answer：Okay, so I have this problem where I'm given two sets, M and N. M is defined as all real numbers x such that x squared is greater than 4, and N is all real numbers x such that x is less than 3. I need to figure out which of the options A, B, C, or D are correct regarding the union and intersection of these sets.First, let me understand what set M is. M is {x | x² > 4}. To solve this inequality, I can take the square root of both sides, but I have to remember that taking square roots in inequalities can sometimes be tricky because of the absolute value. So, x² > 4 implies that |x| > 2, which means x is either greater than 2 or less than -2. So, M is the union of two intervals: (-∞, -2) and (2, ∞).Next, set N is given as {x | x < 3}, which is straightforward. This is just all real numbers less than 3, so the interval (-∞, 3).Now, I need to find the union and intersection of M and N.Starting with the union, M ∪ N. The union of two sets includes all elements that are in either set. So, M is (-∞, -2) ∪ (2, ∞), and N is (-∞, 3). When I take the union of these, I need to see what parts overlap and what parts don't. Looking at M, it includes everything less than -2 and everything greater than 2. N includes everything less than 3. So, combining these, the union would include everything less than 3 and everything greater than 2. But wait, everything greater than 2 is already included in N up to 3, and beyond 3, M continues to infinity. So, actually, M ∪ N would be all real numbers because:- From negative infinity to -2, it's covered by M.- From -2 to 3, it's covered by N.- From 3 onwards, it's covered by M.So, M ∪ N is indeed all real numbers, which is option D.Now, moving on to the intersection, M ∩ N. The intersection includes only the elements that are in both sets. So, M is (-∞, -2) ∪ (2, ∞), and N is (-∞, 3). The overlap between these two sets would be:- For the part of M that's less than -2, since N goes all the way to negative infinity, the intersection here is (-∞, -2).- For the part of M that's greater than 2, N only goes up to 3, so the intersection here is (2, 3).So, putting these together, M ∩ N is (-∞, -2) ∪ (2, 3). Now, looking at the options, option B says M ∩ N is {x | 2 < |x| < 3}, which translates to (-3, -2) ∪ (2, 3). Wait, that's not exactly what I have. I have (-∞, -2) ∪ (2, 3), which is a bit different because it includes all numbers less than -2, not just between -3 and -2.Option C says M ∩ N is {x | 2 < x < 3}, which is only the positive part, (2, 3). That's only half of the intersection. So, neither B nor C fully capture the intersection correctly. However, option B is closer because it includes the absolute value, which accounts for both negative and positive sides, but it restricts the negative side to -3, which isn't necessary because M ∩ N includes all x < -2, not just between -3 and -2.Wait, maybe I made a mistake. Let me double-check. M is x² > 4, which is |x| > 2, so x < -2 or x > 2. N is x < 3. So, the intersection would be x < -2 or 2 < x < 3. So, M ∩ N is (-∞, -2) ∪ (2, 3). Looking back at the options, option B says {x | 2 < |x| < 3}, which is equivalent to (-3, -2) ∪ (2, 3). But in our case, the intersection is (-∞, -2) ∪ (2, 3). So, the negative part is not restricted to -3, it's all the way to negative infinity. Therefore, option B is not entirely correct because it limits the negative side to -3, whereas the actual intersection goes all the way to negative infinity.Option C is {x | 2 < x < 3}, which is only the positive part. So, it's missing the negative part entirely. Therefore, neither B nor C is fully correct. However, if I have to choose between them, option B is closer because it at least considers both sides, but it's still not entirely accurate.Wait, maybe I need to reconsider. The problem is asking for the correct options, and I have to choose from A, B, C, D. I already determined