 50625 + 1000 = 51625, then subtract 272 (since 1728 - 1000 = 728, but wait, no, actually, 1728 is 1000 + 728). Wait, maybe a better way is to add 1728 to 50625 directly.50625 + 1728:- 50625 + 1000 = 51625- 51625 + 700 = 52325- 52325 + 28 = 52353So, 50625 + 1728 = 52353.Now, subtract 7776 from 52353. Let me compute 52353 - 7776.52353 - 7000 = 4535345353 - 776 = ?Wait, 7776 is 7000 + 776, so subtracting 7000 first gives 45353, then subtract 776.45353 - 700 = 4465344653 - 76 = 44577Wait, that doesn't seem right. Let me check again.Wait, 52353 - 7776:Alternatively, I can compute 52353 - 7776 by breaking it down:52353 - 7000 = 4535345353 - 700 = 4465344653 - 76 = 44577Wait, that seems correct. So, 52353 - 7776 = 44577.So, the value of the expression (12^3 + 15^4 - 6^5) is 44577.Now, I need to find the largest prime factor of 44577.To find the prime factors, I can start by checking divisibility starting from the smallest primes.First, check if it's even. 44577 ends with a 7, which is odd, so it's not divisible by 2.Next, check divisibility by 3. To do that, I can sum the digits: 4 + 4 + 5 + 7 + 7 = 27. 27 is divisible by 3, so 44577 is divisible by 3.Let me divide 44577 by 3.3 into 4 is 1 with remainder 1.Bring down the next digit: 14. 3 into 14 is 4, remainder 2.Bring down the next digit: 25. 3 into 25 is 8, remainder 1.Bring down the next digit: 17. 3 into 17 is 5, remainder 2.Bring down the last digit: 27. 3 into 27 is 9, remainder 0.So, 44577 ÷ 3 = 14859.So, 44577 = 3 × 14859.Now, let's factor 14859.Again, check if it's divisible by 3. Sum the digits: 1 + 4 + 8 + 5 + 9 = 27. 27 is divisible by 3, so 14859 ÷ 3.3 into 1 is 0, but let's do it properly.3 into 14 is 4, remainder 2.Bring down the 8: 28. 3 into 28 is 9, remainder 1.Bring down the 5: 15. 3 into 15 is 5, remainder 0.Bring down the 9: 9. 3 into 9 is 3, remainder 0.So, 14859 ÷ 3 = 4953.So, now we have 44577 = 3 × 3 × 4953.Now, factor 4953.Again, check divisibility by 3: 4 + 9 + 5 + 3 = 21, which is divisible by 3.4953 ÷ 3: 3 into 4 is 1, remainder 1.Bring down the 9: 19. 3 into 19 is 6, remainder 1.Bring down the 5: 15. 3 into 15 is 5, remainder 0.Bring down the 3: 3. 3 into 3 is 1, remainder 0.So, 4953 ÷ 3 = 1651.So, now 44577 = 3 × 3 × 3 × 1651.Now, let's factor 1651.Check divisibility by small primes:- 2: It's odd, so no.- 3: Sum of digits: 1 + 6 + 5 + 1 = 13, not divisible by 3.- 5: Ends with 1, so no.- 7: Let's check. 1651 ÷ 7. 7 × 235 = 1645. 1651 - 1645 = 6. So, remainder 6. Not divisible by 7.- 11: Alternating sum: (1 + 5) - (6 + 1) = (6) - (7) = -1, not divisible by 11.- 13: Let's check. 13 × 127 = 1651? Let me compute 13 × 120 = 1560, 13 × 7 = 91, so 1560 + 91 = 1651. Yes! So, 1651 = 13 × 127.So, now we have 44577 = 3 × 3 × 3 × 13 × 127.Now, check if 127 is a prime number. Let me see. 127 is a known prime number because it's not divisible by any prime less than its square root (which is around 11.27). So, primes to check: 2, 3, 5, 7, 11.- 127 ÷ 2: Not divisible.- 127 ÷ 3: 1 + 2 + 7 = 10, not divisible by 3.- 127 ÷ 5: Ends with 7, no.- 127 ÷ 7: 7 × 18 = 126, remainder 1.- 127 ÷ 11: 11 × 11 = 121, remainder 6.So, 127 is prime.Similarly, 13 is a prime number.So, the prime factors of 44577 are 3, 13, and 127.Among these, the largest prime factor is 127.Wait, but let me double-check my calculations to make sure I didn't make any mistakes.Starting from the beginning:(12^3 = 1728), correct.(15^4 = 50625), correct.(6^5 = 7776), correct.Then, 1728 + 50625 = 52353, correct.52353 - 7776 = 44577, correct.Factoring 44577:Divided by 3: 44577 ÷ 3 = 14859.14859 ÷ 3 = 4953.4953 ÷ 3 = 1651.1651 ÷ 13 = 127.Yes, that's correct.So, the prime factors are 3, 13, and 127.Therefore, the largest prime factor is 127.

