💡First, I need to isolate the fractional part of the equation. I'll subtract (7x) from both sides to get:[frac{15x^2 - 40x + 18}{4x - 3} = x - 2]Next, I'll eliminate the denominator by multiplying both sides of the equation by (4x - 3):[15x^2 - 40x + 18 = (x - 2)(4x - 3)]Now, I'll expand the right side of the equation:[(x - 2)(4x - 3) = 4x^2 - 11x + 6]This gives me the equation:[15x^2 - 40x + 18 = 4x^2 - 11x + 6]To solve for (x), I'll move all terms to one side to form a quadratic equation:[11x^2 - 29x + 12 = 0]I'll factor the quadratic equation:[(11x - 3)(x - 4) = 0]Setting each factor equal to zero gives the solutions:[x = frac{3}{11} quad text{and} quad x = 4]Among these, the largest value of (x) is (4).

💡Okay, so I need to provide an example where the conditional probability P(B | G)(ω) is not a measure with probability one. Hmm, let me try to understand what this means. First, I know that conditional probability given a σ-algebra G is a way to assign probabilities to events in F based on the information in G. Usually, for each ω, P(· | G)(ω) is supposed to be a probability measure, meaning it should satisfy the axioms of probability: non-negativity, unitarity, and countable additivity.But the problem says that it's not always assumed that P(· | G)(ω) is a measure with probability one. So, I need to find an example where, for some ω, P(· | G)(ω) doesn't behave like a proper measure.Let me think about how conditional probabilities are defined. If G is a sub-σ-algebra of F, then P(B | G) is a G-measurable function such that for every G ∈ G, the integral of P(B | G) over G is equal to P(B ∩ G). So, P(B | G) is defined almost surely, meaning that for almost every ω, P(B | G)(ω) gives a valid probability.But the issue arises when we consider that for each ω individually, P(· | G)(ω) might not satisfy all the measure properties. Maybe it fails countable additivity for some ω, even though it's a measure almost everywhere.Let me try to construct an example. Suppose we have a simple probability space, like Ω = [0,1] with the Borel σ-algebra and Lebesgue measure. Let G be the σ-algebra consisting of sets of measure 0 or 1. So, G is very coarse; it only distinguishes between events that are almost sure or almost impossible.Now, consider the conditional probability P(B | G)(ω). For any Borel set B, P(B | G)(ω) should be either 0 or 1, depending on whether B is in G or not. Wait, no, that's not quite right. Actually, since G is so coarse, the conditional probability P(B | G) should be a constant function, equal to P(B), except possibly on a set of measure zero.But if I define P(B | G)(ω) = P(B) for all ω, then this is a measure for each ω, right? Because it's just the constant function P(B). So, that doesn't seem to work.Maybe I need a different setup. What if G is not as coarse? Let's say G is generated by some partition of Ω. For example, suppose Ω is [0,1], and G is generated by the partition { [0,1/2), [1/2,1] }. Then, for each ω, P(B | G)(ω) would be either the measure of B in [0,1/2) or in [1/2,1], depending on which partition element ω is in.In this case, for each ω, P(B | G)(ω) is a measure because it's just the restriction of P to the partition element containing ω. So, again, it seems to be a measure.Hmm, maybe I need a more pathological example. What if G is not countably generated or something? Or perhaps if G is not nice in some way.Wait, another thought: conditional probabilities are defined only up to sets of measure zero. So, for each B, P(B | G) is defined almost surely, but for individual ω, it might not be a measure because it could fail on a set of measure zero.But how can I make it fail? Maybe if I define P(B | G)(ω) in a way that it's not countably additive for some ω.Let me think of a specific example. Suppose Ω = [0,1], F = Borel sets, P = Lebesgue measure. Let G be the σ-algebra generated by the intervals [0,1/2) and [1/2,1]. So, G has four elements: ∅, [0,1/2), [1/2,1], and Ω.Now, for any Borel set B, P(B | G)(ω) is equal to P(B ∩ [0,1/2)) / P([0,1/2)) if ω ∈ [0,1/2), and P(B ∩ [1/2,1)) / P([1/2,1)) if ω ∈ [1/2,1). So, for each ω, P(B | G)(ω) is either 2P(B ∩ [0,1/2)) or 2P(B ∩ [1/2,1)).Now, for each ω, is this a measure? Well, for each ω, it's either the measure 2λ restricted to [0,1/2) or 2λ restricted to [1/2,1). Both of these are measures because they are just scaled Lebesgue measures on their respective intervals.So, again, it seems to be a measure for each ω. Hmm, not helpful.Maybe I need a different approach. What if G is not generated by a partition, but by some other sets? Or perhaps if G is not nice in terms of being countably generated.Wait, another idea: suppose G is the σ-algebra generated by a single random variable that is not measurable with respect to F. But I'm not sure.Alternatively, maybe consider a case where G is not complete. So, there are sets in F that are subsets of null sets not in G. Then, perhaps the conditional probability could fail to be a measure on F for some ω.But I'm not sure how to formalize that.Wait, let's think about the definition again. For P(· | G)(ω) to be a measure, it needs to