that D is correct because M ∪ N is all real numbers. For the intersection, since neither B nor C is entirely correct, but B is closer, maybe it's acceptable? Or perhaps the question expects a different interpretation.Alternatively, maybe I should express the intersection in terms of absolute value. Since M is |x| > 2, and N is x < 3, the intersection would be |x| > 2 and x < 3. So, for x < 3, |x| > 2 implies that x < -2 or 2 < x < 3. So, in terms of absolute value, it's 2 < |x| < 3 for x positive, but for x negative, it's |x| > 2 without an upper limit. So, actually, the intersection cannot be neatly expressed as 2 < |x| < 3 because for negative x, it's just |x| > 2, not bounded by 3. Therefore, option B is not entirely correct.So, perhaps none of the options B or C are correct. But since the question provides options, and I have to choose from them, maybe I need to see if the intersection can be expressed differently.Wait, another way to look at it: M ∩ N is (-∞, -2) ∪ (2, 3). So, in terms of absolute value, for the positive side, it's 2 < x < 3, and for the negative side, it's x < -2. So, combining these, it's x < -2 or 2 < x < 3. So, in terms of absolute value, it's |x| > 2 and x < 3. But that's not a standard interval expression.Alternatively, maybe the problem expects the intersection to be expressed as 2 < |x| < 3, which would be (-3, -2) ∪ (2, 3). But in reality, the intersection is (-∞, -2) ∪ (2, 3). So, the negative part is not limited to -3, it's all the way to negative infinity. Therefore, option B is not entirely correct because it restricts the negative side to -3, whereas the actual intersection is unbounded on the negative side.Similarly, option C only considers the positive side, so it's incomplete.Therefore, the only correct option is D: M ∪ N = ℝ.Wait, but the problem asks to determine the correct option(s), so maybe both D and another one? But I don't think so because the intersection isn't correctly described by B or C.Alternatively, maybe I made a mistake in interpreting the intersection. Let me try again.M is x² > 4, which is x < -2 or x > 2.N is x < 3.So, the intersection M ∩ N would be the set of x such that x < -2 or (x > 2 and x < 3). So, that's x < -2 or 2 < x < 3. So, in interval notation, (-∞, -2) ∪ (2, 3).Now, looking at the options:A: M ∪ N = {x | x < 3} → This is incorrect because M ∪ N is all real numbers, not just x < 3.B: M ∩ N = {x | 2 < |x| < 3} → This would be (-3, -2) ∪ (2, 3), which is not the same as (-∞, -2) ∪ (2, 3). So, B is incorrect.C: M ∩ N = {x | 2 < x < 3} → This is only the positive part, missing the negative part. So, C is incorrect.D: M ∪ N = ℝ → This is correct because M ∪ N covers all real numbers.Therefore, the only correct option is D.Wait, but earlier I thought that the intersection could be expressed as 2 < |x| < 3, but that's not accurate because the negative part isn't bounded by -3. So, B is incorrect.So, to summarize:- M ∪ N = ℝ → Correct (Option D)- M ∩ N = (-∞, -2) ∪ (2, 3) → None of the options B or C correctly describe this.Therefore, only option D is correct.

answer：First, I need to determine the number of white marbles. Given that the probability of drawing a white marble is 1/4 and there are 84 marbles in total, I can calculate the number of white marbles by multiplying the total number of marbles by the probability of drawing a white marble.Next, I'll calculate the number of green marbles using the given probability of 2/7. Similar to the previous step, I'll multiply the total number of marbles by the probability of drawing a green marble.After finding the number of white and green marbles, I'll subtract these from the total number of marbles to determine the number of red and blue marbles combined.Finally, to find the probability of drawing either a red or blue marble, I'll divide the number of red and blue marbles by the total number of marbles.

