💡Okay, so I have this problem about something called a "dot fraction." It says that given positive integers ( a, b, c, ldots ) all greater than 1, the dot fraction ( a . b . c . cdots ) is defined as ( frac{1}{a} + frac{1}{ab} + frac{1}{abc} + cdots ). The question is asking which rational numbers can be represented as finite dot fractions.Hmm, first, I need to make sure I understand what a dot fraction is. It seems like it's a sum of reciprocals where each term is the product of the previous integers. So, for example, if I have ( a . b . c ), it would be ( frac{1}{a} + frac{1}{ab} + frac{1}{abc} ). Each term is getting smaller because we're multiplying more integers greater than 1, so each subsequent term is a fraction of the previous one.The problem is asking about finite dot fractions, so it's not an infinite series but a finite sum. So, we're looking at sums like ( frac{1}{a} + frac{1}{ab} + frac{1}{abc} + ldots + frac{1}{a_1 a_2 ldots a_n} ) for some finite ( n ).Now, the question is about which rational numbers can be expressed in this form. So, I need to figure out the conditions under which a rational number can be written as such a sum.First, let's think about the range of possible values. Since each term is positive and decreasing, the sum will be positive. Also, since each term is less than or equal to ( frac{1}{2} ) (because ( a, b, c, ldots ) are all greater than 1), the sum will be less than 1. So, the dot fraction will always be between 0 and 1.Therefore, any rational number that can be expressed as a dot fraction must be between 0 and 1. But not all rational numbers in that interval might be expressible as such. So, I need to find out which ones are.Let me try some examples. Let's say I take ( a = 2 ). Then, the dot fraction ( 2 ) is just ( frac{1}{2} ). If I take ( a = 2 ) and ( b = 3 ), then the dot fraction ( 2 . 3 ) is ( frac{1}{2} + frac{1}{6} = frac{2}{3} ). If I take ( a = 2 ), ( b = 3 ), and ( c = 7 ), then ( 2 . 3 . 7 ) is ( frac{1}{2} + frac{1}{6} + frac{1}{42} = frac{21}{42} + frac{7}{42} + frac{1}{42} = frac{29}{42} ).Wait, so these are all fractions between 0 and 1, but they seem to be specific fractions. Maybe there's a pattern or a way to represent any fraction in that interval.I recall that Egyptian fractions represent numbers as sums of distinct unit fractions, but this is a bit different because each term is a reciprocal of a product of integers greater than 1. So, it's a more structured kind of sum.Let me think about how to approach this. Maybe I can represent any fraction ( frac{p}{q} ) in ( (0,1) ) as a finite dot fraction. To do that, I need to find a sequence of integers ( a_1, a_2, ldots, a_n ) such that:[frac{p}{q} = frac{1}{a_1} + frac{1}{a_1 a_2} + frac{1}{a_1 a_2 a_3} + ldots + frac{1}{a_1 a_2 ldots a_n}]I wonder if this is always possible. Maybe I can use a greedy algorithm approach, similar to how Egyptian fractions are constructed. Start with the largest possible term, subtract it, and then proceed with the remainder.But in this case, each term is not just a unit fraction but a reciprocal of a product. So, maybe I can choose ( a_1 ) such that ( frac{1}{a_1} ) is less than or equal to ( frac{p}{q} ), then subtract ( frac{1}{a_1} ) from ( frac{p}{q} ) and proceed with the remainder.Let me try an example. Suppose I want to represent ( frac{3}{4} ) as a dot fraction.First, choose ( a_1 ). The largest ( a_1 ) such that ( frac{1}{a_1} leq frac{3}{4} ) is 2 because ( frac{1}{2} = 0.5 leq 0.75 ). So, ( a_1 = 2 ). Then, subtract ( frac{1}{2} ) from ( frac{3}{4} ) to get ( frac{1}{4} ).Now, I need to represent ( frac{1}{4} ) as ( frac{1}{a_1 a_2} + frac{1}{a_1 a_2 a_3} + ldots ). Since ( a_1 = 2 ), the next term is ( frac{1}{2 a_2} ). I need ( frac{1}{2 a_2} leq frac{1}{4} ). So, ( a_2 geq 2 ). Let's choose ( a_2 = 2 ). Then, ( frac{1}{2 times 2} = frac{1}{4} ), which is exactly the remainder. So, the dot fraction is ( 2 . 