💡Alright, so I've got this problem here about sequences of positive integers and real numbers. Let me try to unpack it step by step. We have two sequences: ( a_1, a_2, a_3, ldots ) which are positive integers, and ( b_1, b_2, b_3, ldots ) which are real numbers defined by [b_n = dfrac{a_1a_2cdots a_n}{a_1 + a_2 + cdots + a_n}]for each ( n geq 1 ). The problem states that if there's at least one integer among every million consecutive terms of the sequence ( b_1, b_2, b_3, ldots ), then we need to show that there exists some ( k ) such that ( b_k > 2021^{2021} ). Okay, so first, let's get a grip on what ( b_n ) represents. It's the product of the first ( n ) terms of the sequence ( a_i ) divided by their sum. So, ( b_n ) is kind of a ratio of the product to the sum of the first ( n ) terms. Given that, the problem tells us that in any million consecutive ( b_n ) terms, there's at least one integer. So, every block of a million ( b_n )'s has at least one integer value. Our goal is to show that under this condition, there must be some ( b_k ) that's larger than ( 2021^{2021} ). That's a pretty big number, so we need to show that the sequence ( b_n ) can't be bounded above by something smaller than that indefinitely.Hmm. So, maybe we can approach this by contradiction. Let's assume that all ( b_n ) are bounded above by some number ( M ), which is less than or equal to ( 2021^{2021} ). If we can show that this leads to a contradiction, then our assumption must be wrong, and hence there must be some ( b_k ) exceeding ( 2021^{2021} ).So, let's suppose that ( b_n leq M ) for all ( n ), where ( M leq 2021^{2021} ). Given that, let's think about the structure of the ( b_n ) terms. Since each ( b_n ) is the product of the first ( n ) ( a_i )'s divided by their sum, and each ( a_i ) is a positive integer, the product ( a_1a_2cdots a_n ) grows at least exponentially if the ( a_i )'s are all at least 2, but the sum ( a_1 + a_2 + cdots + a_n ) grows linearly if the ( a_i )'s are bounded, or polynomially otherwise. But in our case, since we're assuming ( b_n ) is bounded, the product can't be growing too fast relative to the sum. So, maybe the ( a_i )'s can't be growing too quickly either. But wait, the problem says that every million consecutive terms of ( b_n ) has at least one integer. So, in other words, in any block of a million ( b_n )'s, there's at least one term where ( b_n ) is an integer. So, perhaps we can use this to argue about the structure of the ( a_i )'s. If ( b_n ) is an integer, then the product ( a_1a_2cdots a_n ) must be divisible by the sum ( a_1 + a_2 + cdots + a_n ). That is, when ( b_n ) is an integer, we have:[a_1 + a_2 + cdots + a_n mid a_1a_2cdots a_n]So, the sum divides the product. This is interesting because it's a non-trivial condition. For example, if all the ( a_i )'s are 1, then the sum is ( n ) and the product is 1, so ( b_n = 1/n ), which is never an integer for ( n > 1 ). So, in that case, the condition of having at least one integer in every million terms would fail. But in our problem, it's given that such a condition holds, so the ( a_i )'s must be chosen in such a way that every million terms have at least one ( n ) where the sum divides the product.So, perhaps we can use this to find some structure in the ( a_i )'s, or to show that the ( a_i )'s must grow in a certain way, which would cause ( b_n ) to eventually exceed ( 2021^{2021} ).Alternatively, maybe we can use the fact that if ( b_n ) is bounded, then the product ( a_1a_2cdots a_n ) is bounded by ( M times (a_1 + a_2 + cdots + a_n) ). So, if ( b_n leq M ), then:[a_1a_2cdots a_n leq M(a_1 + a_2 + cdots + a_n)]This inequality must hold for all ( n ).But if we can show that this inequality can't hold indefinitely given the condition on the ( b_n )'s, then we can conclude that ( b_n ) must exceed ( M ) for some ( n ).So, let's try to formalize this.Suppose that ( b_n leq M ) for all ( n ). Then, for each ( n ), we have:[a_1a_2cdots a_n leq M(a_1 + a_2 + cdots + a_n)]Let's denote ( S_n = a_1 + a_2 + cdots + a_n ) and ( P_n = a_1a_2cdots a_n ). So, our inequality becomes:[P_n leq M S_n]Now, since ( P_n = P_{n-1} times a_n ), and ( S_n = S_{n-1} + a_n ), we can write:[P_{n} = P_{n-1} a_n leq M S_n = M(S_{n-1} + a_n)]But since ( P_{n-1} leq M S_{n-1} ), we can substitute:[P_n = P_{n-1} a_n leq M S_{n-1} a_n]So, combining the two inequalities:[M S_{n-1} a_n leq M(S_{n-1} + a_n)]Dividing both sides by ( M ) (since ( M > 0 )):[S_{n-1} a_n leq S_{n-1} + a_n]Let's rearrange this:[S_{n-1} a_n - S_{n-1} - a_n leq 0]Factor:[S_{n-1}(a_n - 1) - a_n leq 0]Hmm, so:[S_{n-1}(a_n - 1) leq a_n]Which can be rewritten as:[S_{n-1} leq frac{a_n}{a_n - 1}]But ( frac{a_n}{a_n - 1} = 1 + frac{1}{a_n - 1} ). Since ( a_n ) is a positive integer greater than or equal to 1, ( a_n - 1 ) is at least 0. But if ( a_n = 1 ), then ( frac{a_n}{a_n - 1} ) is undefined (division by zero). So, ( a_n ) must be at least 2 for this to make sense.Wait, but