answer：Okay, so I have this problem here about a sequence of squares. Let me try to understand it step by step. The problem says there are eleven squares with side lengths 3, 6, 9, 12, and so on up to 33. So, each side length is a multiple of 3, starting from 3 and going up by 3 each time until we reach 33. That makes sense because 3 times 11 is 33, so there are 11 terms in this sequence.Now, each of these squares has an area, right? The area of a square is side length squared. So, for each square, the area would be (3)^2, (6)^2, (9)^2, and so on. So, the areas are 9, 36, 81, etc., up to 33 squared, which is 1089.The problem then says that eleven copies of a single square, each with area A, have the same total area as the total area of these eleven squares. So, if I add up all the areas of the eleven squares in the sequence, that total should be equal to 11 times A. Therefore, I need to find A such that 11A equals the sum of the areas of these squares.Alright, so my first task is to find the sum of the areas of these eleven squares. Let me write down the areas:1. 3^2 = 92. 6^2 = 363. 9^2 = 814. 12^2 = 1445. 15^2 = 2256. 18^2 = 3247. 21^2 = 4418. 24^2 = 5769. 27^2 = 72910. 30^2 = 90011. 33^2 = 1089Now, I need to add all these numbers together. Let me list them again for clarity:9, 36, 81, 144, 225, 324, 441, 576, 729, 900, 1089.Hmm, adding all these up manually might be time-consuming and prone to error. Maybe there's a formula I can use instead. I remember that the sum of the squares of the first n natural numbers is given by n(n + 1)(2n + 1)/6. But in this case, the side lengths are multiples of 3, so the areas are 9 times the squares of the first 11 natural numbers.Wait, let me think about that. If each side length is 3n, where n goes from 1 to 11, then the area of each square is (3n)^2 = 9n^2. So, the total area is the sum from n=1 to n=11 of 9n^2, which is 9 times the sum of the squares of the first 11 natural numbers.Yes, that seems right. So, I can use the formula for the sum of squares:Sum = n(n + 1)(2n + 1)/6.Here, n is 11. So, plugging in:Sum = 11 * 12 * (2*11 + 1)/6.Let me compute that step by step.First, 2*11 + 1 is 22 + 1 = 23.So, the sum becomes 11 * 12 * 23 / 6.Now, let's compute 11 * 12 first. 11 * 12 is 132.Then, 132 * 23. Hmm, let me calculate that.23 times 100 is 2300, 23 times 30 is 690, and 23 times 2 is 46. So, 2300 + 690 is 2990, plus 46 is 3036.So, 11 * 12 * 23 = 3036.Now, divide that by 6: 3036 / 6.Let me do that division. 6 goes into 30 five times, 5*6=30, subtract 30, bring down the 3. 6 goes into 3 zero times, so we have 50 so far. Bring down the 6, making it 36. 6 goes into 36 six times. So, 506.So, the sum of the squares from 1 to 11 is 506.But remember, the total area is 9 times this sum because each area is 9n^2. So, total area is 9 * 506.Let me compute 9 * 506.9 * 500 is 4500, and 9 * 6 is 54, so 4500 + 54 = 4554.So, the total area of all eleven squares is 4554.Now, the problem states that eleven copies of a single square, each with area A, have the same total area. So, 11A = 4554.To find A, I need to divide 4554 by 11.Let me compute that. 11 goes into 45 four times (11*4=44), remainder 1. Bring down the 5, making it 15. 11 goes into 15 once, remainder 4. Bring down the 4, making it 44. 11 goes into 44 four times exactly. So, putting it all together, 4554 divided by 11 is 414.Therefore, A is 414.Wait, let me double-check my calculations to make sure I didn't make any mistakes.First, the sum of the squares formula: n(n + 1)(2n + 1)/6. For n=11, that's 11*12*23/6. 11*12 is 132, 132*23 is 3036, divided by 6 is 506. So, that part is correct.Then, total area is 9*506. 9*500 is 4500, 9*6 is 54, so 4554. That seems right.Then, 4554 divided by 11. Let me do this division again.11 into 45 is 4, remainder 1. 11 into 15 is 1, remainder 4. 11 into 44 is 4, remainder 0. So, 414. Yes, that's correct.So, A is 414. Therefore, the area of each of the eleven identical squares is 414.I think that makes sense. The total area is 4554, and dividing that by 11 gives 414. So, each of the eleven squares has an area of 414.Just to make sure, let me think about whether 414 is a reasonable number. The areas of the original squares range from 9 up to 1089, so 414 is somewhere in the middle. It's not too big or too small, so it seems plausible.Alternatively, I could have approached this problem by recognizing that the side lengths form an arithmetic sequence, and the areas form a sequence of squares of that arithmetic sequence. But since each term is a multiple of 3, the areas are multiples of 9, which simplifies the sum.Another way to think about it is that the areas are 9, 36, 81, etc., which are 9 times 1, 4, 9, 16, etc., so it's 9 times the sum of the squares of the first 11 natural numbers. That's exactly what I did, so that confirms my approach.I don't see any mistakes in my reasoning, so I'm confident that A is 414.