2 ), which is ( frac{1}{2} + frac{1}{4} = frac{3}{4} ).Okay, that worked. Let's try another one. How about ( frac{5}{6} )?Start with ( a_1 ). The largest ( a_1 ) such that ( frac{1}{a_1} leq frac{5}{6} ) is 2, since ( frac{1}{2} = 0.5 leq 0.833... ). Subtract ( frac{1}{2} ) from ( frac{5}{6} ) to get ( frac{1}{3} ).Now, we need to represent ( frac{1}{3} ) as ( frac{1}{2 a_2} + frac{1}{2 a_2 a_3} + ldots ). So, ( frac{1}{2 a_2} leq frac{1}{3} ). Thus, ( a_2 geq 2 ) because ( frac{1}{2 times 2} = frac{1}{4} leq frac{1}{3} ). Let's choose ( a_2 = 3 ). Then, ( frac{1}{2 times 3} = frac{1}{6} ). Subtracting this from ( frac{1}{3} ) gives ( frac{1}{3} - frac{1}{6} = frac{1}{6} ).Now, we need to represent ( frac{1}{6} ) as ( frac{1}{2 times 3 times a_3} + ldots ). So, ( frac{1}{6 a_3} leq frac{1}{6} ). Thus, ( a_3 geq 2 ). Let's choose ( a_3 = 2 ). Then, ( frac{1}{6 times 2} = frac{1}{12} ). Subtracting this from ( frac{1}{6} ) gives ( frac{1}{6} - frac{1}{12} = frac{1}{12} ).Hmm, this seems like it's going into an infinite loop. I need to find a way to terminate this process. Maybe I need to choose ( a_3 ) such that ( frac{1}{6 a_3} = frac{1}{6} ), which would require ( a_3 = 1 ), but ( a_3 ) must be greater than 1. So, that's not possible.Wait, maybe I should have chosen a different ( a_2 ). Instead of ( a_2 = 3 ), what if I choose ( a_2 = 2 )? Then, ( frac{1}{2 times 2} = frac{1}{4} ). Subtracting this from ( frac{1}{3} ) gives ( frac{1}{3} - frac{1}{4} = frac{1}{12} ).Now, we need to represent ( frac{1}{12} ) as ( frac{1}{2 times 2 times a_3} + ldots ). So, ( frac{1}{4 a_3} leq frac{1}{12} ). Thus, ( a_3 geq 4 ). Let's choose ( a_3 = 4 ). Then, ( frac{1}{4 times 4} = frac{1}{16} ). Subtracting this from ( frac{1}{12} ) gives ( frac{1}{12} - frac{1}{16} = frac{1}{48} ).This is getting too small, and I'm not sure if this will terminate. Maybe I need a different approach. Perhaps instead of always choosing the smallest possible ( a_i ), I should choose ( a_i ) such that the remaining fraction can be exactly represented by the next term.Let me try again with ( frac{5}{6} ). Start with ( a_1 = 2 ), subtract ( frac{1}{2} ) to get ( frac{1}{3} ). Now, to represent ( frac{1}{3} ), I need ( frac{1}{2 a_2} + frac{1}{2 a_2 a_3} + ldots = frac{1}{3} ).If I choose ( a_2 = 3 ), then ( frac{1}{2 times 3} = frac{1}{6} ). Subtracting this from ( frac{1}{3} ) leaves ( frac{1}{6} ). Now, to represent ( frac{1}{6} ), I need ( frac{1}{2 times 3 times a_3} + ldots = frac{1}{6} ). If I choose ( a_3 = 2 ), then ( frac{1}{6 times 2} = frac{1}{12} ). Subtracting this leaves ( frac{1}{12} ).Hmm, this isn't working. Maybe I need to choose ( a_3 ) such that ( frac{1}{6 a_3} = frac{1}{6} ), but that would require ( a_3 = 1 ), which is not allowed. So, perhaps I need a different ( a_2 ).Alternatively, maybe I can choose ( a_2 = 4 ). Then, ( frac{1}{2 times 4} = frac{1}{8} ). Subtracting this from ( frac{1}{3} ) gives ( frac{1}{3} - frac{1}{8} = frac{5}{24} ). Now, I need to represent ( frac{5}{24} ) as ( frac{1}{8 a_3} + ldots ). Choosing ( a_3 = 5 ), ( frac{1}{8 times 5} = frac{1}{40} ). Subtracting this gives ( frac{5}{24} - frac{1}{40} = frac{25}{120} - frac{3}{120} = frac{22}{120} = frac{11}{60} ). This is getting more complicated.Maybe there's a better way. Perhaps instead of trying to subtract each term, I can express the fraction as a sum of these reciprocals by choosing appropriate ( a_i )s.Let me think about the structure of the dot fraction. Each term is ( frac{1}{a_1 a_2 ldots a_i} ). So, the denominators are products of the sequence ( a_i ). If I can choose the ( a_i )s such that the sum equals the desired fraction, that would work.Alternatively, maybe I can represent the fraction as a continued product or something. Wait, not sure.Let me try another example. Suppose I want to represent ( frac{1}{2} ). That's easy, just ( a_1 = 2 ). How about ( frac{1}{3} )? Let's see.Start with ( a_1 ). The largest ( a_1 ) such that ( frac{1}{a_1} leq frac{1}{3} ) is 4, because ( frac{1}{4} = 0.25 leq 0.333... ). But wait, ( frac{1}{3} ) is approximately 0.333, so ( a_1 = 3 ) would give ( frac{1}{3} ), which is exactly the fraction. So, ( a_1 = 3 ), and we're done. So, ( 3 ) as a dot fraction is ( frac{1}{3} ).Wait, but ( a_i ) must be greater than 1, so 3 is fine.Another example: ( frac{2}{5} ). Let's try to represent this.Start with ( a_1 ). The largest ( a_1 ) such that ( frac{1}{a_1} leq frac{2}{5} ) is 3 because ( frac{1}{3} approx 0.333 leq 0.4 ). Subtract ( frac{1}{3} ) from ( frac{2}{5} ) to get ( frac{2}{5} - frac{1}{3} = frac{6}{15} - frac{5}{15} = frac{1}{15} ).Now, we need to represent ( frac{1}{15} ) as ( frac{1}{3 a_2} + frac{1}{3 a_2 a_3} + ldots ). So, ( frac{1}{3 a_2} leq frac{1}{15} ). Thus, ( a_2 geq 5 ). Let's choose ( a_2 = 5 ). Then, ( frac{1}{3 times 5} = frac{1}{15} ), which exactly matches the remainder. So, the dot fraction is ( 3 . 