if ( a_n = 1 ), then ( S_{n-1}(1 - 1) = 0 leq 1 ), which is true, but doesn't give us any information. So, perhaps the case ( a_n = 1 ) is trivial, but for ( a_n geq 2 ), we have:[S_{n-1} leq 1 + frac{1}{a_n - 1}]But ( S_{n-1} ) is the sum of the first ( n-1 ) positive integers, each at least 1, so ( S_{n-1} geq n - 1 ). Therefore, for ( a_n geq 2 ):[n - 1 leq S_{n-1} leq 1 + frac{1}{a_n - 1}]But wait, ( 1 + frac{1}{a_n - 1} ) is at most 2, since ( a_n geq 2 ) implies ( frac{1}{a_n - 1} leq 1 ). Therefore, ( S_{n-1} leq 2 ). But ( S_{n-1} geq n - 1 ), so ( n - 1 leq 2 ), which implies ( n leq 3 ).So, this suggests that for ( n geq 4 ), ( a_n ) must be 1, because otherwise, we get a contradiction. But if ( a_n = 1 ), then ( S_n = S_{n-1} + 1 ), and ( P_n = P_{n-1} times 1 = P_{n-1} ). But if ( a_n = 1 ) for all ( n geq 4 ), then ( P_n = P_3 ) for all ( n geq 3 ), and ( S_n = S_3 + (n - 3) ). So, ( b_n = frac{P_3}{S_3 + (n - 3)} ). As ( n ) increases, ( b_n ) tends to 0. So, eventually, ( b_n ) becomes less than 1, which contradicts the condition that every million terms has at least one integer. Because once ( b_n ) is less than 1, it can't be an integer (since ( b_n ) is positive). Therefore, our assumption that ( b_n leq M ) for all ( n ) leads to a contradiction, because it forces ( a_n = 1 ) for all ( n geq 4 ), which in turn causes ( b_n ) to become less than 1, violating the condition that every million terms has at least one integer.Wait, but this seems too quick. Maybe I need to think more carefully. Let me recap. If ( b_n leq M ), then for each ( n ), ( P_n leq M S_n ). Then, for ( n geq 4 ), we derived that ( a_n ) must be 1, leading to ( P_n = P_3 ) and ( S_n = S_3 + (n - 3) ). Therefore, ( b_n = frac{P_3}{S_3 + (n - 3)} ). As ( n ) increases, ( b_n ) decreases, approaching zero. So, for sufficiently large ( n ), ( b_n < 1 ). But ( b_n ) is defined as a ratio of positive integers, so it's a positive real number. However, it can be a fraction. But the problem states that among every million consecutive terms, there is at least one integer. So, if ( b_n ) becomes less than 1, it can't be an integer anymore (since integers are at least 1). Therefore, in the blocks where ( b_n < 1 ), there are no integers, which contradicts the given condition.Therefore, our initial assumption that ( b_n leq M ) for all ( n ) must be false. Hence, there must exist some ( k ) such that ( b_k > M ). Since ( M ) was arbitrary (as long as it's less than or equal to ( 2021^{2021} )), we can choose ( M = 2021^{2021} ), and thus, there exists some ( k ) with ( b_k > 2021^{2021} ).Wait, but in my earlier reasoning, I concluded that ( a_n ) must be 1 for ( n geq 4 ), which leads to ( b_n ) decreasing to zero. But maybe there's a way to have ( a_n ) not equal to 1, but still have ( b_n ) bounded. Let me think.Suppose that ( a_n ) is not 1 for some ( n geq 4 ). Then, from the inequality:[S_{n-1} leq 1 + frac{1}{a_n - 1}]But ( S_{n-1} geq n - 1 ), so:[n - 1 leq 1 + frac{1}{a_n - 1}]Which implies:[n - 2 leq frac{1}{a_n - 1}]Since ( a_n geq 2 ), ( frac{1}{a_n - 1} leq 1 ). Therefore:[n - 2 leq 1 implies n leq 3]Which is a contradiction because ( n geq 4 ). Therefore, our assumption that ( a_n geq 2 ) for some ( n geq 4 ) leads to a contradiction. Hence, all ( a_n ) for ( n geq 4 ) must be 1.But as we saw earlier, this causes ( b_n ) to decrease to zero, which contradicts the condition that every million terms has at least one integer. Therefore, our initial assumption that ( b_n ) is bounded above by ( M ) must be false.Thus, there must exist some ( k ) such that ( b_k > M ). Since ( M ) was arbitrary, we can choose ( M = 2021^{2021} ), and hence, there exists some ( k ) with ( b_k > 2021^{2021} ).So, putting it all together, by assuming that ( b_n ) is bounded above, we derived a contradiction, which means that ( b_n ) must be unbounded, and in particular, must exceed ( 2021^{2021} ) for some ( k ).I think this makes sense. The key idea is that if ( b_n ) were bounded, then the ( a_n )'s would have to eventually become 1, causing ( b_n ) to decrease to zero, which contradicts the condition that every million terms has an integer. Therefore, ( b_n ) can't be bounded, and must exceed any given bound, including ( 2021^{2021} ).**Final Answer**boxed{2021^{2021}}

💡First, I need to determine the relative speed between the two trains since they are moving in the same direction. The relative speed is the difference between their speeds.Next, I'll convert the relative speed from kilometers per hour to meters per second to match the unit of the train lengths.After obtaining the relative speed in meters per second, I'll calculate the total distance that needs to be covered for the faster train to completely pass the slower train. This distance is the sum of the lengths of both trains.Finally, I'll use the formula Time = Distance / Speed to find out how long it takes for the faster train to pass the slower train.