answer：Alright, so I've got this problem here that's asking whether there exist positive integers ( n_1, n_2, dots, n_{2022} ) such that the product [left( n_1^{2020} + n_2^{2019} right)left( n_2^{2020} + n_3^{2019} right) cdots left( n_{2021}^{2020} + n_{2022}^{2019} right)left( n_{2022}^{2020} + n_1^{2019} right)]is a power of 11. Hmm, okay, so the product is supposed to be ( 11^k ) for some integer ( k ). First, I need to understand what it means for a number to be a power of 11. It means that all the prime factors of that number are 11. So, if the product is a power of 11, then each of the terms in the product must also be powers of 11, right? Because if any term had a prime factor other than 11, then the entire product would have that prime factor as well, which would contradict it being a power of 11.So, that suggests that each ( n_i^{2020} + n_{i+1}^{2019} ) must be a power of 11. Let me write that down:For each ( i ), ( n_i^{2020} + n_{i+1}^{2019} = 11^{k_i} ) for some integer ( k_i ).Now, since we're dealing with exponents, maybe I can analyze this modulo 11. Fermat's Little Theorem says that if ( a ) is not divisible by 11, then ( a^{10} equiv 1 mod 11 ). So, perhaps I can use that to simplify the exponents.Let me see, 2020 divided by 10 is 202, so 2020 is a multiple of 10. Similarly, 2019 divided by 10 is 201 with a remainder of 9. So, 2019 is 10*201 + 9. Therefore, ( n_i^{2020} equiv (n_i^{10})^{202} equiv 1^{202} equiv 1 mod 11 ) if ( n_i ) is not divisible by 11. Similarly, ( n_{i+1}^{2019} equiv n_{i+1}^9 mod 11 ).So, putting that together, ( n_i^{2020} + n_{i+1}^{2019} equiv 1 + n_{i+1}^9 mod 11 ). But since ( n_{i+1} ) is not divisible by 11 (because otherwise, the term would be divisible by 11, but the sum would have to be a power of 11, which is only divisible by 11), so ( n_{i+1} ) is coprime to 11. Therefore, ( n_{i+1}^9 equiv (n_{i+1}^{10}) cdot n_{i+1}^{-1} equiv 1 cdot n_{i+1}^{-1} mod 11 ). Wait, that might complicate things. Maybe I should just consider that ( n_{i+1}^9 ) is some number modulo 11.But actually, since ( n_{i+1}^9 ) is congruent to some number between 1 and 10 modulo 11, because ( n_{i+1} ) is not divisible by 11. So, ( n_i^{2020} + n_{i+1}^{2019} equiv 1 + c mod 11 ), where ( c ) is between 1 and 10. Therefore, the entire term is congruent to ( 1 + c mod 11 ), which is between 2 and 11 modulo 11.But since the term is supposed to be a power of 11, it must be congruent to 0 modulo 11. However, ( 1 + c ) is between 2 and 11, so it can only be congruent to 0 modulo 11 if ( 1 + c = 11 ), which would mean ( c = 10 ). So, ( n_{i+1}^9 equiv 10 mod 11 ).Therefore, for each ( i ), ( n_{i+1}^9 equiv 10 mod 11 ). Hmm, so each ( n_{i+1} ) must satisfy ( n_{i+1}^9 equiv 10 mod 11 ). Let me see if such numbers exist.I know that 10 is congruent to -1 modulo 11, so ( n_{i+1}^9 equiv -1 mod 11 ). Let me see what numbers raised to the 9th power modulo 11 give -1.Let me try small numbers:- ( 1^9 equiv 1 mod 11 )- ( 2^9 ): Let's compute ( 2^5 = 32 equiv 10 mod 11 ), then ( 2^9 = 2^5 cdot 2^4 = 10 cdot 5 = 50 equiv 6 mod 11 )- ( 3^9 ): ( 3^5 = 243 equiv 1 mod 11 ), so ( 3^9 = 3^5 cdot 3^4 = 1 cdot 81 equiv 4 mod 11 )- ( 4^9 ): ( 4^5 = 1024 equiv 1 mod 11 ), so ( 4^9 = 4^5 cdot 4^4 = 1 cdot 256 equiv 3 mod 11 )- ( 5^9 ): ( 5^5 = 3125 equiv 10 mod 11 ), so ( 5^9 = 5^5 cdot 5^4 = 10 cdot 625 equiv 10 cdot 4 = 40 equiv 7 mod 11 )- ( 6^9 ): ( 6^5 = 7776 equiv 10 mod 11 ), so ( 6^9 = 6^5 cdot 6^4 = 10 cdot 1296 equiv 10 cdot 9 = 90 equiv 2 mod 11 )- ( 7^9 ): ( 7^5 = 16807 equiv 10 mod 11 ), so ( 7^9 = 7^5 cdot 7^4 = 10 cdot 2401 equiv 10 cdot 5 = 50 equiv 6 mod 11 )- ( 8^9 ): ( 8^5 = 32768 equiv 10 mod 11 ), so ( 8^9 = 8^5 cdot 8^4 = 10 cdot 4096 equiv 10 cdot 1 = 10 mod 11 )- ( 9^9 ): ( 9^5 = 59049 equiv 1 mod 11 ), so ( 9^9 = 9^5 cdot 9^4 = 1 cdot 6561 equiv 1 cdot 4 = 4 mod 11 )- ( 10^9 equiv (-1)^9 equiv -1 mod 11 )Ah, so ( 10^9 equiv -1 mod 11 ). Therefore, ( n_{i+1} equiv 10 mod 11 ). So, each ( n_{i+1} ) must be congruent to 10 modulo 11. That is, ( n_{i+1} = 11m + 10 ) for some integer ( m geq 0 ).So, if each ( n_i equiv 10 mod 11 ), then ( n_i^{2020} + n_{i+1}^{2019} equiv 1 + 10^{2019} mod 11 ). Wait, but earlier I thought ( n_i^{2020} equiv 1 mod 11 ) and ( n_{i+1}^{2019} equiv 10 mod 11 ). Let me verify that.Given ( n_i equiv 10 mod 11 ), then ( n_i^{2020} equiv 10^{2020} mod 11 ). Since ( 10 equiv -1 mod 11 ), so ( 10^{2020} equiv (-1)^{2020} equiv 1 mod 11 ). Similarly, ( n_{i+1}^{2019} equiv 10^{2019} equiv (-1)^{2019} equiv -1 mod 11 ). Therefore, ( n_i^{2020} + n_{i+1}^{2019} equiv 1 + (-1) equiv 0 mod 11 ). So, each term is divisible by 11, which is good because the product is supposed to be a power of 11.But wait, earlier I thought that ( n_{i+1}^9 equiv 10 mod 11 ), but now I see that ( n_{i+1} equiv 10 mod 11 ) suffices to make ( n_{i+1}^{2019} equiv -1 mod 11 ). So, that seems consistent.But now, if each ( n_i equiv 10 mod 11 ), then each term ( n_i^{2020} + n_{i+1}^{2019} equiv 0 mod 11 ), but we need each term to be exactly a power of 11, not just divisible by 11. So, perhaps each term is 11 times something, but we need the entire product to be a power of 11, so each term must be a power of 11.Therefore, each ( n_i^{2020} + n_{i+1}^{2019} ) must be equal to ( 11^{k_i} ) for some ( k_i geq 1 ).Given that, perhaps I can set each ( n_i = 10 ). Let me try that.If ( n_i = 10 ) for all ( i ), then each term becomes ( 10^{2020} + 10^{2019} = 10^{2019}(10 + 1) = 10^{2019} times 11 ). So, each term is ( 11 times 10^{2019} ). Therefore, the product would be ( (11 times 10^{2019})^{2022} = 11^{2022} times 10^{2019 times 2022} ). But this is not a pure power of 11 because of the ( 10^{2019 times 2022} ) factor. So, that doesn't work.Hmm, so setting all ( n_i = 10 ) doesn't work because we get extra factors of 10. Maybe I need to choose ( n_i ) such that ( n_i^{2020} + n_{i+1}^{2019} ) is exactly a power of 11, without any other prime factors.Given that, perhaps each ( n_i ) must be a multiple of 11? Wait, but earlier I thought that if ( n_i ) is divisible by 11, then ( n_i^{2020} ) would be divisible by ( 11^{2020} ), which would make the term ( n_i^{2020} + n_{i+1}^{2019} ) divisible by ( 11^{2020} ), but then the entire product would have a very high power of 11, but we need the product to be a power of 11, which is okay, but then the other terms might not be compatible.Wait, but if ( n_i ) is divisible by 11, then ( n_i^{2020} ) is