assign a value to every B ∈ F such that it's countably additive. But if G is not complete, maybe there are sets in F that are not measurable with respect to G, and thus the conditional probability might not be defined properly.But I'm not sure.Wait, another thought: suppose we have an uncountable partition of Ω, and G is the σ-algebra generated by this partition. Then, for each ω, P(B | G)(ω) would be the probability of B within the partition element containing ω.But if the partition is uncountable, then G is not countably generated, and maybe the conditional probability fails to be a measure for some ω.But I'm not sure.Wait, maybe I need to think about a specific example where the conditional probability is not countably additive for some ω.Let me try to construct such an example.Suppose Ω = [0,1], F = Borel sets, P = Lebesgue measure. Let G be the σ-algebra generated by the singletons {ω} for ω ∈ [0,1]. So, G consists of all countable and co-countable sets.Now, for any Borel set B, P(B | G)(ω) is equal to 1 if ω ∈ B, and 0 otherwise. Because G is generated by singletons, so for each ω, the conditional probability is just the indicator function of B at ω.But wait, is this a measure? For each ω, P(B | G)(ω) is either 0 or 1. So, for each ω, it's a Dirac measure concentrated at ω. So, yes, it is a measure.Hmm, not helpful.Wait, maybe I need to consider a different σ-algebra G where the conditional probability is not a measure for some ω.Let me think of a case where G is not nice, like G is not countably generated.Suppose Ω = [0,1], F = Borel sets, P = Lebesgue measure. Let G be the σ-algebra generated by all intervals [0, q) where q is rational. So, G is the Borel σ-algebra restricted to the rationals.Wait, no, that's still countably generated.Alternatively, let G be the σ-algebra generated by all intervals [0, q) where q is in some uncountable set. But I'm not sure.Wait, maybe consider G as the σ-algebra generated by a non-measurable set. But then G would not be a σ-algebra.Wait, no, G has to be a σ-algebra.Hmm, this is tricky.Wait, another idea: suppose we have Ω = {0,1}^N, the space of infinite binary sequences, with the product σ-algebra and the product measure P = (1/2,1/2)^N.Let G be the σ-algebra generated by the first coordinate, so G consists of sets where the first bit is 0 or 1, and the rest are arbitrary.Now, for any event B, P(B | G)(ω) is the conditional probability given the first bit. So, for ω where the first bit is 0, P(B | G)(ω) is P(B | first bit is 0), and similarly for first bit 1.Now, for each ω, P(B | G)(ω) is a measure because it's just the conditional probability given the first bit.Hmm, still not helpful.Wait, maybe I need to consider a case where the conditional probability is not regular. That is, for some ω, P(· | G)(ω) is not a measure because it's not countably additive.But how?Wait, maybe consider a case where G is not countably generated, so that the conditional probability is not countably additive for some ω.But I'm not sure.Wait, another idea: suppose we have Ω = [0,1], F = Borel sets, P = Lebesgue measure. Let G be the σ-algebra generated by the intervals [0,1/2^n) for n ∈ N.So, G is countably generated, but maybe it's not nice in some way.Now, for any Borel set B, P(B | G)(ω) is the limit of P(B ∩ [0,1/2^n)) / P([0,1/2^n)) as n goes to infinity, but only for ω in [0,1/2^n). Wait, no, that's not quite right.Actually, P(B | G)(ω) is defined as the limit of P(B ∩ G_n) / P(G_n) as n goes to infinity, where G_n is a sequence of sets in G decreasing to {ω}.But in this case, G is generated by [0,1/2^n), so the sequence G_n = [0,1/2^n) decreases to {0}.So, for ω = 0, P(B | G)(0) would be the limit of P(B ∩ [0,1/2^n)) / P([0,1/2^n)) as n→∞, which is just P(B ∩ {0}) / P({0}), but P({0}) = 0, so it's undefined.Wait, but in reality, conditional probabilities are defined only up to sets of measure zero. So, maybe P(B | G)(0) is defined as 1 if 0 ∈ B and 0 otherwise, but that's a Dirac measure.But for other ω ≠ 0, P(B | G)(ω) would be the limit as n→∞ of P(B ∩ [0,1/2^n)) / P([0,1/2^n)). But for ω ≠ 0, eventually ω ∉ [0,1/2^n), so P(B ∩ [0,1/2^n)) / P([0,1/2^n)) would be P(B ∩ [0,1/2^n)) / (1/2^n).But as n→∞, [0,1/2^n) shrinks to {0}, so for ω ≠ 0, this limit would be 0 if B doesn't contain 0, and undefined if B contains 0 because P(B ∩ [0,1/2^n)) would be P([0,1/2^n)) if B contains [0,1/2^n), which would make the limit 1.Wait, this is getting complicated.But in any case, for ω ≠ 0, P(B | G)(ω) would be 1 if B contains 0 and 0 otherwise, which is again a Dirac measure.But for ω = 0, it's undefined because P({0}) = 0.So, in this case, for almost every ω, P(B | G)(ω) is a Dirac measure, which is a measure. For ω = 0, it's undefined, but that's a set of measure zero.So, again, it seems to be a measure almost everywhere.Hmm, I'm struggling to find an example where P(· | G)(ω) is not a measure for some ω with positive probability.Wait, maybe I need to consider a case where G is not nice, like G is not countably generated, or something like that.Alternatively, maybe consider a case where G is not complete, so there are