💡Okay, so I have this problem that says for every natural number N, there exists some k such that N can be expressed as a sum where each term is either 1 or 2 multiplied by a power of 2. Specifically, N equals a0 times 2^0 plus a1 times 2^1 plus ... up to ak times 2^k, and each ai is either 1 or 2. Hmm, let me think about how to approach this. Maybe I can start by understanding what this representation means. Normally, in binary, each digit is either 0 or 1, and you add up powers of 2 multiplied by these digits. Here, instead of 0 or 1, we're allowed to have 1 or 2. So, it's like a modified binary representation where each digit can be 1 or 2 instead of 0 or 1.Let me take a simple example to see how this works. Let's say N is 3. In binary, 3 is 11, which is 1*2^1 + 1*2^0. But according to this problem, I need to express it as a sum where each coefficient is 1 or 2. Well, 3 can be written as 1*2^1 + 1*2^0, which already fits the requirement. So, that's straightforward.What about N=4? In binary, 4 is 100. So, that's 1*2^2. But according to the problem, I need to have coefficients 1 or 2. So, 1*2^2 is already fine because 1 is allowed. So, that works too.Wait, what about N=5? In binary, 5 is 101, which is 1*2^2 + 0*2^1 + 1*2^0. But here, we have a 0 coefficient, which isn't allowed. So, how can I express 5 using only 1s and 2s as coefficients?Maybe I can adjust the binary representation. If I have 1*2^2 + 0*2^1 + 1*2^0, and I need to eliminate the 0, perhaps I can carry over from the higher bit. Let's see: If I take the 2^2 term and split it into two 2^1 terms, that would give me 2*2^1. Then, I still have the 1*2^0. So, 2*2^1 + 1*2^0 equals 4 + 1, which is 5. So, that works. So, 5 can be expressed as 2*2^1 + 1*2^0.Okay, so in this case, I had to carry over from a higher power to eliminate the 0 coefficient. That seems to be a useful strategy. Let me try another number, say N=6. In binary, 6 is 110, which is 1*2^2 + 1*2^1 + 0*2^0. Again, there's a 0 coefficient. So, how can I adjust this?If I take the 2^2 term and split it into two 2^1 terms, that would give me 2*2^1. Then, I already have a 1*2^1, so adding that together would give me 3*2^1, which is 6. But 3 is not allowed as a coefficient. Hmm, so that doesn't work. Maybe I need a different approach.Alternatively, perhaps I can split the 2^1 term into two 2^0 terms. So, 1*2^2 + 1*2^1 + 0*2^0 becomes 1*2^2 + 0*2^1 + 2*2^0. But then, 2*2^0 is 2, and 1*2^2 is 4, so 4 + 2 is 6. So, that works. So, 6 can be expressed as 1*2^2 + 2*2^0.Wait, but in this case, I had to split the 2^1 term into two 2^0 terms, which allowed me to eliminate the 0 coefficient. So, it seems like I can adjust the representation by carrying over or splitting terms to ensure that all coefficients are either 1 or 2.Let me try a larger number, say N=7. In binary, 7 is 111, which is 1*2^2 + 1*2^1 + 1*2^0. That already fits the requirement because all coefficients are 1. So, no adjustment is needed.What about N=8? In binary, 8 is 1000, which is 1*2^3. Since 1 is allowed, that's fine. So, 8 can be expressed as 1*2^3.Now, let's try N=9. In binary, 9 is 1001, which is 1*2^3 + 0*2^2 + 0*2^1 + 1*2^0. There are two 0 coefficients here. How can I adjust this?Maybe I can carry over from the higher bits. If I take the 2^3 term and split it into two 2^2 terms, that would give me 2*2^2. Then, I still have the 1*2^0. So, 2*2^2 + 1*2^0 equals 8 + 1, which is 9. So, that works. So, 9 can be expressed as 2*2^2 + 1*2^0.Alternatively, could I split the 2^3 term differently? If I split it into one 2^2 and one 2^1, that would give me 1*2^2 + 1*2^1. Then, adding the 1*2^0, I get 4 + 2 + 1 = 7, which is less than 9. So, that doesn't help. So, splitting into two 2^2 terms seems necessary.Let me try N=10. In binary, 10 is 1010, which is 1*2^3 + 0*2^2 + 1*2^1 + 0*2^0. There are two 0 coefficients. How can I adjust this?If I take the 2^3 term and split it into two 2^2 terms, that gives me 2*2^2. Then, I have the 1*2^1 term. So, 2*2^2 + 1*2^1 equals 8 + 2 = 10. So, that works. So, 10 can be expressed as 2*2^2 + 1*2^1.Alternatively, could I split the 2^3 term into one 2^2 and one 2^1? That would give me 1*2^2 + 1*2^1. Then, adding the existing 1*2^1, I get 4 + 2 + 2 = 8, which is less than 10. So, that doesn't help. So, splitting into two 2^2 terms seems necessary.Wait, but in this case, I also have a 0*2^0 term. How do I handle that? Well, in the adjusted representation, I have 2*2^2 + 1*2^1, which sums to 10, and there's no need to include the 2^0 