divisible by ( 11^{2020} ), and ( n_{i+1}^{2019} ) is divisible by ( 11^{2019} ) if ( n_{i+1} ) is also divisible by 11. But then, ( n_i^{2020} + n_{i+1}^{2019} ) would be divisible by ( 11^{2019} ), but not necessarily by a higher power, unless both terms are divisible by higher powers.But if we set all ( n_i ) to be multiples of 11, say ( n_i = 11m_i ), then each term becomes ( (11m_i)^{2020} + (11m_{i+1})^{2019} = 11^{2020}m_i^{2020} + 11^{2019}m_{i+1}^{2019} = 11^{2019}(11m_i^{2020} + m_{i+1}^{2019}) ). So, each term is divisible by ( 11^{2019} ), but the product would be ( (11^{2019})^{2022} times prod (11m_i^{2020} + m_{i+1}^{2019}) ). So, the entire product is ( 11^{2019 times 2022} times prod (11m_i^{2020} + m_{i+1}^{2019}) ). For this to be a power of 11, the product ( prod (11m_i^{2020} + m_{i+1}^{2019}) ) must also be a power of 11. But now, each ( 11m_i^{2020} + m_{i+1}^{2019} ) must be a power of 11 as well. Let me denote ( m_i = n_i' ), so we have ( 11n_i'^{2020} + n_{i+1}'^{2019} = 11^{k_i} ). This seems similar to the original problem but with smaller exponents. Maybe I can apply the same logic recursively. If I assume that each ( n_i' ) is also a multiple of 11, then I can factor out another power of 11, but this would lead to an infinite descent unless all ( n_i' ) are zero, which is not allowed since we need positive integers. Therefore, this approach might not work.Alternatively, perhaps not all ( n_i ) are multiples of 11. Maybe some are and some aren't. But if any ( n_i ) is not a multiple of 11, then ( n_i^{2020} equiv 1 mod 11 ), and ( n_{i+1}^{2019} equiv c mod 11 ), so ( n_i^{2020} + n_{i+1}^{2019} equiv 1 + c mod 11 ). For this to be a power of 11, it must be congruent to 0 modulo 11, so ( 1 + c equiv 0 mod 11 ), which means ( c equiv 10 mod 11 ). Therefore, ( n_{i+1}^{2019} equiv 10 mod 11 ), which as before, implies ( n_{i+1} equiv 10 mod 11 ).So, if ( n_i ) is not divisible by 11, then ( n_{i+1} equiv 10 mod 11 ). But then, ( n_{i+1} ) could be 10, 21, 32, etc. If ( n_{i+1} = 10 ), then ( n_{i+1}^{2019} = 10^{2019} ). Then, ( n_i^{2020} + 10^{2019} ) must be a power of 11. Let's see if this is possible.Suppose ( n_i^{2020} + 10^{2019} = 11^k ). Let's try small values. If ( n_i = 1 ), then ( 1 + 10^{2019} ). But ( 10^{2019} ) is a huge number, and adding 1 to it won't make it a power of 11. Similarly, if ( n_i = 2 ), ( 2^{2020} + 10^{2019} ) is also not likely to be a power of 11. It seems difficult to find such ( n_i ).Alternatively, maybe ( n_i ) is also 10. Then, ( 10^{2020} + 10^{2019} = 10^{2019}(10 + 1) = 10^{2019} times 11 ). So, this is ( 11 times 10^{2019} ), which is not a pure power of 11 because of the ( 10^{2019} ) factor. So, that doesn't work either.Hmm, maybe I need to consider that each term ( n_i^{2020} + n_{i+1}^{2019} ) is exactly 11, the smallest power of 11. Let's see if that's possible.So, ( n_i^{2020} + n_{i+1}^{2019} = 11 ). Let's try small values for ( n_i ) and ( n_{i+1} ).If ( n_i = 1 ), then ( 1 + n_{i+1}^{2019} = 11 ), so ( n_{i+1}^{2019} = 10 ). But 10 is not a perfect 2019th power, so no solution here.If ( n_i = 2 ), ( 2^{2020} ) is a huge number, much larger than 11, so ( n_{i+1}^{2019} ) would have to be negative, which is impossible since ( n_{i+1} ) is positive.Similarly, ( n_i = 3 ) would make ( 3^{2020} ) even larger. So, it seems impossible to have ( n_i^{2020} + n_{i+1}^{2019} = 11 ).Therefore, maybe each term must be a higher power of 11, like ( 11^2 = 121 ), ( 11^3 = 1331 ), etc.Let's try ( n_i^{2020} + n_{i+1}^{2019} = 121 ).Again, trying small ( n_i ):- ( n_i = 1 ): ( 1 + n_{i+1}^{2019} = 121 ) → ( n_{i+1}^{2019} = 120 ). Not a perfect power.