answer：Okay, so I have this problem about depositing money in a fixed deposit account. Let me try to understand it step by step. The problem says that the annual interest rate for a three-year fixed deposit is 14%. Additionally, there's a 7% value preservation subsidy every month. But, if I withdraw the money after the maturity period plus an extra half year, the excess period is calculated with the current savings interest rate, which is a monthly interest rate of 2%, and without the value preservation subsidy. I need to find out how much principal and interest I'll receive in total if I deposit 1000 yuan and withdraw it after an additional half year.First, let me break down the components of this problem:1. **Fixed Deposit for 3 Years**: The main deposit is for three years with an annual interest rate of 14%. I think this is a simple interest calculation because it's a fixed deposit, but I need to confirm if it's simple or compound interest. The problem doesn't specify, so I might assume it's simple interest unless stated otherwise.2. **Value Preservation Subsidy**: There's an additional 7% subsidy every month. This seems like an extra interest or benefit provided monthly. I need to figure out how this is applied. Is it 7% of the principal each month, or is it 7% of the current amount? The problem says "value preservation subsidy every month," so it might be a monthly addition to the principal or interest.3. **Excess Period After Maturity**: If I withdraw the money after the three years plus an additional half year (which is six months), the interest for this extra period is calculated with the current savings interest rate of 2% per month, and without the value preservation subsidy. So, for the first three years, I get the fixed deposit interest and the subsidy, and for the next six months, I get only the current savings rate.Let me try to structure the calculations step by step.**Step 1: Calculate the interest for the first three years.**The annual interest rate is 14%, so for three years, the total interest would be:Interest = Principal × Rate × TimeInterest = 1000 × 14% × 3But wait, the problem also mentions a 7% value preservation subsidy every month. So, I need to incorporate that as well. Is this 7% per month or 7% per year? The problem says "every month," so I think it's 7% per month. That seems quite high, but let's go with the information given.So, each month, I get an additional 7% of the principal as a subsidy. Since it's every month, over three years, that would be 36 months.Subsidy = Principal × Subsidy Rate × Number of MonthsSubsidy = 1000 × 7% × 36Wait, but 7% per month is extremely high. 7% per month would translate to 84% per year, which is way higher than the annual interest rate of 14%. That doesn't make sense because the subsidy should be a smaller addition to preserve value, not to exceed the interest rate. Maybe I misinterpreted the subsidy rate.Perhaps the 7% is an annual value preservation subsidy, and it's distributed monthly. So, 7% per year, which is 7% divided by 12 months. Let me check the problem again.The problem says: "a 7% value preservation subsidy every month." So, it's 7% every month. That still seems high, but maybe it's a typo or specific to the problem. I'll proceed with 7% per month.So, the total subsidy over three years would be:Subsidy = 1000 × 7% × 36But let's calculate that:7% of 1000 is 70 yuan per month. Over 36 months, that would be 70 × 36 = 2520 yuan.That seems too high because the interest from the fixed deposit is only 1000 × 14% × 3 = 420 yuan. The subsidy is way higher than the interest. Maybe I'm misunderstanding the subsidy.Alternatively, perhaps the 7% is a simple addition to the interest, not compounded. Or maybe it's 7% of the interest earned each month. Let me think.Wait, the problem says "value preservation subsidy every month." Value preservation usually refers to protecting against inflation, so it's likely a small addition. Maybe 7% per annum, distributed monthly as 7%/12 per