5 ), which is ( frac{1}{3} + frac{1}{15} = frac{5}{15} + frac{1}{15} = frac{6}{15} = frac{2}{5} ).Okay, that worked. So, it seems like with careful selection of ( a_i )s, I can represent certain fractions. But is this always possible?Let me think about the general case. Suppose I have a fraction ( frac{p}{q} ) in ( (0,1) ). I want to find a sequence ( a_1, a_2, ldots, a_n ) such that:[frac{p}{q} = sum_{i=1}^{n} prod_{j=1}^{i} frac{1}{a_j}]This is equivalent to:[frac{p}{q} = frac{1}{a_1} + frac{1}{a_1 a_2} + frac{1}{a_1 a_2 a_3} + ldots + frac{1}{a_1 a_2 ldots a_n}]I need to find such ( a_i )s. Maybe I can use a recursive approach. Start with ( a_1 ), then express the remainder as a new fraction, and repeat the process.Let me formalize this. Suppose I have ( frac{p}{q} ). Choose ( a_1 ) such that ( frac{1}{a_1} leq frac{p}{q} < frac{1}{a_1 - 1} ). Then, subtract ( frac{1}{a_1} ) from ( frac{p}{q} ) to get a new fraction ( frac{p'}{q'} ). Then, repeat the process for ( frac{p'}{q'} ), choosing ( a_2 ), and so on, until the remainder is zero.But will this process always terminate? I'm not sure. It depends on whether the remainders can be expressed as the required reciprocals.Wait, in the examples I tried, it worked. Maybe it's always possible. Let me try another example to see.How about ( frac{4}{7} ). Let's see.Start with ( a_1 ). The largest ( a_1 ) such that ( frac{1}{a_1} leq frac{4}{7} ) is 2, because ( frac{1}{2} = 0.5 leq 0.571... ). Subtract ( frac{1}{2} ) from ( frac{4}{7} ) to get ( frac{4}{7} - frac{1}{2} = frac{8}{14} - frac{7}{14} = frac{1}{14} ).Now, we need to represent ( frac{1}{14} ) as ( frac{1}{2 a_2} + frac{1}{2 a_2 a_3} + ldots ). So, ( frac{1}{2 a_2} leq frac{1}{14} ). Thus, ( a_2 geq 8 ). Let's choose ( a_2 = 8 ). Then, ( frac{1}{2 times 8} = frac{1}{16} ). Subtracting this from ( frac{1}{14} ) gives ( frac{1}{14} - frac{1}{16} = frac{8}{112} - frac{7}{112} = frac{1}{112} ).Now, we need to represent ( frac{1}{112} ) as ( frac{1}{2 times 8 times a_3} + ldots ). So, ( frac{1}{16 a_3} leq frac{1}{112} ). Thus, ( a_3 geq 8 ). Let's choose ( a_3 = 8 ). Then, ( frac{1}{16 times 8} = frac{1}{128} ). Subtracting this gives ( frac{1}{112} - frac{1}{128} = frac{128 - 112}{14336} = frac{16}{14336} = frac{1}{896} ).This is getting too small, and I'm not sure if this will ever terminate. Maybe I need to choose a different ( a_2 ). Instead of ( a_2 = 8 ), perhaps a larger ( a_2 ) such that ( frac{1}{2 a_2} = frac{1}{14} ). That would require ( a_2 = 7 ), because ( frac{1}{2 times 7} = frac{1}{14} ). So, if I choose ( a_2 = 7 ), then ( frac{1}{2 times 7} = frac{1}{14} ), which exactly matches the remainder. So, the dot fraction is ( 2 . 7 ), which is ( frac{1}{2} + frac{1}{14} = frac{7}{14} + frac{1}{14} = frac{8}{14} = frac{4}{7} ).Ah, that worked! So, by choosing ( a_2 = 7 ) instead of 8, I was able to terminate the process. So, the key is to choose ( a_i ) such that the next term exactly matches the remaining fraction, if possible.So, in general, the strategy is:1. Start with the given fraction ( frac{p}{q} ).2. Choose ( a_1 ) such that ( frac{1}{a_1} leq frac{p}{q} < frac{1}{a_1 - 1} ).3. Subtract ( frac{1}{a_1} ) from ( frac{p}{q} ) to get a new fraction ( frac{p'}{q'} ).4. Repeat the process for ( frac{p'}{q'} ), choosing ( a_2 ), and so on.5. If at any step, the remaining fraction can be exactly represented by ( frac{1}{a_1 a_2 ldots a_i} ), then the process terminates.This seems similar to the greedy algorithm for Egyptian fractions, but with the added structure of the denominators being products of the sequence ( a_i ).But is this always possible? For example, can every fraction ( frac{p}{q} ) in ( (0,1) ) be expressed as a finite dot fraction?I think yes, because at each step, we can choose ( a_i ) such that the remaining fraction is reduced, and since we're dealing with fractions with denominators that are products of integers greater than 1, the denominators grow rapidly, allowing the process to terminate.Wait, but in my earlier attempt with ( frac{5}{6} ), I had trouble because I kept getting remainders that didn't seem to terminate. But then I realized that by choosing ( a_2 = 3 ), I could actually terminate the process. So, maybe with careful selection, it's always possible.Another thought: since each term in the dot fraction is a unit fraction with a denominator that is a product of the sequence ( a_i ), and since we can choose the ( a_i )s to make the denominators as large as needed, we can always represent any fraction by choosing