- ( n_i = 2 ): ( 2^{2020} ) is way larger than 121.- Similarly, ( n_i = 3 ) is too big.So, no solution here either.This suggests that it's impossible to have each term be a small power of 11. Maybe the terms have to be larger powers, but then the product would have a very high power of 11, which might not be compatible with the other terms.Alternatively, perhaps all terms are equal to the same power of 11, say ( 11^k ), and then the product would be ( (11^k)^{2022} = 11^{2022k} ), which is a power of 11.But for that, each ( n_i^{2020} + n_{i+1}^{2019} = 11^k ). So, all terms are equal. Maybe setting all ( n_i ) equal to each other. Let me try that.Let ( n_1 = n_2 = dots = n_{2022} = n ). Then, each term becomes ( n^{2020} + n^{2019} = n^{2019}(n + 1) ). So, ( n^{2019}(n + 1) = 11^k ).Since ( 11 ) is prime, both ( n^{2019} ) and ( n + 1 ) must be powers of 11. Let ( n = 11^m ). Then, ( n + 1 = 11^m + 1 ). For ( n + 1 ) to be a power of 11, ( 11^m + 1 = 11^p ) for some ( p > m ). So, ( 11^p - 11^m = 1 ). Let's see if this is possible.If ( m = 0 ), then ( 11^p - 1 = 1 ) → ( 11^p = 2 ), which is not possible since ( 11^p ) is at least 11.If ( m = 1 ), then ( 11^p - 11 = 1 ) → ( 11^p = 12 ), which is not a power of 11.If ( m = 2 ), ( 11^p - 121 = 1 ) → ( 11^p = 122 ), not a power of 11.Similarly, higher ( m ) will give even larger numbers, so no solution here.Therefore, setting all ( n_i ) equal doesn't work either.Hmm, maybe I need to consider that the terms alternate between different powers of 11. For example, some terms are ( 11^a ) and others are ( 11^b ), such that their product is ( 11^{a + b + dots} ). But this seems complicated, and I'm not sure how to proceed.Alternatively, perhaps the only way for the product to be a power of 11 is if each term is a power of 11, and their exponents add up appropriately. But as I saw earlier, it's difficult to find ( n_i ) such that ( n_i^{2020} + n_{i+1}^{2019} ) is a power of 11.Wait, maybe I can consider the exponents modulo something. Let's see, 2020 and 2019 are consecutive exponents, differing by 1. Maybe I can find a relationship between ( n_i ) and ( n_{i+1} ) such that ( n_i^{2020} + n_{i+1}^{2019} ) is a power of 11.But this seems too vague. Maybe I need to think about the prime factorization of each term. Since each term must be a power of 11, they can't have any other prime factors. So, ( n_i^{2020} + n_{i+1}^{2019} ) must be 11-smooth, but actually, it must be exactly a power of 11.Given that, perhaps the only way is if ( n_i ) and ( n_{i+1} ) are both powers of 11. Let me assume ( n_i = 11^{a_i} ) and ( n_{i+1} = 11^{b_i} ). Then, ( n_i^{2020} + n_{i+1}^{2019} = 11^{2020a_i} + 11^{2019b_i} ).For this sum to be a power of 11, say ( 11^k ), we must have that one of the terms is much larger than the other, so that