term because it's zero. So, that's fine.Let me try N=11. In binary, 11 is 1011, which is 1*2^3 + 0*2^2 + 1*2^1 + 1*2^0. There's one 0 coefficient. How can I adjust this?If I take the 2^3 term and split it into two 2^2 terms, that gives me 2*2^2. Then, I have the 1*2^1 and 1*2^0 terms. So, 2*2^2 + 1*2^1 + 1*2^0 equals 8 + 2 + 1 = 11. So, that works. So, 11 can be expressed as 2*2^2 + 1*2^1 + 1*2^0.Alternatively, could I split the 2^3 term into one 2^2 and one 2^1? That would give me 1*2^2 + 1*2^1. Then, adding the existing 1*2^1 and 1*2^0, I get 4 + 2 + 2 + 1 = 9, which is less than 11. So, that doesn't help. So, splitting into two 2^2 terms seems necessary.Let me try N=12. In binary, 12 is 1100, which is 1*2^3 + 1*2^2 + 0*2^1 + 0*2^0. There are two 0 coefficients. How can I adjust this?If I take the 2^3 term and split it into two 2^2 terms, that gives me 2*2^2. Then, I have the 1*2^2 term. So, 2*2^2 + 1*2^2 equals 3*2^2, which is 12. But 3 is not allowed as a coefficient. Hmm, so that doesn't work.Alternatively, could I split the 2^3 term into one 2^2 and one 2^1? That would give me 1*2^2 + 1*2^1. Then, adding the existing 1*2^2 term, I get 2*2^2 + 1*2^1. That equals 8 + 2 = 10, which is less than 12. So, that doesn't help.Wait, maybe I need to split the 2^3 term into two 2^2 terms and then handle the 2^2 term differently. So, 2*2^2 + 1*2^2 equals 3*2^2, which is 12, but again, 3 is not allowed. Hmm.Alternatively, maybe I can split the 2^3 term into one 2^2 and one 2^1, and then split the 2^2 term into two 2^1 terms. So, 1*2^3 becomes 1*2^2 + 1*2^1, and then 1*2^2 becomes 2*2^1. So, altogether, I have 2*2^1 + 1*2^1 + 1*2^1, which is 4 + 2 + 2 = 8, which is less than 12. That doesn't help either.Wait, maybe I need to split the 2^3 term into two 2^2 terms, and then split one of those 2^2 terms into two 2^1 terms. So, 2*2^2 becomes 1*2^2 + 2*2^1. Then, adding the existing 1*2^2 term, I get 2*2^2 + 2*2^1. That equals 8 + 4 = 12. So, that works. So, 12 can be expressed as 2*2^2 + 2*2^1.Wait, but in this case, I had to split one of the 2^2 terms into two 2^1 terms, which allowed me to eliminate the 0 coefficients. So, that seems to be a way to handle it.Let me try N=13. In binary, 13 is 1101, which is 1*2^3 + 1*2^2 + 0*2^1 + 1*2^0. There's one 0 coefficient. How can I adjust this?If I take the 2^3 term and split it into two 2^2 terms, that gives me 2*2^2. Then, I have the 1*2^2 term and the 1*2^0 term. So, 2*2^2 + 1*2^2 + 1*2^0 equals 8 + 4 + 1 = 13. But 2*2^2 + 1*2^2 is 3*2^2, which is 12, plus 1 is 13. But 3 is not allowed as a coefficient. Hmm, so that doesn't work.Alternatively, could I split the 2^3 term into one 2^2 and one 2^1? That would give me 1*2^2 + 1*2^1. Then, adding the existing 1*2^2 term and 1*2^0, I get 2*2^2 + 1*2^1 + 1*2^0, which equals 8 + 2 + 1 = 11, which is less than 13. So, that doesn't help.Wait, maybe I need to split the 2^3 term into two 2^2 terms, and then split one of those 2^2 terms into two 2^1 terms. So, 2*2^2 becomes 1*2^2 + 2*2^1. Then, adding the existing 1*2^2 term and 1*2^0, I get 2*2^2 + 2*2^1 + 1*2^0, which equals 8 + 4 + 1 = 13. So, that works. So, 13 can be expressed as 2*2^2 + 2*2^1 + 1*2^0.Okay, so it seems like the strategy is to carry over from higher bits when there's a 0 coefficient in the binary representation. By splitting higher powers into lower ones, we can adjust the coefficients to be either 1 or 2.Let me try N=14. In binary, 14 is 1110, which is 1*2^3 + 1*2^2 + 1*2^1 + 0*2^0. There's one 0 coefficient. How can I adjust this?If I take the 2^3 term and split it into two 2^2 terms, that gives me 2*2^2. Then, I have the 1*2^2 term and the 1*2^1 term. So, 2*2^2 + 1*2^2 + 1*2^1 equals 8 + 4 + 2 = 14. But 2*2^2 + 1*2^2 is 3*2^2, which is 12, plus 2 is 14. But 3 is not allowed as a coefficient. Hmm, so that doesn't work.Alternatively, could I split the 2^3 term into one 2^2 and one 2^1? That would give me 1*2^2 + 1*2^1. Then, adding the existing 1*2^2 term and 1*2^1 term, I get 2*2^2 + 2*2^1, which equals 8 + 4 = 12, which is less than 14. So, that doesn't help.Wait, maybe I need to split the 2^3 term into two 2^2 terms, and then split one of those 2^2 terms into two 2^1 terms. So, 2*2^2 becomes 1*2^2 + 2*2^1. Then, adding the existing 1*2^2 term and 1*2^1 term, I get 2*2^2 + 3*2^1, which is 8 + 6 = 14. But 3 is not allowed as a coefficient. Hmm, that doesn't work either.Wait, maybe I need to split the 2^3 term into one 2^2 and one 2^1, and then split the 2^2 term into two 2^1 terms. So, 