sets in F that are subsets of null sets not in G, and thus the conditional probability might not be defined properly.But I'm not sure.Wait, another idea: suppose we have Ω = [0,1], F = Borel sets, P = Lebesgue measure. Let G be the σ-algebra generated by the intervals [0, q) where q is rational.So, G is countably generated, but maybe it's not nice in some way.Now, for any Borel set B, P(B | G)(ω) is the limit of P(B ∩ [0, q_n)) / P([0, q_n)) as q_n decreases to ω.But for irrational ω, this limit might not exist because the rationals are dense.Wait, actually, for any ω, the limit as q_n decreases to ω of P(B ∩ [0, q_n)) / q_n exists and is equal to the density of B at ω, which is either 0 or 1 almost everywhere.Wait, no, that's not necessarily true. The density might not exist for some B.Wait, but by the Lebesgue density theorem, for almost every ω, the density of B at ω exists and is equal to 1 if ω is in the essential interior of B, and 0 otherwise.So, in this case, P(B | G)(ω) would be the density of B at ω, which exists almost everywhere and is either 0 or 1.So, for almost every ω, P(B | G)(ω) is either 0 or 1, which is a measure.Wait, but for some B, maybe the density doesn't exist for some ω, but those ω would be a null set.So, again, it seems that P(· | G)(ω) is a measure almost everywhere.Hmm, I'm not getting anywhere. Maybe I need to think differently.Wait, perhaps the issue is that P(· | G)(ω) is only required to be a measure almost everywhere, not necessarily for every ω. So, even if for some ω it's not a measure, as long as it's a measure almost everywhere, it's fine.But the problem says that it's usually not assumed that P(· | G)(ω) is a measure with probability one. So, I need to find an example where it's not a measure for some ω with positive probability.Wait, maybe consider a case where G is not countably generated, so that the conditional probability is not countably additive for some ω.But I'm not sure.Wait, another idea: suppose we have Ω = [0,1], F = Borel sets, P = Lebesgue measure. Let G be the σ-algebra generated by the intervals [0,1/2) and [1/2,1], and also all singletons in [0,1/2).So, G consists of sets that are either in [0,1/2) and are countable or co-countable, and in [1/2,1] they are arbitrary.Now, for any Borel set B, P(B | G)(ω) is equal to P(B ∩ [0,1/2)) / P([0,1/2)) if ω ∈ [0,1/2), and P(B ∩ [1/2,1)) / P([1/2,1)) if ω ∈ [1/2,1).But for ω ∈ [0,1/2), since G includes all singletons, P(B | G)(ω) is 1 if ω ∈ B and 0 otherwise.Wait, no, because G includes all singletons in [0,1/2), so for ω ∈ [0,1/2), P(B | G)(ω) is 1 if ω ∈ B and 0 otherwise.But for ω ∈ [1/2,1), P(B | G)(ω) is P(B ∩ [1/2,1)) / (1/2).Now, for ω ∈ [0,1/2), P(B | G)(ω) is the indicator function of B at ω, which is a measure (Dirac measure). For ω ∈ [1/2,1), it's a constant measure.So, in this case, P(· | G)(ω) is a measure for every ω.Hmm, still not helpful.Wait, maybe I need to consider a case where G is not nice, like G is not countably generated, or something.Wait, another idea: suppose we have Ω = [0,1], F = Borel sets, P = Lebesgue measure. Let G be the σ-algebra generated by the intervals [0,1/2^n) for n ∈ N.So, G is countably generated, but maybe it's not nice in some way.Now, for any Borel set B, P(B | G)(ω) is the limit of P(B ∩ [0,1/2^n)) / P([0,1/2^n)) as n→∞, but only for ω in [0,1/2^n).Wait, but for ω ≠ 0, eventually ω ∉ [0,1/2^n), so P(B | G)(ω) would be 0 if B doesn't contain ω, and undefined if B contains ω because P(B ∩ [0,1/2^n)) / P([0,1/2^n)) would be P([0,1/2^n)) / P([0,1/2^n)) = 1 if B contains [0,1/2^n), which it does if B contains ω.Wait, but as n→∞, [0,1/2^n) shrinks to {0}, so for ω ≠ 0, P(B | G)(ω) would be 1 if B contains 0 and 0 otherwise.But for ω = 0, it's undefined because P({0}) = 0.So, in this case, for ω ≠ 0, P(B | G)(ω) is 1 if 0 ∈ B and 0 otherwise, which is a measure (Dirac measure at 0). For ω = 0, it's undefined.But since ω = 0 is a null set, it's fine. So, P(· | G)(ω) is a measure almost everywhere.Hmm, still not helpful.Wait, maybe I need to consider a case where G is not nice, like G is not countably generated, or something.Wait, another idea: suppose we have Ω = [0,1], F = Borel sets, P = Lebesgue measure. Let G be the σ-algebra generated by the intervals [0, q) where q is in some uncountable set.But I'm not sure.Wait, maybe consider a case where G is not countably generated, so that the conditional probability is not countably additive for some ω.But I'm not sure.Wait, another thought: suppose we have Ω = [0,1], F = Borel sets, P = Lebesgue measure. Let G be the σ-algebra generated by the intervals [0,1/2) and [1/2,1], and also all singletons in [0,1/2).So, G consists of sets that are either in [0,1/2) and are countable or co-countable, and in [1/2,1] they are arbitrary.Now, for any Borel set B, P(B | G)(ω) is equal to P(B ∩ [0,1/2)) / P([0,1/2)) if ω ∈ [0,1/2), and P(B ∩ [1/2,1)) / P([1/2,1)) if ω ∈ [1/2,1).But for ω ∈ [0,1/2), since G includes all singletons, P(B | G)(ω) is 1 if ω ∈ B and 0 otherwise.Wait, no, because G includes all singletons in [0,1/2), so for ω ∈ [0,1/2), P(B | G)(ω) is 1 if ω ∈ B