🧠:[<think>Okay, so I need to figure out this problem where numbers from 1 to 2009 are written on a board. Then, some of them are erased and replaced with the remainder of their sum when divided by 13. This process is repeated several times until only three numbers are left: two of them are 9 and 999, and I need to find the third one. Let me start by understanding the process step by step. The key here is that every time we erase a couple of numbers and replace them with their sum modulo 13. So, each operation reduces the total count of numbers on the board by one. For example, if there are n numbers, after one operation, there will be n-1 numbers. We start with 2009 numbers and end up with 3. So, the number of operations performed is 2009 - 3 = 2006 operations. But maybe I don't need to track each operation. Instead, perhaps I can find an invariant or some quantity that remains constant modulo 13 throughout the operations. That is, something that doesn't change when we replace numbers with their sum modulo 13. What happens when we replace two numbers, say a and b, with (a + b) mod 13? Let's think about the sum of all numbers on the board. Originally, the sum is S. After replacing a and b with (a + b) mod 13, the new sum S' becomes S - a - b + (a + b) mod 13. The difference between S and S' is a + b - (a + b) mod 13. Since (a + b) mod 13 is the remainder when a + b is divided by 13, the difference is a multiple of 13. Therefore, the sum modulo 13 remains the same. Therefore, the invariant here is the total sum of all numbers modulo 13. This sum remains constant throughout all operations. So, if I can compute the initial sum modulo 13 and then compute the final sum modulo 13 (which is 9 + 999 + x mod 13, where x is the third number), then they must be equal. Therefore, I can set up an equation and solve for x. Let me verify this reasoning. Suppose we have numbers a and b, replace them with (a + b) mod 13. Then, the sum before is S = ... + a + b + ... and after the operation, it's S' = ... + (a + b) mod 13 + ... So, S' = S - a - b + (a + b) mod 13. Then, S' ≡ S - a - b + (a + b) mod 13 (mod 13). But (a + b) mod 13 ≡ a + b (mod 13), so S' ≡ S - a - b + a + b ≡ S (mod 13). Therefore, the sum modulo 13 is indeed preserved. Perfect. So, the key is to compute the initial sum of numbers from 1 to 2009 modulo 13, and then equate that to (9 + 999 + x) modulo 13, then solve for x. First, let's compute the initial sum S_initial = 1 + 2 + 3 + ... + 2009. The formula for the sum of the first n natural numbers is n(n + 1)/2. Therefore, S_initial = 2009 * 2010 / 2. Let me compute that:2009 * 2010 = 2009 * (2000 + 10) = 2009*2000 + 2009*10 = 4,018,000 + 20,090 = 4,038,090. Then divide by 2: 4,038,090 / 2 = 2,019,045. So S_initial = 2,019,045. Now, we need to compute this modulo 13. Let's compute 2,019,045 mod 13. Calculating such a large number modulo 13 directly might be tedious, but there are tricks to simplify it. One way is to note that 10 ≡ -3 mod 13, so we can use the method of writing the number in base 10 and substituting powers of 10 with -3. Alternatively, we can break the number into parts. Let's see.Alternatively, since 13 is a prime, maybe we can use properties of modular arithmetic. Let's recall that 10^6 ≡ 1 mod 13 because of Fermat's little theorem, since 13 is prime and 10 not a multiple of 13, so 10^(13-1) ≡ 1 mod 13. However, 13-1 = 12, so 10^12 ≡ 1 mod 13. But 6 is a divisor of 12, so maybe 10^6 ≡ -1 mod 13? Let's check:10^1 ≡ 10 mod 1310^2 ≡ 100 ≡ 9 mod 1310^3 ≡ 10*9 = 90 ≡ 12 mod 1310^4 ≡ 10*12 = 120 ≡ 3 mod 1310^5 ≡ 10*3 = 30 ≡ 4 mod 1310^6 ≡ 10*4 = 40 ≡ 1 mod 13Ah, so 10^6 ≡ 1 mod 13. Therefore, every 6 digits can be cycled. But maybe this approach isn't the most straightforward here. Let's instead compute 2,019,045 mod 13 by breaking it down.First, note that 2,019,045 can be written as:2*10^6 + 0*10^5 + 1*10^4 + 9*10^3 + 0*10^2 + 4*10 + 5But 10^6 mod 13: as above, 10^6 ≡ 1 mod 1310^5 mod 13: 10^5 ≡ 4 mod 13 (from earlier steps: 10^5 ≡ 4)10^4 ≡ 3 mod 1310^3 ≡ 12 mod 1310^2 ≡ 9 mod 1310^1 ≡ 10 mod 1310^0 ≡ 1 mod 13So, using these:2*10^6 ≡ 2*1 = 2 mod 130*10^5 ≡ 0*4 = 0 mod 131*10^4 ≡ 1*3 = 3 mod 139*10^3 ≡ 9*12 = 108 ≡ 108 - 8*13 = 108 - 104 = 4 mod 130*10^2 ≡ 0*9 = 0 mod 134*10 ≡ 4*10 = 40 ≡ 1 mod 13 (since 40 - 3*13 = 40 - 39 = 1)5*1 ≡ 5 mod 13Now, summing all these up:2 + 0 + 3 + 4 + 0 + 1 + 5 = 15. Then 15 mod 13 = 2. So 2,019,045 ≡ 2 mod 13. Alternatively, maybe there's a simpler way. Let's check:We can use the formula for the sum of numbers from 1 to n, which is n(n + 1)/2. So, 2009*2010/2 mod 13. Let's compute 2009 mod 13 and 2010 mod 13 first.2009 divided by 13. Let's compute:13*154 = 2002 (since 13*150=1950, 13*4=52; 1950+52=2002). Therefore, 2009 - 2002 = 7. So, 2009 ≡ 7 mod 13.Similarly, 2010 = 2009 + 1 ≡ 7 + 1 = 8 mod 13.Therefore, the product 2009*2010 ≡ 7*8 = 56 mod 13. 56 divided by 13: 13*4=52, 56-52=4. So 56 ≡ 4 mod 13.Then, divide by 2: 4/2 mod 13. But division by 2 in mod 13 is multiplication by the inverse of 2 mod 13. The inverse of 2 mod 13 is 7, since 2*7=14 ≡1 mod13. Therefore, 4/2 ≡ 4*7 = 28 ≡ 2 mod13. Therefore, the sum S_initial ≡ 2 mod13. So that's consistent with my previous result. Therefore, the initial sum modulo13 is 2. Now, the final sum modulo13 must also be 2. The final numbers are 9, 999, and x. So, 9 + 999 + x ≡ 2 mod13. Let's compute 9 + 999 mod13 first.Compute 999 mod13. Let's divide 999 by13. 13*76=988. 999 - 988=11. So 999≡11 mod13. Therefore, 9 + 999 ≡9 +11=20 mod13. 20 -13=7. So 9 +999 ≡7 mod13. Therefore, 7 + x ≡2 mod13. Solving for x: x ≡2 -7 ≡-5 ≡8 mod13. So x≡8 mod13. Therefore, the third number must be congruent to 8 modulo13. But the problem says that during the process, each time a couple of numbers are erased and replaced with their sum modulo13. Therefore, all numbers on the board at any time are remainders modulo13, except possibly the original numbers. Wait, but when we replace numbers, we write the remainder of their sum modulo13. So, after each operation, the new number is between 0 and 12. However, in the final three numbers, two of them are 9 and 999. But 999 is a number that was originally on the board. Wait, but 999 was part of the original numbers (since numbers from 1 to 2009 are written, and 999 is within that range). So, if 999 was never erased, then it's possible that it's still on the board. Similarly, 9 might have been part of the original numbers or it might be a result of some operations. But the problem states that after several operations, only three numbers are left: two of them are 9 and 999. So, 9 could be original or a result. But wait, the replacement process writes the remainder modulo13, which is between 0 and 12. However, the original numbers are from 1 to 2009, which are all positive integers, so if 9 is left, it could be either original (since 9 is between 1 and 2009) or it could be a result of replacements. Similarly, 999 is original. But the replacement process replaces numbers with their sum modulo13, so if 9 is the result of replacing some numbers, then 9 is between 0 and 12. But since 9 is within 0-12, but the original 9 is also present. However, the problem says that "after a couple of repetitions of this erasing, only 3 numbers are left, of which two are 9 and 999". So, 9 could be either original or a result. However, since 9 is less than 13, if it were a result of replacements, then it's a modulo13 result. But the original 9 is still a single number. Therefore, it's possible that 9 is original or replaced. However, in the final three numbers, they can be a mix of original numbers and replaced numbers (which are between 0 and 12). But the key point is that the third number x is either an original number (so between 1 and 2009) or a replaced number (so between 0 and 12). However, since all operations replace numbers with their sum modulo13, which is between 0 and 12, unless x is an original number that was never erased. But the problem says "after a couple of repetition of this erasing, only 3 numbers are left". So, unless the process stops when three numbers are left, which could include original numbers and replaced numbers. However, since replacements result in numbers between 0 and 12, but original numbers can be up to 2009. So, how come 999 is still there? Because 999 is an original number that was never erased during the process. Similarly, 9 could be an original number or a replaced number. But the problem states "the numbers from 1 to 2009 are written. A couple of them are erased and instead of them, on the board is written the remainder of the sum of the erased numbers divided by 13. After a couple of repetitions of this erasing, only 3 numbers are left, of which two are 9 and 999. Find the third number." So, the third number could be either original or a result of replacements, but the problem is to find what it is. Since modulo13 equivalence is maintained, we can say that the sum of the three numbers modulo13 must equal the original sum modulo13, which is 2. We have 9 + 999 + x ≡2 mod13. Therefore, x ≡2 -9 -999 mod13. As we computed earlier, x ≡8 mod13. Therefore, x is congruent to 8 modulo13. Now, the question is, is x equal to 8? Or is it another number congruent to 8 modulo13. But if x is a result of replacements, then x must be between 0 and 12. Therefore, x=8. However, if x is an original number that was never erased, it could be any number from 1 to 2009 congruent to 8 modulo13. But in that case, the problem doesn't give us enough information to determine which one it is. However, the problem says "Find the third number", implying that the answer is uniquely determined. Hence, the third number must be 8. Because if it were an original number, there are multiple possibilities, but since the process replaces numbers with their sum modulo13, the third number must be the result of such replacements, which is between 0 and 12. Therefore, the third number is 8. Wait, but the original numbers 1 to 2009 include numbers congruent to 8 modulo13. For example, 8, 21, 34, ..., up to 2009. But the problem says that two of the remaining numbers are 9 and 999. If 999 was never erased, it's possible that the third number was also never erased. But how could that happen? If we have 2009 numbers and we perform 2006 operations, each operation reduces the number of numbers by 1, so after 2006 operations, 3 numbers remain. Each operation consists of erasing a couple (i.e., two numbers) and replacing them with one number. Therefore, each operation reduces the count by 1. So, starting with 2009, after 2006 operations, we get 2009 - 2006 = 3 numbers. But in this case, the remaining numbers could be original numbers that were never erased, or numbers that were results of replacements. However, 999 is an original number. So, if 999 was never chosen to be erased in any of the operations, then it remains. Similarly, 9 could be original or a result. But if x is an original number, say 8, then it's possible, but how do we know? The problem states "Find the third number", which is unique. Therefore, likely, the third number must be 8, since modulo13 equivalence requires it, and since the problem is solvable, the answer is 8. But let me check again. Suppose that x is an original number. Then, since x was never erased, then 999 and 9 were also never erased. Therefore, the sum of all numbers is equal to 9 + 999 + x plus the sum of all numbers that were erased and replaced. But since each replacement replaces two numbers a and b with (a + b) mod13. However, the total sum modulo13 is preserved. Therefore, regardless of which numbers are erased, the final sum modulo13 is equal to the initial sum modulo13, which is 2. Therefore, 9 + 999 + x ≡2 mod13, so x≡8 mod13. Therefore, x must be congruent to8 mod13. But x could be 8, 21, 34, etc., up to 2009. But since we need to find the third number, and in the context of replacements, which produce numbers between0 and12, the only possible number between0 and12 congruent to8 mod13 is8. Therefore, the third number must be8. Alternatively, if x is an original number, it must be congruent to8 mod13, but there's no way to know which one. However, since the problem is presented in a way that expects a unique answer, and given the process replaces numbers with their modulo13 remainders, the third number is8. Therefore, the answer is8. Final Answerboxed{8}</think>To solve the problem, we need to determine the third number left on the board after repeatedly erasing pairs of numbers and replacing them with their sum modulo 13. The key insight is that the sum of all numbers on the board modulo 13 remains invariant throughout the process.1. Initial Sum Calculation: - The numbers from 1 to 2009 are written on the board. The sum of these numbers is calculated using the formula for the sum of the first ( n ) natural numbers: ( frac{n(n+1)}{2} ). - For ( n = 2009 ), the sum is ( frac{2009 times 2010}{2} = 2019045 ).2. Modulo 13 Calculation: - We need to find the initial sum modulo 13. First, compute ( 2009 mod 13 ) and ( 2010 mod 13 ): - ( 2009 equiv 7 mod 13 ) - ( 2010 equiv 8 mod 13 ) - The product ( 2009 times 2010 equiv 7 times 8 = 56 mod 13 equiv 4 mod 13 ). - Dividing by 2 (modulo 13), we find the inverse of 2 modulo 13 is 7. Thus, ( 4 times 7 = 28 equiv 2 mod 13 ). Therefore, the initial sum modulo 13 is 2.3. Final Sum Calculation: - The remaining numbers are 9, 999, and the third number ( x ). We need their sum modulo 13 to equal the initial sum modulo 13 (which is 2). - Calculate ( 999 mod 13 ): - ( 999 div 13 = 76 ) remainder 11, so ( 999 equiv 11 mod 13 ). - The sum of 9 and 999 modulo 13 is ( 9 + 11 = 20 equiv 7 mod 13 ). - Therefore, ( 7 + x equiv 2 mod 13 ). Solving for ( x ), we get ( x equiv 2 - 7 equiv -5 equiv 8 mod 13 ).4. Conclusion: - The third number ( x ) must be congruent to 8 modulo 13. Since the replacement process results in numbers between 0 and 12, the only possible value is 8.[boxed{8}],<think>Okay, so I have this problem here where numbers from 1 to 2009 are written on a board. Then, some numbers are erased and replaced with the remainder of their sum divided by 13. After doing this multiple times, only three numbers are left: two of them are 9 and 999, and I need to find the third one. Hmm, interesting. Let me try to break this down step by step.First, let me make sure I understand the process. Initially, the board has all numbers from 1 to 2009. Then, repeatedly, someone erases a couple of numbers, adds them up, divides by 13, takes the remainder, and writes that remainder back on the board. So each time they do this, the total count of numbers on the board decreases by one. For example, if they erase two numbers, they replace them with one, so total numbers reduce by one each operation. They keep doing this until only three numbers are left: 9, 999, and another one which we need to find.So, the key here must be some invariant or property that doesn't change during the operations. Since we're dealing with modulo 13, maybe the sum modulo 13 remains the same throughout the operations? Let me check that.Suppose we have numbers a and b on the board. Their sum is a + b. When we replace them with (a + b) mod 13, the new sum modulo 13 would be (original sum - a - b + (a + b) mod 13) mod 13. Wait, is that equal to the original sum modulo 13?Let's compute:Original sum mod 13: S ≡ (sum of all numbers) mod 13After replacing a and b with (a + b) mod 13, the new sum is (sum - a - b + (a + b mod 13)) mod 13. Let's see:sum_new = sum - a - b + (a + b) mod 13sum_new mod 13 = [sum - a - b + (a + b) mod 13] mod 13But (a + b) mod 13 is congruent to (a + b) mod 13, so we can write:sum_new ≡ sum - a - b + (a + b) mod 13 (mod 13)But sum ≡ sum (mod 13). So:sum_new ≡ sum - (a + b) + ((a + b) mod 13) (mod 13)Now, note that (a + b) mod 13 = (a + b) - 13k for some integer k. Therefore:sum_new ≡ sum - (a + b) + (a + b) - 13k (mod 13)Simplify:sum_new ≡ sum - 13k (mod 13)But since -13k ≡ 0 (mod 13), this simplifies to sum_new ≡ sum (mod 13)Therefore, the total sum modulo 13 remains invariant throughout the operations. That's a crucial point. So, the sum modulo 13 when we start is the same as when we finish with three numbers.Therefore, if we can compute the initial sum modulo 13, then the sum of the three remaining numbers modulo 13 should be equal to that. Given that two of the numbers are 9 and 999, we can find the third number by solving (9 + 999 + x) ≡ initial_sum mod 13. Therefore, x ≡ (initial_sum - 9 - 999) mod 13. So, first, I need to compute initial_sum mod 13.The initial numbers are from 1 to 2009. The sum of the first n integers is n(n + 1)/2. So, sum = 2009 * 2010 / 2. Let me compute that:2009 * 2010 / 2 = (2009 * 2010) / 2First, compute 2009 * 2010. Let me note that 2010 = 2009 + 1, so 2009 * 2010 = 2009^2 + 2009.But maybe it's easier to compute directly:2009 * 2010 = 2009 * (2000 + 10) = 2009*2000 + 2009*10 = 4,018,000 + 20,090 = 4,038,090.Then divide by 2: 4,038,090 / 2 = 2,019,045.So the initial sum is 2,019,045. Now, we need to compute this modulo 13.Calculating 2,019,045 mod 13. Hmm, how to do that?Well, one way is to divide 2,019,045 by 13 and find the remainder, but that might be tedious. Alternatively, we can use the fact that 10 ≡ -3 mod 13, so maybe we can break the number down digit by digit using modular arithmetic.But maybe there's a smarter way. Let's note that 13 is a prime number, and perhaps use properties of modulo operations here.Alternatively, compute 2,019,045 divided by 13:Let me see, 13 * 155,000 = 2,015,000. Then subtract that from 2,019,045: 2,019,045 - 2,015,000 = 4,045.Now, compute 4,045 / 13. Let's see:13 * 311 = 4,043. Because 13 * 300 = 3,900; 13*11=143; 3,900 + 143 = 4,043. Therefore, 4,045 - 4,043 = 2. So, 4,045 = 13*311 + 2. Therefore, 4,045 mod 13 = 2. Hence, 2,019,045 mod 13 is 2.Wait, let me verify:13 * 155,000 = 2,015,000. Then 2,019,045 - 2,015,000 = 4,045. Then 4,045 divided by 13 is 311 with remainder 2. Therefore, total remainder is 2. So initial_sum mod 13 is 2.Therefore, the sum of the three remaining numbers modulo 13 must be 2. The three numbers are 9, 999, and x. So:(9 + 999 + x) ≡ 2 mod 13Compute 9 + 999 = 1008. So 1008 + x ≡ 2 mod 13. Therefore, x ≡ (2 - 1008) mod 13.Compute 1008 mod 13 first. Let's see, 13 * 77 = 1001. So 1008 - 1001 = 7. Therefore, 1008 ≡ 7 mod 13. Therefore:x ≡ (2 - 7) mod 13 ≡ (-5) mod 13 ≡ 8 mod 13.Therefore, x ≡ 8 mod 13. But x is a number on the board. However, in the process, numbers are either original numbers (from 1 to 2009) or numbers that are written as remainders modulo 13. Wait, when you replace numbers with their sum modulo 13, those remainders can be from 0 to 12. But 9 and 999 are two of the remaining numbers. Wait, 999 is an original number, since 999 is less than 2009. Similarly, 9 is an original number. However, x could be either an original number or a remainder.But since we are allowed to replace numbers multiple times, maybe x is a remainder modulo 13? Wait, but if all operations replace numbers with their sum modulo 13, then the possible numbers on the board at any time are either original numbers or numbers that are between 0 and 12 (since modulo 13 is from 0 to 12). But 9 is already between 0 and 12, but 999 is an original number. So the third number could be either an original number or a remainder.But wait, the problem states "after a couple of repetition of this erasing, only 3 numbers are left, of which two are 9 and 999". So 9 and 999 could be either original numbers or remainders. However, 9 is an original number (since 9 is in 1 to 2009), and 999 is also original. However, in the process, when we replace numbers with their sum modulo 13, those remainders could be written as numbers from 0 to 12. So if 9 was obtained as a remainder, but 9 is already an original number. Wait, but maybe 9 is the original 9, and 999 is the original 999, and the third number is the remainder. Alternatively, maybe 9 is a remainder and 999 is original. But how can we know?Wait, but the problem says "the numbers from 1 to 2009 are written. A couple of them are erased and instead of them, on the board is written the remainder of the sum of the erased numbers divided by 13. After a couple of repetition of this erasing, only 3 numbers are left, of which two are 9 and 999. Find the third number."So perhaps during the erasing process, when you erase numbers, you can erase any two numbers, even if they are original or previously replaced remainders. So the 9 could be an original 9 or a replaced remainder. Similarly for 999.But regardless, the key is that the invariant is the sum modulo 13. So regardless of how the numbers are replaced, the total sum modulo 13 remains the same. Therefore, the sum of the three numbers modulo 13 must equal the initial sum modulo 13, which was 2. Therefore, regardless of whether 9 and 999 are original or remainders, their values modulo 13 are fixed. Let's check:9 mod 13 is 9. 999 mod 13: Let's compute 999 divided by 13.13*76 = 988. 999 - 988 = 11. Therefore, 999 ≡ 11 mod 13.So 9 + 999 ≡ 9 + 11 ≡ 20 ≡ 7 mod 13. Then, adding x, we need 7 + x ≡ 2 mod 13. Therefore, x ≡ 2 - 7 ≡ -5 ≡ 8 mod 13. Therefore, x must be congruent to 8 modulo 13.But x is a number on the board. Now, possible options for x are either:1. An original number from 1 to 2009, which is ≡8 mod13.2. A remainder from a previous operation, which would be between 0 and 12, so 8 is possible.But the problem says "Find the third number." So maybe it's 8? But wait, 8 is possible, but maybe the problem expects it to be 8 regardless. But let's check.But wait, the third number could be 8, which is between 0 and 12, but if 8 is an original number (which it is, 8 is in 1-2009), but when you replace numbers, you can end up with 8 as a remainder. However, in the end, the third number is either 8 or another number congruent to 8 mod13. But since after replacements, the numbers can be remainders (i.e., 0-12), or original numbers. But the problem states that two of the numbers left are 9 and 999, which are original numbers. Wait, 9 is an original number, 999 is original. If the third number is a remainder, it would be between 0-12. But if it's an original number, it could be any number from 1-2009 congruent to 8 mod13.But how can we know which one it is? The problem says "Find the third number," so perhaps it's uniquely determined. But according to our reasoning, the third number must be congruent to 8 modulo13. However, since modulo13 could be 8 or 8+13k, but given the operations, if it's a remainder, it's 8. If it's an original number, it's some number like 8, 21, 34, etc., up to 2009.But the problem states "after a couple of repetition of this erasing, only 3 numbers are left, of which two are 9 and 999. Find the third number." The question is, is the third number necessarily 8? Or could it be another number congruent to 8 mod13?Wait, but in the process, every time you replace two numbers with their sum mod13, the numbers on the board can be either original numbers or numbers in 0-12. So it's possible that the third number is 8, but also possible that it's a larger number that wasn't replaced. However, since the problem states that after multiple erasing operations, only three numbers are left. If the erasing operations can be done in such a way that you can choose which numbers to erase, maybe you could leave certain numbers untouched. So if you wanted to leave 9, 999, and another number congruent to 8 mod13, you could. But the key is that the third number must be congruent to 8 mod13. So if 8 is still on the board, you could have 8, but if 8 was erased and replaced earlier, then the third number would have to be something else congruent to 8 mod13.However, the problem states that the numbers are initially from 1 to 2009. So 8 is present initially. But during the operations, you might erase 8 with another number and replace it with (8 + other) mod13. So it's possible that 8 is erased. Therefore, the third number might not be 8. But how can we know?Wait, but the problem is asking for the third number given that two of them are 9 and 999. Since 9 and 999 are original numbers, is it possible that they were never involved in any erasing operations? If so, then the rest of the numbers (from 1 to 2009 except 9, 999, and the third number) were erased in pairs or groups and replaced with their sum mod13. However, the key invariant is that the total sum mod13 remains 2. Therefore, regardless of how the numbers are erased and replaced, the sum of the remaining numbers mod13 is 2. So 9 + 999 + x ≡ 2 mod13, so x ≡ 8 mod13. Therefore, the third number must be congruent to 8 mod13.But since the problem asks to "Find the third number," and the answer is unique, perhaps we need to give 8 as the answer. However, maybe the answer is 8, but let me check if there's a possible mistake here.Wait, let's double-check the initial sum mod13. The initial sum is 2009*2010/2 = 2,019,045. Then, 2,019,045 mod13. As calculated earlier, 2,019,045 divided by 13 is 155,311 with a remainder of 2. Therefore, 2,019,045 ≡2 mod13. So the invariant is 2 mod13.Then, the final three numbers must sum to 2 mod13. So 9 + 999 + x ≡2 mod13. 9 + 999 = 1008. 1008 mod13. As calculated before, 13*77=1001, so 1008 -1001=7. Therefore, 1008≡7 mod13. So 7 + x ≡2 mod13 => x ≡2 -7 ≡-5≡8 mod13.Therefore, x≡8 mod13. So the third number must be congruent to 8 mod13. However, the problem is asking for the third number. If the answer is expected to be a number between 0 and 12, then 8 is the answer. But if it's possible that the third number is an original number, which is 8 mod13, then there could be multiple answers. But the problem states "Find the third number," implying that the answer is uniquely determined.But why would it be uniquely determined? Because even though there are multiple numbers congruent to 8 mod13 in 1-2009, the replacement process replaces numbers with their sum modulo13, so the third number could be 8 or any other number congruent to 8 mod13. However, but if you have two original numbers left, 9 and 999, perhaps the third number is a replaced number, which would be between 0 and 12. Therefore, the answer is 8.But let me think again. Suppose we have 9 and 999 as original numbers. Then, the third number must be a replaced number, which is a remainder modulo13, so 0-12. Therefore, the third number is 8, which is the only number in 0-12 congruent to 8 mod13 (since 8 is already in that range). Therefore, the answer is 8.Alternatively, if the third number is an original number, which was never replaced, then it would be some number congruent to 8 mod13, like 8, 21, 34, etc. But since the problem states that "a couple of them are erased and instead of them...", so the process is repeated until three numbers are left. So perhaps not all numbers except three are erased; rather, pairs are erased and replaced until three remain. But in that case, the three numbers could be any combination of original numbers and replaced remainders.But the problem states "after a couple of repetition of this erasing, only 3 numbers are left, of which two are 9 and 999." So 9 and 999 could be original numbers that were never erased, and the third number could be a remainder. Alternatively, maybe 9 and 999 were themselves results of replacements. However, 9 is in the original set, so even if it was erased and replaced, the remainder would be 9 if the sum was 9 mod13. Similarly, 999 is a large number; replacing it with another number would require that the sum of 999 and another number modulo13 is written, but 999 mod13 is 11, as calculated earlier. So replacing 999 with another number would require that 11 + something ≡ new number mod13.But regardless, since the invariant is the total sum modulo13, the sum of the three numbers must be 2 mod13. Therefore, regardless of the path taken, the third number must be 8 mod13. But how do we know whether it's 8 or another number?Wait, perhaps the answer is 8. Because when you replace numbers, you can write any remainder, so even if 8 was erased, you could get it back by having numbers that sum to 8 mod13. However, the problem states "Find the third number," and given that it's a math competition problem, the answer is likely 8. So I think the answer is 8.But let me check once again. Let's verify with small examples.Suppose we have numbers 1, 2, 3. Sum is 6. Modulo13 is 6. If we erase 1 and 2, replace them with 3 (sum 3). Now numbers are 3 and 3. Sum is 6. Modulo13 is 6. If we erase 3 and 3, replace with 6. Now only 6 is left. Sum is 6, which is 6 mod13. So invariant holds.Another example: numbers 1, 2. Replace them with 3. Sum remains 3. Then replace 3 with 3, but you can't. Wait, but maybe if you have three numbers, you can only replace two at a time. Wait, in the original problem, they start with 2009 numbers, which is an odd number. Each operation reduces the count by 1. So starting from 2009 (odd), after one operation: 2008 (even), then 2007, etc. So to get down to 3 numbers, which is odd, since 2009 - 2*1003 = 2009 - 2006 = 3. So it's possible.But in any case, the key is the invariant. Therefore, regardless of the operations, the sum modulo13 remains the same. Therefore, the third number must be 8 mod13. Since the problem asks for the third number, which is left on the board, and given that replacements can result in any remainder, including 8, and 8 is a possible remainder, the answer is 8.Therefore, I think the answer is 8.Final Answerboxed{8}</think>The numbers from 1 to 2009 are written on a board. Repeatedly, pairs of numbers are erased and replaced with their sum modulo 13. We need to find the third number left on the board when only three numbers remain, two of which are 9 and 999.1. Invariant Check: The sum of the numbers modulo 13 remains invariant throughout the operations. The initial sum of numbers from 1 to 2009 is calculated as follows: [ text{Sum} = frac{2009 times 2010}{2} = 2019045 ] Calculating this sum modulo 13: [ 2019045 div 13 = 155311 text{ remainder } 2 implies 2019045 equiv 2 pmod{13} ]2. Final Sum Calculation: The sum of the remaining three numbers modulo 13 must also be 2. Given two of the remaining numbers are 9 and 999: [ 9 + 999 + x equiv 2 pmod{13} ] Calculating (9 + 999) modulo 13: [ 999 equiv 11 pmod{13} implies 9 + 11 = 20 equiv 7 pmod{13} ] Solving for (x): [ 7 + x equiv 2 pmod{13} implies x equiv 2 - 7 equiv -5 equiv 8 pmod{13} ]Thus, the third number must be congruent to 8 modulo 13. Given the problem constraints, the third number is (boxed{8}).]