1*2^3 becomes 1*2^2 + 1*2^1, and then 1*2^2 becomes 2*2^1. So, altogether, I have 2*2^1 + 1*2^1 + 1*2^1, which is 4 + 2 + 2 = 8, which is less than 14. That doesn't help.Hmm, maybe I'm overcomplicating this. Let's try a different approach. Since N=14 is 1110 in binary, which is 8 + 4 + 2. So, that's 1*2^3 + 1*2^2 + 1*2^1 + 0*2^0. To eliminate the 0, I can try to carry over from the 2^1 term. If I take the 2^1 term and split it into two 2^0 terms, that would give me 2*2^0. Then, adding the existing 0*2^0, I get 2*2^0. So, the total becomes 8 + 4 + 2*2^0, which is 8 + 4 + 2 = 14. So, that works. So, 14 can be expressed as 1*2^3 + 1*2^2 + 2*2^0.Wait, but in this case, I had to split the 2^1 term into two 2^0 terms, which allowed me to eliminate the 0 coefficient. So, that seems to be another way to handle it.Let me try N=15. In binary, 15 is 1111, which is 1*2^3 + 1*2^2 + 1*2^1 + 1*2^0. All coefficients are 1, so that's fine. No adjustment needed.What about N=16? In binary, 16 is 10000, which is 1*2^4. Since 1 is allowed, that's fine. So, 16 can be expressed as 1*2^4.Let me try N=17. In binary, 17 is 10001, which is 1*2^4 + 0*2^3 + 0*2^2 + 0*2^1 + 1*2^0. There are three 0 coefficients. How can I adjust this?If I take the 2^4 term and split it into two 2^3 terms, that gives me 2*2^3. Then, I have the 1*2^0 term. So, 2*2^3 + 1*2^0 equals 16 + 1 = 17. So, that works. So, 17 can be expressed as 2*2^3 + 1*2^0.Alternatively, could I split the 2^4 term into one 2^3 and one 2^2? That would give me 1*2^3 + 1*2^2. Then, adding the 1*2^0 term, I get 8 + 4 + 1 = 13, which is less than 17. So, that doesn't help. So, splitting into two 2^3 terms seems necessary.Wait, but in this case, I also have multiple 0 coefficients. How do I handle that? Well, in the adjusted representation, I have 2*2^3 + 1*2^0, which sums to 17, and there's no need to include the other terms because they're zero. So, that's fine.Let me try N=18. In binary, 18 is 10010, which is 1*2^4 + 0*2^3 + 0*2^2 + 1*2^1 + 0*2^0. There are two 0 coefficients. How can I adjust this?If I take the 2^4 term and split it into two 2^3 terms, that gives me 2*2^3. Then, I have the 1*2^1 term. So, 2*2^3 + 1*2^1 equals 16 + 2 = 18. So, that works. So, 18 can be expressed as 2*2^3 + 1*2^1.Alternatively, could I split the 2^4 term into one 2^3 and one 2^2? That would give me 1*2^3 + 1*2^2. Then, adding the 1*2^1 term, I get 8 + 4 + 2 = 14, which is less than 18. So, that doesn't help. So, splitting into two 2^3 terms seems necessary.Let me try N=19. In binary, 19 is 10011, which is 1*2^4 + 0*2^3 + 0*2^2 + 1*2^1 + 1*2^0. There are two 0 coefficients. How can I adjust this?If I take the 2^4 term and split it into two 2^3 terms, that gives me 2*2^3. Then, I have the 1*2^1 and 1*2^0 terms. So, 2*2^3 + 1*2^1 + 1*2^0 equals 16 + 2 + 1 = 19. So, that works. So, 19 can be expressed as 2*2^3 + 1*2^1 + 1*2^0.Alternatively, could I split the 2^4 term into one 2^3 and one 2^2? That would give me 1*2^3 + 1*2^2. Then, adding the 1*2^1 and 1*2^0 terms, I get 8 + 4 + 2 + 1 = 15, which is less than 19. So, that doesn't help. So, splitting into two 2^3 terms seems necessary.Let me try N=20. In binary, 20 is 10100, which is 1*2^4 + 0*2^3 + 1*2^2 + 0*2^1 + 0*2^0. There are two 0 coefficients. How can I adjust this?If I take the 2^4 term and split it into two 2^3 terms, that gives me 2*2^3. Then, I have the 1*2^2 term. So, 2*2^3 + 1*2^2 equals 16 + 4 = 20. So, that works. So, 20 can be expressed as 2*2^3 + 1*2^2.Alternatively, could I split the 2^4 term into one 2^3 and one 2^2? That would give me 1*2^3 + 1*2^2. Then, adding the existing 1*2^2 term, I get 1*2^3 + 2*2^2, which equals 8 + 8 = 16, which is less than 20. So, that doesn't help. So, splitting into two 2^3 terms seems necessary.Wait, but in this case, I also have a 0*2^1 and 0*2^0 term. How do I handle that? Well, in the adjusted representation, I have 2*2^3 + 1*2^2, which sums to 20, and there's no need to include the other terms because they're zero. So, that's fine.Let me try N=21. In binary, 21 is 10101, which is 1*2^4 + 0*2^3 + 1*2^2 + 0*2^1 + 1*2^0. There are two 0 coefficients. How can I adjust this?If I take the 2^4 term and split it into two 2^3 terms, that gives me 2*2^3. Then, I have the 1*2^2 and 1*2^0 terms. So, 2*2^3 + 1*2^2 + 1*2^0 equals 16 + 4 + 1 = 21. So, that works. So, 21 can be expressed as 2*2^3 + 1*2^2 + 1*2^0.Alternatively, could I split the 2^4 term into one 2^3 and one 2^2? That would give me 1*2^3 + 1*2^2. Then, adding the existing 1*2^2 