appropriate ( a_i )s.But I need to make sure that the process doesn't get stuck in an infinite loop. Since each step reduces the remaining fraction, and the denominators grow exponentially, the number of terms needed is finite.Wait, but in the case of ( frac{5}{6} ), I initially chose ( a_2 = 3 ), which led to a remainder that required another term. But then I realized that choosing ( a_2 = 3 ) actually allowed me to terminate because ( frac{1}{2 times 3} = frac{1}{6} ), which was exactly the remainder after subtracting ( frac{1}{2} ).So, in general, if I can choose ( a_i ) such that the next term exactly matches the remaining fraction, then the process terminates. If not, I can always choose a larger ( a_i ) to make the next term smaller, but then I have to continue the process.But since the denominators grow rapidly, the number of terms needed is finite. Therefore, every fraction ( frac{p}{q} ) in ( (0,1) ) can be expressed as a finite dot fraction.Wait, but I need to be careful. For example, what about ( frac{1}{1} )? It's not in ( (0,1) ), so it's excluded. What about ( frac{0}{1} )? It's also excluded. So, we're only considering fractions strictly between 0 and 1.Another thought: the dot fraction is a sum of distinct unit fractions, but with denominators that are products of the sequence ( a_i ). So, it's a more restricted form than Egyptian fractions, but still, it seems flexible enough to represent any fraction in ( (0,1) ).Wait, but in the examples I tried, it worked. Let me try a more complicated fraction, say ( frac{11}{13} ).Start with ( a_1 ). The largest ( a_1 ) such that ( frac{1}{a_1} leq frac{11}{13} ) is 2, because ( frac{1}{2} = 0.5 leq 0.846... ). Subtract ( frac{1}{2} ) from ( frac{11}{13} ) to get ( frac{11}{13} - frac{1}{2} = frac{22}{26} - frac{13}{26} = frac{9}{26} ).Now, represent ( frac{9}{26} ) as ( frac{1}{2 a_2} + ldots ). So, ( frac{1}{2 a_2} leq frac{9}{26} ). Thus, ( a_2 geq 2 ) because ( frac{1}{2 times 2} = frac{1}{4} = 0.25 leq 0.346... ). Let's choose ( a_2 = 3 ). Then, ( frac{1}{2 times 3} = frac{1}{6} approx 0.1667 ). Subtracting this from ( frac{9}{26} ) gives ( frac{9}{26} - frac{1}{6} = frac{27}{78} - frac{13}{78} = frac{14}{78} = frac{7}{39} ).Now, represent ( frac{7}{39} ) as ( frac{1}{2 times 3 times a_3} + ldots ). So, ( frac{1}{6 a_3} leq frac{7}{39} approx 0.1795 ). Thus, ( a_3 geq 2 ) because ( frac{1}{6 times 2} = frac{1}{12} approx 0.0833 leq 0.1795 ). Let's choose ( a_3 = 4 ). Then, ( frac{1}{6 times 4} = frac{1}{24} approx 0.0417 ). Subtracting this gives ( frac{7}{39} - frac{1}{24} = frac{56}{312} - frac{13}{312} = frac{43}{312} approx 0.1378 ).Now, represent ( frac{43}{312} ) as ( frac{1}{24 a_4} + ldots ). So, ( frac{1}{24 a_4} leq frac{43}{312} approx 0.1378 ). Thus, ( a_4 geq 2 ) because ( frac{1}{24 times 2} = frac{1}{48} approx 0.0208 leq 0.1378 ). Let's choose ( a_4 = 3 ). Then, ( frac{1}{24 times 3} = frac{1}{72} approx 0.0139 ). Subtracting this gives ( frac{43}{312} - frac{1}{72} = frac{129}{936} - frac{13}{936} = frac{116}{936} = frac{29}{234} approx 0.1239 ).This is getting tedious, but I can see that each step reduces the remaining fraction, though slowly. However, since the denominators are growing exponentially, the process must terminate after a finite number of steps. Therefore, even though it might take many terms, it's still a finite dot fraction.But wait, in practice, it's taking a lot of terms, but the question is about finite dot fractions, regardless of the number of terms. So, as long as the process can terminate after some finite ( n ), it's acceptable.Therefore, it seems that every rational number in ( (0,1) ) can be expressed as a finite dot fraction by appropriately choosing the sequence ( a_1, a_2, ldots, a_n ).To summarize, the key steps are:1. Recognize that a dot fraction is a sum of reciprocals of products of integers greater than 1.2. Observe that such sums are always between 0 and 1.3. Use a greedy algorithm approach to construct the sequence ( a_i ) by choosing each ( a_i ) to be the smallest integer such that the next term does not exceed the remaining fraction.4. Show that this process must terminate after a finite number of steps because the denominators grow rapidly, ensuring that the remaining fraction can eventually be represented exactly by a term in the sequence.Therefore, the conclusion is that all rational numbers in the interval ( (0,1) ) can be represented as finite dot fractions.