and 0 otherwise.But for ω ∈ [1/2,1), P(B | G)(ω) is P(B ∩ [1/2,1)) / (1/2).Now, for ω ∈ [0,1/2), P(B | G)(ω) is the indicator function of B at ω, which is a measure (Dirac measure). For ω ∈ [1/2,1), it's a constant measure.So, in this case, P(· | G)(ω) is a measure for every ω.Hmm, still not helpful.Wait, maybe I need to think about a different probability space. Suppose Ω is uncountable, F is the σ-algebra of countable and co-countable sets, and P is the probability measure where P(A) = 0 if A is countable and 1 otherwise.Now, let G be the σ-algebra generated by a single non-measurable set. Wait, but G has to be a σ-algebra.Wait, no, G has to be a sub-σ-algebra of F. So, in this case, F is the σ-algebra of countable and co-countable sets.Let G be the σ-algebra generated by a single co-countable set A. So, G consists of ∅, A, A^c, and Ω.Now, for any B ∈ F, P(B | G)(ω) is equal to P(B ∩ A)/P(A) if ω ∈ A, and P(B ∩ A^c)/P(A^c) if ω ∈ A^c.But since P(A) = 1 if A is co-countable, and P(A^c) = 0.Wait, but P(A^c) = 0 because A is co-countable, so A^c is countable.So, for ω ∈ A, P(B | G)(ω) = P(B ∩ A)/1 = P(B ∩ A).For ω ∈ A^c, P(B | G)(ω) = P(B ∩ A^c)/0, which is undefined.But since A^c is countable, and thus P(A^c) = 0, we can ignore it.So, for ω ∈ A, P(B | G)(ω) = P(B ∩ A).Now, is this a measure for each ω ∈ A?Well, for each ω ∈ A, P(B | G)(ω) = P(B ∩ A). But P(B ∩ A) is just P(B) if B is co-countable, and 0 if B is countable.Wait, no, P(B ∩ A) is P(B) if B is co-countable, because A is co-countable, so B ∩ A is co-countable if B is co-countable.If B is countable, then B ∩ A is countable, so P(B ∩ A) = 0.So, for each ω ∈ A, P(B | G)(ω) is 1 if B is co-countable and 0 if B is countable.But is this a measure? Let's see.A measure needs to assign a value to every B ∈ F such that it's countably additive.But in this case, for each ω ∈ A, P(B | G)(ω) is 1 if B is co-countable and 0 otherwise.But this is not a measure because it's not countably additive. For example, consider a countable union of countable sets, which is countable, so P(B | G)(ω) = 0. But the sum of P(B_n | G)(ω) for countable B_n would be 0, which matches. However, if we take a countable union of co-countable sets, which is co-countable, so P(B | G)(ω) = 1, but the sum of P(B_n | G)(ω) would be 1 for each n, which would diverge to infinity, violating countable additivity.Wait, no, because in this case, each B_n is co-countable, so P(B_n | G)(ω) = 1 for each n. So, the sum would be infinite, but P(∪B_n | G)(ω) = 1, which is not equal to the sum. So, countable additivity fails.Therefore, in this case, P(· | G)(ω) is not a measure for ω ∈ A, because it's not countably additive.But wait, is this correct? Because in this specific setup, F is the σ-algebra of countable and co-countable sets, so any countable union of co-countable sets is co-countable, and any countable union of countable sets is countable.So, for each ω ∈ A, P(B | G)(ω) is 1 if B is co-countable and 0 otherwise. Now, if we take a countable union of co-countable sets, it's co-countable, so P(∪B_n | G)(ω) = 1. But the sum of P(B_n | G)(ω) is the sum of 1's, which is infinite. So, 1 ≠ ∞, which violates countable additivity.Therefore, P(· | G)(ω) is not a measure for ω ∈ A.But in this setup, G is the σ-algebra generated by a single co-countable set A, and P is the probability measure where P(A) = 1 if A is co-countable and 0 otherwise.So, in this case, for ω ∈ A, P(· | G)(ω) is not a measure because it's not countably additive.Therefore, this serves as an example where P(· | G)(ω) is not a measure with probability one, because for ω ∈ A, which has probability 1, P(· | G)(ω) is not a measure.So, the example is:Let Ω be an uncountable set, F be the σ-algebra of countable and co-countable sets, P be the probability measure where P(A) = 0 if A is countable and 1 otherwise. Let G be the σ-algebra generated by a single co-countable set A. Then, for ω ∈ A (which has probability 1), P(· | G)(ω) is not a measure because it's not countably additive.

💡Okay, so I have this problem where I need to find all prime numbers ( p ) and ( q ) such that ( pq ) divides ( 2^p + 2^q ). Hmm, that sounds interesting. Let me try to break it down step by step.First, I know that ( p ) and ( q ) are primes, so they can be 2, 3, 5, 7, etc. I need to find pairs ( (p, q) ) where the product ( pq ) divides the sum ( 2^p + 2^q ). That means when I divide ( 2^p + 2^q ) by ( pq ), there should be no remainder.Let me start by considering small primes because they are easier to handle. Maybe I can test some small values of ( p ) and ( q ) to see if they satisfy the condition.Let's try ( p = 2 ). Then, ( pq = 2q ). So, I need ( 2q ) to divide ( 2^2 + 2^q = 4 + 2^q ). Let me see what happens for different ( q ).If ( q = 2 ), then ( pq = 4 ) and ( 2^p + 2^q = 4 + 4 = 8 ). Since 4 divides 8, that works. So, ( (2, 2) ) is a solution.What if ( q = 3 )? Then, ( pq = 6 ) and ( 2^p + 2^q = 4 + 8 = 12 ). 6 divides 12, so ( (2, 3) ) is also a solution.Let me check ( q = 5 ). Then, ( pq = 10 ) and ( 2^p + 2^q = 4 + 32 = 36 ). 10 divides 36? 