term and 1*2^0 term, I get 1*2^3 + 2*2^2 + 1*2^0, which equals 8 + 8 + 1 = 17, which is less than 21. So, that doesn't help. So, splitting into two 2^3 terms seems necessary.Let me try N=22. In binary, 22 is 10110, which is 1*2^4 + 0*2^3 + 1*2^2 + 1*2^1 + 0*2^0. There are two 0 coefficients. How can I adjust this?If I take the 2^4 term and split it into two 2^3 terms, that gives me 2*2^3. Then, I have the 1*2^2 and 1*2^1 terms. So, 2*2^3 + 1*2^2 + 1*2^1 equals 16 + 4 + 2 = 22. So, that works. So, 22 can be expressed as 2*2^3 + 1*2^2 + 1*2^1.Alternatively, could I split the 2^4 term into one 2^3 and one 2^2? That would give me 1*2^3 + 1*2^2. Then, adding the existing 1*2^2 term and 1*2^1 term, I get 1*2^3 + 2*2^2 + 1*2^1, which equals 8 + 8 + 2 = 18, which is less than 22. So, that doesn't help. So, splitting into two 2^3 terms seems necessary.Let me try N=23. In binary, 23 is 10111, which is 1*2^4 + 0*2^3 + 1*2^2 + 1*2^1 + 1*2^0. There's one 0 coefficient. How can I adjust this?If I take the 2^4 term and split it into two 2^3 terms, that gives me 2*2^3. Then, I have the 1*2^2, 1*2^1, and 1*2^0 terms. So, 2*2^3 + 1*2^2 + 1*2^1 + 1*2^0 equals 16 + 4 + 2 + 1 = 23. So, that works. So, 23 can be expressed as 2*2^3 + 1*2^2 + 1*2^1 + 1*2^0.Alternatively, could I split the 2^4 term into one 2^3 and one 2^2? That would give me 1*2^3 + 1*2^2. Then, adding the existing 1*2^2 term, 1*2^1 term, and 1*2^0 term, I get 1*2^3 + 2*2^2 + 1*2^1 + 1*2^0, which equals 8 + 8 + 2 + 1 = 19, which is less than 23. So, that doesn't help. So, splitting into two 2^3 terms seems necessary.Let me try N=24. In binary, 24 is 11000, which is 1*2^4 + 1*2^3 + 0*2^2 + 0*2^1 + 0*2^0. There are three 0 coefficients. How can I adjust this?If I take the 2^4 term and split it into two 2^3 terms, that gives me 2*2^3. Then, I have the 1*2^3 term. So, 2*2^3 + 1*2^3 equals 3*2^3, which is 24. But 3 is not allowed as a coefficient. Hmm, so that doesn't work.Alternatively, could I split the 2^4 term into one 2^3 and one 2^2? That would give me 1*2^3 + 1*2^2. Then, adding the existing 1*2^3 term, I get 2*2^3 + 1*2^2. That equals 16 + 4 = 20, which is less than 24. So, that doesn't help.Wait, maybe I need to split the 2^4 term into two 2^3 terms, and then split one of those 2^3 terms into two 2^2 terms. So, 2*2^3 becomes 1*2^3 + 2*2^2. Then, adding the existing 1*2^3 term, I get 2*2^3 + 2*2^2, which equals 16 + 8 = 24. So, that works. So, 24 can be expressed as 2*2^3 + 2*2^2.Wait, but in this case, I had to split one of the 2^3 terms into two 2^2 terms, which allowed me to eliminate the 0 coefficients. So, that seems to be a way to handle it.Let me try N=25. In binary, 25 is 11001, which is 1*2^4 + 1*2^3 + 0*2^2 + 0*2^1 + 1*2^0. There are two 0 coefficients. How can I adjust this?If I take the 2^4 term and split it into two 2^3 terms, that gives me 2*2^3. Then, I have the 1*2^3 term and the 1*2^0 term. So, 2*2^3 + 1*2^3 + 1*2^0 equals 16 + 8 + 1 = 25. But 2*2^3 + 1*2^3 is 3*2^3, which is 24, plus 1 is 25. But 3 is not allowed as a coefficient. Hmm, so that doesn't work.Alternatively, could I split the 2^4 term into one 2^3 and one 2^2? That would give me 1*2^3 + 1*2^2. Then, adding the existing 1*2^3 term and 1*2^0 term, I get 2*2^3 + 1*2^2 + 1*2^0, which equals 16 + 4 + 1 = 21, which is less than 25. So, that doesn't help.Wait, maybe I need to split the 2^4 term into two 2^3 terms, and then split one of those 2^3 terms into two 2^2 terms. So, 2*2^3 becomes 1*2^3 + 2*2^2. Then, adding the existing 1*2^3 term and 1*2^0 term, I get 2*2^3 + 2*2^2 + 1*2^0, which equals 16 + 8 + 1 = 25. So, that works. So, 25 can be expressed as 2*2^3 + 2*2^2 + 1*2^0.Okay, so it seems like the strategy is to carry over from higher bits when there's a 0 coefficient in the binary representation. By splitting higher powers into lower ones, we can adjust the coefficients to be either 1 or 2.From these examples, it looks like for any natural number N, we can express it as a sum of powers of 2 with coefficients 1 or 2 by appropriately splitting higher powers into lower ones whenever there's a 0 coefficient in the binary representation. This process ensures that all coefficients are either 1 or 2, thus proving the statement.I think this approach works because binary representation covers all natural numbers, and by adjusting the coefficients through splitting, we can handle any 0 coefficients by redistributing the value into lower powers. This way, we ensure that each coefficient is either 1 or 2, as required.