💡Alright, so I have this probability problem to solve. It involves random variables, sums, and expectations. Let me try to unpack it step by step.First, we have these independent and identically distributed (i.i.d.) random variables ξ₁, ξ₂, ..., ξ_{2n}. Each ξ_i can be either 1 or -1, each with probability 1/2. That means each step is like a coin flip—either you go up by 1 or down by 1. So, S_{2n} is just the sum of these 2n steps. Essentially, S_{2n} represents the position after 2n steps in a symmetric random walk.Then, M_{2n} is defined as the maximum of all the partial sums S₁, S₂, ..., S_{2n}. So, M_{2n} is the highest point the random walk reaches in those 2n steps.The problem asks me to show two things:1. For k ≤ n, the probability that M_{2n} is at least k and S_{2n} is 0 is equal to the probability that S_{2n} is 2k. In symbols: [ mathrm{P}left{M_{2n} geq k, S_{2n}=0right} = mathrm{P}left{S_{2n}=2kright} ] 2. From this, it follows that: [ mathrm{P}left(M_{2n} geq k mid S_{2n}=0right)=frac{mathrm{P}left{S_{2n}=2kright}}{mathrm{P}left{S_{2n}=0right}} = frac{C_{2n}^{n+k}}{C_{2n}^{n}} ] where C_{2n}^{n+k} is the binomial coefficient.Finally, using this result, I need to derive the expectation: [ mathrm{E}left(M_{2n} mid S_{2n}=0right)=frac{1}{2}left[frac{1}{mathrm{P}left{S_{2n}=0right}}-1right] ]Okay, let's tackle the first part. I need to show that P{M_{2n} ≥ k, S_{2n}=0} = P{S_{2n}=2k}.Hmm, so this is saying that the probability that the maximum of the partial sums is at least k and that the final sum is 0 is equal to the probability that the final sum is 2k.I remember that in random walks, there's something called the reflection principle which is used to calculate probabilities related to maxima and minima. Maybe that can help here.The reflection principle states that the number of paths that reach a certain level can be reflected to calculate probabilities. So, if I consider the paths that reach k and end at 0, maybe reflecting them can relate to paths that end at 2k.Let me think. If a path reaches k at some point and ends at 0, then reflecting the part of the path after it first reaches k would result in a path that ends at 2k. Is that right?Yes, that seems familiar. So, the number of paths that reach k and end at 0 is equal to the number of paths that end at 2k. Therefore, their probabilities should be equal.So, P{M_{2n} ≥ k, S_{2n}=0} = P{S_{2n}=2k}.Okay, that makes sense. So, that's the first part.Now, moving on to the conditional probability. We have:P(M_{2n} ≥ k | S_{2n}=0) = P(S_{2n}=2k) / P(S_{2n}=0)And this simplifies to C_{2n}^{n+k} / C_{2n}^{n}.Alright, so let's recall that for a symmetric random walk, the probability that S_{2n} = 2k is equal to the number of ways to have (n + k) steps up and (n - k) steps down, divided by the total number of paths, which is 2^{2n}.So, P(S_{2n}=2k) = C_{2n}^{n+k} / 2^{2n}Similarly, P(S_{2n}=0) = C_{2n}^{n} / 2^{2n}Therefore, the ratio P(S_{2n}=2k) / P(S_{2n}=0) = C_{2n}^{n+k} / C_{2n}^{n}So, that's straightforward.Now, the last part is to derive the expectation E(M_{2n} | S_{2n}=0).So, expectation is the sum over k of k * P(M_{2n}=k | S_{2n}=0).But from the previous result, we have P(M_{2n} ≥ k | S_{2n}=0) = C_{2n}^{n+k} / C_{2n}^{n}Hmm, but expectation involves P(M_{2n}=k | S_{2n}=0), not P(M_{2n} ≥ k | S_{2n}=0).Wait, maybe I can express the expectation in terms of the survival function, which is P(M_{2n} ≥ k | S_{2n}=0).I recall that for non-negative integer-valued random variables, the expectation can be written as the sum over k ≥ 1 of P(X ≥ k).So, E(M_{2n} | S_{2n}=0) = sum_{k=1}^{n} P(M_{2n} ≥ k | S_{2n}=0)Therefore, substituting the expression we