🔑:1. Define the invariant: Let ( I ) be the remainder when the sum of all the integers on the board is divided by 13. Initially, the numbers from 1 to 2009 are written on the board. The sum of these numbers is: [ S = sum_{k=1}^{2009} k = frac{2009 times 2010}{2} = 2009 times 1005 ] We need to find ( S mod 13 ).2. Calculate ( 2009 mod 13 ) and ( 1005 mod 13 ): [ 2009 div 13 = 154 quad text{remainder} quad 7 quad Rightarrow quad 2009 equiv 7 pmod{13} ] [ 1005 div 13 = 77 quad text{remainder} quad 4 quad Rightarrow quad 1005 equiv 4 pmod{13} ]3. Calculate ( S mod 13 ): [ S = 2009 times 1005 equiv 7 times 4 = 28 equiv 2 pmod{13} ] Therefore, ( I = 2 ).4. Determine the third number ( a ): After several operations, only three numbers are left on the board: 9, 999, and ( a ). The sum of these three numbers modulo 13 must be equal to the invariant ( I ): [ a + 9 + 999 equiv 2 pmod{13} ]5. Calculate ( 999 mod 13 ): [ 999 div 13 = 76 quad text{remainder} quad 11 quad Rightarrow quad 999 equiv 11 pmod{13} ]6. Set up the equation: [ a + 9 + 11 equiv 2 pmod{13} ] Simplify the equation: [ a + 20 equiv 2 pmod{13} ] Subtract 20 from both sides: [ a equiv 2 - 20 pmod{13} ] [ a equiv -18 pmod{13} ] Since (-18 equiv -18 + 26 equiv 8 pmod{13}), we have: [ a equiv 8 pmod{13} ]7. Find the possible values of ( a ) between 1 and 2009: The possible values of ( a ) are of the form: [ a = 8 + 13k quad text{for integer} quad k ] We need ( 1 leq a leq 2009 ): [ 1 leq 8 + 13k leq 2009 ] [ -7 leq 13k leq 2001 ] [ 0 leq k leq 154 ] Therefore, the possible values of ( a ) are: [ a = 8, 21, 34, ldots, 2002 ]The final answer is ( boxed{8} ).

🧠:[<think>Okay, let's tackle this problem step by step. It's about an arithmetic sequence and a geometric sequence. Hmm. First, let's make sure I understand the problem correctly.We have an arithmetic sequence {a_n} with a non-zero common difference. That means each term is the previous term plus some constant d, which isn't zero. So, the general term should be a_n = a_1 + (n-1)d, right?Now, the problem states that the terms a_1, a_4, and a_8 form a geometric sequence. A geometric sequence has each term multiplied by a constant ratio r. So, if a_1, a_4, a_8 are in geometric progression, then the ratio between a_4 and a_1 should be the same as the ratio between a_8 and a_4. In other words, (a_4)/(a_1) = (a_8)/(a_4). That should give us an equation to work with.Also, the sum of the first 10 terms of the arithmetic sequence is 45. The formula for the sum of the first n terms of an arithmetic sequence is S_n = n/2 * (2a_1 + (n - 1)d). So, plugging in n=10, we have S_10 = 45 = 10/2 * (2a_1 + 9d). That simplifies to 45 = 5*(2a_1 + 9d), so dividing both sides by 5, we get 9 = 2a_1 + 9d. So that's another equation.So, we have two equations here: one from the geometric sequence condition and one from the sum. We need to solve these two equations to find a_1 and d, then write the general term formula. That should be part (1). Let's start with that.First, let's write expressions for a_1, a_4, a_8 in terms of a_1 and d.Since it's an arithmetic sequence:a_1 = a_1 (obviously)a_4 = a_1 + 3da_8 = a_1 + 7dNow, since these three terms form a geometric sequence, the square of the middle term (a_4)^2 should be equal to the product of the first and third terms (a_1 * a_8). So, (a_4)^2 = a_1 * a_8.Substituting the expressions in terms of a_1 and d:(a_1 + 3d)^2 = a_1 * (a_1 + 7d)Let's expand both sides.Left side: (a_1)^2 + 6a_1d + 9d^2Right side: (a_1)^2 + 7a_1dSubtract the right side from the left side to set the equation to zero:(a_1^2 + 6a_1d + 9d^2) - (a_1^2 + 7a_1d) = 0Simplify:0 + (-a_1d) + 9d^2 = 0So:- a_1d + 9d^2 = 0Factor out d:d(-a_1 + 9d) = 0Since the common difference d is non-zero, we can divide both sides by d:- a_1 + 9d = 0Therefore:a_1 = 9dOkay, so we have a relationship between a_1 and d: a_1 is 9 times d.Now, we also have the sum equation: 2a_1 + 9d = 9.But since a_1 = 9d, substitute that into the equation:2*(9d) + 9d = 918d + 9d = 927d = 9Therefore, d = 9 / 27 = 1/3.Then, since a_1 = 9d = 9*(1/3) = 3.So, the first term a_1 is 3, and the common difference d is 1/3.Therefore, the general term formula is:a_n = a_1 + (n - 1)d = 3 + (n - 1)*(1/3)Simplify that:a_n = 3 + (n/3 - 1/3) = (9/3) + (n/3 - 1/3) = (8/3) + n/3 = (n + 8)/3Wait, let's check that again:a_1 = 3a_2 = 3 + 1/3 = 10/3a_3 = 3 + 2*(1/3) = 11/3Wait, but if a_n = (n + 8)/3, then a_1 would be (1 + 8)/3 = 9/3 = 3, which is correct. a_4 would be (4 + 8)/3 = 12/3 = 4. Let's check if that's consistent with the arithmetic sequence.Original a_1 is 3, d=1/3, so a_4 = 3 + 3*(1/3) = 3 + 1 = 4, which matches. Similarly, a_8 = 3 + 7*(1/3) = 3 + 7/3 = 16/3. Wait, but according to the formula (n + 8)/3, a_8 would be (8 + 8)/3 = 16/3, which is the same. So that works.But let's verify the geometric sequence part. a_1 = 3, a_4 = 4, a_8 = 16/3. Let's check if these are in geometric progression.First ratio: a_4 / a_1 = 4 / 3Second ratio: a_8 / a_4 = (16/3) / 4 = (16/3)*(1/4) = 4/3So both ratios are 4/3. Perfect, that's a geometric sequence with common ratio 4/3. So that checks out.Also, the sum of the first 10 terms: Let's compute S_10. Using the formula S_n = n/2*(2a_1 + (n - 1)d)S_10 = 10/2*(2*3 + 9*(1/3)) = 5*(6 + 3) = 5*9 = 45, which matches the given condition. Great, so part (1) is solved. The general term is a_n = (n + 8)/3. Wait, but let's write it properly. The arithmetic sequence formula is a_1 + (n - 1)d.Given a_1 = 3 and d = 1/3, then:a_n = 3 + (n - 1)*(1/3) = 3 + (n/3 - 1/3) = (9/3 + n/3 - 1/3) = (n + 8)/3. Yes, that's correct. So, answer to part (1) is a_n = (n + 8)/3.Now part (2): If b_n = 1/(a_n a_{n+1}), and T_n is the sum of the first n terms of {b_n}, find T_n.So, we need to find T_n = sum_{k=1}^n b_k = sum_{k=1}^n [1/(a_k a_{k+1})]Given that a_k is an arithmetic sequence, perhaps we can use the concept of telescoping series here. Because when you have terms like 1/(a_k a_{k+1}), which can be expressed as a difference of reciprocals.Since a_k is an arithmetic sequence, the difference a_{k+1} - a_k is equal to the common difference d. Which we found to be 1/3.So, let's recall that for an arithmetic sequence, a_{k+1} - a_k = d.So, perhaps we can write 1/(a_k a_{k+1}) as (1/d)(1/a_k - 1/a_{k+1})). Let's check that.Let's compute (1/a_k - 1/a_{k+1}):= (a_{k+1} - a_k)/(a_k a_{k+1}))= d/(a_k a_{k+1})Therefore, 1/(a_k a_{k+1}) ) = (1/d)(1/a_k - 1/a_{k+1})So, this is a telescoping series. Therefore, the sum T_n = sum_{k=1}^n [1/(a_k a_{k+1})] = (1/d) sum_{k=1}^n [1/a_k - 1/a_{k+1}]Which telescopes to (1/d)(1/a_1 - 1/a_{n+1})Given that d = 1/3, a_1 = 3, and a_{n+1} = (n + 1 + 8)/3 = (n + 9)/3Therefore,T_n = (1/(1/3)) [1/3 - 1/((n + 9)/3)] = 3 [1/3 - 3/(n + 9)] = 3*(1/3) - 3*(3)/(n + 9) = 1 - 9/(n + 9) = (n + 9 - 9)/(n + 9) = n/(n + 9)Wait, let me verify step by step.First, 1/a_1 = 1/3.1/a_{n+1} = 1/[(n + 9)/3] = 3/(n + 9).Therefore, 1/a_1 - 1/a_{n+1} = 1/3 - 3/(n + 9)Multiply by 1/d = 1/(1/3) = 3:So, T_n = 3*(1/3 - 3/(n + 9)) = 3*(1/3) - 3*(3/(n + 9)) = 1 - 9/(n + 9)Combine the terms:1 can be written as (n + 9)/(n + 9), so:(n + 9)/(n + 9) - 9/(n + 9) = (n + 9 - 9)/ (n + 9) = n/(n + 9)Hence, T_n = n/(n + 9)Let me test this with n=1. If n=1, then T_1 = b_1 = 1/(a_1 a_2). Given a_1 = 3, a_2 = (2 + 8)/3 = 10/3. So, b_1 = 1/(3 * 10/3) = 1/10. According to the formula, T_1 = 1/(1 + 9) = 1/10. Correct.Another test with n=2. T_2 = b_1 + b_2 = 1/10 + 1/(a_2 a_3). a_3 = (3 + 8)/3 = 11/3. So, a_2 a_3 = (10/3)(11/3) = 110/9. Thus, b_2 = 9/110. Then T_2 = 1/10 + 9/110 = (11 + 9)/110 = 20/110 = 2/11. According to the formula, T_2 = 2/(2 + 9) = 2/11. Correct. Good.Another test: n=3. T_3 = 1/(a1 a2) + 1/(a2 a3) + 1/(a3 a4). We have a1=3, a2=10/3, a3=11/3, a4=12/3=4.So, T3 = 1/(3*10/3) + 1/(10/3 *11/3) + 1/(11/3 *4) = 1/10 + 9/110 + 3/44.Compute:1/10 = 0.19/110 ≈ 0.08183/44 ≈ 0.06818Adding together: ≈0.1 + 0.0818 + 0.06818 ≈0.25.Formula gives T3 = 3/(3 + 9) = 3/12 = 0.25. Perfect. So the formula works.Therefore, part (2) answer is T_n = n/(n + 9).So summarizing:(1) The general term is a_n = (n + 8)/3.(2) The sum T_n = n/(n + 9).Final Answer(1) The general term formula is boxed{a_n = dfrac{n + 8}{3}}.(2) The sum ( T_n ) is boxed{T_n = dfrac{n}{n + 9}}.</think>Given a non-zero common difference arithmetic sequence ({a_n}), the terms (a_1), (a_4), and (a_8) form a geometric sequence. The sum of the first 10 terms of the sequence ({a_n}) is 45.(1) To find the general term formula for the sequence ({a_n}):1. The terms (a_1), (a_4), and (a_8) form a geometric sequence, so ((a_4)^2 = a_1 cdot a_8).2. Expressing these terms in terms of (a_1) and common difference (d): - (a_1 = a_1) - (a_4 = a_1 + 3d) - (a_8 = a_1 + 7d)3. Substituting these into the geometric sequence condition: [ (a_1 + 3d)^2 = a_1(a_1 + 7d) ] Expanding and simplifying: [ a_1^2 + 6a_1d + 9d^2 = a_1^2 + 7a_1d implies -a_1d + 9d^2 = 0 implies a_1 = 9d ]4. Using the sum of the first 10 terms: [ S_{10} = frac{10}{2} (2a_1 + 9d) = 45 implies 2a_1 + 9d = 9 ]5. Substituting (a_1 = 9d) into the sum equation: [ 2(9d) + 9d = 9 implies 27d = 9 implies d = frac{1}{3} ]6. Therefore, (a_1 = 9d = 3), and the general term formula is: [ a_n = a_1 + (n - 1)d = 3 + (n - 1)left(frac{1}{3}right) = frac{n + 8}{3} ](2) To find (T_n), the sum of the first (n) terms of the sequence ({b_n}) where (b_n = frac{1}{a_n a_{n+1}}):1. Recognize that (b_n) can be expressed as a telescoping series: [ b_n = frac{1}{a_n a_{n+1}} = frac{1}{d} left( frac{1}{a_n} - frac{1}{a_{n+1}} right) ] where (d = frac{1}{3}).2. Summing the first (n) terms: [ T_n = sum_{k=1}^n b_k = frac{1}{d} left( frac{1}{a_1} - frac{1}{a_{n+1}} right) ]3. Substituting (a_1 = 3), (a_{n+1} = frac{n + 9}{3}), and (d = frac{1}{3}): [ T_n = 3 left( frac{1}{3} - frac{3}{n + 9} right) = 1 - frac{9}{n + 9} = frac{n}{n + 9} ] Final Answer(1) The general term formula is boxed{a_n = dfrac{n + 8}{3}}.(2) The sum (T_n) is boxed{T_n = dfrac{n}{n + 9}}.,<think>Okay, let's tackle this problem step by step. It's an arithmetic sequence with some conditions, and then a related question about another sequence. Let me start with part (1).First, we know that in an arithmetic sequence, each term is the previous term plus a common difference, d. So, the general term formula for an arithmetic sequence is a_n = a_1 + (n - 1)d. Since it's a non-zero common difference, d ≠ 0.The problem states that the terms a₁, a₄, and a₈ form a geometric sequence. In a geometric sequence, each term is multiplied by a constant ratio. So, if these three terms are in a geometric sequence, the square of the middle term (a₄) should equal the product of the first and third terms (a₁ and a₈). So, let's write that down:a₄² = a₁ * a₈Now, substituting the arithmetic sequence formula into this equation:a₄ = a₁ + 3da₈ = a₁ + 7dSo, the equation becomes:(a₁ + 3d)² = a₁ * (a₁ + 7d)Let me expand the left side:(a₁)² + 6a₁d + 9d² = a₁² + 7a₁dSubtracting the right side from both sides:(a₁² + 6a₁d + 9d²) - (a₁² + 7a₁d) = 0Simplify:- a₁d + 9d² = 0Factor out d:d(-a₁ + 9d) = 0Since d is non-zero (given), the other factor must be zero:- a₁ + 9d = 0 ⇒ a₁ = 9dOkay, so we have a relationship between a₁ and d: a₁ = 9d. That's useful.Next, the problem also states that the sum of the first 10 terms of the arithmetic sequence is 45. The sum of the first n terms of an arithmetic sequence is given by S_n = (n/2)(2a₁ + (n - 1)d). So, plugging n = 10:S₁₀ = (10/2)(2a₁ + 9d) = 5(2a₁ + 9d) = 45Simplify:5(2a₁ + 9d) = 45 ⇒ 2a₁ + 9d = 9But we already know that a₁ = 9d. Let's substitute that into the equation:2*(9d) + 9d = 9 ⇒ 18d + 9d = 9 ⇒ 27d = 9 ⇒ d = 9/27 = 1/3So, d = 1/3. Then, a₁ = 9d = 9*(1/3) = 3.Therefore, the arithmetic sequence has a first term of 3 and common difference of 1/3. Let me confirm that.The terms would be:a₁ = 3a₂ = 3 + 1/3 = 10/3a₃ = 10/3 + 1/3 = 11/3a₄ = 11/3 + 1/3 = 12/3 = 4Wait, a₄ is 4. Let's check if a₁, a₄, a₈ form a geometric sequence.a₁ = 3a₄ = 4a₈ = a₁ + 7d = 3 + 7*(1/3) = 3 + 7/3 = 16/3 ≈ 5.333...So, the geometric sequence would be 3, 4, 16/3. Let's check if 4² = 3*(16/3).4² = 163*(16/3) = 16Yes, that's equal. Good, that checks out.Also, the sum of the first 10 terms. Let's compute it:S₁₀ = (10/2)*(2*3 + 9*(1/3)) = 5*(6 + 3) = 5*9 = 45. Correct. That matches the given condition.So part (1) is solved. The general term formula is a_n = a₁ + (n - 1)d = 3 + (n - 1)*(1/3). Let me write that more neatly:a_n = 3 + (n - 1)/3Alternatively, we can write that as:a_n = (9 + n - 1)/3 = (n + 8)/3Wait, let me compute that:3 is 9/3, so 9/3 + (n - 1)/3 = (9 + n - 1)/3 = (n + 8)/3Yes, that's correct. So the general term is (n + 8)/3. Let me check with n=1: (1 + 8)/3 = 9/3 = 3. Correct. For n=4: (4 + 8)/3 = 12/3 = 4. Correct. For n=8: (8 + 8)/3 = 16/3. Correct. So, yes, a_n = (n + 8)/3.Okay, moving on to part (2). Given b_n = 1/(a_n * a_{n+1}), and T_n is the sum of the first n terms of b_n. We need to find T_n.First, let's write down what b_n is. Since a_n = (n + 8)/3, then a_{n+1} = (n + 1 + 8)/3 = (n + 9)/3.So, b_n = 1/(a_n * a_{n+1}) = 1/[ ( (n + 8)/3 ) * ( (n + 9)/3 ) ] = 1/[ (n + 8)(n + 9)/9 ] = 9 / [ (n + 8)(n + 9) ]So, b_n = 9 / [ (n + 8)(n + 9) ]This looks like a telescoping series. To find the sum T_n, we can use partial fractions.Let me decompose 9 / [ (n + 8)(n + 9) ] into partial fractions.Assume 9 / [ (n + 8)(n + 9) ] = A/(n + 8) + B/(n + 9)Multiply both sides by (n + 8)(n + 9):9 = A(n + 9) + B(n + 8)To find A and B, set up equations by choosing suitable n.Let n = -9: 9 = A(0) + B(-1) ⇒ 9 = -B ⇒ B = -9Let n = -8: 9 = A(1) + B(0) ⇒ 9 = A ⇒ A = 9Therefore, the partial fractions are:9/(n + 8) - 9/(n + 9)Therefore, b_n = 9/(n + 8) - 9/(n + 9)Therefore, the sum T_n = sum_{k=1}^n b_k = sum_{k=1}^n [9/(k + 8) - 9/(k + 9)]This is a telescoping sum. Let's write out the terms:For k = 1: 9/(1 + 8) - 9/(1 + 9) = 9/9 - 9/10 = 1 - 9/10For k = 2: 9/10 - 9/11For k = 3: 9/11 - 9/12...For k = n: 9/(n + 8) - 9/(n + 9)When we add all these terms, most terms will cancel. The sum T_n is:[1 - 9/10] + [9/10 - 9/11] + [9/11 - 9/12] + ... + [9/(n + 8) - 9/(n + 9)]All the intermediate terms cancel out. The remaining terms are the first term of the first expression and the last term of the last expression.So, T_n = 1 - 9/(n + 9)Simplify that:T_n = 1 - 9/(n + 9) = (n + 9)/(n + 9) - 9/(n + 9) = (n + 9 - 9)/(n + 9) = n/(n + 9)Therefore, T_n = n/(n + 9)Let me check with n=1. If n=1, T₁ = b₁ = 1/(a₁a₂) = 1/(3*(10/3)) = 1/(10). According to the formula, T₁ = 1/(1 + 9) = 1/10. Correct.For n=2, T₂ = 1/10 + 1/(a₂a₃) = 1/10 + 1/( (10/3)*(11/3) ) = 1/10 + 1/(110/9) = 1/10 + 9/110 = 11/110 + 9/110 = 20/110 = 2/11. According to the formula, T₂ = 2/(2 + 9) = 2/11. Correct.So the formula seems to work. Therefore, the answer for part (2) is T_n = n/(n + 9).Final Answer(1) The general term is boxed{a_n = dfrac{n + 8}{3}}.(2) The sum ( T_n ) is boxed{T_n = dfrac{n}{n + 9}}.</think>Given a non-zero common difference arithmetic sequence ({a_n}), the terms (a_1), (a_4), and (a_8) form a geometric sequence. The sum of the first 10 terms of the sequence ({a_n}) is 45.(1) To find the general term formula for the sequence ({a_n}):1. The terms (a_1), (a_4), and (a_8) form a geometric sequence, so we have: [ (a_1 + 3d)^2 = a_1 cdot (a_1 + 7d) ]2. Expanding and simplifying: [ a_1^2 + 6a_1d + 9d^2 = a_1^2 + 7a_1d implies -a_1d + 9d^2 = 0 implies a_1 = 9d ]3. The sum of the first 10 terms is 45: [ S_{10} = frac{10}{2} left(2a_1 + 9dright) = 45 implies 5(2a_1 + 9d) = 45 implies 2a_1 + 9d = 9 ]4. Substituting (a_1 = 9d) into the equation: [ 2(9d) + 9d = 9 implies 27d = 9 implies d = frac{1}{3} ]5. Therefore, (a_1 = 9d = 3). The general term formula is: [ a_n = 3 + (n - 1) cdot frac{1}{3} = frac{n + 8}{3} ]The general term formula for the sequence ({a_n}) is (boxed{a_n = dfrac{n + 8}{3}}).(2) To find the sum (T_n) of the first (n) terms of the sequence ({b_n}) where (b_n = frac{1}{a_n a_{n+1}}):1. Express (b_n) using the general term formula of ({a_n}): [ b_n = frac{1}{left(frac{n + 8}{3}right)left(frac{n + 9}{3}right)} = frac{9}{(n + 8)(n + 9)} ]2. Decompose (b_n) using partial fractions: [ frac{9}{(n + 8)(n + 9)} = frac{9}{n + 8} - frac{9}{n + 9} ]3. The sum (T_n) is a telescoping series: [ T_n = sum_{k=1}^n left(frac{9}{k + 8} - frac{9}{k + 9}right) ]4. The series simplifies to: [ T_n = left(1 - frac{9}{10}right) + left(frac{9}{10} - frac{9}{11}right) + cdots + left(frac{9}{n + 8} - frac{9}{n + 9}right) ]5. Most terms cancel out, leaving: [ T_n = 1 - frac{9}{n + 9} = frac{n}{n + 9} ]The sum (T_n) is (boxed{T_n = dfrac{n}{n + 9}}).]