the smaller term doesn't affect the exponent. For example, if ( 2020a_i > 2019b_i ), then ( 11^{2020a_i} + 11^{2019b_i} approx 11^{2020a_i} ), but to be exactly ( 11^k ), we need ( 11^{2020a_i} + 11^{2019b_i} = 11^k ).This can only happen if ( 2020a_i = 2019b_i ), because otherwise, the smaller term would not be a multiple of the larger term, making the sum not a pure power of 11. So, ( 2020a_i = 2019b_i ). Let me write this as ( frac{a_i}{b_i} = frac{2019}{2020} ). Since 2019 and 2020 are coprime (their GCD is 1), this implies that ( a_i = 2019m ) and ( b_i = 2020m ) for some integer ( m ).Therefore, ( n_i = 11^{2019m} ) and ( n_{i+1} = 11^{2020m} ). Then, ( n_i^{2020} + n_{i+1}^{2019} = 11^{2019m times 2020} + 11^{2020m times 2019} = 11^{2019 times 2020 m} + 11^{2019 times 2020 m} = 2 times 11^{2019 times 2020 m} ).But this sum is ( 2 times 11^{k} ), which is not a pure power of 11 unless ( 2 times 11^{k} = 11^{k'} ), which would require ( 2 = 11^{k' - k} ), which is impossible since 2 is not a power of 11.Therefore, this approach also doesn't work. It seems that no matter how I try to set ( n_i ) as powers of 11, the sum ( n_i^{2020} + n_{i+1}^{2019} ) ends up being twice a power of 11, which can't be a pure power of 11.Given all these attempts, it seems that it's impossible to find positive integers ( n_1, n_2, dots, n_{2022} ) such that the given product is a power of 11. Each approach either leads to a contradiction or an impossibility, suggesting that such integers do not exist.

answer：First, I need to determine the total mass of the 100 bags of lemons. Since each bag weighs 8 kilograms, multiplying the number of bags by the weight per bag will give the total mass.Next, I'll calculate the total mass of the lemons by multiplying 100 bags by 8 kilograms per bag, which equals 800 kilograms.The truck has a maximum load capacity of 900 kilograms. To find out how many more kilograms can still be loaded, I'll subtract the total mass of the lemons from the truck's maximum load capacity.Subtracting 800 kilograms from 900 kilograms gives 100 kilograms.Therefore, the farmer can still load an additional 100 kilograms into the truck.

answer：To convert the fraction (frac{3}{40}) into a decimal, we can use a method similar to the one described in the original problem. Here's a step-by-step explanation:1. **Identify the Denominator**: The denominator here is 40. To make the conversion to a decimal easier, we aim to find a multiplier that will convert the denominator into a power of 10 (such as 10, 100, 1000, etc.).2. **Choose an Appropriate Multiplier**: Since 40 multiplied by 2.5 equals 100, we can use 2.5 as our multiplier. This is because: [ 40 times 2.5 = 100 ]3. **Apply the Multiplier to Both Numerator and Denominator**: To maintain the value of the fraction, we multiply both the numerator and the denominator by 2.5: [ frac{3}{40} = frac{3 times 2.5}{40 times 2.5} = frac{7.5}{100} ]4. **Convert the Fraction to a Decimal**: The fraction (frac{7.5}{100}) can be easily converted to a decimal by moving the decimal point two places to the left: [ frac{7.5}{100} = 0.075 ]5. **Conclusion**: Therefore, the decimal representation of (frac{3}{40}) is 0.075.The correct answer is (boxed{text{B}}).