have:E(M_{2n} | S_{2n}=0) = sum_{k=1}^{n} C_{2n}^{n+k} / C_{2n}^{n}Hmm, that seems a bit complicated. Maybe there's a way to simplify this sum.Alternatively, perhaps there's a combinatorial identity or a generating function approach that can help here.Wait, another thought: since we're dealing with symmetric random walks, maybe there's a relationship between the maximum and the final position.Given that S_{2n}=0, the walk returns to the origin after 2n steps. The maximum M_{2n} is the highest point reached during this walk.I remember that for such walks, there's a formula called the range or the maximum displacement, and it relates to the number of paths that reach a certain maximum.But I'm not sure about the exact formula. Maybe I can think about it in terms of the reflection principle again.Alternatively, perhaps I can use generating functions or recursive relations.Wait, another idea: since we have P(M_{2n} ≥ k | S_{2n}=0) = C_{2n}^{n+k} / C_{2n}^{n}, maybe we can express the expectation as a sum over k of C_{2n}^{n+k} / C_{2n}^{n}.But sum_{k=1}^{n} C_{2n}^{n+k} / C_{2n}^{n} = sum_{k=1}^{n} C_{2n}^{n+k} / C_{2n}^{n}But C_{2n}^{n+k} = C_{2n}^{n - k}, since C_{n}^{k} = C_{n}^{n - k}.Wait, no, C_{2n}^{n+k} is not equal to C_{2n}^{n - k} because n + k and n - k are different.Wait, unless k is less than or equal to n, which it is, since k ≤ n.But I don't see an immediate simplification.Alternatively, perhaps I can write the sum as:sum_{k=1}^{n} C_{2n}^{n+k} / C_{2n}^{n} = sum_{k=1}^{n} [C_{2n}^{n+k} / C_{2n}^{n}]But C_{2n}^{n+k} / C_{2n}^{n} = [ (2n)! / ( (n + k)! (n - k)! ) ] / [ (2n)! / (n! n!) ) ] = [n! n!] / [ (n + k)! (n - k)! ) ]Simplifying, this is equal to [n! / (n + k)!)] * [n! / (n - k)!)] = [1 / ( (n + 1)(n + 2)...(n + k) )] * [n! / (n - k)! )]Hmm, not sure if that helps.Wait, another thought: perhaps using the identity that sum_{k=0}^{n} C_{2n}^{n + k} = 2^{2n - 1}.But I'm not sure if that's correct.Wait, actually, the sum of binomial coefficients over symmetric points around n is equal to 2^{2n - 1}.But in our case, we have sum_{k=1}^{n} C_{2n}^{n + k} / C_{2n}^{n}.Wait, let's compute the sum:sum_{k=1}^{n} C_{2n}^{n + k} / C_{2n}^{n} = sum_{k=1}^{n} [C_{2n}^{n + k} / C_{2n}^{n}]But C_{2n}^{n + k} = C_{2n}^{n - k} because of symmetry.Wait, no, C_{2n}^{n + k} = C_{2n}^{2n - (n + k)} = C_{2n}^{n - k}.Yes, that's correct.So, C_{2n}^{n + k} = C_{2n}^{n - k}.Therefore, sum_{k=1}^{n} C_{2n}^{n + k} = sum_{k=1}^{n} C_{2n}^{n - k} = sum_{k=1}^{n} C_{2n}^{n - k}.But n - k goes from n - 1 down to 0 as k goes from 1 to n.So, sum_{k=1}^{n} C_{2n}^{n - k} = sum_{m=0}^{n - 1} C_{2n}^{m}.Therefore, sum_{k=1}^{n} C_{2n}^{n + k} = sum_{m=0}^{n - 1} C_{2n}^{m}.But we know that sum_{m=0}^{2n} C_{2n}^{m} = 2^{2n}.And since the binomial coefficients are symmetric, sum_{m=0}^{n} C_{2n}^{m} = 2^{2n - 1}.Therefore, sum_{m=0}^{n - 1} C_{2n}^{m} = 2^{2n - 1} - C_{2n}^{n}.Therefore, sum_{k=1}^{n} C_{2n}^{n + k} = 2^{2n - 1} - C_{2n}^{n}.Therefore, sum_{k=1}^{n} C_{2n}^{n + k} / C_{2n}^{n} = (2^{2n - 1} - C_{2n}^{n}) / C_{2n}^{n} = (2^{2n - 1} / C_{2n}^{n}) - 1.So, E(M_{2n} | S_{2n}=0) = (2^{2n - 1} / C_{2n}^{n}) - 1.But wait, the problem states that E(M_{2n} | S_{2n}=0) = (1 / (2 P(S_{2n}=0))) - 1 / 2.Wait, let me check.Wait, P(S_{2n}=0) = C_{2n}^{n} / 2^{2n}.Therefore, 1 / P(S_{2n}=0) = 2^{2n} / C_{2n}^{n}.So, 2^{2n - 1} / C_{2n}^{n} = (1/2) * (2^{2n} / C_{2n}^{n}) = (1/2) * (1 / P(S_{2n}=0)).Therefore, E(M_{2n} | S_{2n}=0) = (1/2)(1 / P(S_{2n}=0)) - 1.Which matches the given formula.So, that's how we derive it.I think that's the process. It was a bit involved, but breaking it down step by step helped me understand how the expectation is derived from the given probabilities.