🔑:(1) In a non-zero common difference arithmetic sequence {a_n}, the terms a_1, a_4, and a_8 form a geometric sequence. This implies that a_1a_8 = a_4^2. Substituting the arithmetic sequence terms, we get a_1(a_1+7d)=(a_1+3d)^2. Solving this equation, we find a_1=9d.The sum of the first 10 terms of the sequence is given as 45, so we have 10a_1 + 45d = 45. Solving this equation, we get a_1=3 and d=frac{1}{3}. Thus, the general term formula for the sequence {a_n} is:a_n = 3 + frac{1}{3}(n-1) = frac{n+8}{3}(2) For the sequence {b_n} defined as b_n = frac{1}{a_n a_{n+1}}, substituting the general term formula for {a_n}, we obtain:b_n = frac{9}{(n+8)(n+9)} = 9left(frac{1}{n+8} - frac{1}{n+9}right)Now, to find the sum T_n of the first n terms of the sequence {b_n}, we use telescoping series:T_n = 9left(frac{1}{9} - frac{1}{10} + frac{1}{10} - frac{1}{11} + ldots + frac{1}{n+8} - frac{1}{n+9}right) = 9left(frac{1}{9} - frac{1}{n+9}right)Simplifying this expression, we get:boxed{T_n = frac{n}{n+9}}

🧠:[<think>Alright, so I need to figure out how many integers n satisfy the inequality x² + nx + 100 > 0 for all real numbers x. Hmm, okay. Let me start by recalling some algebra. First off, this is a quadratic inequality. The general form of a quadratic is ax² + bx + c. In this case, a is 1, b is n, and c is 100. The inequality says that this quadratic is always positive, meaning it never touches or crosses the x-axis. For a quadratic to be always positive, the discriminant must be negative. Right? Because if the discriminant is positive, there are two real roots, which means the quadratic crosses the x-axis twice. If it's zero, there's one real root (a repeated root), so the vertex touches the x-axis. But if the discriminant is negative, there are no real roots, which means the quadratic is always above or below the x-axis. Since the coefficient of x² here is 1, which is positive, the parabola opens upwards. Therefore, if the discriminant is negative, the quadratic is always positive. So, the discriminant D is given by b² - 4ac. Plugging in the values, that's n² - 4*1*100. So D = n² - 400. For the quadratic to be always positive, we need D < 0. Therefore, n² - 400 < 0. Solving that inequality:n² < 400Taking square roots on both sides, we get |n| < 20. Since n is an integer, this means that n can be any integer from -19 up to 19 inclusive. Let's check that. If |n| < 20, then n must be greater than -20 and less than 20. So n ∈ (-20, 20). But since n is an integer, the integers in this interval are -19, -18, ..., -1, 0, 1, ..., 18, 19. Now, let's count how many integers that is. From -19 to 19 inclusive. The number of integers from a to b inclusive is b - a + 1. But here, a is -19 and b is 19. So 19 - (-19) + 1 = 19 + 19 + 1 = 39. Wait, is that right? Wait, no. Wait, if you count from -19 to 19, how does that work? Let's think. From 1 to n, the count is n. From -m to n, it's m + n + 1? Hmm, maybe another way. Let's break it into negative integers, zero, and positive integers. Negative integers: from -1 to -19. That's 19 numbers. Then zero, which is 1 number. Positive integers: from 1 to 19. That's 19 numbers. So total is 19 + 1 + 19 = 39. Yeah, so 39 integers. But hold on, wait. Let me verify this. Suppose n = 20. Then discriminant would be 400 - 400 = 0, so the quadratic would have a repeated root. Therefore, at x = -n/2 = -10, the quadratic touches the x-axis, so the inequality x² + 20x + 100 > 0 would not hold for x = -10, since it equals zero there. Similarly for n = -20, discriminant is (-20)^2 - 400 = 400 - 400 = 0, same thing. So n cannot be 20 or -20. So the permissible integers are strictly between -20 and 20. So that's n from -19 to 19 inclusive, which is 39 integers. Therefore, the number of integers n is 39. Hmm. But let me think again. Wait, perhaps I made a mistake here. Let me double-check. So the discriminant is n² - 400 < 0. So n² < 400. Which means that n is between -20 and 20. So n must be greater than -20 and less than 20. So if n is an integer, then the integers n satisfying -20 < n < 20 are from -19 up to 19 inclusive. That's correct. So how many numbers is that? Let's see: from 1 to 19, that's 19 numbers. From -1 to -19, that's 19 numbers. Plus zero. 19 + 19 + 1 = 39. Yes, 39 integers. Wait, but hold on. Let me take a small example. Suppose instead of 400, the discriminant was 4, so n² < 4. Then n can be -1, 0, 1. So 3 integers. Which is from -1 to 1 inclusive. So yes, 1 - (-1) + 1 = 3. Wait, that formula: if the numbers are from a to b inclusive, the count is b - a + 1. So in the original problem, from -19 to 19 inclusive: 19 - (-19) + 1 = 19 + 19 +1 = 39. So that's correct. Therefore, the answer should be 39. But let me make sure that there isn't any mistake in reasoning here. The key steps are: quadratic is always positive when discriminant is negative. Discriminant is n² - 400 < 0, so n must be between -20 and 20. Since n must be an integer, the integers in that interval are -19, -18,...,19, which is 39 integers. Therefore, I think that's right. So the number of integers n is 39.Final AnswerThe number of integers ( n ) is boxed{39}.</think>To determine how many integers ( n ) satisfy the inequality ( x^2 + nx + 100 > 0 ) for all real numbers ( x ), we need to ensure that the quadratic expression is always positive. This occurs when the discriminant of the quadratic equation is negative, as the quadratic opens upwards (coefficient of ( x^2 ) is positive).The quadratic expression ( x^2 + nx + 100 ) has coefficients ( a = 1 ), ( b = n ), and ( c = 100 ). The discriminant ( D ) is given by:[D = b^2 - 4ac = n^2 - 4 cdot 1 cdot 100 = n^2 - 400]For the quadratic to be always positive, the discriminant must be negative:[n^2 - 400 < 0 implies n^2 < 400]Taking the square root of both sides, we get:[|n| < 20]This implies ( -20 < n < 20 ). Since ( n ) must be an integer, the possible values of ( n ) are the integers from -19 to 19 inclusive. To count the number of integers in this range:- The integers from -19 to -1 are 19 numbers.- The integer 0 is 1 number.- The integers from 1 to 19 are 19 numbers.Adding these together, we get:[19 + 1 + 19 = 39]Thus, the number of integers ( n ) is (boxed{39}).,<think>Alright, let's see. I need to find how many integers n make the quadratic inequality x² + nx + 100 > 0 true for all real numbers x. Hmm, okay. So, this is a quadratic in x, and I want it to be positive no matter what x is. That means the quadratic never touches or crosses the x-axis, right? So, the graph of the quadratic is a parabola that opens upwards and stays entirely above the x-axis. First, I remember that for a quadratic equation ax² + bx + c, the discriminant is b² - 4ac. If the discriminant is negative, the quadratic doesn't have real roots, which means it doesn't cross the x-axis. Since the coefficient of x² here is 1, which is positive, the parabola opens upwards. So, if the discriminant is negative, the quadratic is always positive. That's exactly what we need here. So, the condition should be that the discriminant is less than zero.Let me write that down. The discriminant D is n² - 4*1*100. So, D = n² - 400. We want D < 0. So, n² - 400 < 0. Solving for n, that would be n² < 400. Taking square roots on both sides, |n| < √400. √400 is 20, so |n| < 20. Which means that n is between -20 and 20, but not including -20 and 20 themselves. But wait, the question asks for integers n. So, n must be an integer. Therefore, the integers between -20 and 20, not including -20 and 20. Let's see, the integers from -19 up to 19. How many integers is that?Well, from -19 to 19 inclusive. Let's count. If we consider from -19 to 0, that's 20 numbers (including 0), and from 1 to 19, that's 19 numbers. Wait, no: from -19 to -1 is 19 numbers, plus 0 is 1, then 1 to 19 is 19. So total is 19 + 1 + 19 = 39. Wait, is that right?Alternatively, the number of integers between a and b inclusive is b - a + 1. But here, from -19 to 19. So, 19 - (-19) + 1 = 19 + 19 + 1 = 39. Yes, that's 39 integers. But wait, if n is allowed to be -19 up to 19, but not including -20 and 20. Wait, but |n| < 20, so n must be greater than -20 and less than 20. So, n can be -19, -18, ..., 18, 19. So, how many integers is that?Let me list them out. Starting from -19, the next integer is -18, ..., up to 19. So, from -19 to 19 inclusive. The number of integers in this range is (19 - (-19)) + 1 = 39 + 1 = 40? Wait, no. Wait, let me think again. The number of integers from a to b inclusive is b - a + 1. So, from -19 to 19, that's 19 - (-19) + 1 = 19 + 19 + 1 = 39. Hmm. Wait, but if I count from 1 to n, there are n numbers, but when dealing with negative numbers, the formula still holds. For example, from -2 to 2 inclusive: 2 - (-2) +1 = 5, which is correct: -2, -1, 0, 1, 2. So, yes, 5 numbers. Similarly, from -19 to 19: 19 - (-19) +1 = 39. So, 39 integers.But wait, let's check with a smaller example. Suppose the discriminant has to be less than 4, so |n| < 2. Then n can be -1, 0, 1. That's 3 integers. Which is (1 - (-1)) +1 = 3. So, yes, the formula works. So, in this case, n ranges from -19 to 19 inclusive, which is 39 integers. So, the answer should be 39. But wait, let me confirm again. The inequality is n² < 400. So, n must satisfy -20 < n < 20. Since n must be an integer, the integers in this interval are from -19 up to 19. So, how many? Let's count: starting at -19, each step increases by 1. The total number of integers is 19 (negative numbers) + 1 (zero) + 19 (positive numbers) = 39. Yes, that's 39. So, the answer is 39 integers. Therefore, the number of integers n is 39. But wait, hold on a second. Let me check once more. If n is an integer, and n must be strictly between -20 and 20. So, n can be -19, -18,...,18,19. So, that's 19 negative integers, 19 positive integers, and zero. So, 19 + 19 + 1 = 39. Yes. So, 39 integers. Therefore, the final answer should be 39. I think that's it. So, the number of integers n is 39.Final AnswerThe number of integers ( n ) is boxed{39}.</think>To determine how many integers ( n ) make the inequality ( x^2 + nx + 100 > 0 ) hold for all real numbers ( x ), we need the quadratic to be positive for all ( x ). This occurs when the quadratic has no real roots, meaning its discriminant must be negative.The discriminant ( D ) of the quadratic ( x^2 + nx + 100 ) is given by:[D = n^2 - 4 cdot 1 cdot 100 = n^2 - 400]We require ( D < 0 ), which leads to:[n^2 - 400 < 0 implies n^2 < 400]Taking the square root of both sides, we get:[|n| < 20]This implies ( -20 < n < 20 ). Since ( n ) must be an integer, the possible values of ( n ) are the integers from -19 to 19 inclusive.To count the number of integers in this range, we note that the integers from -19 to 19 inclusive are:- From -19 to -1: 19 integers- 0: 1 integer- From 1 to 19: 19 integersAdding these together, we get ( 19 + 1 + 19 = 39 ) integers.Thus, the number of integers ( n ) is (boxed{39}).]

🔑:## Solution.The discriminant of the quadratic function f(x)=x^{2}+n x+100 is D=n^{2}-400.Since the leading coefficient is positive, for the desired inequality to hold, it is necessary that D<0.Thus, the inequality n^{2}<400 must hold, i.e., |n|<20.Finally, the integers -19,-18, ldots,-1,0,1,2, ldots, 19 satisfy the condition of the problem, which totals 39.Note: For the last 1 point in the scoring scheme, it is not necessary to list the numbers n.

🧠:[<think>Okay, let's try to solve this problem step by step. It's about an ellipse, so I need to remember the standard equation and properties of an ellipse. Let me start with part (1), finding the equation of ellipse C. The problem states that the ellipse has the equation (frac{x^2}{a^2} + frac{y^2}{b^2} = 1) where (a > b > 0), with an eccentricity of (frac{sqrt{3}}{2}), and it passes through the point ((1, frac{sqrt{3}}{2})). I need to find (a) and (b). First, I recall that the eccentricity (e) of an ellipse is given by (e = frac{c}{a}), where (c) is the distance from the center to each focus. Also, the relationship between (a), (b), and (c) is (c^2 = a^2 - b^2). Given that the eccentricity (e = frac{sqrt{3}}{2}), we can write:[frac{c}{a} = frac{sqrt{3}}{2} implies c = frac{sqrt{3}}{2} a]Then, substituting (c) into the equation (c^2 = a^2 - b^2):[left( frac{sqrt{3}}{2} a right)^2 = a^2 - b^2 frac{3}{4} a^2 = a^2 - b^2 ]Subtracting (frac{3}{4}a^2) from both sides:[0 = a^2 - b^2 - frac{3}{4}a^2 0 = frac{1}{4}a^2 - b^2 implies b^2 = frac{1}{4}a^2 implies b = frac{a}{2}]So, (b) is half of (a). Now, the ellipse passes through the point ((1, frac{sqrt{3}}{2})). Let's plug this point into the ellipse equation to find (a) and (b).The standard ellipse equation is (frac{x^2}{a^2} + frac{y^2}{b^2} = 1). Substituting (x = 1) and (y = frac{sqrt{3}}{2}):[frac{1^2}{a^2} + frac{left( frac{sqrt{3}}{2} right)^2}{b^2} = 1 frac{1}{a^2} + frac{3/4}{b^2} = 1]But we already know that (b = frac{a}{2}), so (b^2 = frac{a^2}{4}). Substitute (b^2) into the equation:[frac{1}{a^2} + frac{3/4}{a^2/4} = 1 frac{1}{a^2} + frac{3/4 times 4}{a^2} = 1 frac{1}{a^2} + frac{3}{a^2} = 1 frac{4}{a^2} = 1 implies a^2 = 4 implies a = 2]Since (a > 0), we take the positive root. Then, (b = frac{a}{2} = frac{2}{2} = 1), so (b = 1).Therefore, the equation of the ellipse is:[frac{x^2}{4} + frac{y^2}{1} = 1]So that's part (1). Let me verify if this makes sense. The eccentricity (e = c/a), and (c = sqrt{a^2 - b^2} = sqrt{4 - 1} = sqrt{3}), so (e = sqrt{3}/2), which matches the given eccentricity. Also, plugging the point ((1, sqrt{3}/2)) into the equation:[frac{1}{4} + frac{3/4}{1} = frac{1}{4} + frac{3}{4} = 1]Yes, that works. So part (1) is done.Now, moving on to part (2): Point (P) moves on the ellipse, find the maximum value of (|PF_1| cdot |PF_2|).Hmm, (F_1) and (F_2) are the left and right focal points. Since the ellipse is centered at the origin (as the standard equation is given with x and y terms only), the foci are located at ((pm c, 0)). From part (1), we found (c = sqrt{3}), so the foci are at ((-sqrt{3}, 0)) and ((sqrt{3}, 0)).We need to find the maximum value of the product of the distances from a point (P) on the ellipse to each of the foci. Let me denote (d_1 = |PF_1|) and (d_2 = |PF_2|). We need to maximize (d_1 cdot d_2).First, let me recall that for any point on the ellipse, the sum of the distances to the foci is constant and equal to (2a). In this case, (a = 2), so (d_1 + d_2 = 4).But we need the product (d_1 d_2). To maximize this product given that the sum is constant, I can use the AM-GM inequality, which states that the product is maximized when the two numbers are equal. However, in this case, can (d_1) and (d_2) be equal?If (d_1 = d_2), then since (d_1 + d_2 = 4), each would be 2. But for an ellipse, the only points where the distances to the foci are equal are the points along the minor axis. Let me check.Wait, no. The minor axis points are equidistant from both foci. Let me see: the endpoints of the minor axis are at ((0, pm b)), which in this case is ((0, pm 1)). Let's compute (d_1) and (d_2) for point ((0,1)).Distance to (F_1 = (-sqrt{3}, 0)):[sqrt{(0 + sqrt{3})^2 + (1 - 0)^2} = sqrt{3 + 1} = sqrt{4} = 2]Similarly, distance to (F_2 = (sqrt{3}, 0)) is also 2. So indeed, at point ((0,1)), (d_1 = d_2 = 2), so the product is 4. But is this the maximum?Wait, let's test another point. For example, the right vertex at ((2,0)). Then, distance to (F_1) is (|2 - (-sqrt{3})| = 2 + sqrt{3}), and distance to (F_2) is (|2 - sqrt{3}|). The product would be:[(2 + sqrt{3})(2 - sqrt{3}) = 4 - 3 = 1]Which is less than 4. Similarly, at the left vertex ((-2,0)), distances would be the same, product is 1. What about another point, say ((1, sqrt{3}/2)), which is the given point.Compute (d_1) and (d_2):Coordinates of (P): (1, (sqrt{3}/2))Distance to (F_1 = (-sqrt{3}, 0)):[sqrt{(1 + sqrt{3})^2 + left( frac{sqrt{3}}{2} - 0 right)^2}]Let's compute this:First, expand ((1 + sqrt{3})^2 = 1 + 2sqrt{3} + 3 = 4 + 2sqrt{3})Then, (left( frac{sqrt{3}}{2} right)^2 = frac{3}{4})So total distance squared: (4 + 2sqrt{3} + frac{3}{4} = frac{16}{4} + frac{8sqrt{3}}{4} + frac{3}{4} = frac{19}{4} + 2sqrt{3})Similarly, distance to (F_2 = (sqrt{3}, 0)):[sqrt{(1 - sqrt{3})^2 + left( frac{sqrt{3}}{2} right)^2}]Compute ((1 - sqrt{3})^2 = 1 - 2sqrt{3} + 3 = 4 - 2sqrt{3})Add (frac{3}{4}): total distance squared is (4 - 2sqrt{3} + frac{3}{4} = frac{19}{4} - 2sqrt{3})So, the distances squared are (frac{19}{4} + 2sqrt{3}) and (frac{19}{4} - 2sqrt{3}). Let me compute the product:[left( sqrt{frac{19}{4} + 2sqrt{3}} right) cdot left( sqrt{frac{19}{4} - 2sqrt{3}} right) = sqrt{ left( frac{19}{4} right)^2 - (2sqrt{3})^2 } ]Calculating the expression under the square root:[left( frac{19}{4} right)^2 - (2sqrt{3})^2 = frac{361}{16} - 12 = frac{361}{16} - frac{192}{16} = frac{169}{16}]Therefore, the product is (sqrt{frac{169}{16}} = frac{13}{4} = 3.25), which is 3.25, less than 4. So the product at that point is indeed less than 4. So maybe the maximum occurs at the minor axis points. But let me check another point to be sure. Let's take a general point on the ellipse.Parametrize the ellipse. Since it's an ellipse with major axis along the x-axis, we can write parametric equations as:[x = a cos theta = 2 cos theta y = b sin theta = sin theta]Where (theta) is the parameter. Then, for any point (P) on the ellipse, its coordinates are ((2 cos theta, sin theta)).Compute (d_1 = |PF_1|) and (d_2 = |PF_2|):(F_1 = (-sqrt{3}, 0)), (F_2 = (sqrt{3}, 0)).Distance from (P) to (F_1):[sqrt{(2 cos theta + sqrt{3})^2 + (sin theta - 0)^2}]Similarly, distance to (F_2):[sqrt{(2 cos theta - sqrt{3})^2 + (sin theta)^2}]So, the product (d_1 d_2) is:[sqrt{[(2 cos theta + sqrt{3})^2 + sin^2 theta][(2 cos theta - sqrt{3})^2 + sin^2 theta]}]This looks complicated. Let's try to simplify the expressions inside the square roots first.First, expand the terms inside each distance squared:For (d_1^2):[(2 cos theta + sqrt{3})^2 + sin^2 theta = 4 cos^2 theta + 4 sqrt{3} cos theta + 3 + sin^2 theta]Similarly, (d_2^2):[(2 cos theta - sqrt{3})^2 + sin^2 theta = 4 cos^2 theta - 4 sqrt{3} cos theta + 3 + sin^2 theta]Notice that these expressions are similar except for the middle term. Let me denote (A = 4 cos^2 theta + 3 + sin^2 theta), then:(d_1^2 = A + 4 sqrt{3} cos theta)(d_2^2 = A - 4 sqrt{3} cos theta)Therefore, the product (d_1^2 d_2^2 = (A + 4 sqrt{3} cos theta)(A - 4 sqrt{3} cos theta) = A^2 - (4 sqrt{3} cos theta)^2)So, (d_1 d_2 = sqrt{A^2 - (4 sqrt{3} cos theta)^2})First, compute (A):(A = 4 cos^2 theta + 3 + sin^2 theta)Combine the cosine and sine terms:(4 cos^2 theta + sin^2 theta = 3 cos^2 theta + (cos^2 theta + sin^2 theta) = 3 cos^2 theta + 1)Therefore, (A = 3 cos^2 theta + 1 + 3 = 3 cos^2 theta + 4). Wait, wait, let's check that again.Wait, original expression:(A = 4 cos^2 theta + 3 + sin^2 theta)Break down:4 cos²θ + sin²θ + 3= 3 cos²θ + (cos²θ + sin²θ) + 3= 3 cos²θ + 1 + 3= 3 cos²θ + 4Yes, correct. So (A = 3 cos^2 theta + 4). Then,(d_1 d_2 = sqrt{(3 cos^2 theta + 4)^2 - (4 sqrt{3} cos theta)^2})Compute the expression under the square root:First, expand ((3 cos^2 theta + 4)^2):= 9 cos⁴θ + 24 cos²θ + 16Then subtract ((4 sqrt{3} cos theta)^2 = 16 * 3 cos²θ = 48 cos²θSo:9 cos⁴θ + 24 cos²θ + 16 - 48 cos²θ= 9 cos⁴θ - 24 cos²θ + 16Therefore, (d_1 d_2 = sqrt{9 cos^4 theta - 24 cos^2 theta + 16})Hmm, this is still a bit complicated. Let me see if I can factor this expression or find a maximum in another way.Alternatively, maybe instead of parametrizing by theta, I can use coordinates. Let me consider using coordinates (x, y) on the ellipse. Then, the product |PF1| * |PF2| can be written as:[sqrt{(x + c)^2 + y^2} cdot sqrt{(x - c)^2 + y^2}]Where (c = sqrt{3}). Let me square the product to simplify:[[(x + c)^2 + y^2][(x - c)^2 + y^2]]Expanding each term:First, expand ((x + c)^2 + y^2 = x^2 + 2xc + c^2 + y^2)Similarly, ((x - c)^2 + y^2 = x^2 - 2xc + c^2 + y^2)Multiply the two:[(x^2 + 2xc + c^2 + y^2)(x^2 - 2xc + c^2 + y^2)]This is of the form ((A + B)(A - B) = A^2 - B^2), where (A = x^2 + c^2 + y^2) and (B = 2xc)Therefore, the product becomes:[(x^2 + c^2 + y^2)^2 - (2xc)^2]Compute each part:First, (x^2 + y^2 + c^2). Then square it:((x^2 + y^2 + c^2)^2 = x^4 + 2x^2 y^2 + y^4 + 2x^2 c^2 + 2y^2 c^2 + c^4)Subtract ((2xc)^2 = 4x^2 c^2):So, the entire expression is:(x^4 + 2x^2 y^2 + y^4 + 2x^2 c^2 + 2y^2 c^2 + c^4 - 4x^2 c^2)Simplify:Combine the (x^2 c^2) terms: (2x^2 c^2 - 4x^2 c^2 = -2x^2 c^2)So,(x^4 + 2x^2 y^2 + y^4 - 2x^2 c^2 + 2y^2 c^2 + c^4)Hmm, not sure if this helps. Maybe a better approach.Since the ellipse equation is (frac{x^2}{4} + y^2 = 1), so (y^2 = 1 - frac{x^2}{4}). Let's substitute (y^2) into the expression for the product.The product squared is:[[(x + sqrt{3})^2 + y^2][(x - sqrt{3})^2 + y^2]]Substitute (y^2 = 1 - frac{x^2}{4}):First, compute each bracket:For the first bracket:[(x + sqrt{3})^2 + y^2 = x^2 + 2xsqrt{3} + 3 + 1 - frac{x^2}{4} = x^2 + 2xsqrt{3} + 4 - frac{x^2}{4}]Simplify:Combine x² terms: (x^2 - frac{x^2}{4} = frac{3x^2}{4})So, first bracket becomes: (frac{3x^2}{4} + 2xsqrt{3} + 4)Similarly, the second bracket:[(x - sqrt{3})^2 + y^2 = x^2 - 2xsqrt{3} + 3 + 1 - frac{x^2}{4} = x^2 - 2xsqrt{3} + 4 - frac{x^2}{4}]Simplify similarly: (frac{3x^2}{4} - 2xsqrt{3} + 4)Now, multiply these two expressions:[left( frac{3x^2}{4} + 2xsqrt{3} + 4 right) left( frac{3x^2}{4} - 2xsqrt{3} + 4 right)]Again, this is of the form ((M + N)(M - N) = M^2 - N^2), where (M = frac{3x^2}{4} + 4) and (N = 2xsqrt{3})Therefore, the product becomes:[left( frac{3x^2}{4} + 4 right)^2 - (2xsqrt{3})^2]Compute each part:First, (left( frac{3x^2}{4} + 4 right)^2 = frac{9x^4}{16} + 2 * frac{3x^2}{4} * 4 + 16 = frac{9x^4}{16} + 6x^2 + 16)Then, subtract ((2xsqrt{3})^2 = 4x^2 * 3 = 12x^2)So:[frac{9x^4}{16} + 6x^2 + 16 - 12x^2 = frac{9x^4}{16} - 6x^2 + 16]Therefore, the square of the product (d_1 d_2) is (frac{9x^4}{16} - 6x^2 + 16), so:[d_1 d_2 = sqrt{ frac{9x^4}{16} - 6x^2 + 16 }]To find the maximum of this expression, we can find the maximum of the expression inside the square root, let's denote it as (Q(x)):[Q(x) = frac{9x^4}{16} - 6x^2 + 16]To maximize (Q(x)), take derivative with respect to x and set to zero.First, compute derivative:(Q'(x) = frac{9 * 4 x^3}{16} - 12x = frac{36x^3}{16} - 12x = frac{9x^3}{4} - 12x)Set equal to zero:[frac{9x^3}{4} - 12x = 0 x left( frac{9x^2}{4} - 12 right) = 0 ]Solutions are (x = 0) or (frac{9x^2}{4} - 12 = 0 implies frac{9x^2}{4} = 12 implies x^2 = frac{48}{9} = frac{16}{3} implies x = pm frac{4}{sqrt{3}})But wait, the ellipse has (x^2 leq 4) (since (a = 2)), so (x = pm frac{4}{sqrt{3}} approx pm 2.309), which is outside the ellipse's x-range. Therefore, those solutions are extraneous. Therefore, the critical points within the ellipse are only at (x = 0).Therefore, maximum and minima occur either at (x = 0) or at the endpoints (x = pm 2). Let's evaluate (Q(x)) at these points.First, at (x = 0):[Q(0) = 0 - 0 + 16 = 16]Therefore, (d_1 d_2 = sqrt{16} = 4), which matches our earlier calculation for the minor axis points.At (x = pm 2):[Q(2) = frac{9*(16)}{16} - 6*(4) + 16 = 9 - 24 + 16 = 1]So, (d_1 d_2 = sqrt{1} = 1), which is the product at the vertices, as we saw before.Therefore, the maximum value of (d_1 d_2) is 4, achieved at (x = 0), i.e., points ((0, pm 1)).But wait, just to make sure, let's check if the critical point at (x = pm frac{4}{sqrt{3}}) is outside the ellipse. Since (x^2 = frac{16}{3} approx 5.333), which is greater than (a^2 = 4), so yes, outside the ellipse. Therefore, the maximum within the ellipse is indeed at (x = 0).Alternatively, since we parametrized with theta earlier, perhaps I can use calculus with theta.Recall that we had:(d_1 d_2 = sqrt{9 cos^4 theta - 24 cos^2 theta + 16})Let me denote (u = cos^2 theta), so that (u in [0,1]), since (theta) is real.Then the expression under the square root becomes:(9u^2 - 24u + 16)We need to find the maximum of this quadratic in (u) over (u in [0,1]).But since the quadratic is (9u^2 -24u +16), which is a parabola opening upwards (coefficient of (u^2) is positive). The vertex is at (u = -b/(2a) = 24/(2*9) = 24/18 = 4/3 ≈ 1.333). But (u) is only in [0,1], so the minimum occurs at the vertex, but since the parabola is increasing on [4/3, ∞), but decreasing on (-∞, 4/3). Therefore, on [0,1], the maximum occurs at the endpoints.Compute the value at u=0:(9*0 -24*0 +16 = 16)At u=1:(9*1 -24*1 +16 = 9 -24 +16 = 1)Therefore, the maximum is 16 at u=0, which corresponds to (cos^2 theta = 0 implies theta = pi/2) or (3pi/2), i.e., the points (0,1) and (0,-1). So indeed, the maximum value is 16 under the square root, hence 4. Therefore, confirming our previous result.Thus, the maximum value of (|PF_1| cdot |PF_2|) is 4.But wait, let me check once more. If the product of distances is maximized at the minor axis, but intuitively, when you move towards a focus, the distance to that focus decreases while the other increases. The product might have a maximum somewhere in between. But according to the calculations, the maximum is at the minor axis. Let me think.Wait, but if we think of the product (d_1 d_2), and we know that (d_1 + d_2 = 4), then using AM ≥ GM:[frac{d_1 + d_2}{2} geq sqrt{d_1 d_2} 2 geq sqrt{d_1 d_2} implies d_1 d_2 leq 4]Equality holds when (d_1 = d_2 = 2). Therefore, by AM-GM inequality, the maximum product is 4, achieved when (d_1 = d_2 = 2). Therefore, this confirms our result. So even though it seems counterintuitive, the maximum product occurs when the two distances are equal, which happens at the top and bottom of the ellipse.Therefore, the answer to part (2) is 4.Final Answer(1) The equation of ellipse (C) is (boxed{dfrac{x^2}{4} + y^2 = 1}).(2) The maximum value of (|PF_{1}| cdot |PF_{2}|) is (boxed{4}).</think>Given an ellipse (C): (frac{x^2}{a^2} + frac{y^2}{b^2} = 1) with an eccentricity of (frac{sqrt{3}}{2}) and passing through the point ((1, frac{sqrt{3}}{2})):1. Finding the equation of ellipse (C): - The eccentricity (e = frac{sqrt{3}}{2}) gives (frac{c}{a} = frac{sqrt{3}}{2}), leading to (c = frac{sqrt{3}}{2}a). - Using the relationship (c^2 = a^2 - b^2), we substitute (c) to get (left(frac{sqrt{3}}{2}aright)^2 = a^2 - b^2), which simplifies to (frac{3}{4}a^2 = a^2 - b^2). Solving this gives (b^2 = frac{1}{4}a^2), so (b = frac{a}{2}). - Substituting the point ((1, frac{sqrt{3}}{2})) into the ellipse equation, we solve for (a) and (b). This results in (a = 2) and (b = 1). Thus, the equation of the ellipse is (frac{x^2}{4} + y^2 = 1).2. Finding the maximum value of (|PF_1| cdot |PF_2|): - The foci (F_1) and (F_2) are at ((-sqrt{3}, 0)) and ((sqrt{3}, 0)) respectively. - For any point (P) on the ellipse, the sum of distances to the foci is (2a = 4). Using the AM-GM inequality, the product (|PF_1| cdot |PF_2|) is maximized when (|PF_1| = |PF_2| = 2). - This occurs at the points ((0, pm 1)) on the minor axis, leading to the maximum product value of (4). Final Answer(1) The equation of ellipse (C) is (boxed{dfrac{x^2}{4} + y^2 = 1}).(2) The maximum value of (|PF_{1}| cdot |PF_{2}|) is (boxed{4}).,<think>Alright, so I have this problem about an ellipse. Let me try to tackle it step by step. First, part (1) is asking to find the equation of ellipse C. The given equation is the standard form of an ellipse: x²/a² + y²/b² = 1, where a > b > 0. The ellipse has an eccentricity of √3/2 and passes through the point (1, √3/2). Okay, so let's recall that the eccentricity e of an ellipse is given by e = c/a, where c is the distance from the center to each focus, and c² = a² - b². Here, the eccentricity is √3/2, so that gives us a relationship between a and c. Since e = c/a = √3/2, that means c = (√3/2)a. Then, using the relationship c² = a² - b², we can substitute c:c² = (3/4)a² = a² - b². Therefore, solving for b², we get b² = a² - (3/4)a² = (1/4)a². So, b = (1/2)a. Now, we also know that the ellipse passes through the point (1, √3/2). So we can substitute x=1 and y=√3/2 into the ellipse equation to find a² and b². Let's do that:(1)²/a² + (√3/2)²/b² = 1. Simplify:1/a² + (3/4)/b² = 1.But we already found that b² = (1/4)a², so substitute that into the equation:1/a² + (3/4)/( (1/4)a² ) = 1. Let's compute the second term:(3/4) divided by (1/4)a² is equal to (3/4) * (4/a²) = 3/a². Therefore, the equation becomes:1/a² + 3/a² = 4/a² = 1. Therefore, 4/a² = 1 => a² = 4 => a = 2. Since a must be positive, a = 2.Then, since b² = (1/4)a² = (1/4)*4 = 1. Therefore, b² = 1, so b = 1. Therefore, the equation of the ellipse is x²/4 + y²/1 = 1, or written more neatly, x²/4 + y² = 1. Let me check if that works with the given point (1, √3/2). Plugging in x=1 and y=√3/2:1²/4 + ( (√3)/2 )² = 1/4 + (3/4) = 1. Yes, that works out. And the eccentricity: c = √3/2 * a = √3/2 * 2 = √3. Then c² = 3, and a² - b² = 4 - 1 = 3, which checks out. So part (1) is done. The equation is x²/4 + y² = 1.Now part (2): Point P moves on the ellipse, find the maximum value of |PF1| · |PF2|. Hmm. So F1 and F2 are the left and right focal points. Since the ellipse is centered at the origin (because the equation is x²/a² + y²/b² = 1), the foci are at (-c, 0) and (c, 0). From part (1), we found c = √3, so F1 is (-√3, 0) and F2 is (√3, 0). We need to find the maximum value of the product of the distances from a moving point P on the ellipse to each of the foci. Let me think about how to approach this. Maybe parametrize the ellipse and express the distances in terms of the parameter, then find the maximum. Alternatively, use coordinates (x,y) on the ellipse and express the product in terms of x and y, then maximize it.First, let's recall that any point P on the ellipse can be represented parametrically as (2 cos θ, sin θ), since a = 2 and b = 1. So x = 2 cos θ, y = sin θ. Then the distances |PF1| and |PF2| can be written in terms of θ. Let me compute those distances.First, F1 is (-√3, 0), F2 is (√3, 0). For a point P = (2 cos θ, sin θ), the distance to F1 is sqrt[ (2 cos θ + √3)^2 + (sin θ - 0)^2 ], and similarly for F2: sqrt[ (2 cos θ - √3)^2 + sin² θ ].But multiplying these two distances would be complicated. Maybe we can square the product? Wait, but the problem asks for the maximum value of |PF1| · |PF2|. Alternatively, maybe there's a smarter way.Alternatively, since P is on the ellipse, perhaps we can use properties of ellipses. Remember that for an ellipse, the sum of distances from any point on the ellipse to the two foci is constant: |PF1| + |PF2| = 2a. But here, we need the product. Hmm. Maybe express the product in terms of the sum and something else?Alternatively, think in coordinates. Let me consider |PF1| · |PF2| as expressions in terms of x and y. Let's write that down.Let P = (x, y). Then |PF1| = sqrt( (x + √3)^2 + y^2 ), |PF2| = sqrt( (x - √3)^2 + y^2 ). So their product is sqrt[ ( (x + √3)^2 + y^2 ) ( (x - √3)^2 + y^2 ) ].But that seems messy. Maybe square the product to make it easier: (|PF1| · |PF2|)^2 = [ (x + √3)^2 + y^2 ][ (x - √3)^2 + y^2 ].Let me compute this expression. Let's denote A = (x + √3)^2 + y^2 and B = (x - √3)^2 + y^2. Then AB = [ (x + √3)^2 + y^2 ][ (x - √3)^2 + y^2 ].Alternatively, expand the product. Let's compute:First, expand (x + √3)^2 (x - √3)^2. That's [ (x^2 - (√3)^2 ) ]² = (x² - 3)^2.Then, cross terms. Wait, maybe it's better to compute AB directly:A = (x + √3)^2 + y^2 = x² + 2√3 x + 3 + y²B = (x - √3)^2 + y^2 = x² - 2√3 x + 3 + y²Multiply A and B:= (x² + 3 + y² + 2√3 x)(x² + 3 + y² - 2√3 x)= [ (x² + y² + 3 ) + 2√3 x ] [ (x² + y² + 3 ) - 2√3 x ]= (x² + y² + 3 )² - (2√3 x )²= (x² + y² + 3 )² - 12 x²Now, let's compute that.First, expand (x² + y² + 3)^2:= x^4 + 2x² y² + y^4 + 6x² + 6y² + 9Subtract 12x²:= x^4 + 2x² y² + y^4 + 6x² + 6y² + 9 - 12x²= x^4 + 2x² y² + y^4 - 6x² + 6y² + 9Hmm, not sure if this helps. Maybe substitute the ellipse equation into this expression.Given that the ellipse is x²/4 + y² = 1, so x² = 4(1 - y²) or y² = 1 - x²/4.Alternatively, maybe express x² in terms of y² or vice versa. Let's try substituting y² = 1 - x²/4 into the expression (x² + y² + 3)^2 - 12x².First, compute x² + y² + 3:x² + y² + 3 = x² + (1 - x²/4) + 3 = x² + 1 - x²/4 + 3 = (3x²/4) + 4.So, (x² + y² + 3)^2 = ( (3x²/4) + 4 )² = (3x²/4 + 4)^2.Then subtract 12x²:So AB = (3x²/4 + 4)^2 - 12x².Let's compute that:First expand (3x²/4 + 4)^2:= (3x²/4)^2 + 2*(3x²/4)*4 + 4^2= 9x^4/16 + 24x²/4 + 16= 9x^4/16 + 6x² + 16Subtract 12x²:= 9x^4/16 + 6x² + 16 - 12x²= 9x^4/16 - 6x² + 16So AB = 9x^4/16 - 6x² + 16Therefore, (|PF1| · |PF2|)^2 = 9x^4/16 - 6x² + 16.But we need to maximize |PF1| · |PF2|, which is equivalent to maximizing its square. So perhaps we can maximize AB = 9x^4/16 - 6x² + 16 over x such that P is on the ellipse, i.e., x²/4 + y² = 1. Since x² must be ≤ 4 (since a² =4), x ranges from -2 to 2.Alternatively, we can express AB in terms of x², let me let t = x². Then since x² ∈ [0,4], t ∈ [0,4].So AB becomes 9t²/16 - 6t + 16. Let's denote f(t) = 9t²/16 - 6t + 16. We need to find the maximum of f(t) for t ∈ [0,4], then take the square root to get the maximum |PF1| · |PF2|.To find the maximum of f(t), since it's a quadratic function in t, opening upwards (since the coefficient of t² is positive), its minimum is at the vertex, and the maximum occurs at one of the endpoints of the interval [0,4].Wait, that can't be. If the quadratic opens upwards, then it has a minimum at the vertex, and the maximum would be at the endpoints. Let me check.Compute f(0): 0 - 0 +16 =16f(4): 9*(16)/16 -6*4 +16 = 9 -24 +16= 1f(t) is a quadratic with vertex at t = -B/(2A) where A=9/16, B=-6.So t_vertex = 6/(2*(9/16)) = 6/(9/8) = 6*(8/9)= 48/9=16/3≈5.333, which is outside the interval [0,4]. Therefore, on the interval [0,4], the function f(t) is decreasing from t=0 to t=4 because the vertex is at t=16/3 >4. Therefore, the maximum of f(t) is at t=0, which is 16, and the minimum at t=4, which is 1. Wait, but when t=0, x²=0, so x=0, then y²=1. So point P is (0,1) or (0,-1). Let's compute |PF1|·|PF2| at (0,1):Distance to F1 (-√3,0): sqrt( (0 + √3)^2 + (1 -0)^2 ) = sqrt(3 +1)=2Distance to F2 (√3,0): same, sqrt( (0 - √3)^2 +1 )=sqrt(3 +1)=2Product: 2*2=4. Then (|PF1|·|PF2|)^2=16, which matches f(0)=16. So the square is 16, so the product is 4.At t=4, x²=4, so x=2 or -2, y=0. Then point P is (2,0). Compute |PF1|: distance to (-√3,0): |2 - (-√3)|=2 + √3. Similarly, |PF2|: |2 - √3|. Then product is (2 + √3)(2 - √3)=4 -3=1. Which matches f(4)=1. So sqrt(1)=1. So |PF1|·|PF2|=1.But we need the maximum value, which is 4. Wait, but this contradicts the quadratic analysis. Wait, but according to this, the maximum of AB is 16, so |PF1|·|PF2|=4. But when I check at point (0,1), the product is 4, but maybe there are other points where the product is higher?Wait, but according to the function f(t) = 9t²/16 -6t +16, which is (|PF1|·|PF2|)^2, the maximum is 16 at t=0, so the maximum product is 4. However, is this correct? Let me check another point.Suppose P is at (sqrt(2), sqrt(2)/2). Wait, does this point lie on the ellipse? Let's check: x²/4 + y² = (2)/4 + ( (2)/4 )= 0.5 +0.5=1. Yes. So this point is on the ellipse.Compute distances to F1 and F2:x=√2, y=√2/2|PF1| = sqrt( (√2 + √3)^2 + (√2/2)^2 )= sqrt( (√2 + √3)^2 + 0.5 )Similarly, |PF2| = sqrt( (√2 - √3)^2 + 0.5 )Compute the product:sqrt( ( (√2 + √3)^2 + 0.5 ) ( (√2 - √3)^2 + 0.5 ) )This seems complicated, but let's compute numerically.First, compute (√2 + √3)^2: 2 + 2√6 +3 =5 +2√6 ≈5 +4.898≈9.898. Then +0.5≈10.398. So |PF1|≈sqrt(10.398)≈3.225Similarly, (√2 - √3)^2: 2 - 2√6 +3=5 -4.898≈0.102. Then +0.5≈0.602. So |PF2|≈sqrt(0.602)≈0.776. Then the product is ≈3.225*0.776≈2.5. Which is less than 4. So maybe 4 is indeed the maximum. But wait, let's check another point. Maybe at some angle θ, the product is larger.Wait, but according to the previous analysis, the maximum is at t=0, which corresponds to x=0, y=±1. Let me check another point. Take θ=π/3. Then x=2 cos π/3=2*(1/2)=1, y=sin π/3=√3/2. Which is the given point (1, √3/2). Let's compute |PF1| and |PF2|.F1=(-√3,0), F2=(√3,0). Point P=(1,√3/2).|PF1|=sqrt( (1 + √3)^2 + (√3/2)^2 )= sqrt(1 + 2√3 + 3 + 3/4 )= sqrt(4 + 2√3 + 3/4 )= sqrt(4.75 + 2√3 )≈ sqrt(4.75 +3.464)=sqrt(8.214)≈2.866|PF2|=sqrt( (1 - √3)^2 + (√3/2)^2 )= sqrt(1 - 2√3 +3 +3/4 )= sqrt(4 -2√3 + 3/4 )≈ sqrt(4 -3.464 +0.75)=sqrt(1.286)≈1.134Product≈2.866*1.134≈3.25. Still less than 4.Wait, but according to the previous calculation, the maximum product is 4 at (0,1). Let me verify that.At (0,1), |PF1|=sqrt( (0 + √3)^2 + (1 -0)^2 )=sqrt(3 +1)=2. Similarly, |PF2|=sqrt( (0 - √3)^2 +1 )=sqrt(3 +1)=2. So product is 4.Is there any other point where the product is higher?Wait, but according to the function f(t) = 9t²/16 -6t +16, which is the square of the product, the maximum occurs at t=0, giving 16, so the product is 4. The next thought is, maybe 4 is indeed the maximum.But let me think again. Maybe there's another approach. Remember that in an ellipse, the product |PF1|·|PF2| can be expressed in terms of the sum and difference.We know that |PF1| + |PF2| = 2a =4 (since a=2). Let me denote S = |PF1| + |PF2| =4, and let D = |PF1| - |PF2|. Then, |PF1|·|PF2| = [(S)^2 - D^2]/4 = [16 - D^2]/4 =4 - D²/4.So to maximize |PF1|·|PF2|, we need to minimize D², i.e., minimize | |PF1| - |PF2| |.The minimal value of | |PF1| - |PF2| | occurs when |PF1| = |PF2|, which would be at the points where P is on the minor axis. Wait, but in an ellipse, the points where |PF1| = |PF2| are the points on the minor axis. Because at the minor axis, the distance to both foci is the same. Let's check that.At point (0,1), which is on the minor axis, |PF1| = |PF2|=2, as computed earlier, so product is 4. If P is on the major axis, say at (2,0), then |PF1|=2 + √3, |PF2|=2 - √3, product is (2 +√3)(2 -√3)=4 -3=1. So that's minimal. So as per the formula |PF1|·|PF2|=4 - D²/4, where D is | |PF1| - |PF2| |. So to maximize |PF1|·|PF2|, D should be as small as possible. The minimal D is 0, achieved at the minor axis, giving |PF1|·|PF2|=4. So the maximum value is 4.Therefore, the answer to part (2) is 4.But let me check another approach. Using parametric equations. Let me parametrize the ellipse as x=2 cosθ, y=sinθ. Then compute |PF1|·|PF2|.Compute |PF1|: sqrt( (2 cosθ + √3)^2 + (sinθ)^2 )Similarly, |PF2|: sqrt( (2 cosθ - √3)^2 + (sinθ)^2 )Multiply them:sqrt[ ( (2 cosθ + √3)^2 + sin²θ ) ( (2 cosθ - √3)^2 + sin²θ ) ]Let me compute the product inside the square root.Let me compute A = (2 cosθ + √3)^2 + sin²θ and B = (2 cosθ - √3)^2 + sin²θ.Compute A = 4 cos²θ + 4√3 cosθ +3 + sin²θ= (4 cos²θ + sin²θ) +4√3 cosθ +3Similarly, B = 4 cos²θ -4√3 cosθ +3 + sin²θ= (4 cos²θ + sin²θ) -4√3 cosθ +3Now, note that 4 cos²θ + sin²θ = 3 cos²θ + (cos²θ + sin²θ) = 3 cos²θ +1Therefore, A = 3 cos²θ +1 +4√3 cosθ +3 =3 cos²θ +4√3 cosθ +4Similarly, B =3 cos²θ +1 -4√3 cosθ +3 =3 cos²θ -4√3 cosθ +4Therefore, AB = (3 cos²θ +4√3 cosθ +4)(3 cos²θ -4√3 cosθ +4)This looks like (M + N)(M - N) where M=3 cos²θ +4 and N=4√3 cosθTherefore, AB = (3 cos²θ +4)^2 - (4√3 cosθ)^2Compute that:=9 cos^4θ +24 cos²θ +16 - 16*3 cos²θ=9 cos^4θ +24 cos²θ +16 -48 cos²θ=9 cos^4θ -24 cos²θ +16Therefore, (|PF1|·|PF2|)^2 =9 cos^4θ -24 cos²θ +16Let me denote t = cos²θ. Then t ∈ [0,1] since cosθ ranges between -1 and 1. So the expression becomes 9t² -24t +16. Let's call this f(t)=9t² -24t +16.We need to find the maximum of sqrt(f(t)) for t ∈ [0,1]. The maximum of sqrt(f(t)) occurs at the maximum of f(t). So let's find the maximum of f(t) on [0,1].f(t)=9t² -24t +16. This is a quadratic function in t, opening upwards (since coefficient of t² is positive). The vertex is at t=24/(2*9)=24/18=4/3≈1.333. But the interval is [0,1], so the vertex is outside the interval. Therefore, the function is decreasing on [0,1]. Therefore, maximum at t=0: f(0)=0 -0 +16=16. Minimum at t=1: f(1)=9 -24 +16=1.Therefore, maximum value of |PF1|·|PF2| is sqrt(16)=4, which occurs when t=0, i.e., cosθ=0, so θ=π/2 or 3π/2, which corresponds to points (0,1) and (0,-1). This matches our previous conclusion.Therefore, the maximum value is 4.So part (2)'s answer is 4.Final Answer(1) The equation of ellipse ( C ) is boxed{dfrac{x^2}{4} + y^2 = 1}.(2) The maximum value of ( |PF_{1}| cdot |PF_{2}| ) is boxed{4}.</think>Given an ellipse ( C ): (frac{x^{2}}{a^{2}} + frac{y^{2}}{b^{2}} = 1 ) with an eccentricity of (frac{sqrt{3}}{2}) and passing through the point ((1, frac{sqrt{3}}{2})):(1) To find the equation of the ellipse ( C ):- The eccentricity ( e = frac{sqrt{3}}{2} ) gives ( c = frac{sqrt{3}}{2}a ).- Using ( c^2 = a^2 - b^2 ), we substitute ( c ) to get ( left(frac{sqrt{3}}{2}aright)^2 = a^2 - b^2 ), which simplifies to ( frac{3}{4}a^2 = a^2 - b^2 ). Solving for ( b^2 ), we find ( b^2 = frac{1}{4}a^2 ).- The ellipse passes through the point ((1, frac{sqrt{3}}{2})). Substituting ( x = 1 ) and ( y = frac{sqrt{3}}{2} ) into the ellipse equation, we get ( frac{1}{a^2} + frac{3}{4b^2} = 1 ). Substituting ( b^2 = frac{1}{4}a^2 ), we solve to find ( a^2 = 4 ) and ( b^2 = 1 ).- Therefore, the equation of the ellipse is (frac{x^2}{4} + y^2 = 1).(2) To find the maximum value of ( |PF_1| cdot |PF_2| ):- The foci ( F_1 ) and ( F_2 ) are at ((- sqrt{3}, 0)) and ((sqrt{3}, 0)) respectively.- Using the parametric form of the ellipse ( x = 2 cos theta ) and ( y = sin theta ), we compute the distances ( |PF_1| ) and ( |PF_2| ).- The product ( |PF_1| cdot |PF_2| ) is maximized when the point ( P ) is on the minor axis, leading to the maximum product of 4. Final Answer(1) The equation of ellipse ( C ) is (boxed{dfrac{x^2}{4} + y^2 = 1}).(2) The maximum value of ( |PF_{1}| cdot |PF_{2}| ) is (boxed{4}).]

🔑:(1) According to the given conditions, we have the system of equations:begin{cases}frac{c}{a} = frac{sqrt{3}}{2} frac{1}{a^{2}} + frac{3}{4b^{2}} = 1 a^{2} = b^{2} + c^{2}end{cases}Solving the system, we obtain:begin{cases}a = 2 b = 1 c = sqrt{3}end{cases}Hence, the equation of ellipse C is frac{x^{2}}{4} + y^{2} = 1.(2) Since point P moves on the ellipse,|PF_{1}| + |PF_{2}| = 2a = 4.Applying the AM-GM inequality, we have:|PF_{1}| cdot |PF_{2}| leqslant left( frac{|PF_{1}| + |PF_{2}|}{2} right)^{2} = left( frac{4}{2} right)^{2} = 4.The equality holds if and only if |PF_{1}| = |PF_{2}|.Hence, the maximum value of |PF_{1}| cdot |PF_{2}| is boxed{4}.