36 divided by 10 is 3.6, which is not an integer. So, that doesn't work.Hmm, maybe ( q = 7 )? Then, ( pq = 14 ) and ( 2^p + 2^q = 4 + 128 = 132 ). 132 divided by 14 is approximately 9.428... Not an integer. So, that doesn't work either.It seems like for ( p = 2 ), only ( q = 2 ) and ( q = 3 ) work. Let me check ( q = 5 ) again. Wait, 4 + 32 is 36, and 10 doesn't divide 36. Yeah, that's correct.Now, let me try ( p = 3 ). Then, ( pq = 3q ). So, I need ( 3q ) to divide ( 2^3 + 2^q = 8 + 2^q ).If ( q = 2 ), then ( pq = 6 ) and ( 8 + 4 = 12 ). 6 divides 12, so ( (3, 2) ) is a solution.What about ( q = 3 )? Then, ( pq = 9 ) and ( 8 + 8 = 16 ). 9 doesn't divide 16 because 16 divided by 9 is about 1.777... So, that doesn't work.How about ( q = 5 )? Then, ( pq = 15 ) and ( 8 + 32 = 40 ). 15 divides 40? 40 divided by 15 is approximately 2.666... Not an integer. So, nope.( q = 7 ): ( pq = 21 ) and ( 8 + 128 = 136 ). 136 divided by 21 is about 6.476... Not an integer. Doesn't work.So, for ( p = 3 ), only ( q = 2 ) works.Let me try ( p = 5 ). Then, ( pq = 5q ). So, ( 2^5 + 2^q = 32 + 2^q ). I need ( 5q ) to divide ( 32 + 2^q ).Testing ( q = 2 ): ( 5*2 = 10 ) and ( 32 + 4 = 36 ). 10 divides 36? No, as before.( q = 3 ): ( 5*3 = 15 ) and ( 32 + 8 = 40 ). 15 divides 40? No.( q = 5 ): ( 5*5 = 25 ) and ( 32 + 32 = 64 ). 25 divides 64? No.( q = 7 ): ( 5*7 = 35 ) and ( 32 + 128 = 160 ). 35 divides 160? 160 divided by 35 is approximately 4.571... Not an integer.Hmm, seems like ( p = 5 ) doesn't give any solutions.Let me try ( p = 7 ). Then, ( pq = 7q ) and ( 2^7 + 2^q = 128 + 2^q ).Testing ( q = 2 ): ( 7*2 = 14 ) and ( 128 + 4 = 132 ). 14 divides 132? 132 divided by 14 is about 9.428... Not an integer.( q = 3 ): ( 7*3 = 21 ) and ( 128 + 8 = 136 ). 21 divides 136? 136 divided by 21 is approximately 6.476... No.( q = 5 ): ( 7*5 = 35 ) and ( 128 + 32 = 160 ). 35 divides 160? 160 divided by 35 is about 4.571... No.( q = 7 ): ( 7*7 = 49 ) and ( 128 + 128 = 256 ). 49 divides 256? 256 divided by 49 is approximately 5.224... No.So, ( p = 7 ) doesn't give any solutions either.I'm noticing a pattern here. For primes ( p ) and ( q ) greater than 3, it's getting harder to find solutions. Maybe the only solutions are when one of the primes is 2 or 3.Wait, let me think about this more systematically. Maybe I can use some number theory here.If ( pq ) divides ( 2^p + 2^q ), then both ( p ) and ( q ) must divide ( 2^p + 2^q ). So, ( p ) divides ( 2^p + 2^q ) and ( q ) divides ( 2^p + 2^q ).Let me consider ( p ) dividing ( 2^p + 2^q ). Since ( p ) is a prime, by Fermat's Little Theorem, ( 2^{p-1} equiv 1 mod p ). So, ( 2^p equiv 2 mod p ).Similarly, ( 2^q equiv 2^{q mod (p-1)} mod p ). So, ( 2^p + 2^q equiv 2 + 2^{q mod (p-1)} mod p ). For ( p ) to divide ( 2^p + 2^q ), this sum must be congruent to 0 modulo ( p ).So, ( 2 + 2^{q mod (p-1)} equiv 0 mod p ). That implies ( 2^{q mod (p-1)} equiv -2 mod p ).Similarly, for ( q ) dividing ( 2^p + 2^q ), we can write ( 2 + 2^{p mod (q-1)} equiv 0 mod q ), so ( 2^{p mod (q-1)} equiv -2 mod q ).This seems a bit abstract, but maybe I can apply it to specific cases.Let me take ( p = 2 ). Then, ( p - 1 = 1 ), so ( q mod 1 = 0 ). So, ( 2^{0} = 1 ). Then, ( 1 equiv -2 mod 2 ). But ( -2 mod 2 = 0 ), and 1 is not congruent to 0 mod 2. Wait, that doesn't make sense. Maybe my earlier step was wrong.Wait, when ( p = 2 ), ( 2^p = 4 ), and ( 2^q ) is as is. So, ( 2^p + 2^q = 4 + 2^q ). For ( p = 2 ), we need ( 2q ) to divide ( 4 + 2^q ). So, ( q ) must divide ( 4 + 2^q ).But ( q ) is an odd prime (since we already considered ( q = 2 )). So, ( 2^q equiv 2 mod q ) by Fermat's Little Theorem. Therefore, ( 4 + 2^q equiv 4 + 2 = 6 mod q ). So, ( q ) must divide 6. The primes dividing 6 are 2 and 3. So, ( q = 2 ) or ( q = 3 ). We already saw that ( q = 2 ) and ( q = 3 ) work.Similarly, if ( q = 2 ), then ( pq = 2p ) must divide ( 2^p + 4 ). So, ( p ) must divide ( 2^p + 4 ). Again, using Fermat's Little Theorem, ( 2^{p-1} equiv 1 mod p ), so ( 2^p equiv 2 mod p ). Therefore, ( 2^p + 4 equiv 2 + 4 = 6 mod p ). So, ( p ) must divide 6, which means ( p = 2 ) or ( p = 3 ). We already saw that ( p = 2 ) and ( p = 3 ) work.So, that explains why when one of the primes is 2, the other can only be 2 or 3.What about when both ( p ) and ( q ) are odd primes? Let's assume ( p ) and ( q ) are both odd primes greater than 2.Then, ( pq ) divides ( 2^p + 2^q ). So, both ( p ) and ( q ) must divide ( 2^p + 2^q ).Let me consider ( p ) dividing ( 2^p + 2^q ). As before, ( 2^p equiv 2 mod p ), so ( 2^p + 2^q equiv 2 + 2^q mod p ). Therefore, ( 2 + 2^q equiv 0 mod p ), which implies ( 2^q equiv -2 mod p ).Similarly, ( q ) divides ( 2^p + 2^q ), so ( 2^p + 2^q equiv 0 mod q ). By Fermat's Little Theorem, ( 2^{q-1} equiv 1 mod q ), so ( 2^q equiv 2 mod q ). Therefore, ( 2^p + 2^q equiv 2^p + 2 equiv 0 mod q ), which implies ( 2^p equiv -2 mod q ).So, we have:1. ( 2^q equiv -2 mod p )2. ( 2^p equiv -2 mod q )This seems like a system of congruences. Maybe I can find some relationship between ( p ) and ( q ).Let me see if I can express one in terms of the other. From the first equation, ( 2^q equiv -2 mod p ). Let me write this as ( 2^{q} + 2 equiv 0 mod p ). Similarly, from the second equation, ( 2^{p} + 2 equiv 0 mod q ).So, both ( p ) and ( q ) divide ( 2^{k} + 2 ) where ( k ) is the other prime.This seems symmetric. Maybe I can assume without loss of generality that ( p < q ) and see if I can find such primes.Let me try ( p = 3 ). Then, ( 2^3 + 2 = 8 + 2 = 10 ). So, ( q ) must divide 10. The primes dividing 10 are 2 and 5. We already considered ( q = 2 ), which works. What about ( q = 5 )?Let me check if ( p = 3 ) and ( q = 5 ) satisfy the original condition. So, ( pq = 15 ) and ( 2^3 + 2^5 = 8 + 32 = 40 ). Does 15 divide 40? 