💡Alright, so I've got this problem where I need to prove that the quadrilateral ABCD is a rectangle. The points given are A(-2, 0), B(1, 6), C(5, 4), and D(2, -2). I remember that a rectangle is a four-sided figure with opposite sides equal and parallel, and all angles are right angles. So, to prove that ABCD is a rectangle, I need to show that the opposite sides are equal and that the adjacent sides are perpendicular to each other.First, I think I should plot these points on a coordinate plane to get a visual idea of how the quadrilateral looks. Let me see, point A is at (-2, 0), which is two units to the left on the x-axis. Point B is at (1, 6), so that's one unit to the right and six units up. Point C is at (5, 4), which is five units to the right and four units up. Point D is at (2, -2), so that's two units to the right and two units down.Now, connecting these points in order: A to B, B to C, C to D, and D back to A. Hmm, it's a bit hard to visualize, but I think it might form a rectangle. To be sure, I need to calculate the lengths of the sides and check the slopes to see if the sides are perpendicular.Let me start by finding the lengths of the sides. I can use the distance formula, which is sqrt[(x2 - x1)^2 + (y2 - y1)^2].First, the length of AB:A(-2, 0) to B(1, 6):Distance AB = sqrt[(1 - (-2))^2 + (6 - 0)^2] = sqrt[(3)^2 + (6)^2] = sqrt[9 + 36] = sqrt[45] = 3*sqrt(5).Next, the length of BC:B(1, 6) to C(5, 4):Distance BC = sqrt[(5 - 1)^2 + (4 - 6)^2] = sqrt[(4)^2 + (-2)^2] = sqrt[16 + 4] = sqrt[20] = 2*sqrt(5).Then, the length of CD:C(5, 4) to D(2, -2):Distance CD = sqrt[(2 - 5)^2 + (-2 - 4)^2] = sqrt[(-3)^2 + (-6)^2] = sqrt[9 + 36] = sqrt[45] = 3*sqrt(5).Finally, the length of DA:D(2, -2) to A(-2, 0):Distance DA = sqrt[(-2 - 2)^2 + (0 - (-2))^2] = sqrt[(-4)^2 + (2)^2] = sqrt[16 + 4] = sqrt[20] = 2*sqrt(5).So, the lengths are AB = 3*sqrt(5), BC = 2*sqrt(5), CD = 3*sqrt(5), and DA = 2*sqrt(5). This shows that opposite sides are equal: AB = CD and BC = DA. That's one property of a rectangle satisfied.Now, I need to check if the adjacent sides are perpendicular. For that, I'll calculate the slopes of the sides and see if the product of the slopes of adjacent sides is -1, which indicates perpendicular lines.Let's find the slope of AB:Slope of AB = (y2 - y1)/(x2 - x1) = (6 - 0)/(1 - (-2)) = 6/3 = 2.Slope of BC:Slope of BC = (4 - 6)/(5 - 1) = (-2)/4 = -0.5.Slope of CD:Slope of CD = (-2 - 4)/(2 - 5) = (-6)/(-3) = 2.Slope of DA:Slope of DA = (0 - (-2))/(-2 - 2) = 2/(-4) = -0.5.Now, let's check the product of the slopes of adjacent sides:Slope of AB * Slope of BC = 2 * (-0.5) = -1.Slope of BC * Slope of CD = (-0.5) * 2 = -1.Slope of CD * Slope of DA = 2 * (-0.5) = -1.Slope of DA * Slope of AB = (-0.5) * 2 = -1.All adjacent sides have slopes whose product is -1, which means they are perpendicular to each other. Therefore, all angles in the quadrilateral are right angles.Since we've established that opposite sides are equal and all adjacent sides are perpendicular, quadrilateral ABCD satisfies the properties of a rectangle. Therefore, ABCD is a rectangle.I think that's a solid proof. I calculated the lengths and slopes methodically, checked each step, and made sure that all properties align with those of a rectangle. It all adds up, so I'm confident in the conclusion.

💡Okay, so I have this problem where I need to find the value of ( x ) that minimizes the function ( y = x + frac{5}{x + 1} ), and ( x ) has to be greater than or equal to 2. The options given are A) ( sqrt{5} - 1 ), B) 2, C) ( sqrt{5} ), and D) ( sqrt{5} + 1 ). First, I remember that to find the minimum of a function, especially when it's a single-variable function like this, calculus is usually the way to go. So, I think I should take the derivative of ( y ) with respect to ( x ) and set it equal to zero to find the critical points. Let me write down the function again: ( y = x + frac{5}{x + 1} ). Taking the derivative, ( y' ), with respect to ( x ), I get:( y' = 1 - frac{5}{(x + 1)^2} ).Okay, so I set this derivative equal to zero to find the critical points:( 1 - frac{5}{(x + 1)^2} = 0 ).Solving for ( x ), I can rearrange this equation:( frac{5}{(x + 1)^2} = 1 ).Multiplying both sides by ( (x + 1)^2 ):( 5 = (x + 1)^2 ).Taking the square root of both sides:( x + 1 = sqrt{5} ) or ( x + 1 = -sqrt{5} ).Since ( x ) is greater than or equal to 2, ( x + 1 ) must be positive, so we can ignore the negative root. Therefore:( x + 1 = sqrt{5} ).Subtracting 1 from both sides:( x = sqrt{5} - 1 ).Hmm, that's one of the options, option A. But wait, the problem specifies that ( x geq 2 ). Is ( sqrt{5} - 1 ) greater than or equal to 2?Let me calculate ( sqrt{5} ). I know that ( sqrt{4} = 2 ) and ( sqrt{9} = 3 ), so ( sqrt{5} ) is approximately 2.236. Therefore, ( sqrt{5} - 1 ) is approximately 1.236, which is less than 2. So, this critical point is actually not in the domain we're considering, which is ( x geq 2 ).That means the minimum can't occur at ( x = sqrt{5} - 1 ) because that's outside our domain. So, I need to check the behavior of the function at the boundary of our domain, which is ( x = 2 ), and see if the function is increasing or decreasing beyond that point.To determine if the function is increasing or decreasing for ( x geq 2 ), I can look at the derivative again. Let's plug in a value greater than 2, say ( x = 3 ), into the derivative:( y' = 