🧠:[<think>Okay, let me try to work through this problem step by step. The question is about the angle formed by a straight line and a plane. The options are given as (A) acute, (B) a first quadrant angle, (C) 0° ≤ α ≤ 90°, and (D) 0° < α ≤ 90°. I need to figure out which one is correct.First, I remember that when dealing with angles between a line and a plane, there's a specific way to define that angle. If I recall correctly, the angle between a line and a plane is the complement of the angle between the line and the normal to the plane. Wait, let me make sure. So, if we have a straight line intersecting a plane, the angle between them is actually the smallest angle that the line makes with any line on the plane. But maybe that's not exactly right. Let me think.Another approach: In 3D geometry, the angle between a line and a plane is defined as the angle between the line and its orthogonal projection onto the plane. This angle would be the smallest angle between the original line and any line lying on the plane. Since this is the case, the angle should be between 0° and 90°, inclusive. Because if the line is lying on the plane, the angle would be 0°, right? And if the line is perpendicular to the plane, then the angle between the line and the plane would be 90°. So, the angle can't be more than 90°, and it can be as low as 0°, which would mean the line is on the plane.But wait, the options given are (C) 0° ≤ α ≤ 90° and (D) 0° < α ≤ 90°. The difference here is whether the angle can be exactly 0° or not. If the line is lying on the plane, then the angle between the line and the plane is 0°, so that would make (C) correct. However, maybe there's a definition that excludes 0°, but I don't think so. In standard definitions, the angle between a line and a plane can be 0°, which occurs when the line is on the plane. So, perhaps (C) is the correct answer.But let me check again. Let me recall the precise definition. The angle between a line and a plane is the complement of the angle between the line and the normal to the plane. So, if θ is the angle between the line and the normal, then the angle φ between the line and the plane is 90° - θ. Since θ can range from 0° to 90°, φ would range from 90° down to 0°, right? Wait, but that might not be the case. Let me think again.If θ is the angle between the line and the normal, then θ can be between 0° and 90°, so φ = 90° - θ would be between 0° and 90° as well. For example, if the line is perpendicular to the normal (i.e., lying on the plane), then θ = 90°, so φ = 0°. If the line is parallel to the normal (i.e., perpendicular to the plane), then θ = 0°, so φ = 90°. Therefore, φ, the angle between the line and the plane, is indeed between 0° and 90°, inclusive. So that would make option (C) correct.But why is option (D) there? Maybe some sources define the angle between a line and a plane as strictly greater than 0°, but that would be if they consider only lines that are not lying on the plane. But in standard definitions, if the line is on the plane, the angle is 0°, so (C) should include 0°. Therefore, (C) is the correct answer.Wait, but let me check a textbook reference. For example, according to some geometry references, the angle between a line and a plane is defined as the smallest angle between the line and its projection onto the plane. If the line is on the plane, its projection is itself, so the angle is 0°. If the line is perpendicular to the plane, the projection is a single point, so the angle is 90°. Therefore, 0° ≤ α ≤ 90°, so (C) is correct.So, the answer should be (C). But let me see the options again. The options are (A) acute; which would mean between 0° and 90°, but not including 0° or 90°, but that's not necessarily true because the angle can be 0° or 90°. (B) a first quadrant angle; first quadrant angles are between 0° and 90°, but not including 0° and 90°, so that's similar to acute, but in quadrant terms. However, in terms of angles, first quadrant angle might not be a standard term here. Then (C) and (D) specify the ranges, with (C) including 0° and (D) excluding it. Since we established that 0° is possible when the line is on the plane, (C) is the correct answer.Therefore, I think the correct answer is (C).Final Answerboxed{C}</think>To determine the angle formed by a straight line and a plane, we need to consider the standard definition in 3D geometry. The angle between a line and a plane is defined as the angle between the line and its orthogonal projection onto the plane. This angle is the smallest angle between the line and any line on the plane. If the line is lying on the plane, the angle is 0°. If the line is perpendicular to the plane, the angle is 90°. Therefore, the angle must be between 0° and 90°, inclusive. To verify, we also consider the complement of the angle between the line and the normal to the plane. If θ is the angle between the line and the normal, then the angle φ between the line and the plane is 90° - θ. Since θ can range from 0° to 90°, φ ranges from 0° to 90°, inclusive. Thus, the correct answer is (0^circ leqslant alpha leqslant 90^circ), which corresponds to option (C).boxed{C},<think>Okay, so I have this problem here about the angle formed by a straight line and a plane. The question is asking which of the options A, B, C, or D is correct. Let me try to recall what I know about angles between lines and planes.First, I remember that when you have a straight line and a plane, the angle between them isn't just any angle. If the line is lying on the plane, then the angle should be zero because they're kind of "touching" each other completely. But if the line is intersecting the plane at some other angle, how is that measured?Wait, the angle between a line and a plane is defined as the complement of the angle between the line and the normal to the plane. Hmm, so if the line is perpendicular to the plane, then the angle between the line and the plane would be 90 degrees minus 90 degrees, which is 0 degrees. But that seems contradictory because if the line is perpendicular, the angle should be 90 degrees. Maybe I got that mixed up.Let me check again. The angle between a line and a plane is the smallest angle between the line and its projection onto the plane. Oh, right! So if the line is perpendicular to the plane, its projection onto the plane is just a point, so the angle between the line and the plane is 90 degrees. If the line is lying on the plane, then the angle is 0 degrees. So the angle should range from 0 degrees up to 90 degrees. So that would mean the angle is between 0 and 90 degrees inclusive.Looking back at the options, option C says 0° ≤ α ≤ 90°, and option D says 0° < α ≤ 90°. The difference is whether it includes 0° or not. If the line lies on the plane, then the angle is exactly 0°, so 0° should be included. Therefore, option C would be correct.But wait, let me make sure. If the line is in the plane, is the angle considered 0°, or is there a different interpretation? I think according to the definition, since the angle is measured between the line and its projection on the plane. If the line is already on the plane, its projection is itself, so the angle between them is 0°. So 0° is possible, which makes option C correct. However, some sources might argue that the angle is measured when the line is not on the plane, so maybe 0° is excluded. But according to standard definitions, I believe 0° is included when the line is lying on the plane. So the correct answer should be C.But wait, let me check another perspective. In 3D geometry, when they talk about the angle between a line and a plane, they usually consider the smallest angle between the line and any line on the plane. If the line is on the plane, that angle is 0°, which is allowed. So yes, 0° is a valid measure here. So option C is the right answer.But wait, the options are (C) 0° ≤ α ≤ 90°, (D) 0° < α ≤ 90°. So the difference is whether 0° is included. Since the angle can be 0° when the line is in the plane, option C must be correct. Therefore, the answer is C.Final Answerboxed{C}</think>The angle formed by a straight line and a plane is defined as the smallest angle between the line and its projection onto the plane. If the line is lying on the plane, its projection is itself, resulting in an angle of (0^circ). If the line is perpendicular to the plane, the angle is (90^circ). Therefore, the angle must be between (0^circ) and (90^circ), inclusive. Given the options:- (C) (0^circ leqslant alpha leqslant 90^circ)- (D) (0^circ < alpha leqslant 90^circ)Since (0^circ) is included when the line lies on the plane, the correct answer is (boxed{C}).,<think>Okay, let's try to figure out this problem. The question is about the angle α formed by a straight line and a plane. The options are different ranges for α. Hmm, I need to recall how the angle between a line and a plane is defined.First, I remember that when you have a line and a plane, the angle between them isn't the same as the angle between two lines. For two lines, the angle is between 0° and 180°, but for a line and a plane, it's different. Let me think... I think the angle between a line and a plane is defined as the complement of the angle between the line and the normal to the plane. Wait, no, maybe it's the angle between the line and its projection onto the plane. Let me verify that.So, if you have a straight line intersecting a plane, the angle α between them is the smallest angle between the line and any line on the plane. That makes sense. To find this angle, you can project the line onto the plane and then measure the angle between the original line and its projection. Since the projection is on the plane, the angle between the original line and the projection must be acute or right. Because if it were obtuse, the smaller angle on the other side would be considered instead.Alternatively, another way to define it is by using the normal vector of the plane. The angle between the line and the normal is β, then the angle α between the line and the plane is 90° - β. Since β can range from 0° to 90°, α would also range from 0° to 90°. Wait, no. If β is the angle between the line and the normal, then β can be between 0° and 90°, so α = 90° - β would be between 0° and 90° as well. But if the line is parallel to the plane, then the angle α should be 0°, right? And if the line is perpendicular to the plane, then α would be 90°.Wait, hold on. If the line is lying on the plane, then the angle between the line and the plane should be 0°, since it's parallel. If the line is perpendicular to the plane, then the angle is 90°. So the angle α should range from 0° to 90°, inclusive. Let me check that.Suppose the line is on the plane. Then, the angle between the line and the plane is 0°, which is the minimum possible. If the line is perpendicular, then the angle is 90°, which is the maximum. If the line is somewhere in between, the angle is acute. So that would mean the angle α must be between 0° and 90°, including both endpoints.Looking at the options, option C is 0° ≤ α ≤ 90°, and option D is 0° < α ≤ 90°. The difference is whether 0° is included or not. If the line lies on the plane, the angle is 0°, so 0° should be possible. Therefore, option C includes 0°, which is correct. But wait, some definitions might exclude the case where the line is on the plane. Let me verify.In some contexts, the angle between a line and a plane is considered only when the line is not lying on the plane, but that's more for the dihedral angle or something else. However, according to standard definitions, when the line lies on the plane, the angle is indeed 0°, which is acceptable. So the correct range should include 0°.So option C: 0° ≤ α ≤ 90°, is correct. Option D excludes 0°, which is wrong because the line can lie on the plane. Therefore, the answer should be C.But wait, let me make sure. Suppose the line is in the plane. Then the angle between the line and the plane is 0°. If the line is not in the plane, then the angle is acute. If the line is perpendicular, it's 90°. So yes, the angle can be 0°, so the correct answer is C. Therefore, the answer is C.Final Answerboxed{C}</think>To determine the angle (alpha) formed by a straight line and a plane, we need to consider the definition of this angle. The angle between a line and a plane is defined as the smallest angle between the line and any line on the plane. This angle can be found by projecting the line onto the plane and measuring the angle between the original line and its projection. If the line lies on the plane, the angle (alpha) is (0^circ). If the line is perpendicular to the plane, the angle (alpha) is (90^circ). For any other orientation, the angle (alpha) is acute. Therefore, the angle (alpha) must range from (0^circ) to (90^circ), inclusive.The correct answer is (boxed{C}).,<think>Okay, let's see. The problem is about the angle formed by a straight line and a plane. The options are given, and I need to figure out which one is correct. Hmm, angles between lines and planes... I remember there's a specific way to define that angle. Let me recall.First, when we talk about the angle between a line and a plane, it's not the same as the angle between two lines or between two planes. From what I remember, if a line intersects a plane, the angle between them is the smallest angle between the line and its projection onto the plane. Or maybe it's the complement of that angle? Wait, let me make sure.So, suppose we have a straight line intersecting a plane. The angle α between them is defined as the angle between the line and its orthogonal projection onto the plane. This angle should be acute, right? Because the projection would form a right angle with the line's direction vector perpendicular to the plane. So the angle between the line and the plane would be the acute angle between them. But then again, if the line is lying on the plane, the angle would be zero. If the line is perpendicular to the plane, the angle would be 90 degrees. So the angle should range from 0 to 90 degrees, inclusive. That seems right.Looking at the options: (C) is 0° ≤ α ≤ 90°, and (D) is 0° < α ≤ 90°. The difference is whether the angle can be exactly zero. If the line lies on the plane, then the angle between them is zero. So that's possible. Therefore, the correct interval should include 0°, right? So option (C) would include that, whereas (D) excludes zero. But wait, let me verify the exact definition to be sure.In some textbooks, the angle between a line and a plane is defined as the angle between the line and the normal to the plane. Wait, no, that would be a different angle. Let me check. If θ is the angle between the line and the normal, then the angle between the line and the plane would be 90° - θ. Since θ can range from 0° to 90°, then 90° - θ would also range from 0° to 90°. But in that case, if the line is in the plane, the angle with the normal is 90°, so the angle with the plane is 0°. If the line is perpendicular to the plane, the angle with the normal is 0°, so the angle with the plane is 90°. So yes, the angle α between the line and the plane is between 0° and 90°, inclusive. Therefore, option (C) seems correct.But wait, some sources might argue that the angle between a line and a plane is considered to be the smallest angle greater than zero. But that doesn't seem right. If the line is in the plane, the angle should be zero. For example, in geometry, when a line lies on a plane, they are said to be coplanar, and the angle between them is zero. So 0° should be included.Therefore, option (C) is the correct answer. However, let me check the options again. (C) says 0° ≤ α ≤ 90°, and (D) is 0° < α ≤ 90°. So the only difference is whether zero is included. Since the line can lie on the plane, zero is possible, so (C) should be right.Wait, but maybe there's a convention where the angle between a line and a plane is considered to be the smallest positive angle, hence excluding zero. But in standard definitions, when the line is on the plane, the angle is zero. For instance, in vector terms, if the line is on the plane, the angle between the line's direction vector and the plane's normal is 90°, so the angle between the line and the plane would be 90° - 90° = 0°. So yes, zero is possible.Alternatively, maybe in some contexts, they consider only lines that are not lying on the plane, but the problem statement here says "a straight line and a plane", so it could be any line, including those lying on the plane. So the answer should include 0°, making (C) the correct choice. Therefore, I think the answer is (C).Final Answerboxed{C}</think>To determine the angle alpha formed by a straight line and a plane, we need to consider the definition of this angle. The angle between a line and a plane is defined as the smallest angle between the line and its orthogonal projection onto the plane. This angle can be zero if the line lies on the plane, and it can be 90 degrees if the line is perpendicular to the plane. Key points:1. If the line lies on the plane, the angle alpha is 0^circ.2. If the line is perpendicular to the plane, the angle alpha is 90^circ.3. For any other position, the angle alpha is acute, ranging between 0^circ and 90^circ.Thus, the angle alpha must satisfy 0^circ leq alpha leq 90^circ. This includes both the possibility of the line lying on the plane (0°) and being perpendicular to the plane (90°). Therefore, the correct answer is (C).boxed{C}]

🔑:This question is knowledge-intensive, as the angle formed by a line and a plane should include four cases: oblique intersection, perpendicular intersection, parallel, and coincident. If not considered comprehensively, it's easy to answer (A),(D). However, the first quadrant angle should be 2 k pi<alpha<2 k pi+frac{pi}{2}, so the correct choice should be (C).

🧠:[<think>Okay, let's try to solve this equation: [x]^4 + {x}^4 + x^4 = 2048. Hmm, first I need to remember what [x] and {x} mean. [x] is the greatest integer not exceeding x, so that's the floor function, right? And {x} is x minus [x], which is the fractional part. So any real number x can be written as [x] + {x}, where [x] is an integer and {x} is between 0 (inclusive) and 1 (exclusive). So let's denote [x] as n, which is an integer, and {x} as f, which is 0 ≤ f < 1. Then x = n + f. The equation becomes n^4 + f^4 + (n + f)^4 = 2048. Let me expand (n + f)^4. Using the binomial theorem, that would be n^4 + 4n^3f + 6n^2f^2 + 4nf^3 + f^4. So substituting back into the equation: n^4 + f^4 + [n^4 + 4n^3f + 6n^2f^2 + 4nf^3 + f^4] = 2048. Combining like terms, we have n^4 + f^4 + n^4 + 4n^3f + 6n^2f^2 + 4nf^3 + f^4. So that's 2n^4 + 2f^4 + 4n^3f + 6n^2f^2 + 4nf^3 = 2048. Hmm, that's a bit complicated. Maybe there's a way to factor this expression or simplify it. Let me see. Let's factor terms step by step. First, 2n^4 + 2f^4 can be written as 2(n^4 + f^4). Then, the remaining terms are 4n^3f + 6n^2f^2 + 4nf^3. Let's factor out a common factor from these. Each term has at least nf, so factoring out 2nf:Wait, 4n^3f + 6n^2f^2 + 4nf^3. Let me check: 4n^3f is 4n^3f, 6n^2f^2 is 6n^2f^2, and 4nf^3 is 4nf^3. Hmm, maybe these terms form part of a binomial expansion. Let's see: (n + f)^4 is n^4 + 4n^3f + 6n^2f^2 + 4nf^3 + f^4. So the middle terms here are exactly 4n^3f + 6n^2f^2 + 4nf^3. So our equation is 2n^4 + 2f^4 + [(n + f)^4 - n^4 - f^4] = 2048. Let me verify that. The (n + f)^4 expands to n^4 + 4n^3f + 6n^2f^2 + 4nf^3 + f^4. Therefore, 4n^3f + 6n^2f^2 + 4nf^3 is equal to (n + f)^4 - n^4 - f^4. So substituting back into the original equation:2n^4 + 2f^4 + (n + f)^4 - n^4 - f^4 = 2048. Simplify that: 2n^4 - n^4 + 2f^4 - f^4 + (n + f)^4 = 2048. Which becomes n^4 + f^4 + (n + f)^4 = 2048. Wait, but that's exactly the original equation. So that didn't help. Maybe another approach.Alternatively, since x = n + f, and x^4 = (n + f)^4. So the equation is [x]^4 + {x}^4 + x^4 = 2048, which is n^4 + f^4 + (n + f)^4 = 2048. Let me compute (n + f)^4 + n^4 + f^4. Let's see:(n + f)^4 + n^4 + f^4 = 2n^4 + 4n^3f + 6n^2f^2 + 4nf^3 + 2f^4. Wait, but that's the same as before. Hmm. Maybe instead of expanding, we can find possible integer values of n first, since n is an integer, and then solve for f.Given that x = n + f, and n is integer, f ∈ [0,1). Then x^4 is between n^4 and (n + 1)^4. Similarly, {x}^4 is between 0 and 1. So the total sum [x]^4 + {x}^4 + x^4 is between n^4 + 0 + n^4 = 2n^4 and n^4 + 1 + (n + 1)^4. So 2n^4 ≤ 2048 ≤ n^4 + 1 + (n + 1)^4.So maybe we can bound n first. Let's find n such that 2n^4 ≤ 2048 ≤ n^4 + (n + 1)^4 + 1.First, solve 2n^4 ≤ 2048. So n^4 ≤ 1024. Then n ≤ (1024)^(1/4). Let's compute that. 1024 is 2^10, so (2^10)^(1/4) = 2^(10/4) = 2^2.5 ≈ 5.656. So n ≤ 5.656, so n can be at most 5. Similarly, the upper bound: 2048 ≤ n^4 + (n + 1)^4 + 1. Let's check for n=5: n^4 + (n +1)^4 +1 = 5^4 + 6^4 +1 = 625 + 1296 +1=1922. Which is less than 2048. So that's not sufficient. Wait, perhaps n=6?Wait, but n was supposed to be less than or equal to 5.656. So n is at most 5. Wait, but if n=5, the upper bound would be 5^4 + (6)^4 +1=625 +1296 +1=1922 <2048. So that suggests that even for n=5, the maximum possible value is 1922, which is less than 2048. But the equation is 2048. So maybe my initial approach is wrong.Wait, perhaps there's an error in the bounds. Let me check again. Let's suppose that x is in [n, n +1), so x = n + f with 0 ≤ f <1. Then [x] = n, {x} = f, and x^4 = (n + f)^4. The equation is n^4 + f^4 + (n + f)^4 = 2048. The maximum possible value when x is in [n, n +1) is when f approaches 1. So (n +1)^4. So the sum would be n^4 + (almost 1) + (n +1)^4. Therefore, for each n, the maximum value of the sum is approximately n^4 +1 + (n +1)^4. So if we can find n such that n^4 + (n +1)^4 +1 is greater than or equal to 2048, then possible n's are those where n^4 is around 2048.Wait, let's compute n^4 for various n:n=5: 625n=6: 1296n=7: 2401So n=7 gives 2401 which is over 2048, but n=6 is 1296. So if n=6, then x is in [6,7). Then, the sum [x]^4 + {x}^4 + x^4 =6^4 + f^4 + (6 +f)^4. Let's compute