40 divided by 15 is approximately 2.666... So, no, it doesn't. Therefore, ( q = 5 ) doesn't work.Wait, but earlier, I thought ( q ) must divide ( 2^p + 2 ), which is 10 in this case. So, ( q = 5 ) divides 10, but when I plug it back into the original condition, it doesn't satisfy. So, maybe there's more to it.Perhaps the fact that ( p ) divides ( 2^q + 2 ) and ( q ) divides ( 2^p + 2 ) is necessary but not sufficient. There might be additional constraints.Let me try another approach. Suppose both ( p ) and ( q ) are odd primes greater than 3. Then, ( p ) and ( q ) are both congruent to 1 or 2 modulo 3, but since they are primes greater than 3, they must be congruent to 1 or 2 modulo 3.Wait, actually, primes greater than 3 are congruent to 1 or 2 modulo 3, but not necessarily. For example, 5 is 2 mod 3, 7 is 1 mod 3, 11 is 2 mod 3, etc.I'm not sure if this helps. Maybe I can look at the exponents.From ( 2^q equiv -2 mod p ), I can write ( 2^{q} equiv -2 mod p ). Let me divide both sides by 2: ( 2^{q - 1} equiv -1 mod p ). Similarly, from ( 2^p equiv -2 mod q ), dividing both sides by 2: ( 2^{p - 1} equiv -1 mod q ).So, now I have:1. ( 2^{q - 1} equiv -1 mod p )2. ( 2^{p - 1} equiv -1 mod q )This is interesting. So, the order of 2 modulo ( p ) must divide ( 2(q - 1) ) because ( 2^{2(q - 1)} equiv 1 mod p ). Similarly, the order of 2 modulo ( q ) must divide ( 2(p - 1) ).Let me denote ( d_p ) as the order of 2 modulo ( p ), and ( d_q ) as the order of 2 modulo ( q ). Then, ( d_p ) divides ( 2(q - 1) ) and ( d_q ) divides ( 2(p - 1) ).Also, since ( 2^{q - 1} equiv -1 mod p ), squaring both sides gives ( 2^{2(q - 1)} equiv 1 mod p ). So, the order ( d_p ) divides ( 2(q - 1) ) but does not divide ( q - 1 ), because ( 2^{q - 1} equiv -1 mod p neq 1 mod p ). Therefore, ( d_p ) must be exactly ( 2(q - 1) ) or a divisor that is twice an odd divisor of ( q - 1 ).Similarly, ( d_q ) must be exactly ( 2(p - 1) ) or a divisor that is twice an odd divisor of ( p - 1 ).This seems complicated. Maybe I can consider specific cases where ( p ) and ( q ) are small odd primes.Let me try ( p = 5 ). Then, ( 2^{5 - 1} = 16 equiv -1 mod q ). So, ( 16 equiv -1 mod q ), which implies ( q ) divides ( 17 ). Since 17 is prime, ( q = 17 ).Let me check if ( p = 5 ) and ( q = 17 ) satisfy the original condition. So, ( pq = 85 ) and ( 2^5 + 2^{17} = 32 + 131072 = 131104 ). Does 85 divide 131104?Let me compute ( 131104 div 85 ). 85 times 1542 is 130, 85*1542 = 85*(1500 + 42) = 127500 + 3570 = 131070. Then, 131104 - 131070 = 34. So, 131104 = 85*1542 + 34. Therefore, 85 does not divide 131104. So, ( q = 17 ) doesn't work.Hmm, so even though ( q = 17 ) divides ( 2^4 + 2 = 18 ), it doesn't satisfy the original condition. Maybe my earlier reasoning was incomplete.Let me try ( p = 7 ). Then, ( 2^{7 - 1} = 64 equiv -1 mod q ). So, ( 64 equiv -1 mod q ), which implies ( q ) divides ( 65 ). The primes dividing 65 are 5 and 13.Let me check ( q = 5 ). Then, ( pq = 35 ) and ( 2^7 + 2^5 = 128 + 32 = 160 ). Does 35 divide 160? 160 divided by 35 is approximately 4.571... So, no.How about ( q = 13 )? Then, ( pq = 91 ) and ( 2^7 + 2^{13} = 128 + 8192 = 8320 ). Does 91 divide 8320? Let's compute 91*91 = 8281. 8320 - 8281 = 39. So, 8320 = 91*91 + 39. Therefore, 91 does not divide 8320.So, ( q = 13 ) doesn't work either.This is getting frustrating. Maybe there are no solutions where both ( p ) and ( q ) are odd primes greater than 3.Wait, let me try ( p = 3 ) again. If ( p = 3 ), then ( 2^{3 - 1} = 4 equiv -1 mod q ). So, ( 4 equiv -1 mod q ), which implies ( q ) divides ( 5 ). So, ( q = 5 ).But earlier, when I checked ( p = 3 ) and ( q = 5 ), it didn't work because 15 doesn't divide 40. So, that doesn't help.Wait, but maybe I made a mistake in my earlier reasoning. Let me double-check.If ( p = 3 ), then ( 2^{p - 1} = 4 equiv -1 mod q ). So, ( 4 equiv -1 mod q ) implies ( q ) divides ( 4 + 1 = 5 ). So, ( q = 5 ).But ( pq = 15 ) and ( 2^3 + 2^5 = 8 + 32 = 40 ). 