month. Let me try that.So, monthly subsidy rate = 7% / 12 ≈ 0.5833% per month.Then, total subsidy over three years:Subsidy = 1000 × (7%/12) × 36= 1000 × 7% × 3= 1000 × 0.07 × 3= 210 yuan.That makes more sense because 210 yuan is a reasonable addition to the 420 yuan interest.But the problem says "7% value preservation subsidy every month," which could mean 7% per month. However, given the context, it's more likely to be 7% per annum, distributed monthly. Otherwise, the numbers don't add up.I think the problem might have intended the subsidy to be 7% per annum, so I'll proceed with that interpretation.**Step 2: Calculate the total amount after three years.**Principal = 1000 yuanInterest from fixed deposit = 1000 × 14% × 3 = 420 yuanSubsidy = 1000 × 7% × 3 = 210 yuanTotal amount after three years = Principal + Interest + Subsidy= 1000 + 420 + 210= 1630 yuanWait, but the initial calculation in the problem statement was different. It had:(1000 + 1000 × 14% × 3 + 1000 × 7% × 12 × 3) × (1 + 2% × 6)= 3920 × 1.12= 4390.4 yuanHmm, so in the initial solution, they calculated the interest as 14% × 3 = 42%, and the subsidy as 7% × 12 × 3 = 252%, which is way too high. That would make the total amount after three years as 1000 + 420 + 2520 = 3920, which seems incorrect because the subsidy is 7% per month, which is 84% per year, leading to 252% over three years.But that would mean the subsidy is 7% of the principal each month, which is 70 yuan per month, totaling 2520 yuan over three years. That seems excessive, but perhaps that's how it's intended.Alternatively, maybe the subsidy is 7% of the interest earned each month. Let me check.If the fixed deposit interest is 14% per annum, then monthly interest is 14%/12 ≈ 1.1667% per month.If the subsidy is 7% of that interest, then monthly subsidy = 1.1667% × 7% ≈ 0.081667% of the principal. That would be negligible.Alternatively, maybe the subsidy is 7% of the principal each month, but that would be 70 yuan per month, which is 2520 yuan over three years, as before.Given that the initial solution used 7% × 12 × 3, which is 252%, leading to 2520 yuan, I think that's how it's intended, even though it seems high.So, perhaps the problem is designed that way, with a very high subsidy, so I'll go with that.Therefore, total amount after three years:= 1000 + (1000 × 14% × 3) + (1000 × 7% × 12 × 3)= 1000 + 420 + 2520= 3920 yuanNow, after three years, I have 3920 yuan. Then, I withdraw it after an additional half year, which is six months. For this extra period, the interest is calculated with the current savings interest rate of 2% per month, without the value preservation subsidy.So, for the next six months, the interest is:Interest = Principal × Rate × Time= 3920 × 2% × 6But wait, is it simple interest or compound interest? The problem doesn't specify, but since it's a savings interest rate, it's likely simple interest unless stated otherwise.So, simple interest for six months:Interest = 3920 × 0.02 × 6= 3920 × 0.12= 470.4 yuanTherefore, total amount after three years and six months:= 3920 + 470.4= 4390.4 yuanSo, the total amount received is 4390.4 yuan.But let me double-check the calculations.First, fixed deposit interest:14% per annum for three years: 1000 × 0.14 × 3 = 420 yuanValue preservation subsidy: 7% per month for 36 months: 1000 × 0.07 × 36 = 2520 yuanTotal after three years: 1000 + 420 + 2520 = 3920 yuanThen, for the next six months, interest at 2% per month:3920 × 0.02 × 6 = 470.4 yuanTotal amount: 3920 + 470.4 = 4390.4 yuanYes, that seems correct.Alternatively, if the interest for the extra six months is compounded monthly, the calculation would be:Amount = Principal × (1 + Rate)^Time= 3920 × (1 + 0.02)^6Let me calculate that:(1.02)^6 ≈ 1.12616So, Amount ≈ 3920 × 1.12616 ≈ 4410.7 yuanBut the initial solution used simple interest, so perhaps that's the intended method.Given that the initial solution used simple interest for the extra period, I'll stick with that.Therefore, the total amount is 4390.4 yuan.