that:6^4 =1296. Then f^4 is between 0 and 1. (6 +f)^4 =6^4 + 4*6^3 f + 6*6^2 f^2 +4*6 f^3 +f^4. So that's 1296 + 864f + 216f^2 +24f^3 +f^4. So total sum: 1296 + f^4 + 1296 +864f +216f^2 +24f^3 +f^4=2592 + 2f^4 +864f +216f^2 +24f^3.Set this equal to 2048: 2592 +2f^4 +864f +216f^2 +24f^3=2048. Subtract 2048: 544 +2f^4 +864f +216f^2 +24f^3=0. But 544 is positive, and all other terms are non-negative because f is between 0 and1. So this is impossible. Hence, n=6 gives a sum that's too big. Wait, but when n=6, the minimal sum is when f=0: 6^4 +0 +6^4=2*1296=2592, which is already more than 2048. So n cannot be 6. Then n must be less than 6, but earlier we saw that for n=5, the maximum sum is 1922, which is less than 2048. So this is a problem.Wait, so according to this, there's a contradiction. If n ≤5, maximum sum is 1922, which is less than 2048. If n=6, minimum sum is 2592, which is more than 2048. So there's no solution? But the problem says "the solution to the equation", implying that there is a solution. So maybe I made a mistake in my reasoning.Alternatively, maybe n is negative? Let's check. If x is negative, then [x] is the greatest integer not exceeding x. For example, if x is -1.2, then [x] is -2, because -2 is the greatest integer not exceeding -1.2. Then {x}=x - [x] = -1.2 - (-2)=0.8. So even for negative x, {x} is in [0,1). So maybe x is negative.Let's check n negative. Let's try n=-1. Then x is in [-1,0). So [x]=-1, {x}=x +1 (since x = -1 + f, where f= x +1, so 0 ≤f <1). Then the equation becomes (-1)^4 + f^4 + (-1 +f)^4 =1 +f^4 + (-1 +f)^4. Let's compute (-1 +f)^4. That would be 1 -4f +6f^2 -4f^3 +f^4. So total sum:1 +f^4 +1 -4f +6f^2 -4f^3 +f^4=2 +2f^4 -4f +6f^2 -4f^3. Set equal to 2048: 2 +2f^4 -4f +6f^2 -4f^3=2048. Then 2f^4 -4f^3 +6f^2 -4f -2046=0. This seems impossible because f is between 0 and1, so 2f^4 is at most 2, -4f^3 is between -4 and0, 6f^2 is between0 and6, -4f is between -4 and0, so total is between (2 -4 +0 -4)= -6 and (2 +0 +6 +0)=8. So the left-hand side is between -6 and8, but we need it to be 2048. Not possible. So n=-1 is invalid.Similarly, n=-2: x in [-2,-1), so [x]=-2, {x}=x +2, 0≤f<1. Then equation: (-2)^4 +f^4 + (-2 +f)^4=16 +f^4 +(-2 +f)^4. Compute (-2 +f)^4. Let's expand that:(-2 +f)^4 = (f -2)^4 = f^4 -8f^3 +24f^2 -32f +16. So total sum:16 +f^4 +f^4 -8f^3 +24f^2 -32f +16=32 +2f^4 -8f^3 +24f^2 -32f. Set equal to 2048: 32 +2f^4 -8f^3 +24f^2 -32f=2048. Then 2f^4 -8f^3 +24f^2 -32f -2016=0. Again, f in [0,1). Let's see approximate values. For f=0: -2016. For f=1: 2 -8 +24 -32 -2016= (2 -8)= -6; (-6 +24)=18; (18 -32)= -14; (-14 -2016)= -2030. So the left-hand side is negative throughout, so no solution here.n=-3: x in [-3,-2). [x]=-3, {x}=x +3. Equation: (-3)^4 +f^4 + (-3 +f)^4=81 +f^4 +(-3 +f)^4. Compute (-3 +f)^4:(f -3)^4 =f^4 -12f^3 +54f^2 -108f +81. So total sum:81 +f^4 +f^4 -12f^3 +54f^2 -108f +81=162 +2f^4 -12f^3 +54f^2 -108f. Set equal to 2048: 162 +2f^4 -12f^3 +54f^2 -108f=2048. Then 2f^4 -12f^3 +54f^2 -108f -1886=0. Again, f in [0,1). At f=0: -1886. At f=1:2 -12 +54 -108 -1886= (2 -12)= -10; (-10 +54)=44; (44 -108)= -64; (-64 -1886)= -1950. Still negative. So no solution.It's clear that negative n are not working. Let's check n=0. Then x in [0,1). [x]=0, {x}=x. Then equation becomes 0 +x^4 +x^4=2x^4=2048. So x^4=1024. So x= (1024)^(1/4)= (2^10)^(1/4)=2^(10/4)=2^2.5= sqrt(2^5)=sqrt(32)=about 5.656. But x is in [0,1). So 5.656 is not in [0,1). So no solution here.n=1: x in [1,2). [x]=1, {x}=x -1. Equation:1^4 + (x -1)^4 +x^4=1 +f^4 +x^4=1 +f^4 + (1 +f)^4. Let's compute that. (1 +f)^4=1 +4f +6f^2 +4f^3 +f^4. So total sum:1 +f^4 +1 +4f +6f^2 +4f^3 +f^4=2 +2f^4 +4f +6f^2 +4f^3. Set equal to 2048:2 +2f^4 +4f +6f^2 +4f^3=2048. Then 2f^4 +4f^3 +6f^2 +4f -2046=0. Again, f ∈ [0,1). Plugging f=0: -2046. f=1:2 +4 +6 +4 -2046= -2030. So no solution.n=2: x in [2,3). Equation:2^4 + (x -2)^4 +x^4=16 +f^4 +x^4=16 +f^4 + (2 +f)^4. Expand (2 +f)^4=16 +32f +24f^2 +8f^3 +f^4. So total sum:16 +f^4 +16 +32f +24f^2 +8f^3 +f^4=32 +2f^4 +32f +24f^2 +8f^3. Set equal to 2048:32 +2f^4 +32f +24f^2 +8f^3=2048. Then 2f^4 +8f^3 +24f^2 +32f -2016=0. Again, f in [0,1). Plugging f=0: -2016. f=1:2 +8 +24 +32 -2016= -1950. No solution.n=3: x in [3,4). Equation:3^4 + (x -3)^4 +x^4=81 +f^4 +x^4=81 +f^4 + (3 +f)^4. Expand (3 +f)^4=81 +108f +54f^2 +12f^3 +f^4. Total sum:81 +f^4 +81 +108f +54f^2 +12f^3 +f^4=162 +2f^4 +108f +54f^2 +12f^3. Set equal to 2048:162 +2f^4 +108f +54f^2 +12f^3=2048. Then 2f^4 +12f^3 +54f^2 +108f -1886=0. Again, plugging f=0: -1886. f=1:2 +12 +54 +108 -1886= -1712. No solution.n=4: x in [4,5). Equation:4^4 + (x -4)^4 +x^4=256 +f^4 + (4 +f)^4. Expand (4 +f)^4=256 +256f +96f^2 +16f^3 +f^4. Total sum:256 +f^4 +256 +256f +96f^2 +16f^3 +f^4=512 +2f^4 +256f +96f^2 +16f^3. Set equal to 2048:512 +2f^4 +256f +96f^2 +16f^3=2048. Then 2f^4 +16f^3 +96f^2 +256f -1536=0. Divide equation by 2: f^4 +8f^3 +48f^2 +128f -768=0. For f ∈ [0,1). Let's check f=0: -768. f=1:1 +8 +48 +128 -768=185 -768= -583. Still negative.n=5: x in [5,6). Equation:5^4 + (x -5)^4 +x^4=625 +f^4 + (5 +f)^4. Expand (5 +f)^4=625 +500f +150f^2 +20f^3 +f^4. Total sum:625 +f^4 +625 +500f +150f^2 +20f^3 +f^4=1250 +2f^4 +500f +150f^2 +20f^3. Set equal to 2048:1250 +2f^4 +500f +150f^2 +20f^3=2048. Then 2f^4 +20f^3 +150f^2 +500f -798=0. Divide by 2: f^4 +10f^3 +75f^2 +250f -399=0. Let's check if this equation has a solution in [0,1). Let's compute left-hand side at f=0: -399. At f=1:1 +10 +75 +250 -399= (1+10)=11; 11+75=86; 86+250=336; 336-399= -63. Still negative. At f=0.5: (0.5)^4 +10*(0.5)^3 +75*(0.5)^2 +250*(0.5) -399=0.0625 +10*0.125 +75*0.25 +250*0.5 -399=0.0625 +1.25 +18.75 +125 -399= (0.0625 +1.25)=1.3125; (1.3125 +18.75)=20.0625; (20.0625 +125)=145.0625; (145.0625 -399)= -253.9375. Still negative. At f=0.8: f^4=0.4096; 10f^3=10*(0.512)=5.12;75f^2=75*0.64=48;250f=200. So total:0.4096 +5.12 +48 +200 -399= (0.4096 +5.12)=5.5296; (5.5296 +48)=53.5296; (53.5296 +200)=253.5296; (253.5296 -399)= -145.4704. Still negative. At f=0.9: f^4=0.6561;10f^3=10*0.729=7.29;75f^2=75*0.81=60.75;250f=225. Total:0.6561 +7.29 +60.75 +225 -399= (0.6561 +7.29)=7.9461; (7.9461 +60.75)=68.6961; (68.6961 +225)=293.6961; (293.6961 -399)= -105.3039. Still negative. So no solution in n=5.Wait, but according to this, there is no solution. But the problem states "the solution to the equation". Did I make a miscalculation? Let me check again for n=6. Wait, n=6 gives x in [6,7). Then, the equation is 6^4 + f^4 + (6 +f)^4=1296 +f^4 + (6 +f)^4. Let's compute (6 +f)^4=1296 + 864f + 216f^2 +24f^3 +f^4. So total sum:1296 +f^4 +1296 +864f +216f^2 +24f^3 +f^4=2592 +2f^4 +864f +216f^2 +24f^3. Setting this equal to 2048 gives 2592 + ... =2048. Which is impossible because 2592 is already more than 2048. So even if f is negative (which it can't be), the left-hand side is too big.Wait, so is there a mistake here? Because the problem states that there is a solution. Maybe I need to check higher n? But for n=7, x is in [7,8). Let's compute n=7: [x]^4=7^4=2401. Then {x}^4 is less than1. x^4 is at least7^4=2401. So total sum would be at least2401 +0 +2401=4802, which is way over 2048. So n=7 and above are too big.But according to this, there is no solution. But the problem says "the solution to the equation". Maybe fractional parts can be larger? Wait, no, {x} is always in [0,1). Wait, unless x is an integer. If x is an integer, then {x}=0. So the equation becomes [x]^4 +0 +x^4=2[x]^4=2048. So 2n^4=2048 =>n^4=1024 =>n=1024^(1/4)=5.656..., but n must be integer. So no integer n satisfies this. Therefore, if x is integer, no solution. If x is not integer, then we have [x]=n, {x}=f∈(0,1), and x=n +f. Then n^4 + f^4 + (n +f)^4=2048.But as per previous calculations, for n=5, the maximum possible value is around 1922, which is less than 2048, and for n=6, the minimum is 2592, which is over 2048. Hence, no solution. But the problem says there is a solution. Maybe I made a mistake in calculating the maximum for n=5?Wait, let's recalculate for n=5. x=5 +f, f∈[0,1). Then the equation is 5^4 +f^4 + (5 +f)^4=625 +f^4 + (625 +500f +150f^2 +20f^3 +f^4)=625 +f^4 +625 +500f +150f^2 +20f^3 +f^4=1250 +2f^4 +500f +150f^2 +20f^3. Setting equal to 2048:20f^3 +150f^2 +500f +2f^4 = 2048 -1250=798. So 2f^4 +20f^3 +150f^2 +500f=798. Let's write this as f^4 +10f^3 +75f^2 +250f=399. Let's check at f=0.8:f=0.8: 0.4096 +10*(0.512) +75*(0.64) +250*(0.8)=0.4096 +5.12 +48 +200=253.5296. Which is 253.5296, less than399. At f=0.9: 0.6561 +10*0.729 +75*0.81 +250*0.9=0.6561 +7.29 +60.75 +225=293.6961. Still less. At f=0.95: 0.81450625 +10*(0.857375) +75*(0.9025) +250*0.95≈0.8145 +8.57375 +67.6875 +237.5≈0.8145+8.57375=9.38825+67.6875=77.07575+237.5≈314.57575. Still less than399. Wait, but how can this reach 399? It's increasing as f approaches1. Let's plug f approaching1. At f=1, the left-hand side would be1 +10 +75 +250=336. But since f<1, the maximum approaches336, which is still less than399. So even as f approaches1, the left-hand side approaches336, which is less than399. So no solution in n=5. Therefore, according to this, the equation has no solution. But the problem states there is a solution. That suggests that I made a mistake.Wait, let me check if I copied the equation correctly. The problem says [x]^4 + {x}^4 +x^4=2048. Yes. Wait, maybe I have miscalculated the expansion somewhere. Let me check n=5 again:[x]^4 =5^4=625.{x}^4 =f^4.x^4=(5 +f)^4=5^4 +4*5^3 f +6*5^2 f^2 +4*5 f^3 +f^4=625 +500f +150f^2 +20f^3 +f^4.So total sum:625 +f^4 +625 +500f +150f^2 +20f^3 +f^4=1250 +2f^4 +500f +150f^2 +20f^3=2048. Then 2f^4 +20f^3 +150f^2 +500f=2048 -1250=798. So 2f^4 +20f^3 +150f^2 +500f -798=0. Divide by2: f^4 +10f^3 +75f^2 +250f -399=0. Wait, but maybe I can factor this polynomial? Let's try rational roots. The possible rational roots are factors of399 over factors of1, so ±1, ±3, ±7, ±19, ±21, etc. Test f=1:1 +10 +75 +250 -399=336 -399=-63≠0. f=3:81 +270 + 675 +750 -399= (81+270)=351; 351+675=1026; 1026+750=1776; 1776-399=1377≠0. Not useful. Maybe there's an irrational root. Let's check the function at f=0.9: approximately, as before, 0.6561 +7.29 +60.75 +225 -399≈293.696 -399≈-105.3. At f=1:336 -399=-63. So from f=0.9 to f=1, function goes from -105.3 to -63. Still negative. So no root in n=5.Wait, then the problem must have a typo? Or maybe I miscalculated somewhere. Alternatively, maybe the equation is [x]^4 + {x}^4 +x^2=2048? Because 2048 is 2^11, which might be related to powers of 2, but the equation as stated is with x^4. Alternatively, maybe 2048 is a typo. Alternatively, maybe I need to consider that [x] can be negative, but even so, as we saw earlier, negative n don't give high enough sums.Alternatively, maybe x is not a real number? No, the problem states x is a real number. Hmm. Wait, perhaps multiple solutions? But given the way the problem is phrased, "the solution", implying a unique solution. Wait, perhaps the problem is in the Chinese or another language original, and "the solution" might be plural. But the user wrote "the solution", maybe there is one solution. Hmm.Alternatively, maybe I messed up the floor function. Let me confirm: [x] is the greatest integer not exceeding x, so for positive x, it's the integer part, like floor(x). For negative x, e.g., x=-1.2, [x]=-2. So {x}=x - [x]= -1.2 - (-2)=0.8. Yes.Wait, what if x is a non-integer but [x]^4 + {x}^4 +x^4=2048. Wait, maybe x is a number like 6. So [x]=6, {x}=0, x=6. Then equation:6^4 +0 +6^4=2*1296=2592≠2048. Not equal. If x is 5.5, then [x]=5, {x}=0.5. Then equation:5^4 +0.5^4 +5.5^4=625 +0.0625 +915.0625≈625 +0.0625=625.0625 +915.0625≈1540.125≠2048. Still too low. If x=6. Even if f=0. So x=6, sum is2592. Which is over. But there's no x between5 and6 that can reach2048. As we saw, n=5 can't reach it. n=6 is over. Therefore, maybe the problem is a trick question with no solution? But the user says "the solution to the equation", so there must be one. Wait, maybe I made a mistake in calculations for n=5. Let me check with f=0.8:n=5, f=0.8. Then [x]^4=625, {x}^4=0.8^4=0.4096, x^4=(5.8)^4. Let's compute5.8^4:5.8^2=33.64. Then5.8^4=(33.64)^2=1131.6496. So sum:625 +0.4096 +1131.6496≈625 +0.4096=625.4096 +1131.6496≈1757.0592. Still way below2048. For x=5.9:5.9^4:5.9^2=34.81, then34.81^2≈1211.3361. Sum:625 +0.9^4 +1211.3361=625 +0.6561 +1211.3361≈625 +0.6561=625.6561 +1211.3361≈1836.9922. Still under. x=5.99:5.99^4. First,5.99^2≈35.8801, then squared:35.8801^2≈1287.3. Sum:625 +0.99^4 +1287.3≈625 +0.9606 +1287.3≈625 +0.9606=625.9606 +1287.3≈1913.2606. Still under. x approaching6 from below:At x=6-, f approaching1. x^4 approaching6^4=1296. So sum:5^4 +f^4 +x^4≈625 +1 +1296=1922. Which is less than2048. So no solution.Wait, so according to all this, there is no real solution to the equation. But the problem says there is. Maybe the original problem was different? Let me check the problem again:"Let [x] denote the greatest integer not exceeding the real number x, and {x}=x−[x]. Then the solution to the equation [x]^4 + {x}^4 +x^4=2048 is…"Hmm. Maybe there's a typo and the equation is [x]^3 + {x}^3 +x^3=2048? Or maybe the exponents are different? Alternatively, maybe the equation is [x]^2 + {x}^2 +x^2=2048. But the user wrote the exponent as4.Alternatively, maybe I need to consider that [x] and {x} can be negative. Wait, [x] can be negative, but {x} is always non-negative because {x}=x - [x], and [x] is the greatest integer not exceeding x. So even for negative x, {x} is in[0,1). So if x is negative, {x}=x - [x], which is x - (floor(x)). For example, x=-1.3, floor(x)=-2, so {x}= -1.3 - (-2)=0.7.But even with negative [x], we saw that the sums are much lower. Let's check with n=-4. x in [-4,-3). [x]=-4, {x}=x +4. Then equation: (-4)^4 + (x +4)^4 +x^4=256 +f^4 + (-4 +f)^4, where f=x +4, 0≤f<1. Compute (-4 +f)^4=(f -4)^4= f^4 -16f^3 +96f^2 -256f +256. So total sum:256 +f^4 +f^4 -16f^3 +96f^2 -256f +256=512 +2f^4 -16f^3 +96f^2 -256f. Set equal to2048:512 +2f^4 -16f^3 +96f^2 -256f=2048. Then 2f^4 -16f^3 +96f^2 -256f -1536=0. Divide by2: f^4 -8f^3 +48f^2 -128f -768=0. For f in [0,1). Plug f=0: -768. f=1:1 -8 +48 -128 -768= (1 -8)= -7; (-7 +48)=41; (41,<think>Okay, so I need to solve the equation [x]^4 + {x}^4 + x^4 = 2048. Let me recall that [x] is the greatest integer less than or equal to x, which is also called the floor function. And {x} is the fractional part of x, so {x} = x - [x]. That means any real number x can be written as [x] + {x}, right? So x is split into its integer part and the fractional part.Given this equation, I need to find all real numbers x such that when you take the floor of x to the fourth power, the fractional part of x to the fourth power, and x itself to the fourth power, and add them all together, you get 2048.Hmm. Let me think. Let me denote n = [x], so n is an integer, and f = {x}, which is between 0 (inclusive) and 1 (exclusive). Then x = n + f. So substituting into the equation, we have:n^4 + f^4 + (n + f)^4 = 2048.Let me expand (n + f)^4. Using the binomial theorem, that would be n^4 + 4n^3f + 6n^2f^2 + 4nf^3 + f^4. So adding that to the existing n^4 and f^4:Original equation becomes:n^4 + f^4 + [n^4 + 4n^3f + 6n^2f^2 + 4nf^3 + f^4] = 2048Combine like terms:n^4 + f^4 + n^4 + 4n^3f + 6n^2f^2 + 4nf^3 + f^4 = 2048So that simplifies to:2n^4 + 2f^4 + 4n^3f + 6n^2f^2 + 4nf^3 = 2048.Hmm, that seems complicated. Maybe there's a better way to approach this. Let me check again.Wait, the original equation is [x]^4 + {x}^4 + x^4 = 2048. Since x = n + f, then x^4 = (n + f)^4. So substituting into the equation:n^4 + f^4 + (n + f)^4 = 2048.Which is exactly what I had before. So yes, expanding that gives:n^4 + f^4 + n^4 + 4n^3f + 6n^2f^2 + 4nf^3 + f^4 = 2048Simplify:2n^4 + 2f^4 + 4n^3f + 6n^2f^2 + 4nf^3 = 2048.Hmm, this is a quartic equation in variables n (integer) and f (real between 0 and 1). That seems difficult to solve directly. Maybe I can factor it or find some way to simplify.Alternatively, perhaps we can note that 2048 is 2^11, which is 2^11 = 2048. So maybe n and f are such that each term contributes a certain power of 2? Alternatively, maybe n is an integer such that n^4 is close to 2048, but let's check.Compute some fourth powers:3^4 = 814^4 = 2565^4 = 6256^4 = 12967^4 = 2401Ah, so 6^4 is 1296, 7^4 is 2401. 2048 is between 6^4 and 7^4. Wait, but our equation is [x]^4 + {x}^4 + x^4 = 2048.Since [x] is n, an integer, and {x} is f between 0 and 1, so {x}^4 is between 0 and 1. Then x^4 = (n + f)^4. So the total of n^4 + f^4 + (n + f)^4. Since n is an integer, and f is fractional part.Given that 6^4 is 1296 and 7^4 is 2401, which is already more than 2048. So perhaps n must be 6, because if n=7, then even [x]^4 = 2401, which is already larger than 2048, so that's impossible. Similarly, if n is 5, 5^4 is 625, which is much less. So n=6 seems plausible.Let me test n=6. Then n=6, so x = 6 + f, where 0 ≤ f < 1. Then the equation becomes:6^4 + f^4 + (6 + f)^4 = 2048.Compute 6^4: 6*6=36, 36*6=216, 216*6=1296. So 6^4=1296.So the equation is 1296 + f^4 + (6 + f)^4 = 2048.Subtract 1296: f^4 + (6 + f)^4 = 752.Now, compute (6 + f)^4. Let's expand that:(6 + f)^4 = 6^4 + 4*6^3*f + 6*6^2*f^2 + 4*6*f^3 + f^4Which is 1296 + 4*216*f + 6*36*f^2 + 24*f^3 + f^4Compute each term:4*216 = 864, so 864f6*36 = 216, so 216f^224f^3 and f^4.So (6 + f)^4 = 1296 + 864f + 216f^2 + 24f^3 + f^4Therefore, substituting back into the equation:f^4 + [1296 + 864f + 216f^2 + 24f^3 + f^4] = 752Combine like terms:f^4 + 1296 + 864f + 216f^2 + 24f^3 + f^4 = 752So that's 2f^4 + 24f^3 + 216f^2 + 864f + 1296 = 752Subtract 752:2f^4 + 24f^3 + 216f^2 + 864f + 1296 - 752 = 0Calculate 1296 - 752: 1296 - 700 = 596, 596 - 52 = 544So:2f^4 + 24f^3 + 216f^2 + 864f + 544 = 0Hmm, this quartic equation. Let's see if we can factor out common terms. All coefficients are even, so divide by 2:f^4 + 12f^3 + 108f^2 + 432f + 272 = 0Hmm, this still seems complicated. Let me check the coefficients. Wait, 12f^3, 108f^2, 432f. These coefficients are multiples of 12. Let me check:12f^3 = 12*f^3108f^2 = 9*12*f^2432f = 36*12*fBut 272 divided by 12 is not an integer. Hmm, maybe not helpful.Alternatively, perhaps factor the polynomial. Let me try rational roots. The possible rational roots would be factors of 272 over factors of 1, so ±1, ±2, ±4, ±8, ±16, ±17, etc.Let me test f=-1:(-1)^4 + 12*(-1)^3 + 108*(-1)^2 + 432*(-1) + 272 =1 -12 + 108 - 432 + 272 = (1 -12) + (108 -432) + 272 =(-11) + (-324) + 272 = (-335) + 272 = -63 ≠ 0f=-2:16 + 12*(-8) + 108*4 + 432*(-2) + 272 =16 -96 + 432 -864 + 272 =(16 -96) + (432 -864) + 272 =(-80) + (-432) + 272 = (-512) + 272 = -240 ≠0f=-4:256 + 12*(-64) + 108*16 + 432*(-4) +272 =256 -768 + 1728 -1728 +272 =(256 -768) + (1728 -1728) +272 =(-512) + 0 +272 = -240 ≠0f=-8:4096 +12*(-512) +108*64 +432*(-8) +272 =4096 -6144 + 6912 -3456 +272= (4096 -6144) + (6912 -3456) +272= (-2048) + 3456 +272 = (-2048 +3456)=1408 +272=1680≠0Hmm, none of these seem to work. Maybe positive roots? But f is supposed to be between 0 and 1, so if we plug f=0:0 +0 +0 +0 +272=272≠0f=1:1 +12 +108 +432 +272=1+12=13, 13+108=121, 121+432=553, 553+272=825≠0So no root at f=1. Since f is between 0 and 1, but all the potential roots here are negative or outside the range. That suggests that the equation f^4 + 12f^3 + 108f^2 + 432f + 272 = 0 has no real roots in the interval [0,1). But wait, this is a problem because when n=6, we need a solution f in [0,1). If there is no solution, then n=6 is invalid?But that can't be right. Maybe I made a mistake in my calculation.Wait, let's double-check the expansion of (6 + f)^4. Let's compute it step by step.(6 + f)^4:First, (6 + f)^2 = 36 + 12f + f^2Then, (6 + f)^4 = (36 + 12f + f^2)^2Expanding that:= 36^2 + 2*36*12f + 2*36*f^2 + (12f)^2 + 2*12f*f^2 + f^4Wait, that seems more complicated. Alternatively, use the binomial coefficients.(6 + f)^4 = 6^4 + 4*6^3 f + 6*6^2 f^2 + 4*6 f^3 + f^4Yes, which is 1296 + 4*216 f + 6*36 f^2 + 24 f^3 + f^4Which is 1296 + 864f + 216f^2 + 24f^3 + f^4. That's correct.Then, the original equation when n=6 is:6^4 + f^4 + (6 + f)^4 = 20481296 + f^4 + (1296 + 864f + 216f^2 + 24f^3 + f^4) = 2048So adding them:1296 + f^4 + 1296 + 864f + 216f^2 + 24f^3 + f^4 = 2048Combine like terms:2*1296 + 2f^4 + 864f + 216f^2 + 24f^3 = 2048Wait a second, here is a mistake! Previously, I thought that 1296 + 1296 is 2592, but 2592 + other terms would be way over 2048. Wait, but 1296 + f^4 + 1296 + ... = 2*1296 + ... which is 2592 + ... So 2592 plus other terms equals 2048? That's impossible because 2592 is already larger than 2048. Therefore, my previous step must be wrong.Wait, hold on. Wait, the original equation is [x]^4 + {x}^4 + x^4 = 2048. When n=6 and f is the fractional part, [x]^4 = 6^4 = 1296, {x}^4 = f^4, and x^4 = (6 + f)^4. So adding these: 1296 + f^4 + (6 + f)^4 = 2048. But (6 + f)^4 is already 1296 + ... So 1296 + f^4 + 1296 + ... is going to be 2592 + ... which is way more than 2048.But 2592 is already 2592, which is way larger than 2048. So this is a contradiction. Therefore, n cannot be 6? But 7^4 is 2401, which is even larger. Wait, this is confusing.Wait, maybe I messed up the equation. Let me check again.Original equation: [x]^4 + {x}^4 + x^4 = 2048.But x = n + f, so [x] = n, {x} = f, and x^4 = (n + f)^4. So substituting, the equation becomes:n^4 + f^4 + (n + f)^4 = 2048.But when n=6, (n + f)^4 is (6 + f)^4, which when f is between 0 and 1, (6 + f)^4 is between 6^4 = 1296 and 7^4 = 2401. So adding n^4 (1296) and f^4 (between 0 and 1), the total would be between 1296 + 0 + 1296 = 2592 and 1296 + 1 + 2401 = 3698. Both of which are way over 2048. So n=6 is impossible. Similarly, n=5 would have x between 5 and 6.Wait, so let's test n=5. Let me recast.Let n=5. Then x=5 + f, 0 ≤ f <1. Then the equation becomes:5^4 + f^4 + (5 + f)^4 = 2048Compute 5^4: 625.So equation: 625 + f^4 + (5 + f)^4 = 2048Subtract 625: f^4 + (5 + f)^4 = 1423Expand (5 + f)^4:5^4 + 4*5^3 f + 6*5^2 f^2 + 4*5 f^3 + f^4= 625 + 500f + 150f^2 + 20f^3 + f^4Therefore, the equation becomes:f^4 + 625 + 500f + 150f^2 + 20f^3 + f^4 = 1423Combine terms:2f^4 + 20f^3 + 150f^2 + 500f + 625 = 1423Subtract 1423:2f^4 + 20f^3 + 150f^2 + 500f + 625 - 1423 = 0625 -1423= -798So:2f^4 + 20f^3 + 150f^2 + 500f -798 =0Divide by 2:f^4 + 10f^3 + 75f^2 + 250f - 399 = 0Hmm, still a quartic. Let's check possible roots. f is between 0 and 1.Try f=0: 0 +0 +0 +0 -399= -399 ≠0f=1: 1 +10 +75 +250 -399= (1+10)=11, 11+75=86, 86+250=336, 336-399= -63 ≠0So somewhere between 0 and1, the function goes from -399 to -63. Since it's continuous, maybe crosses zero? Wait, but from f=0 to f=1, the value goes from -399 to -63, which is increasing, but remains negative. So no root between 0 and1. Hence, no solution for n=5.Hmm, same problem as with n=6. Wait, then maybe n=7? But [x]^4 for n=7 is 2401, which is already more than 2048. So n must be less than 7. But n=5 and n=6 give equations that don't have solutions. Maybe n=4?Let me test n=4.Then x=4 + f, 0 ≤f <1. Equation becomes:4^4 + f^4 + (4 + f)^4 = 2048.4^4 = 256. So:256 + f^4 + (256 + 4*64f +6*16f^2 +4*4f^3 +f^4) = 2048Wait, let's expand (4 + f)^4 properly.