40 divided by 15 is not an integer. So, ( q = 5 ) doesn't work.Hmm, maybe there's a problem with my approach. Let me try to think differently.Suppose both ( p ) and ( q ) are odd primes. Then, ( 2^p + 2^q ) is even, so ( pq ) must also be even. But since ( p ) and ( q ) are odd, ( pq ) is odd. Wait, that's a contradiction because ( 2^p + 2^q ) is even, and ( pq ) is odd, so an odd number cannot divide an even number unless the even number is a multiple of the odd number. But ( 2^p + 2^q ) is even, so ( pq ) must divide it. But ( pq ) is odd, so ( pq ) must divide ( (2^p + 2^q)/2 ).Wait, that's a good point. Let me write it down.Since ( pq ) divides ( 2^p + 2^q ), and ( 2^p + 2^q ) is even, ( pq ) must divide ( (2^p + 2^q)/2 ). So, ( pq ) divides ( 2^{p - 1} + 2^{q - 1} ).Therefore, ( p ) divides ( 2^{p - 1} + 2^{q - 1} ) and ( q ) divides ( 2^{p - 1} + 2^{q - 1} ).But ( p ) divides ( 2^{p - 1} + 2^{q - 1} ). By Fermat's Little Theorem, ( 2^{p - 1} equiv 1 mod p ). So, ( 1 + 2^{q - 1} equiv 0 mod p ). Therefore, ( 2^{q - 1} equiv -1 mod p ).Similarly, ( q ) divides ( 2^{p - 1} + 2^{q - 1} ). By Fermat's Little Theorem, ( 2^{q - 1} equiv 1 mod q ). So, ( 2^{p - 1} + 1 equiv 0 mod q ). Therefore, ( 2^{p - 1} equiv -1 mod q ).So, we have:1. ( 2^{q - 1} equiv -1 mod p )2. ( 2^{p - 1} equiv -1 mod q )This is similar to what I had before, but now it's in terms of ( p - 1 ) and ( q - 1 ).Let me denote ( a = p - 1 ) and ( b = q - 1 ). Then, the equations become:1. ( 2^{b} equiv -1 mod (a + 1) )2. ( 2^{a} equiv -1 mod (b + 1) )This seems like a system of equations where ( a ) and ( b ) are related through these congruences.I wonder if there are small values of ( a ) and ( b ) that satisfy this.Let me try ( a = 2 ). Then, ( p = 3 ). From equation 2, ( 2^{2} = 4 equiv -1 mod (b + 1) ). So, ( 4 equiv -1 mod (b + 1) ), which implies ( b + 1 ) divides ( 5 ). So, ( b + 1 = 5 ), hence ( b = 4 ). Therefore, ( q = b + 1 = 5 ).So, ( p = 3 ) and ( q = 5 ). Let me check if this works. ( pq = 15 ) and ( 2^3 + 2^5 = 8 + 32 = 40 ). 15 doesn't divide 40, as before. So, this doesn't work.Wait, but according to the earlier reasoning, ( p = 3 ) and ( q = 5 ) should satisfy the conditions. Maybe I'm missing something.Wait, no. The equations ( 2^{b} equiv -1 mod (a + 1) ) and ( 2^{a} equiv -1 mod (b + 1) ) are necessary conditions, but they might not be sufficient. So, even if they are satisfied, the original condition might not hold.Let me try another small ( a ). Let ( a = 4 ), so ( p = 5 ). From equation 2, ( 2^{4} = 16 equiv -1 mod (b + 1) ). So, ( 16 equiv -1 mod (b + 1) ), which implies ( b + 1 ) divides ( 17 ). So, ( b + 1 = 17 ), hence ( b = 16 ). Therefore, ( q = 17 ).Let me check ( p = 5 ) and ( q = 17 ). ( pq = 85 ) and ( 2^5 + 2^{17} = 32 + 131072 = 131104 ). 131104 divided by 85 is approximately 1542.4, which is not an integer. So, it doesn't work.Hmm, maybe ( a = 6 ). Then, ( p = 7 ). From equation 2, ( 2^{6} = 64 equiv -1 mod (b + 1) ). So, ( 64 equiv -1 mod (b + 1) ), which implies ( b + 1 ) divides ( 65 ). So, ( b + 1 = 5 ) or ( 13 ). Therefore, ( b = 4 ) or ( 12 ). Hence, ( q = 5 ) or ( 13 ).Let me check ( q = 5 ). ( pq = 35 ) and ( 2^7 + 2^5 = 128 + 32 = 160 ). 35 doesn't divide 160.How about ( q = 13 )? ( pq = 91 ) and ( 2^7 + 2^{13} = 128 + 8192 = 8320 ). 8320 divided by 91 is approximately 91.428... Not an integer. So, doesn't work.This is getting me nowhere. Maybe there are no solutions where both ( p ) and ( q ) are odd primes greater than 3.Wait, let me think about the original problem again. We found that when one of the primes is 2, the other can be 2 or 3. When both are 2 or 3, we have solutions. But when both are odd primes greater than 3, we can't find any solutions. Maybe those are the only solutions.Let me try to see if there's a theoretical reason why there are no solutions for odd primes greater than 3.Suppose both ( p ) and ( q ) are odd primes greater than 3. Then, ( p ) and ( q ) are both congruent to 1 or 2 modulo 4, but since they are odd, they are congruent to 1 or 3 modulo 4.Wait, actually, primes greater than 2 are odd, so they are congruent to 1 or 3 modulo 4.But I'm not sure if that helps.Let me consider the equation ( 2^{q - 1} equiv -1 mod p ). Squaring both sides, we get ( 2^{2(q - 1)} equiv 1 mod p ). So, the order of 2 modulo ( p ) divides ( 2(q - 1) ) but does not divide ( q - 1 ), as before.Similarly, the order of 2 modulo ( q ) divides ( 2(p - 1) ) but does not divide ( p - 1 ).This suggests that the orders are even numbers, specifically twice an odd number.But I'm not sure how to proceed from here.Maybe I can consider the multiplicative order of 2 modulo ( p ) and ( q ). Let me denote ( d_p ) as the order of 2 modulo ( p ), and ( d_q ) as the order of 2 modulo ( q ).From ( 2^{q - 1} equiv -1 mod p ), we have ( 2^{2(q - 1)} equiv 1 mod p ). So, ( d_p ) divides ( 2(q - 1) ). Also, since ( 2^{q - 1} equiv -1 mod p ), ( d_p ) does not divide ( q - 1 ), so ( d_p ) must be exactly ( 2k ) where ( k ) divides ( q - 1 ).Similarly, ( d_q ) must be exactly ( 2m ) where ( m ) divides ( p - 1 ).This seems like a recursive relationship, but I'm not sure how to exploit it.Maybe I can look for primes where the order of 2 is twice an odd number.For example, let me check ( p = 7 ). The order of 2 modulo 7 is 3 because ( 2^3 = 8 equiv 1 mod 7 ). Wait, no, 2^3 = 8 ≡ 1 mod 7, so the order is 3, which is odd. But we need the order to be even because ( d_p ) must be twice an odd number.Wait, no. If ( d_p ) is the order, and ( d_p ) divides ( 2(q - 1) ), but ( d_p ) doesn't divide ( q - 1 ), then ( d_p ) must be even.So, the order of 2 modulo ( p ) must be even. Therefore, ( p ) must be such that 2 is a quadratic residue modulo ( p ), but I'm not sure.Wait, actually, if the order of 2 modulo ( p ) is even, then 2 is a quadratic residue modulo ( p ) if and only if the order divides ( (p - 1)/2 ). Hmm, maybe not directly useful.Let me check some small primes:- ( p = 5 ): Order of 2 is 4 (since 2^4 = 16 ≡ 1 mod 5). So, even order.