answer：Okay, so I have this problem about a parabola and its reflection. Let me try to understand what it's asking. The original parabola is given by the equation y = ax² + bx + c. Then, it says the parabola is reflected about the x-axis. Hmm, reflecting a graph about the x-axis usually means that we change the sign of the y-values, right? So, if the original equation is y = ax² + bx + c, then the reflection should be y = -ax² - bx - c. That makes sense because reflecting over the x-axis flips the graph upside down.Next, both the original parabola and its reflection are translated vertically, but in different directions by 3 units. I need to figure out what that means. Translating vertically means moving the graph up or down along the y-axis. Since it's in different directions, one will go up and the other will go down. The problem doesn't specify which one goes which way, but I think it might not matter because we're going to subtract their equations later. Let me just assume that the original parabola is translated upwards and the reflection is translated downwards. So, for the original parabola, adding 3 to the equation should move it up, making it y = ax² + bx + c + 3. For the reflection, subtracting 3 should move it down, making it y = -ax² - bx - c - 3.Now, the problem introduces two new functions: f(x) for the original translated parabola and g(x) for the reflection. So, f(x) = ax² + bx + c + 3 and g(x) = -ax² - bx - c - 3. The question asks for the equation of y = (f - g)(x). That means I need to subtract g(x) from f(x). Let me write that out:f(x) - g(x) = (ax² + bx + c + 3) - (-ax² - bx - c - 3)Hmm, subtracting a negative is like adding, so this should simplify. Let me distribute the negative sign inside the parentheses:= ax² + bx + c + 3 + ax² + bx + c + 3Now, combine like terms. The ax² and ax² make 2ax². The bx and bx make 2bx. The c and c make 2c. The 3 and 3 make 6. So, putting it all together:= 2ax² + 2bx + 2c + 6So, the equation of y = (f - g)(x) is 2ax² + 2bx + 2c + 6. Let me just double-check my steps to make sure I didn't make a mistake. I reflected the parabola correctly by changing the signs, translated them in opposite directions by 3 units, subtracted the reflection from the original, and combined like terms. It all seems to add up. I think that's the correct answer.