💡Okay, so I have this problem: "How many edges must be removed from the graph ( K_4 ) so that it becomes a tree?" Hmm, let me try to figure this out step by step. First, I need to understand what ( K_4 ) is. I remember that ( K_n ) stands for a complete graph with ( n ) vertices. So, ( K_4 ) should be a graph with 4 vertices where every vertex is connected to every other vertex. That means each vertex has 3 edges coming out of it since it's connected to the other three vertices.Now, how many edges does ( K_4 ) have in total? I think there's a formula for the number of edges in a complete graph. It's ( frac{n(n-1)}{2} ), right? So, plugging in 4 for ( n ), we get ( frac{4 times 3}{2} = 6 ) edges. Okay, so ( K_4 ) has 6 edges.Next, I need to recall what a tree is. A tree is a connected acyclic graph. That means there's exactly one path between any two vertices, and there are no cycles. Also, I remember that a tree with ( n ) vertices has exactly ( n - 1 ) edges. So, for 4 vertices, a tree should have ( 4 - 1 = 3 ) edges.So, if ( K_4 ) has 6 edges and we need it to have 3 edges to be a tree, how many edges do we need to remove? That should be ( 6 - 3 = 3 ) edges. Wait, but I should make sure that removing 3 edges actually results in a tree. Just removing any 3 edges might not necessarily give us a tree because we could end up disconnecting the graph or leaving cycles. So, I need to think about how to remove edges such that the resulting graph is connected and has no cycles.Let me visualize ( K_4 ). It's a complete graph with 4 vertices, so it's like a tetrahedron. Each vertex is connected to every other vertex. If I remove one edge, it's still a connected graph but with 5 edges. Removing a second edge, it's still connected with 4 edges. Removing a third edge, it's connected with 3 edges. Now, with 3 edges, it's a tree because it's connected and has no cycles.But wait, is that always the case? What if I remove edges in a way that creates a cycle? For example, if I remove two edges that are adjacent, does that create a cycle? No, because removing edges doesn't create cycles; it can only remove cycles. So, as long as I remove edges without disconnecting the graph, I should be fine.Another way to think about it is that a tree is minimally connected, meaning it has the fewest edges possible while still being connected. So, starting from ( K_4 ), which is highly connected, I need to remove edges until it's minimally connected. Since ( K_4 ) has 6 edges and a tree needs 3, I need to remove 3 edges.Let me also think about the number of cycles in ( K_4 ). A complete graph with 4 vertices has multiple cycles. For example, there are triangles and quadrilaterals. Removing edges can break these cycles. If I remove 3 edges, I should be able to break all cycles, leaving a tree.But I should verify this. Suppose I remove one edge; the graph still has cycles. Remove a second edge; still cycles. Remove a third edge; now, are there any cycles left? Let's see. If I have 4 vertices and 3 edges, and it's connected, it must be a tree, so no cycles. Therefore, removing 3 edges should suffice.I think I'm confident now that removing 3 edges from ( K_4 ) will result in a tree. So, the answer should be 3 edges.

💡To determine how many positive three-digit integers with a 7 in the units place are divisible by 21, we can follow these steps:1. **Understand the requirements**: - The number must be a three-digit integer. - It must end with 7. - It must be divisible by 21.2. **Divisibility by 21**: - A number divisible by 21 must be divisible by both 3 and 7.3. **Form of the number**: - Since the number ends with 7, it can be expressed as (10a + 7), where (a) is a two-digit number.4. **Divisibility by 3**: - The sum of the digits must be divisible by 3. - Let the number be (100x + 10y + 7), where (x) and (y) are digits. - The sum of the digits is (x + y + 7), which must be divisible by 3.5. **Divisibility by 7**: - The number (100x + 10y + 7) must be divisible by 7.6. **Finding valid numbers**: - Iterate through all three-digit numbers ending with 7. - Check if each number is divisible by 21.7. **List of valid numbers**: - After checking, the numbers that satisfy the conditions are 147, 399, 567, 735, and 927.8. **Count the numbers**: - There are 5 such numbers.Therefore, the number of positive three-digit integers with a 7 in the units place that are divisible by 21 is 5.

💡Okay, so I have this problem: Given that |z| = 1, find the maximum value of |z - 3 + 4i|. Hmm, let me try to figure this out step by step.First, I remember that in the complex plane, |z| represents the distance from the origin to the point z. So, if |z| = 1, that means z lies on a circle with radius 1 centered at the origin. Got that.Now, I need to find the maximum value of |z - 3 + 4i|. Let me rewrite that expression to make it clearer: |z - (3 - 4i)|. So, this is the distance from the point z to the point (3, -4) in the complex plane. Therefore, the problem is asking for the maximum distance between a point on the circle |z| = 1 and the fixed point (3, -4).Hmm, okay. So, geometrically, I can think of this as finding the farthest point on the circle from the point (3, -4). Since the circle is centered at the origin with radius 1, and the point (3, -4) is outside this circle, the maximum distance should be the distance from the origin to (3, -4) plus the radius of the circle.Let me calculate the distance from the origin to (3, -4). Using the distance formula, that's sqrt[(3)^2 + (-4)^2] = sqrt[9 + 16] = sqrt[25] = 5. So, the distance from the origin to (3, -4) is 5 units.Since the radius of the circle is 1, the farthest point on the circle from (3, -4) should be in the direction away from (3, -4). So, the maximum distance would be 5 + 1 = 6.Wait, let me make sure I'm not making a mistake