1 - frac{5}{(3 + 1)^2} = 1 - frac{5}{16} = frac{11}{16} ), which is positive.Since the derivative is positive for ( x = 3 ), the function is increasing at that point. If the function is increasing for ( x geq 2 ), then the minimum value should occur at the smallest ( x ) in the domain, which is ( x = 2 ).Let me verify this by calculating the function at ( x = 2 ):( y = 2 + frac{5}{2 + 1} = 2 + frac{5}{3} = frac{6}{3} + frac{5}{3} = frac{11}{3} ).So, the value of ( y ) at ( x = 2 ) is ( frac{11}{3} ). Just to be thorough, let me check another point beyond ( x = 2 ), say ( x = 4 ):( y = 4 + frac{5}{4 + 1} = 4 + 1 = 5 ).Which is greater than ( frac{11}{3} ) (since ( frac{11}{3} approx 3.666 )), confirming that the function is indeed increasing beyond ( x = 2 ).Therefore, the minimum value of ( y ) occurs at ( x = 2 ), which is option B.Wait, but just to make sure I didn't make any mistakes, let me recap:1. Took the derivative correctly: ( y' = 1 - frac{5}{(x + 1)^2} ).2. Set derivative to zero: ( 1 - frac{5}{(x + 1)^2} = 0 ) leading to ( x = sqrt{5} - 1 approx 1.236 ).3. Since ( x geq 2 ), this critical point is irrelevant.4. Checked the derivative at ( x = 3 ) and found it positive, meaning function is increasing for ( x geq 2 ).5. Therefore, the minimum is at ( x = 2 ).Yes, that all seems correct. So, I think the answer is B) 2.**Final Answer**boxed{B}

💡Okay, so I need to show two things about this function ( f: mathbb{R} rightarrow mathbb{R} ). First, that it's continuous, and second, that it's strictly monotone. The function has limits at every point and no local extrema. Hmm, let me try to unpack this step by step.Starting with part (a): showing that ( f ) is continuous. I remember that for a function to be continuous at a point ( a ), the limit as ( x ) approaches ( a ) must equal ( f(a) ). The problem says that ( f ) has limits at any point, so for every ( a ), ( lim_{x to a} f(x) ) exists. But does that necessarily mean it's equal to ( f(a) )? Not always, right? So maybe I need to use the fact that ( f ) has no local extrema to show that the limit must equal ( f(a) ).Wait, if ( f ) has no local extrema, that means there are no points where ( f ) changes direction, like from increasing to decreasing or vice versa. So maybe ( f ) is either always increasing or always decreasing? That might help with both parts (a) and (b). But let me focus on part (a) first.Suppose, for contradiction, that ( f ) is not continuous at some point ( a ). Then ( lim_{x to a} f(x) ) exists but is not equal to ( f(a) ). Let's denote ( L = lim_{x to a} f(x) ). So ( L neq f(a) ). Now, since ( f ) has no local extrema, what does that tell me about the behavior of ( f ) near ( a )?If ( f ) is not continuous at ( a ), then there's a jump or a removable discontinuity. But if ( f ) has no local extrema, it can't have a sudden peak or valley. So if there's a jump, does that create a local extremum? Maybe. If ( f(a) ) is higher than the limit from the left and lower than the limit from the right, or vice versa, then ( f(a) ) could be a local extremum. But the problem states there are no local extrema, so that can't happen. Therefore, the limit must equal ( f(a) ), making ( f ) continuous.Okay, that seems to make sense. So part (a) is done by contradiction, using the fact that a discontinuity would create a local extremum, which isn't allowed.Now, moving on to part (b): showing that ( f ) is strictly monotone. That means ( f ) is either strictly increasing or strictly decreasing everywhere. Since ( f ) has no local extrema, it can't have any points where it changes direction. So if it's increasing at some point, it must keep increasing, and similarly for decreasing.But how do I formalize this? Maybe by assuming that ( f ) is not strictly monotone and then showing that leads to a contradiction with the no local extrema condition.Suppose ( f ) is not strictly monotone. Then there exist points ( a < b < c ) such that ( f(a) leq f(b) ) and ( f(b) geq f(c) ) (if decreasing) or ( f(a) geq f(b) ) and ( f(b) leq f(c) ) (if increasing). Wait, but that might not necessarily create a local extremum. Hmm.Alternatively, if ( f ) is not strictly monotone, there must be some interval where it's not increasing or decreasing. But since ( f ) is continuous (from part (a)), by the Intermediate Value Theorem, it must attain every value between any two points. If it's not strictly monotone, it must have a peak or a valley somewhere, which would be a local extremum. But the problem says there are no local extrema, so ( f ) must be strictly monotone.Yeah, that seems right. So by assuming ( f ) isn't strictly monotone, we end up implying there's a local extremum, which contradicts the given condition. Therefore, ( f ) must be strictly monotone.Wait, let me make sure I didn't skip any steps. I used the fact that ( f ) is continuous from part (a) to apply the Intermediate Value Theorem. Without continuity, the IVT doesn't hold, but since we've already shown ( f ) is continuous, it's valid. So if ( f ) weren't strictly monotone, it would have to attain a maximum or minimum in some interval, which is a local extremum, contradicting the given. So yes, ( f ) must be strictly monotone.I think that covers both parts. I used contradiction in both cases, leveraging the given conditions about limits and the absence of local extrema. It all ties together nicely.