answer：Okay, so I have this problem here: Given the set A = {x | ax² - 2x + 1 = 0}, the set A has at most two subsets. I need to determine the range of values for a. Hmm, let me try to figure this out step by step.First, I remember that the number of subsets of a set depends on the number of elements in the set. Specifically, if a set has n elements, it has 2ⁿ subsets. The problem says that A has at most two subsets. So, 2ⁿ ≤ 2. That means n ≤ 1 because 2¹ = 2. So, the set A can have at most one element. In other words, the equation ax² - 2x + 1 = 0 can have at most one solution.Wait, but hold on. If A has at most two subsets, that means the number of subsets is either 1 or 2. If there's only one subset, that would mean the set is empty because the only subset of an empty set is itself. If there are two subsets, that means the set has exactly one element because the subsets would be the empty set and the set itself.So, A can either be empty or have exactly one element. Therefore, the equation ax² - 2x + 1 = 0 must either have no real solutions or exactly one real solution.Now, let's analyze the equation ax² - 2x + 1 = 0. This is a quadratic equation in terms of x. For a quadratic equation, the discriminant determines the nature of the roots. The discriminant D is given by D = b² - 4ac.In this equation, a is the coefficient of x², which is 'a'. The coefficient of x is -2, so b = -2, and the constant term is 1, so c = 1.Calculating the discriminant: D = (-2)² - 4*a*1 = 4 - 4a.Now, for the quadratic equation to have exactly one real solution, the discriminant must be zero. So, 4 - 4a = 0. Solving for a: 4a = 4 => a = 1.For the quadratic equation to have no real solutions, the discriminant must be negative. So, 4 - 4a < 0 => 4a > 4 => a > 1.Wait, but hold on. If a is greater than 1, the discriminant is negative, so there are no real solutions. That means the set A is empty, which only has one subset, the empty set itself. So, that's acceptable because the problem says "at most two subsets," which includes the case where there's only one subset.But what if a is less than 1? Let's see. If a is less than 1, then 4 - 4a would be positive if a is less than 1 but greater than -infinity. Wait, no. Let me think again.Wait, the discriminant is 4 - 4a. So, if a is positive, then 4a is positive, so 4 - 4a is less than 4. If a is negative, then 4a is negative, so 4 - 4a is greater than 4. So, if a is negative, the discriminant becomes larger than 4, which is positive. So, for a < 1, the discriminant is positive, meaning two real solutions.But the problem says that A can have at most two subsets, which would mean that A can have at most one element. So, if a is less than 1, the equation has two real solutions, meaning A has two elements, which would result in four subsets. That's more than two, so that's not acceptable.Wait, hold on. Let me clarify. If a is not equal to zero, then the equation is quadratic. If a is zero, then the equation becomes linear: -2x + 1 = 0, which has exactly one solution, x = 1/2. So, in that case, A has one element, so it has two subsets, which is acceptable.But if a is not zero, then the equation is quadratic. So, if a ≠ 0, then for A to have at most one element, the quadratic equation must have either one real solution or no real solutions.So, when a ≠ 0, the discriminant D = 4 - 4a must be less than or equal to zero. So, 4 - 4a ≤ 0 => 4a ≥ 4 => a ≥ 1.Wait, but earlier I thought that if a is negative, the discriminant becomes positive, which would give two real solutions, which is not acceptable because that would make A have two elements, leading to four subsets. So, to prevent that, a must be such that the discriminant is non-positive.So, 4 - 4a ≤ 0 => a ≥ 1.But wait, if a is negative, the discriminant is positive, which is bad because it leads to two solutions. So, to ensure that the discriminant is non-positive, a must be greater than or equal to 1.But what about a = 0? When a = 0, the equation becomes linear, which has exactly one solution, so A has one element, leading to two subsets, which is acceptable.So, putting it all together, the values of a are a = 0 or a ≥ 1.Wait, but earlier I thought that if a is negative, the discriminant is positive, leading to two solutions, which is not acceptable. So, a cannot be negative. So, a must be either 0 or a ≥ 1.But wait, let me double-check. If a is negative, say a = -1, then the equation becomes -x² - 2x + 1 = 0, which is -x² - 2x + 1 = 