(4 + f)^4 = 4^4 + 4*4^3f +6*4^2f^2 +4*4f^3 +f^4= 256 + 4*64f +6*16f^2 +16f^3 +f^4=256 +256f +96f^2 +16f^3 +f^4Therefore, equation becomes:256 + f^4 +256 +256f +96f^2 +16f^3 +f^4 =2048Combine terms:2f^4 +16f^3 +96f^2 +256f +512 =2048Subtract 2048:2f^4 +16f^3 +96f^2 +256f +512 -2048=0512 -2048 = -1536So:2f^4 +16f^3 +96f^2 +256f -1536 =0Divide by 2:f^4 +8f^3 +48f^2 +128f -768=0Check possible roots. Again, f in [0,1).f=0: -768f=1:1 +8 +48 +128 -768= (1+8)=9, 9+48=57,57+128=185,185-768=-583Still negative. So no solution here either.Hmm, this is confusing. Maybe n is negative? Let's check negative integers.Let's try n= -1. Then x= -1 +f, 0 ≤f <1. So [x]=-1, {x}=f, since x is between -1 and 0. Wait, {x} is defined as x - [x], so if x is negative, [x] is the greatest integer less than or equal to x. For example, if x is -0.5, [x]=-1, and {x}= -0.5 - (-1) =0.5.So {x} is always between 0 and 1, even for negative x.So let's take n=-1. Then x= -1 +f, 0 ≤f <1. Then the equation becomes:(-1)^4 + f^4 + (-1 +f)^4 = 2048.Compute each term:(-1)^4=1f^4 is as before.(-1 +f)^4. Let's expand that.(-1 +f)^4 = (-1)^4 +4*(-1)^3f +6*(-1)^2f^2 +4*(-1)f^3 +f^4=1 -4f +6f^2 -4f^3 +f^4Thus, the equation:1 + f^4 + [1 -4f +6f^2 -4f^3 +f^4] =2048Combine terms:1 +f^4 +1 -4f +6f^2 -4f^3 +f^4=2048So:2f^4 -4f^3 +6f^2 -4f +2 =2048Subtract 2048:2f^4 -4f^3 +6f^2 -4f +2 -2048=0Simplify:2f^4 -4f^3 +6f^2 -4f -2046=0Divide by 2:f^4 -2f^3 +3f^2 -2f -1023=0Testing f in [0,1):f=0: -1023f=1:1 -2 +3 -2 -1023= (1-2)= -1, (-1+3)=2, (2-2)=0, (0 -1023)= -1023So no solution here. Similarly, for other negative n, say n=-2.x=-2 +f, [x]=-2, {x}=f.Equation becomes (-2)^4 +f^4 + (-2 +f)^4=2048.Compute:16 +f^4 + ( -2 +f)^4Expand (-2 +f)^4:= (-2)^4 +4*(-2)^3f +6*(-2)^2f^2 +4*(-2)f^3 +f^4=16 -32f +24f^2 -8f^3 +f^4So equation:16 +f^4 +16 -32f +24f^2 -8f^3 +f^4=2048Combine terms:2f^4 -8f^3 +24f^2 -32f +32 =2048Subtract 2048:2f^4 -8f^3 +24f^2 -32f -2016=0Divide by 2:f^4 -4f^3 +12f^2 -16f -1008=0Testing f=0: -1008f=1:1 -4 +12 -16 -1008= (1-4)= -3, (-3 +12)=9, (9 -16)= -7, (-7 -1008)= -1015No roots. So even negative n don't work. Hmm.Wait, so far, all n tried (4,5,6,7, -1, -2) lead to equations with no solution. Maybe I need to check other n?Wait, but let's check n=3. Let me try n=3. x=3 +f, 0 ≤f <1.Equation:3^4 +f^4 + (3 +f)^4=20483^4=81. So:81 +f^4 + (81 + 4*27f +6*9f^2 +4*3f^3 +f^4)=2048Expand (3 +f)^4:81 +108f +54f^2 +12f^3 +f^4So equation:81 +f^4 +81 +108f +54f^2 +12f^3 +f^4=2048Combine:2f^4 +12f^3 +54f^2 +108f +162=2048Subtract 2048:2f^4 +12f^3 +54f^2 +108f -1886=0Divide by 2:f^4 +6f^3 +27f^2 +54f -943=0Testing f in [0,1):f=0: -943f=1:1 +6 +27 +54 -943= (1+6)=7, (7+27)=34, (34 +54)=88, (88 -943)= -855Still negative. No solution.Hmm. What's going on here? Maybe the solution is with a larger n? But 7^4=2401, which is already over 2048. Wait, but [x]^4 is n^4, {x}^4 is up to 1, and x^4 is (n +f)^4. So for n=7, [x]^4=2401, which is more than 2048, so even with f=0, the total would be 2401 +0 +2401=4802, way over. So n cannot be 7 or higher.n=6 gives total terms over 2592. But 2048 is between 6^4=1296 and 7^4=2401. Wait, but the sum [x]^4 + {x}^4 +x^4 for n=6 and f approaching 1 would approach 1296 +1 + (6 +1)^4=1296 +1 +2401= 3698. Which is still more than 2048. So maybe n=6 is too big.But then, n=5, as above, but n=5 also leads to equations that sum to over 2592? Wait, no. Wait, when n=5, x is between 5 and 6, so x^4 is between 625 and 1296. So [x]^4 + {x}^4 +x^4 would be 625 + (something up to 1) + (something up to 1296). So total up to 625 +1 +1296=1922, which is less than 2048. Wait, so maybe n=5 is possible?Wait, let's recast. When n=5, x=5 +f, so x is between 5 and 6, so x^4 is between 5^4=625 and 6^4=1296.Thus, [x]^4=625, {x}^4 up to 1, and x^4 up to 1296. So total maximum possible sum is 625 +1 +1296=1922, which is less than 2048. Therefore, n=5 cannot reach 2048. Similarly, n=6 gives minimum sum 1296 +0 +1296=2592, which is more than 2048. Therefore, there's a contradiction here.Wait, so how is this possible? The equation [x]^4 + {x}^4 +x^4=2048 must have a solution, but according to this, for n=5, the maximum total is 1922, and for n=6, the minimum total is 2592. So there is a gap between 1922 and 2592. 2048 lies in that gap. Therefore, there is no solution?But the problem states "Then the solution to the equation...". So it must have a solution.Wait, maybe I made a mistake in my analysis. Let me double-check.Wait, when n=5, x=5+f, so x^4=(5+f)^4. Let's compute (5 +f)^4 when f approaches 1. So when f=1, x=6, so x^4=1296. So [x]^4=5^4=625, {x}=1, {x}^4=1, so total=625 +1 +1296=1922. But 1922 <2048. When n=6, x=6 +f, but x=6 +f must be less than 7. So x^4 is less than 7^4=2401. So [x]^4=6^4=1296, {x}^4 up to 1, so total=1296 +1 +2401=3698. But 3698 is way over 2048.But the required total is 2048. So how can this happen? The totals jump from 1922 (n=5, f approaching 1) to 2592 (n=6, f=0). So 2048 lies in between, but there is no n where the total can cross 2048. Hence, there is no solution? But the problem says "the solution to the equation". Maybe the problem has a typo, or maybe I made a mistake in my reasoning.Alternatively, perhaps fractional parts can be negative? No, {x} is defined as x - [x], so even for negative x, {x} is in [0,1). So that's not possible.Wait, wait, but maybe x is negative. Let's check. Suppose x is negative. For example, let n=-2, so x=-2 +f, 0 ≤f <1. Then [x]=-3? Wait, no. Wait, [x] is the greatest integer less than or equal to x. If x is, say, -2.5, then [x]=-3, and {x}=0.5. So {x}=x - [x]. So if x is negative, [x] is more negative than x, so {x}=x - [x] is positive.Wait, but let's take x=-k +f, where k is a positive integer and 0 ≤f <1. Then [x]=-k -1, and {x}=x - [x]= (-k +f) - (-k -1)=1 +f. Wait, but {x} must be in [0,1). So this can't be. Therefore, if x is negative, {x}=x - [x], where [x] is the floor of x. For example, x=-0.5, [x]=-1, {x}=x - [x]= -0.5 - (-1)=0.5. If x=-1.5, [x]=-2, {x}=x - [x]= -1.5 - (-2)=0.5. So {x} is always in [0,1).Therefore, even for negative x, {x} is in [0,1). Therefore, [x]^4 + {x}^4 +x^4 is still positive. Let's take x=-0.5. Then [x]=-1, {x}=0.5. Then [x]^4 + {x}^4 +x^4= (-1)^4 + (0.5)^4 + (-0.5)^4=1 +0.0625 +0.0625=1.125.But 2048 is a large number, so maybe x is positive? Wait, but for positive x, as we saw, n=6 gives totals starting at 2592, which is already over 2048, but n=5 gives up to 1922. So there is a gap. So, perhaps the problem is designed such that the solution is x=8? Wait, but 8 is an integer. Then [x]=8, {x}=0, x^4=4096. Then 8^4 +0 +4096=4096+4096=8192, which is way over. Hmm.Alternatively, maybe the equation is [x]^4 + {x}^4 +x^4=2048. Let me test x=6.5. Then [x]=6, {x}=0.5, x=6.5. Then 6^4 +0.5^4 +6.5^4=1296 +0.0625 +1785.0625=1296 +1785.0625=3081.0625 +0.0625=3081.125, which is way over 2048.Wait, maybe x is a smaller number. Let's try x=4. Let's see, x=4, [x]=4, {x}=0, x^4=256. Then total=256+0+256=512. Not enough. x=5: total=625 +0 +625=1250. x=6:1296+0+1296=2592. So between x=5 and x=6, the total goes from 1250 to2592. 2048 is in between, but according to our earlier analysis, when x is between5 and6, n=5, and the total is [x]^4 + {x}^4 +x^4=625 +f^4 + (5 +f)^4. When f=0, 625 +0 +625=1250. When f approaches1, 625 +1 +1296=1922. So 1250 to1922, still less than2048. Then when x reaches6, it's2592. So there's a jump from1922 to2592. So 2048 is in between. Therefore, there is no x such that the total is2048. But the problem says "the solution to the equation", implying that there is a solution.Wait, maybe I have a miscalculation in the expansion. Let me re-examine the case when n=6.Wait, when n=6, the equation is 6^4 +f^4 + (6 +f)^4=2048. As before, 6^4=1296, so:1296 +f^4 + (6 +f)^4=2048.But (6 +f)^4=1296 +864f +216f^2 +24f^3 +f^4.Therefore, total equation:1296 +f^4 +1296 +864f +216f^2 +24f^3 +f^4=2048.So 2*1296=2592. 2f^4 +24f^3 +216f^2 +864f +2592=2048.Wait, hold on, that would be 2592 + ...=2048. But 2592 is already larger than2048, so adding positive terms makes it even larger. Therefore, impossible. So no solution for n=6. Similarly, n=5 gives maximum total 1922. Therefore, the equation has no solution?But the problem says "the solution to the equation". Maybe the problem is in the negative numbers. Let me try n=-3.x=-3 +f, 0 ≤f <1. Then [x]=-4, {x}=x - [x]= (-3 +f) - (-4)=1 +f. Wait, but {x} must be less than1. So here, {x}=1 +f -1= f? Wait, no. Wait, if x=-3 +f, then [x]=-4 when x is between -4 and -3. Wait, no. If x is between -3 and -2, then [x]=-3. Wait, my mistake. Let me correct.If x is between n and n+1, where n is an integer, then [x]=n. For negative numbers, for example, x between -3 and -2: [x]=-3, {x}=x - (-3)=x +3, which is between0 and1.Similarly, x between -4 and -3: [x]=-4, {x}=x +4.So if x=-3 +f, where 0 ≤f <1, then x is between -3 and -2, [x]=-3, {x}=x - (-3)=f. So {x}=f. Then the equation becomes:(-3)^4 +f^4 + (-3 +f)^4=2048.Compute each term:(-3)^4=81.(-3 +f)^4= (f -3)^4= 81 - 108f +54f^2 -12f^3 +f^4.Therefore, equation:81 +f^4 +81 -108f +54f^2 -12f^3 +f^4=2048.Combine like terms:2f^4 -12f^3 +54f^2 -108f +162=2048.Subtract 2048:2f^4 -12f^3 +54f^2 -108f -1886=0.Divide by 2:f^4 -6f^3 +27f^2 -54f -943=0.Testing f in [0,1):f=0: -943f=1:1 -6 +27 -54 -943= (1-6)=-5, (-5+27)=22, (22-54)=-32, (-32 -943)=-975No solution.Hmm. This is perplexing. According to this analysis, there is no real number x that satisfies the equation [x]^4 + {x}^4 +x^4=2048. But the problem states "the solution to the equation", implying that there is a solution. Did I make a mistake somewhere?Wait, let's check the problem statement again. It says: Let [x] denote the greatest integer not exceeding the real number x, and {x}=x - [x]. Then the solution to the equation [x]^4 + {x}^4 +x^4=2048 is...Wait, perhaps the problem is from an Olympiad or similar, and the solution is an integer. But as we saw, if,<think>Okay, let's tackle this equation: [x]^4 + {x}^4 + x^4 = 2048. Hmm, first I need to recall what [x] and {x} mean. [x] is the greatest integer not exceeding x, so that's the floor function, right? And {x} is the fractional part, which is x minus [x]. So any real number x can be written as [x] + {x}, where [x] is an integer and 0 ≤ {x} < 1.So, if I let n = [x], which is an integer, and f = {x}, which is a fractional part between 0 (inclusive) and 1 (exclusive), then x = n + f. Then the equation becomes n^4 + f^4 + (n + f)^4 = 2048.Let me write that out:n^4 + f^4 + (n + f)^4 = 2048.Hmm, maybe I can expand (n + f)^4. Let's do that. The expansion is n^4 + 4n^3f + 6n^2f^2 + 4nf^3 + f^4. So adding the other terms:Original equation becomes:n^4 + f^4 + n^4 + 4n^3f + 6n^2f^2 + 4nf^3 + f^4 = 2048.Combine like terms:2n^4 + 2f^4 + 4n^3f + 6n^2f^2 + 4nf^3 = 2048.Hmm, not sure if that helps. Maybe another approach. Since n is integer and f is fractional between 0 and 1, maybe I can bound the possible values of n and f.Let's think about the possible range of x. If x is positive or negative? Let's see. Since all terms are raised to the 4th power, which is even, so they are non-negative. So x can be positive or negative, but since [x] is the floor function, if x is negative, then [x] is the next lower integer. For example, if x is -1.2, then [x] is -2, and {x} is 0.8. So the fractional part is still between 0 and 1 even for negative x.But let's consider the magnitude. The total sum is 2048. Let's see. The terms are [x]^4, {x}^4, and x^4. Since x = n + f, then x^4 is (n + f)^4, which is going to be the largest term here, especially if n is large. So maybe n isn't too big. Let's see.Suppose n is a positive integer. Then n^4 plus something else. Let's try to estimate. If n is 6, 6^4 is 1296. Then x = 6 + f, so x^4 is (6 + f)^4. If f is close to 1, then (7)^4 is 2401, which is already over 2048. So maybe n is 6, but then x^4 could be up to 2401, which is too big. Wait, but the total sum would be 6^4 + f^4 + (6 + f)^4. Let's compute for n=6:6^4 = 1296. Then f is between 0 and 1. Let's take f=0. Then x=6, so x^4=1296. The total sum would be 1296 + 0 + 1296 = 2592, which is more than 2048. If f increases, x^4 increases, so the total sum would be even larger. So n=6 seems too big.Wait, but maybe n is negative. Let's check. If n is negative, say n=-5. Then [x] = -5, so x is in [-5, -4). Then x = -5 + f, where f is between 0 and 1. So x^4 would be (-5 + f)^4. Let's compute for f=0: (-5)^4=625. Then the sum would be (-5)^4 + 0 + (-5)^4 = 625 + 625 = 1250. If f increases, x approaches -4, so x^4 becomes (-4)^4=256. So x^4 decreases as f increases. So the sum would be 625 + f^4 + ( -5 + f )^4. Let's try f=1 (but it's not allowed, since f<1). As f approaches 1, x approaches -4, so x^4 approaches 256. Then the sum approaches 625 + 1 + 256 = 882, which is still less than 2048.Wait, so maybe n can't be negative because the total sum would be too small. Let's check n=5. If n=5, then x is in [5,6). Let's compute for f=0: 5^4 + 0 +5^4= 625 + 625=1250. If f approaches 1, x approaches 6, so x^4 approaches 1296. Then the sum would approach 625 +1 +1296=1922, which is still less than 2048. Hmm. So n=6 gives sum over 2048 even at f=0, but n=5 with f approaching 1 gives 1922. So maybe there's no solution in n=5. What about n=7? But n=7 would make x in [7,8), so x^4 is at least 7^4=2401, which already added to n^4=2401, the sum would be 2401 + f^4 + (7 + f)^4. Even at f=0, it's 2401 + 0 +2401=4802, which is way over 2048. So n=7 is too big. So maybe the solution is between n=5 and n=6? Wait, but n has to be integer. So x is in [n, n+1). So n must be integer. So possible n values are from negative up to 6? But n=6 gives sum too big even at f=0. So maybe there is no solution? But the problem says "the solution", implying there is a solution. Maybe I made a mistake here.Wait, let's check n=4. For n=4, x in [4,5). Then n^4=256. At f=0, x=4, sum=256 +0 +256=512. At f approaching 1, x approaches 5, x^4=625. So sum approaches 256 +1 +625=882. Still too low. n=5 gives sum up to 1922, as above. n=6 gives sum starting at 2592. So 2048 is between n=5 and n=6. Wait, but n must be integer. So maybe the solution is in n=6, but with a negative f? No, f is fractional part, so must be between 0 and 1. Wait, unless x is negative. Wait, perhaps x is negative. Let's think again.Suppose x is negative. Let's take n=-6, so x is in [-6, -5). Then [x]=-6, {x}=x - (-6)=x +6, which is between 0 and1. Then x = -6 + f, where f is in [0,1). So x^4 = (-6 + f)^4. Let's compute n^4 + {x}^4 +x^4.n^4= (-6)^4=1296. {x}^4= f^4. x^4= (-6 + f)^4. Let's compute when f=0: 1296 +0 +1296=2592. When f approaches 1: x approaches -5, so x^4=625. Then sum approaches 1296 +1 +625=1922. So similar to n=5 positive. Hmm, so for n=-6, the sum goes from 2592 down to 1922 as f increases. So 2048 would lie somewhere in between. Let's see, maybe there is a value of f in (0,1) where 1296 + f^4 + (-6 + f)^4=2048.Wait, so solving for f in [0,1):1296 + f^4 + (-6 + f)^4=2048Compute (-6 + f)^4. Let me expand that:(-6 + f)^4 = (f -6)^4 = f^4 - 24f^3 + 216f^2 - 864f + 1296So plugging into the equation:1296 + f^4 + [f^4 -24f^3 +216f^2 -864f +1296] =2048Combine terms:1296 + f^4 +f^4 -24f^3 +216f^2 -864f +1296=2048Combine like terms:2f^4 -24f^3 +216f^2 -864f +1296 +1296=2048Which is:2f^4 -24f^3 +216f^2 -864f +2592=2048Subtract 2048:2f^4 -24f^3 +216f^2 -864f +544=0Divide both sides by 2:f^4 -12f^3 +108f^2 -432f +272=0Hmm, quartic equation. This seems difficult. Maybe factor?Let me check if f=1 is a root:1 -12 +108 -432 +272 =1 -12= -11; -11 +108=97; 97 -432= -335; -335 +272= -63 ≠0f=2: 16 -96 + 432 -864 +272= 16-96=-80; -80+432=352; 352-864=-512; -512+272=-240≠0f=4: 256 -768 + 1728 -1728 +272=256-768=-512; -512+1728=1216;1216-1728=-512; -512+272=-240≠0Hmm, maybe rational roots. The possible rational roots are factors of 272 over factors of 1, so ±1, ±2, ±4, ±8, ±16, ±17, etc. Let's try f=4 as above. Hmm, not working.Alternatively, maybe substitute y = f - a to eliminate the cubic term? Not sure. Alternatively, maybe factor as quadratics.Alternatively, notice that coefficients are multiples of 4? Let's see:Original equation after division by 2:f^4 -12f^3 +108f^2 -432f +272=0Hmm, 108 is 12*9, 432 is 12*36. Maybe factor as (f^2 + a f + b)(f^2 + c f + d)=0Multiply out:f^4 + (a + c)f^3 + (b + d + a c)f^2 + (a d + b c)f + b d =0Compare with coefficients:f^4 -12f^3 +108f^2 -432f +272=0So:a + c = -12b + d + a c =108a d + b c = -432b d =272Looking for integers a,b,c,d. Since b d=272. Let's factor 272: 16*17, 8*34, 4*68, 2*136, etc. Let's try b=16, d=17. Then:a + c =-1216 +17 +a c=108 => a c=108 -33=75a d +b c= a*17 +16*c=-432So with a + c=-12, so c= -12 -a.Then a*(-12 -a)=75 => -12a -a²=75 => a² +12a +75=0. Discriminant 144 -300= -156, no real roots. Not good.Try b=8, d=34:Then a + c=-128 +34 +a c=108 => a c=108 -42=66a*34 +8*c=-432Again, c=-12 -a.So a*(-12 -a)=66 => -12a -a²=66 => a² +12a +66=0. Discriminant 144 -264= -120. Nope.Try b=4, d=68:a +c=-124 +68 +a c=108 => a c=108 -72=36a*68 +4*c=-432Again, c=-12 -a.Thus, a*(-12 -a)=36 => -12a -a²=36 => a² +12a +36=0 => (a +6)^2=0 => a=-6. Then c=-12 -(-6)=-6.Check a d + b c= (-6)*68 +4*(-6)= -408 -24= -432. Which matches. So that works!Therefore, the quartic factors into (f² -6f +4)(f² -6f +68)=0Wait, let's check:(f² + a f + b)(f² +c f +d)= (f² -6f +4)(f² -6f +68)Multiply out:First terms: f^4 -6f³ +4f² -6f³ +36f² -24f +68f² -408f +272Combine like terms:f^4 + (-6 -6)f³ + (4 +36 +68)f² + (-24 -408)f +272Which is:f^4 -12f³ +108f² -432f +272. Perfect!So the equation factors into (f² -6f +4)(f² -6f +68)=0Set each quadratic to zero:First quadratic: f² -6f +4=0Solutions: f=(6 ±√(36 -16))/2=(6 ±√20)/2=(6 ±2√5)/2=3 ±√5≈3 ±2.236Which is approximately 5.236 or 0.764. But f must be in [0,1), so 0.764 is a possible solution.Second quadratic: f² -6f +68=0Solutions: f=(6 ±√(36 -272))/2=(6 ±√(-236))/2. Complex solutions, so discard.Therefore, the only real solution in [0,1) is f=3 -√5≈0.764.So f≈0.764, which is 3 -√5≈3 -2.236=0.764. So exact value is 3 -√5.Therefore, x = n + f = -6 + (3 -√5)= -3 -√5≈-3 -2.236≈-5.236.Wait, but let's check this. If n=-6, then x is in [-6, -5). So x= -6 + f, where f is in [0,1). But we have f=3 -√5≈0.764, so x≈-6 +0.764≈-5.236, which is in [-6, -5). Correct.Now, let's verify if this x satisfies the original equation.Compute [x]^4 + {x}^4 +x^4.[x]=-6, {x}=x - [x]=-5.236 - (-6)=0.764=3 -√5.So [x]^4=(-6)^4=1296.{x}^4=(3 -√5)^4. Let's compute that.First compute (3 -√5)^2=9 -6√5 +5=14 -6√5.Then (14 -6√5)^2=196 - 168√5 + 180=376 -168√5.So {x}^4=376 -168√5.x^4=(-6 + f)^4=(-6 +3 -√5)^4=(-3 -√5)^4.Compute (-3 -√5)^4. Since it's positive, same as (3 +√5)^4.(3 +√5)^2=9 +6√5 +5=14 +6√5.(14 +6√5)^2=196 + 168√5 + 180=376 +168√5.So x^4=376 +168√5.Now, sum them up:1296 + (376 -168√5) + (376 +168√5)=1296 +376 +376=1296 +752=2048.Perfect! The terms with √5 cancel out, and the total is indeed 2048. So x= -3 -√5 is the solution.Wait, but x= -6 +f= -6 +3 -√5= -3 -√5≈-5.236, which is correct. Therefore, the solution is x= -3 -√5.But let me check if there are other possible n's. For example, n=-6 gives a solution. Is there another n where this could happen?Suppose n=5: Then x is in [5,6). Let's check if the equation can be satisfied here. For n=5, the equation becomes:5^4 + f^4 + (5 +f)^4=2048.Compute 5^4=625, (5 +f)^4. Let's expand:(5 + f)^4 = 625 + 500f + 150f² +20f³ +f^4.So the equation becomes:625 + f^4 +625 +500f +150f² +20f³ +f^4=2048Combine terms:1250 + 2f^4 +500f +150f² +20f³=2048Subtract 2048:2f^4 +20f³ +150f² +500f -798=0Divide by 2:f^4 +10f³ +75f² +250f -399=0Trying to find real roots in [0,1). Let's test f=0.5:0.5^4 +10*(0.5)^3 +75*(0.5)^2 +250*(0.5) -399=0.0625 +1.25 +18.75 +125 -399= approx 0.0625+1.25=1.3125+18.75=20.0625+125=145.0625-399≈-253.9375 <0At f=1:1 +10 +75 +250 -399= 336 -399= -63 <0So the equation is negative at f=1. Since it's a quartic leading term positive, it tends to infinity as f increases, but between f=1 and higher, but f is limited to [0,1). Thus, no solution in n=5.Similarly, check n= -5. Let me see:For n=-5, x is in [-5, -4). Then [x]=-5, {x}=x +5, which is in [0,1). Then equation is (-5)^4 + f^4 + (-5 +f)^4=2048.Compute (-5)^4=625. (-5 +f)^4=(f -5)^4.So equation:625 + f^4 + (f -5)^4=2048.Expand (f -5)^4:f^4 -20f³ +150f² -500f +625.So equation becomes:625 +f^4 +f^4 -20f³ +150f² -500f +625=2048.Combine terms:2f^4 -20f³ +150f² -500f +1250=2048.Subtract 2048:2f^4 -20f³ +150f² -500f -798=0.Divide by 2:f^4 -10f³ +75f² -250f -399=0.Check for roots in [0,1). At f=0: -399 <0. At f=1:1 -10 +75 -250 -399= (1-10)=-9; (-9 +75)=66; (66 -250)=-184; (-184 -399)=-583 <0. No solution here.Similarly for other n's. So only n=-6 gives a valid solution.Wait, but let me check n=6. Wait, n=6 would mean x is in [6,7). Then equation is 6^4 + f^4 + (6 + f)^4=2048. Compute 6^4=1296. (6 + f)^4=1296 + 864f + 216f² +24f³ +f^4. Then sum:1296 + f^4 +1296 +864f +216f² +24f³ +f^4= 2592 + 2f^4 +24f³ +216f² +864f=2048. So 2592 -2048=544. Thus, 2f^4 +24f³ +216f² +864f +544=0. Divide by 2: f^4 +12f³ +108f² +432f +272=0. All coefficients positive, so no real roots in f≥0. So no solution here.Therefore, only solution is x= -3 -√5. Let me check if it's positive. -3 -√5≈-5.236, which is negative. Wait, the problem didn't specify x needs to be positive, so that's okay.But let me see, maybe there is a positive solution. Let's check n=5 again. Wait, already checked n=5 and n=6. What about n=0? Let's check n=0. Then x is in [0,1). So [x]=0, {x}=x. Then equation becomes 0 +x^4 +x^4=2x^4=2048. Thus x^4=1024. x= (1024)^(1/4)= 5.656, but x is in [0,1). So no solution here.Similarly, n=1. x in [1,2). Then equation:1^4 +f^4 + (1 +f)^4=1 +f^4 + (1 +f)^4=2048. Let's see (1 +f)^4=1 +4f +6f² +4f³ +f^4. So sum:1 +f^4 +1 +4f +6f² +4f³ +f^4=2 +2f^4 +4f +6f² +4f³=2048. Which is way too big even for f=0: 2 +0 +0 +0 +0=2<<2048. So no.So only solution is x=-3 -√5. Wait, but let me verify once more.Wait, x= -3 -√5≈-5.236. Then [x]=-6, since -6 ≤-5.236 < -5. Then {x}=x - [x]= -5.236 - (-6)=0.764. Which is 3 -√5≈0.764. So yes. Then [x]^4=(-6)^4=1296. {x}^4=(3 -√5)^4. We computed that earlier as 376 -168√5. x^4=(-3 -√5)^4=(3 +√5)^4=376 +168√5. So total sum:1296 + (376 -168√5) + (376 +168√5)=1296 +376 +376=2048. Correct.Therefore, the solution is x= -3 -√5.But let me check if there are other solutions for different n. For example, n=-7. Then x in [-7, -6). Let's see:n=-7, so [x]=-7. {x}=x +7 in [0,1). Then equation: (-7)^4 + {x}^4 +x^4=2048. So 2401 +f^4 + (-7 +f)^4=2048. Wait, 2401 is already more than 2048. So even at f=0, sum is 2401 +0 +2401=4802 >2048. So no solution here.Similarly, n= -4. x in [-4, -3). Then [x]=-4. {x}=x +4. Equation: (-4)^4 + (x +4)^4 +x^4=256 +f^4 + (-4 +f)^4=2048.Compute (-4 +f)^4= (f -4)^4= f^4 -16f³ +96f² -256f +256.So equation becomes 256 +f^4 +f^4 -16f³ +96f² -256f +256=2048.Combine terms: 2f^4 -16f³ +96f² -256f +512=2048.Subtract 2048:2f^4 -16f³ +96f² -256f -1536=0. Divide by 2: f^4 -8f³ +48f² -128f -768=0. Test f in [0,1). At f=0: -768 <0. At f=1:1 -8 +48 -128 -768= (1-8)= -7; (-7 +48)=41; (41 -128)=-87; (-87 -768)=-855 <0. No solution here.Thus, only solution is x= -3 -√5. Hence, the answer is boxed{-3 - sqrt{5}}.Final AnswerThe solution to the equation is boxed{-3 - sqrt{5}}.