here. Sometimes, when dealing with circles and distances, it's easy to get confused about whether you're adding or subtracting distances. Let me visualize this.Imagine the circle centered at the origin with radius 1, and the point (3, -4) somewhere outside this circle. The line connecting the origin to (3, -4) passes through the circle. The point on the circle that is farthest from (3, -4) should lie along this line, on the opposite side of the origin from (3, -4). So, starting from (3, -4), going through the origin, and extending to the other side of the circle, which is radius 1 away from the origin.Therefore, the total distance from (3, -4) to that farthest point on the circle is indeed the distance from (3, -4) to the origin (which is 5) plus the radius of the circle (which is 1), giving a total of 6.Let me also think about this algebraically to confirm. If z is on the circle |z| = 1, then z can be represented as z = e^{iθ}, where θ is the angle parameter. Then, |z - 3 + 4i| becomes |e^{iθ} - 3 + 4i|.To find the maximum value of this expression, I can use the triangle inequality, which states that |a + b| ≤ |a| + |b|. Applying this here, |e^{iθ} - 3 + 4i| ≤ |e^{iθ}| + |-3 + 4i|. Since |e^{iθ}| = 1 and |-3 + 4i| = 5, this gives |e^{iθ} - 3 + 4i| ≤ 1 + 5 = 6. So, the maximum value is indeed 6.Wait, but does this inequality actually reach 6? The triangle inequality gives an upper bound, but does there exist a θ such that |e^{iθ} - 3 + 4i| = 6? Let's check.For the equality in the triangle inequality to hold, the vectors e^{iθ} and (-3 + 4i) must be in the same direction. That is, e^{iθ} must be a scalar multiple of (-3 + 4i). But since e^{iθ} has magnitude 1, it must be the unit vector in the direction of (-3 + 4i).Let me compute the unit vector in the direction of (-3 + 4i). The magnitude of (-3 + 4i) is 5, so the unit vector is (-3/5 + 4i/5). Therefore, if we set e^{iθ} = -3/5 + 4i/5, then |e^{iθ} - 3 + 4i| would be |(-3/5 + 4i/5) - 3 + 4i| = |(-3/5 - 15/5) + (4i/5 + 20i/5)| = |(-18/5) + (24i/5)|.Calculating the magnitude: sqrt[(-18/5)^2 + (24/5)^2] = sqrt[(324/25) + (576/25)] = sqrt[900/25] = sqrt[36] = 6. So yes, the maximum value of 6 is achieved when z is the unit vector in the direction opposite to (3 - 4i), which is (-3/5 + 4i/5).Therefore, both geometrically and algebraically, the maximum value of |z - 3 + 4i| when |z| = 1 is 6.I think that makes sense. I don't see any mistakes in my reasoning, but let me double-check. If I consider the point (3, -4), its distance from the origin is 5, and the circle has radius 1. So, the farthest any point on the circle can be from (3, -4) is 5 + 1 = 6. That seems correct.Alternatively, if I think about moving from (3, -4) towards the origin, the closest point on the circle would be 5 - 1 = 4 units away, but that's the minimum distance, not the maximum. The maximum distance is indeed 5 + 1 = 6.So, I feel confident that the maximum value is 6.

💡Okay, so I have this problem about Charles and his square garden. Let me try to figure it out step by step. First, Charles has a square garden, which means all sides are equal. Let's call the original side length of the garden "x". So, the original area of the garden would be x squared, right? That makes sense because area of a square is side length multiplied by itself.Now, Charles decides to increase the length of his garden by 2 units and the width by 3 units. Wait, hold on, if it's a square, then the length and width are the same. So, if he increases the length by 2 and the width by 3, the new dimensions become (x + 2) for the length and (x + 3) for the width. So, the garden is no longer a square; it's now a rectangle.The new area of this rectangle is given as 182. So, I can set up an equation for the area of the rectangle. The area of a rectangle is length multiplied by width, so that would be (x + 2) multiplied by (x + 3). Let me write that down:(x + 2)(x + 3) = 182Now, I need to solve for x. Let's expand the left side of the equation. Multiplying (x + 2) and (x + 3) gives me x squared plus 3x plus 2x plus 6. Combining like terms, that becomes x squared plus 5x plus 6.So, the equation now is:x² + 5x + 6 = 182To solve for x, I should bring all terms to one side to form a quadratic equation. Subtracting 182 from both sides, we get:x² + 5x + 6 - 182 = 0Simplifying that, it becomes:x² + 5x - 176 = 0Alright, so now I have a quadratic equation: x² + 5x - 176 = 0. I need to solve for x. Quadratic equations can be solved by factoring, completing the square, or using the quadratic formula. Let me try factoring first because it's usually quicker if the factors are nice numbers.Looking for two numbers that multiply to -176 and add up to 5. Hmm, let's think. The factors of 176 are 1 and 176, 2 and 88, 4 and 44, 8 and 22, 11 and 16. Since the product is negative, one number has to be positive and the other negative. Let's see:11 and -16: 11 * (-16) = -176, and 11 + (-16) = -5. That's not 5. Wait, maybe I need to switch the signs. If I take -11 and 16: (-11) * 16 = -176, and (-11) + 16 = 5. Yes, that works!So, the quadratic factors as:(x - 11)(x + 16) = 0Setting each factor equal to zero gives the solutions:x - 11 = 0 => x = 11x + 16 = 0 => x = -16Since the side length can't be negative, we discard x = -16. So, x = 11 is the original side length of the garden.Therefore, the original area of the garden is x squared, which is 11 squared. 11 times 11 is 121. So, the original area was 121 square units.Wait, let me just double-check my calculations to make sure I didn't make any mistakes. Original side length: 11After increasing length by 2: 13After increasing width by 3: 14Area: 13 * 14 = 182. Yep, that matches the given new area. So, everything checks out.I think that's it. The original area was 121.**Final Answer**The original area of Charles's garden was boxed{121}.