0. Multiplying both sides by -1, we get x² + 2x - 1 = 0. The discriminant here is 4 + 4 = 8, which is positive, so two real solutions. So, A would have two elements, leading to four subsets, which is more than two. So, that's not acceptable.Therefore, a cannot be negative. So, a must be either 0 or a ≥ 1.But wait, when a = 1, the discriminant is zero, so there's exactly one real solution. So, A has one element, leading to two subsets, which is acceptable.When a > 1, the discriminant is negative, so no real solutions, meaning A is empty, which has one subset, the empty set, which is acceptable because "at most two subsets" includes one subset.When a = 0, the equation is linear, with one solution, so A has one element, leading to two subsets, which is acceptable.Therefore, the range of values for a is a = 0 or a ≥ 1.Wait, but in the initial problem, the equation is ax² - 2x + 1 = 0. So, if a = 0, it's a linear equation, which is fine. If a ≠ 0, it's quadratic.So, to summarize:- If a = 0: A has one element, so two subsets. Acceptable.- If a ≠ 0: To have A with at most one element, the quadratic must have at most one real solution. So, discriminant ≤ 0.Discriminant D = 4 - 4a ≤ 0 => 4a ≥ 4 => a ≥ 1.Therefore, a must be either 0 or a ≥ 1.Wait, but earlier I thought about a being negative, but that leads to two solutions, which is not acceptable. So, a cannot be negative.So, the range of a is a = 0 or a ≥ 1.But let me check if a can be equal to 1. When a = 1, the equation is x² - 2x + 1 = 0, which factors to (x - 1)² = 0, so x = 1 is a double root. So, A = {1}, which has two subsets: {} and {1}. That's acceptable.If a > 1, say a = 2, then the equation is 2x² - 2x + 1 = 0. The discriminant is 4 - 8 = -4, which is negative, so no real solutions. Therefore, A is empty, which has one subset, the empty set. That's acceptable.If a = 0, as we saw, A has one element, so two subsets.But what if a is between 0 and 1? For example, a = 0.5. Then the equation is 0.5x² - 2x + 1 = 0. The discriminant is 4 - 4*(0.5)*1 = 4 - 2 = 2, which is positive. So, two real solutions. Therefore, A has two elements, leading to four subsets, which is more than two. So, that's not acceptable.Therefore, a cannot be between 0 and 1.Similarly, if a is negative, as we saw earlier, the discriminant is positive, leading to two solutions, which is not acceptable.Therefore, the only acceptable values for a are a = 0 or a ≥ 1.Wait, but let me think again. If a is greater than 1, the equation has no real solutions, so A is empty, which has one subset. If a = 1, A has one element, so two subsets. If a = 0, A has one element, so two subsets. If a is between 0 and 1, A has two elements, which is too many subsets. If a is negative, A has two elements, which is also too many subsets.Therefore, the range of a is a = 0 or a ≥ 1.But wait, the problem says "at most two subsets." So, A can have either one or two subsets. If A is empty, it has one subset. If A has one element, it has two subsets. If A has two elements, it has four subsets, which is more than two, so that's not allowed.Therefore, to ensure that A has at most two subsets, A must be either empty or have exactly one element. So, the equation must have either no real solutions or exactly one real solution.Therefore, the discriminant must be less than or equal to zero.So, D = 4 - 4a ≤ 0 => 4a ≥ 4 => a ≥ 1.But wait, when a = 0, the equation is linear, which has exactly one solution, so A has one element, leading to two subsets, which is acceptable. So, a = 0 is also acceptable.Therefore, the range of a is a = 0 or a ≥ 1.But wait, in the quadratic case, when a ≠ 0, we require that D ≤ 0, which gives a ≥ 1. So, combining with a = 0, the range is a = 0 or a ≥ 1.Therefore, the range of values for a is a = 0 or a ≥ 1.But let me write this in interval notation. So, a ∈ {0} ∪ [1, ∞).But the problem might expect a different notation. Alternatively, we can write it as a ≥ 1 or a = 0.Wait, but in the initial problem, the equation is ax² - 2x + 1 = 0. So, if a = 0, it's a linear equation, which is fine. If a ≠ 0, it's quadratic.So, to ensure that A has at most two subsets, A must have at most one element. Therefore, the equation must have at most one real solution.Therefore, for a ≠ 0, the discriminant must be ≤ 0, which gives a ≥ 1. For a = 0, the equation has exactly one solution, so that's acceptable.Therefore, the range of a is a = 0 or a ≥ 1.So, in conclusion, the range of values for a is a = 0 or a ≥ 1.