</think>Given the equation ([x]^4 + {x}^4 + x^4 = 2048), where ([x]) denotes the greatest integer not exceeding (x) and ({x} = x - [x]) is the fractional part of (x), we start by expressing (x) as (n + f) where (n = [x]) is an integer and (0 leq f < 1). Substituting (x = n + f) into the equation, we get:[n^4 + f^4 + (n + f)^4 = 2048]Expanding ((n + f)^4) and combining terms, we derive:[2n^4 + 2f^4 + 4n^3f + 6n^2f^2 + 4nf^3 = 2048]We analyze possible values of (n) and (f). Testing various integer values for (n) and solving for (f) within the interval ([0, 1)), we find that (n = -6) and (f = 3 - sqrt{5}) satisfy the equation. For (n = -6), (x = -6 + f) where (f = 3 - sqrt{5}). This gives:[x = -6 + (3 - sqrt{5}) = -3 - sqrt{5}]Verification shows that ([x] = -6), ({x} = 3 - sqrt{5}), and (x^4 = (-3 - sqrt{5})^4). Calculating each term:[[-3 - sqrt{5}]^4 = (-6)^4 = 1296][{-3 - sqrt{5}}^4 = (3 - sqrt{5})^4 = 376 - 168sqrt{5}][(-3 - sqrt{5})^4 = 376 + 168sqrt{5}]Summing these terms:[1296 + (376 - 168sqrt{5}) + (376 + 168sqrt{5}) = 1296 + 376 + 376 = 2048]Thus, the solution to the equation is (boxed{-3 - sqrt{5}}).,<think>Okay, so I need to solve the equation [x]^4 + {x}^4 + x^4 = 2048. Hmm, let's start by recalling what [x] and {x} mean. [x] is the greatest integer less than or equal to x, also known as the floor function. {x} is the fractional part of x, which is x minus [x]. So, for example, if x is 3.5, then [x] is 3 and {x} is 0.5. Got that.So the equation involves [x]^4, {x}^4, and x^4 adding up to 2048. Since [x] is an integer and {x} is between 0 (inclusive) and 1 (exclusive), maybe I can consider x as n + f, where n is an integer and f is a fractional part between 0 and 1. Then [x] = n and {x} = f. So substituting into the equation, we get n^4 + f^4 + (n + f)^4 = 2048.Let me write that out:n⁴ + f⁴ + (n + f)⁴ = 2048Hmm, maybe expand (n + f)^4 to see if that helps. Let me recall the binomial expansion:(n + f)^4 = n⁴ + 4n³f + 6n²f² + 4nf³ + f⁴So substituting back into the equation:n⁴ + f⁴ + [n⁴ + 4n³f + 6n²f² + 4nf³ + f⁴] = 2048Combine like terms:n⁴ + f⁴ + n⁴ + 4n³f + 6n²f² + 4nf³ + f⁴ = 2048Which simplifies to:2n⁴ + 2f⁴ + 4n³f + 6n²f² + 4nf³ = 2048Hmm, that's a bit complicated. Maybe there's a smarter way to approach this. Let's see.Alternatively, note that x = n + f, so x^4 = (n + f)^4. Then the original equation is:n⁴ + f⁴ + (n + f)^4 = 2048But maybe instead of expanding, think about possible integer values of n such that n^4 is not too large. Since 2048 is 2^11, which is 2048. Let me check the fourth powers of integers to see which could be plausible.Let's see, 6^4 is 1296, 7^4 is 2401. 2401 is already larger than 2048, so n can't be 7 or higher. Similarly, negative integers: (-7)^4 is 2401, (-6)^4 is 1296, so same as positive. But if x is negative, then [x] would be the next lower integer. For example, if x is -3.5, [x] is -4 and {x} is 0.5. Wait, fractional part is always non-negative and less than 1. So even for negative numbers, {x} is between 0 and 1. So n can be negative. Hmm. Let's check.But let's first think about positive n. Since n^4 is part of the sum, and 7^4 is 2401 which is over 2048, n can't be 7. So possible n values are from -6 to 6. Wait, but if n is negative, then x = n + f, which is between n and n + 1. If n is negative, say n = -1, then x is between -1 and 0. Let's see. Let's check if negative n could contribute.But let's try positive n first. Let's test n = 6. Then n^4 = 1296. Then the remaining terms would be f^4 + x^4. But x = 6 + f, so x^4 = (6 + f)^4. Then total equation is 1296 + f^4 + (6 + f)^4 = 2048.Compute (6 + f)^4. Let's compute (6 + f)^4:6^4 + 4*6^3*f + 6*6^2*f^2 + 4*6*f^3 + f^4 = 1296 + 864f + 216f² + 24f³ + f⁴So the equation becomes:1296 + f^4 + 1296 + 864f + 216f² + 24f³ + f⁴ = 2048Combine terms:2*1296 + 2f^4 + 864f + 216f² + 24f³ = 2048Calculate 2*1296: 2592So 2592 + 2f^4 + 864f + 216f² + 24f³ = 2048But 2592 is already larger than 2048, so this would imply negative terms, which is impossible because f is between 0 and 1, so all terms are positive. Therefore, n = 6 is too large. So n can't be 6? Wait, but x^4 would be (6 + f)^4, which is 6^4 plus something. So adding n^4 and x^4 would be 6^4 + (6 + f)^4, which is 1296 + (1296 + ...) as above, leading to over 2592, which is more than 2048. Therefore, n=6 is impossible. So n must be less than 6. Let's check n=5.n=5. Then n^4=625. Then x =5 + f, so x^4=(5 + f)^4. Let's compute that:(5 + f)^4 = 625 + 500f + 150f² + 20f³ + f⁴So the equation becomes:625 + f^4 + 625 + 500f + 150f² + 20f³ + f⁴ = 2048Combine terms:1250 + 2f^4 + 500f + 150f² + 20f³ = 2048Subtract 1250:2f⁴ + 20f³ + 150f² + 500f = 2048 - 1250 = 798So 2f⁴ + 20f³ + 150f² + 500f = 798Hmm, let's factor 2:2(f⁴ + 10f³ + 75f² + 250f) = 798Divide both sides by 2:f⁴ + 10f³ + 75f² + 250f = 399But f is between 0 and 1. Let's check f=1: Plugging f=1 gives 1 + 10 + 75 + 250 = 336, which is less than 399. So even at f=1, the left side is 336, which is less than 399. Therefore, there's no solution for n=5. So n=5 is impossible. Hmm. Maybe I made a miscalculation.Wait, let's verify the expansion of (5 + f)^4:(5 + f)^4 = 5^4 + 4*5^3 f + 6*5^2 f² + 4*5 f³ + f^4Which is 625 + 500f + 150f² + 20f³ + f^4. That's correct. Then the equation becomes n^4 + f^4 + x^4 = 625 + f^4 + 625 + 500f + 150f² + 20f³ + f^4. Wait, that would be 625 + f^4 + (625 + 500f + 150f² + 20f³ + f^4) = 1250 + 2f^4 + 500f + 150f² + 20f³. So 1250 + ... = 2048. So 2048 - 1250 = 798. So 2f^4 + 20f³ + 150f² + 500f = 798. Then dividing by 2: f^4 + 10f³ + 75f² + 250f = 399. But f is between 0 and 1. Let's check f=0.8:f=0.8:0.8^4 + 10*(0.8)^3 + 75*(0.8)^2 + 250*(0.8)0.4096 + 10*0.512 + 75*0.64 + 250*0.80.4096 + 5.12 + 48 + 200 = 0.4096 + 5.12 is 5.5296 + 48 is 53.5296 + 200 is 253.5296. That's way less than 399. Even at f=1, it's 336. So even f=1 gives 336. So no solution for n=5.Wait, so maybe n needs to be higher? But n=6 gives over 2592. So maybe n is negative?Wait, let's check n= -6. Then [x] = -6, and {x} = f, so x = -6 + f, where f is between 0 and 1. So x is between -6 and -5.Compute n^4 + f^4 + x^4 = (-6)^4 + f^4 + (-6 + f)^4 = 1296 + f^4 + ( (-6 + f) )^4.But (-6 + f) is negative because f is less than 1. So (-6 + f)^4 is the same as (6 - f)^4. So (6 - f)^4.So expanding (6 - f)^4:6^4 - 4*6^3 f + 6*6^2 f² - 4*6 f³ + f^4 = 1296 - 864f + 216f² - 24f³ + f^4So the equation becomes:1296 + f^4 + 1296 - 864f + 216f² - 24f³ + f^4 = 2048Combine terms:2*1296 + 2f^4 -864f + 216f² -24f³ = 2048Which is 2592 + 2f^4 -864f + 216f² -24f³ = 2048Subtract 2048:2f^4 -24f³ +216f² -864f +2592 -2048=0Which is 2f^4 -24f³ +216f² -864f +544=0Divide both sides by 2:f^4 -12f³ +108f² -432f +272=0Hmm, this seems complicated. Let's check if f is between 0 and 1. Let's try f=0: 0 -0 +0 -0 +272=272≠0. f=1: 1 -12 +108 -432 +272= (1 -12)= -11; (-11 +108)=97; (97 -432)= -335; (-335 +272)= -63≠0. So no solution at f=1. Maybe somewhere between 0 and 1? Let's see.But this equation is quartic and might not have roots in [0,1). Let's check f=0.5:f^4 = 0.0625-12f³ = -12*(0.125) = -1.5108f²=108*(0.25)=27-432f= -432*0.5= -216+272So total: 0.0625 -1.5 +27 -216 +272= 0.0625 -1.5= -1.4375; -1.4375 +27=25.5625; 25.5625 -216= -190.4375; -190.4375 +272=81.5625≠0. Still positive. At f=0.8:f=0.8:f^4=0.4096-12f³= -12*(0.512)= -6.144108f²=108*(0.64)=69.12-432f= -432*0.8= -345.6+272Total: 0.4096 -6.144= -5.7344; -5.7344 +69.12=63.3856; 63.3856 -345.6= -282.2144; -282.2144 +272= -10.2144. So it's negative at f=0.8, so somewhere between 0.5 and 0.8, the equation crosses zero. But since f is supposed to be between 0 and 1, but {x} is between 0 and 1. Wait, but x is between -6 and -5 in this case. So f is in [0,1), but even if there is a solution here, it's going to be complicated. Maybe this is not the right approach. Let's check n=4.n=4. Then [x]=4, x=4 + f, so {x}=f.Compute n^4 + f^4 + x^4 = 256 + f^4 + (4 + f)^4.Expand (4 + f)^4:256 + 256f + 96f² + 16f³ + f^4So the equation becomes:256 + f^4 + 256 + 256f + 96f² + 16f³ + f^4 = 2048Combine terms:512 + 2f^4 +256f +96f² +16f³=2048Subtract 512:2f^4 +16f³ +96f² +256f=1536Divide by 2:f^4 +8f³ +48f² +128f=768But f is between 0 and 1. Let's check f=1: 1 +8 +48 +128=185, which is much less than 768. So no solution here. Similarly, even if f approaches 1, left side is 185. So n=4 is too low.Hmm. Maybe n= -5?Let me try n= -5. Then x= -5 + f, where f is between 0 and1. So x is between -5 and -4.Compute n^4 + f^4 + x^4 = (-5)^4 + f^4 + (-5 + f)^4 = 625 + f^4 + ( -5 + f )^4. Again, since (-5 + f) is negative (because f <1), (-5 + f)^4 = (5 - f)^4. Let's expand (5 - f)^4:625 - 500f + 150f² -20f³ + f^4So the equation becomes:625 + f^4 + 625 -500f +150f² -20f³ +f^4 = 2048Combine terms:1250 +2f^4 -500f +150f² -20f³ =2048Subtract 1250:2f^4 -20f³ +150f² -500f=798Divide by 2:f^4 -10f³ +75f² -250f=399Again, f is between 0 and1. Check f=1: 1 -10 +75 -250= -184. Not 399. So no solution here. Maybe negative n is not the way to go.Wait, maybe n=3? Let's check n=3.n=3, x=3 + f.n^4=81. x^4=(3 + f)^4.Compute (3 + f)^4=81 + 108f + 54f² +12f³ +f^4Equation becomes 81 + f^4 +81 +108f +54f² +12f³ +f^4=2048Combine terms:162 +2f^4 +108f +54f² +12f³=2048Subtract 162:2f^4 +12f³ +54f² +108f=1886Divide by 2:f^4 +6f³ +27f² +54f=943But f is between 0 and1. At f=1:1 +6 +27 +54=88, which is way less than 943. So impossible.Similarly n=2:n=2, x=2 + f. n^4=16. x^4=(2 + f)^4=16 +32f +24f² +8f³ +f^4.Equation:16 +f^4 +16 +32f +24f² +8f³ +f^4=2048Combine:32 +2f^4 +32f +24f² +8f³=2048Subtract 32:2f^4 +8f³ +24f² +32f=2016Divide by 2:f^4 +4f³ +12f² +16f=1008f between 0 and1. f=1:1 +4 +12 +16=33 <1008. Nope.n=1: n^4=1. x=1 +f. x^4=(1 +f)^4=1 +4f +6f² +4f³ +f^4.Equation:1 +f^4 +1 +4f +6f² +4f³ +f^4=2048Combine:2 +2f^4 +4f +6f² +4f³=2048Subtract 2:2f^4 +4f³ +6f² +4f=2046Divide by 2:f^4 +2f³ +3f² +2f=1023But f is between 0 and1. At f=1:1 +2 +3 +2=8 <1023. Impossible.n=0: x=f. Then [x]=0, {x}=f, x=f. So equation:0 +f^4 +f^4=2f^4=2048. So f^4=1024. f= (1024)^(1/4). 1024 is 2^10, so fourth root is 2^(10/4)=2^2.5=~5.656. But f is between 0 and1. So no solution.So, n=0 gives no solution. n=1,2,3,4,5,6 all don't work. Negative n also seems not to work. Wait, but maybe I made a mistake with n= -1?Let me check n= -1. Then x= -1 + f, with f in [0,1), so x is between -1 and 0. So [x]= -1, {x}=x - [x]= (-1 +f) - (-1)=f. So equation:(-1)^4 +f^4 +x^4=1 +f^4 + (-1 +f)^4=2048Compute (-1 + f)^4. Since x is negative here, (-1 + f) is between -1 and 0. So (-1 +f)^4 is (1 -f)^4. Wait, actually, (-1 + f)^4 is the same as (f -1)^4, which is same as (1 - f)^4. So expand (1 -f)^4:1 -4f +6f² -4f³ +f^4So equation becomes:1 +f^4 +1 -4f +6f² -4f³ +f^4=2048Combine terms:2 +2f^4 -4f +6f² -4f³=2048Subtract 2:2f^4 -4f³ +6f² -4f=2046Divide by 2:f^4 -2f³ +3f² -2f=1023Again, f is between 0 and1. At f=1:1 -2 +3 -2=0. So left side is 0. Not 1023. No solution here.Similarly, n= -2. x= -2 + f, f in [0,1). [x]=-2, {x}=f. Equation:(-2)^4 +f^4 + (-2 + f)^4=16 +f^4 + ( (-2 +f) )^4. Since (-2 + f) is between -2 and -1, so the fourth power is positive. Let's compute it as (2 - f)^4.(2 - f)^4=16 -32f +24f² -8f³ +f^4So equation becomes:16 +f^4 +16 -32f +24f² -8f³ +f^4=2048Combine:32 +2f^4 -32f +24f² -8f³=2048Subtract 32:2f^4 -8f³ +24f² -32f=2016Divide by 2:f^4 -4f³ +12f² -16f=1008Again, f in [0,1). At f=1:1 -4 +12 -16= -7. Not 1008. So no solution.Wait, so maybe there's no solution? But the problem says "Then the solution to the equation...", implying there is a solution. Maybe I missed something.Wait, maybe n= -4? Let's check n= -4. Then x= -4 + f. [x]=-4, {x}=f. Equation:(-4)^4 +f^4 + (-4 +f)^4=256 +f^4 + (-4 +f)^4.Again, (-4 +f)^4=(4 -f)^4=256 -256f +96f² -16f³ +f^4.So equation becomes:256 +f^4 +256 -256f +96f² -16f³ +f^4=2048Combine terms:512 +2f^4 -256f +96f² -16f³=2048Subtract 512:2f^4 -16f³ +96f² -256f=1536Divide by 2:f^4 -8f³ +48f² -128f=768Again, f in [0,1). At f=1:1 -8 +48 -128= -87. No.Wait, maybe n= -3. Let's check:n= -3. x= -3 + f. [x]=-3, {x}=f. Equation:(-3)^4 +f^4 + (-3 +f)^4=81 +f^4 + (-3 +f)^4.(-3 +f) is between -3 and -2, so (-3 +f)^4=(3 -f)^4=81 -108f +54f² -12f³ +f^4Equation becomes:81 +f^4 +81 -108f +54f² -12f³ +f^4=2048Combine terms:162 +2f^4 -108f +54f² -12f³=2048Subtract 162:2f^4 -12f³ +54f² -108f=1886Divide by 2:f^4 -6f³ +27f² -54f=943At f=1:1 -6 +27 -54= -32. Not 943. No.This is getting frustrating. Maybe the solution is not an integer n. Wait, but [x] is always integer. So x must be a real number with integer part n and fractional part f. So n has to be integer, f between 0 and1.Wait, maybe n is a larger integer? Wait, earlier n=6 gave over 2592, which is too big. But n=7 is even bigger. So n can't be 7. Wait, but the problem says "greatest integer not exceeding x", so if x is, say, 6.999, then [x]=6. So n is always less than or equal to x. So for positive x, n is floor(x). If x is negative, say, -6.5, then [x]=-7, so n=-7. But in that case, {x}=x - [x]= -6.5 - (-7)=0.5.Wait, hold on! Wait, maybe I made a mistake in negative numbers. Let me check:If x is negative, say x = -k + f where k is a positive integer and 0 <= f <1, then [x] = -k -1, right? Wait, no. For example, x = -1.5. Then [x] is -2, since -2 is the greatest integer not exceeding -1.5. Then {x}=x - [x]= -1.5 - (-2)=0.5.So in general, for x negative, x = n + f, where n is integer (negative) and f is in [0,1). For example, x = -1.5 can be written as -2 + 0.5. So n = -2, f = 0.5.Therefore, in that case, [x] = n and {x} = f. So even for negative x, we can write x = n + f where n = [x] (which is less than or equal to x) and f = {x} = x - n (which is in [0,1)).So regardless of x being positive or negative, x = n + f, with n integer and 0 <= f <1.So in the equation, [x]^4 + {x}^4 + x^4 = n^4 + f^4 + (n + f)^4 =2048.So maybe I need to check for possible n beyond positive 6? But 7^4 is 2401, which is already bigger than 2048. So even if n=7, n^4=2401>2048. So even if x=7.0, then [x]=7, {x}=0, x=7.0. Then equation becomes 7^4 +0 +7^4= 2401 +0 +2401=4802, way over. So n=7 is impossible.So maybe there's no solution? But the problem says "the solution to the equation". Hmm. Wait, perhaps x is not an integer, but when combining n and f, maybe the sum works out. But in all the cases I checked, even with n=6 and f=1, which is not allowed, but gives 2*1296 + ... which is over 2048. For n=5, even f=1 gives 1250 + 336=1586 <2048. So maybe the solution is between n=5 and n=6? But n must be integer.Wait, perhaps x is not an integer, but n is integer, and f is fractional. So n can't be 5.5, for example. So n must be integer. So perhaps the equation has no solution? But the problem states "the solution to the equation". So maybe I made an error in the expansion.Wait, let me go back to the original equation:[x]^4 + {x}^4 +x^4 =2048.But x = [x] + {x}, so x =n + f. Then x^4=(n + f)^4. So the equation is n^4 + f^4 + (n + f)^4 =2048.Alternatively, perhaps combining terms:n^4 + (n + f)^4 + f^4 =2048.But maybe there's a better way to write (n + f)^4 +n^4 +f^4. Let's see.Wait, (n + f)^4 + n^4 + f^4. Let me think if this can be factored or simplified.Alternatively, maybe factor out something. Let's see:(n + f)^4 +n^4 +f^4.Let me compute this for specific n and f. For example, suppose n=4, f=0.5.But this might not help. Alternatively, maybe there's a case where {x}=0, so f=0. Then the equation becomes n^4 +0 +n^4=2n^4=2048. Then n^4=1024. So n=1024^(1/4). 1024=2^10, so 1024^(1/4)=2^(10/4)=2^2.5= sqrt(2^5)=sqrt(32)=~5.656. But n must be integer. So no solution here.Similarly, if f=1, but f can't be 1. So even if f approaches 1, for n=5, as we saw earlier, the maximum left-hand side is 1250 +336=1586<2048. So maybe there's no solution? But the problem says there is.Wait, maybe I made a miscalculation when n= -6. Let me check again.For n= -6, x= -6 +f, 0<=f<1.Equation: (-6)^4 +f^4 + (-6 +f)^4=1296 +f^4 + ( -6 +f )^4.But (-6 +f)^4= (f -6)^4= (6 -f)^4=1296 - 864f + 216f² -24f³ +f^4.So equation becomes:1296 +f^4 +1296 -864f +216f² -24f³ +f^4=2048Combine terms:2592 +2f^4 -864f +216f² -24f³=2048Subtract 2048:2f^4 -24f³ +216f² -864f +544=0Divide by 2:f^4 -12f³ +108f² -432f +272=0Now, let's try to factor this equation. Maybe possible rational roots are factors of 272 over 1. So possible roots: ±1, ±2, ±4, ±8, ±16, ±17, etc.Test f=2: 16 - 96 + 432 -864 +272= (16-96)= -80; (-80+432)=352; (352 -864)= -512; (-512 +272)= -240≠0.f=4: 256 - 768 + 1728 -1728 +272= (256 -768)= -512; (-512 +1728)=1216; (1216 -1728)= -512; (-512 +272)= -240≠0.f=1:1 -12 +108 -432 +272= (1-12)= -11; (-11 +108)=97; (97 -432)= -335; (-335 +272)= -63≠0.f=0.5: (0.5)^4 -12*(0.5)^3 +108*(0.5)^2 -432*(0.5) +272=0.0625 -12*(0.125) +108*(0.25) -432*0.5 +272=0.0625 -1.5 +27 -216 +272=0.0625 -1.5= -1.4375 +27=25.5625 -216= -190.4375 +272=81.5625≠0.Hmm, so no obvious roots here. Maybe there are real roots between 0 and1? Let's check the value at f=0: 0 -0 +0 -0 +272=272>0.At f=1: -63<0. So by Intermediate Value Theorem, there is a root between f=0.8 and f=1. Let's check f=0.9:f=0.9:0.9^4 -12*(0.9)^3 +108*(0.9)^2 -432*(0.9) +272=0.6561 -12*(0.729) +108*(0.81) -388.8 +272=0.6561 -8.748 +87.48 -388.8 +272Calculate step by step:0.6561 -8.748= -8.0919-8.0919 +87.48=79.388179.3881 -388.8= -309.4119-309.4119 +272= -37.4119 <0At f=0.8:We calculated before as 81.5625>0At f=0.85:f=0.85:0.85^4=0.52200625-12*(0.85)^3= -12*(0.614125)= -7.3695108*(0.85)^2=108*0.7225=78.03-432*0.85= -367.2+272Total:0.52200625 -7.3695= -6.84749375 +78.03=71.18250625 -367.2= -296.01749375 +272= -24.01749375 <0At f=0.75:0.75^4=0.31640625-12*(0.421875)= -5.0625108*(0.5625)=60.75-432*0.75= -324+272Total:0.31640625 -5.0625= -4.74609375 +60.75=56.00390625 -324= -267.99609375 +272=4.00390625>0So between f=0.75 and f=0.85, it crosses from positive to negative. Let's try f=0.775:f=0.775:0.775^4≈0.775^2=0.600625; squared≈0.600625^2≈0.3607509-12*(0.775)^3≈-12*(0.775*0.775=0.600625*0.775≈0.4662890625)≈-12*0.4662890625≈-5.59546875108*(0.775)^2≈108*0.600625≈64.8675-432*0.775≈-335.4+272Total≈0.3607509 -5.59546875≈-5.23471785 +64.8675≈59.63278215 -335.4≈-275.76721785 +272≈-3.76721785 <0At f=0.76:f=0.76:0.76^4≈0.33362176-12*(0.76)^3≈-12*(0.438976)= -5.267712108*(0.76)^2≈108*(0.5776)=62.5008-432*0.76≈-328.32+272Total≈0.33362176 -5.267712≈-4.93409024 +62.5008≈57.56670976 -328.32≈-270.75329024 +272≈1.24670976>0So between f=0.76 and f=0.775, the function crosses zero. So there is a solution for f in that interval. But since f is in [0,1), then x= n + f= -6 +0.76…≈-5.23… So x≈-5.23… is a solution.Wait, but the problem says "the solution", implying there is one solution. But maybe this quartic equation has only one real root in [0,1). But since the problem is likely expecting an exact answer, not a numerical approximation, perhaps there's a specific x that works.Alternatively, maybe there's a clever substitution or identity that can simplify the equation.Wait, let's consider that n and f are parts of x. Maybe if x is an integer, f=0. Then equation is 2n^4=2048 =>n^4=1024, which is not an integer fourth power. So no solution.If f=0.5, then {x}=0.5, x=n+0.5. Then the equation is n^4 + (0.5)^4 + (n+0.5)^4=2048.Let's compute (n +0.5)^4= n^4 + 2n³*(0.5) + 3n²*(0.5)^2 + 2n*(0.5)^3 +0.5^4= n^4 +n³ + (3/4)n² + (1]

🔑:9. -3-sqrt{5}.begin{array}{l}text { Let } a=[x], b={x} text {. Then } x=[x]+{x}=a+b . text { Hence } x^{4}+[x]^{4}+{x}^{4} =(a+b)^{4}+a^{4}+b^{4} =2left(a^{4}+2 a^{3} b+3 a^{2} b^{2}+2 a b^{3}+b^{4}right) =2left(a^{2}+a b+b^{2}right)^{2} Rightarrow a^{2}+a b+b^{2}=32 .end{array}If a>0, thenbegin{array}{l}a^{2} leqslant a^{2}+a b+b^{2}<a^{2}+a+1<(a+1)^{2} Rightarrow a=5 .end{array}Thus, the equation (1) becomesb^{2}+5 b-7=0 Rightarrow b=frac{-5 pm sqrt{53}}{2} text {, }but 0 leqslant b<1, no solution.If a<0, thenbegin{array}{l}a^{2}+a<a^{2}+a b+b^{2}<a^{2}+1 Rightarrow a^{2}+2 a+1<a^{2}+a b+b^{2} leqslant a^{2} Rightarrow a=-6 .end{array}Thus, the equation (1) becomesb^{2}-6 b+4=0 Rightarrow b=3 pm sqrt{5} text {. }Since 0 leqslant b<1, hence b=3-sqrt{5}.Therefore, x=a+b=-3-sqrt{5}.