- ( p = 7 ): Order of 2 is 3 (odd).- ( p = 11 ): Order of 2 is 10 (even).- ( p = 13 ): Order of 2 is 12 (even).- ( p = 17 ): Order of 2 is 8 (even).- ( p = 19 ): Order of 2 is 18 (even).So, primes like 5, 11, 13, 17, 19 have even orders for 2, while primes like 3, 7 have odd orders.Wait, ( p = 3 ): Order of 2 is 2 (even). Hmm, 2^2 = 4 ≡ 1 mod 3. So, order is 2, which is even.Wait, earlier I thought ( p = 3 ) had order 2, which is even. So, maybe ( p = 3 ) is a candidate.But when I tried ( p = 3 ) and ( q = 5 ), it didn't work. Maybe I need to find another prime ( q ) such that ( 2^{q - 1} equiv -1 mod 3 ).Wait, ( 2^{q - 1} equiv -1 mod 3 ). Since 2 ≡ -1 mod 3, so ( (-1)^{q - 1} equiv -1 mod 3 ). Therefore, ( (-1)^{q - 1} = -1 ), which implies that ( q - 1 ) is odd, so ( q ) is even. But the only even prime is 2. So, ( q = 2 ).But we already considered ( q = 2 ) with ( p = 3 ), which works. So, that's consistent.So, for ( p = 3 ), the only possible ( q ) is 2.Similarly, for ( p = 5 ), let's see. The order of 2 modulo 5 is 4, which is even. So, ( 2^{q - 1} equiv -1 mod 5 ). Let's compute ( 2^{q - 1} mod 5 ).We know that ( 2^4 ≡ 1 mod 5 ), so the powers cycle every 4.We need ( 2^{q - 1} ≡ -1 ≡ 4 mod 5 ). So, ( 2^{q - 1} ≡ 4 mod 5 ).Looking at powers of 2 modulo 5:- ( 2^1 ≡ 2 )- ( 2^2 ≡ 4 )- ( 2^3 ≡ 3 )- ( 2^4 ≡ 1 )- ( 2^5 ≡ 2 ), etc.So, ( 2^{q - 1} ≡ 4 mod 5 ) when ( q - 1 ≡ 2 mod 4 ). Therefore, ( q - 1 ≡ 2 mod 4 ), so ( q ≡ 3 mod 4 ).So, ( q ) must be a prime congruent to 3 modulo 4.Let me try ( q = 7 ). Then, ( pq = 35 ) and ( 2^5 + 2^7 = 32 + 128 = 160 ). 35 divides 160? 160 divided by 35 is approximately 4.571... No.Next, ( q = 11 ). ( pq = 55 ) and ( 2^5 + 2^{11} = 32 + 2048 = 2080 ). 2080 divided by 55 is approximately 37.818... Not an integer.Next, ( q = 19 ). ( pq = 95 ) and ( 2^5 + 2^{19} = 32 + 524288 = 524320 ). 524320 divided by 95 is approximately 5519.157... Not an integer.Hmm, not working.Wait, maybe I need to check if ( 2^{p - 1} ≡ -1 mod q ). For ( p = 5 ), ( 2^{4} = 16 ≡ -1 mod q ). So, ( 16 ≡ -1 mod q ), which implies ( q ) divides ( 17 ). So, ( q = 17 ).But earlier, ( p = 5 ) and ( q = 17 ) didn't work because 85 doesn't divide 131104.Wait, but according to the earlier reasoning, ( q ) must divide ( 17 ), so ( q = 17 ). But when I plug it back, it doesn't satisfy the original condition. So, maybe there's a mistake in my reasoning.Alternatively, maybe there are no solutions where both ( p ) and ( q ) are odd primes greater than 3.Given that I've tried several small primes and haven't found any solutions, and considering the complexity of the conditions, I'm starting to think that the only solutions are when one of the primes is 2 or 3.So, summarizing:- When ( p = 2 ), ( q ) can be 2 or 3.- When ( q = 2 ), ( p ) can be 2 or 3.- When both ( p ) and ( q ) are odd primes greater than 3, no solutions seem to exist.Therefore, the only solutions are ( (2, 2) ), ( (2, 3) ), and ( (3, 2) ).I think that's it. I don't see any other pairs of primes that satisfy the condition.

💡Okay, so I have this problem where I need to find the median of a list where each integer n appears n times, starting from 1 up to 300. The list looks like this: 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, ..., 300, 300, ..., 300. The options given are (A) 200, (B) 211, (C) 212, (D) 213, and (E) 250.First, I need to understand what the median is. The median is the middle value of an ordered list of numbers. If there's an odd number of observations, it's the middle one. If there's an even number, it's the average of the two middle numbers.So, my first step is to figure out how many numbers are in this list. Since each integer n appears n times, the total number of elements is the sum from 1 to 300. I remember that the sum of the first k integers is given by the formula k(k + 1)/2. So, plugging in 300 for k, the total number of elements N is:N = 300 × 301 / 2 = (300 × 301) / 2Let me calculate that:300 × 301 = 90,300Divide by 2: 90,300 / 2 = 45,150So, there are 45,150 numbers in the list. Since this is an even number, the median will be the average of the 22,575th and 22,576th numbers.Now, I need to find which number corresponds to these positions. To do this, I can calculate the cumulative count of numbers up to each integer n until I reach or surpass 22,575.The cumulative count up to integer n is the sum from 1 to n, which is n(n + 1)/2. I need to find the smallest n such that n(n + 1)/2 is greater than or equal to 22,575.So, let's set up the inequality:n(n + 1)/2 ≥ 22,575Multiply both sides by 2:n(n + 1) ≥ 45,150This simplifies to:n² + n - 45,150 ≥ 0This is a quadratic equation in the form of ax² + bx + c = 0, where a = 1, b = 1, and c = -45,150.To solve for n, I can use the quadratic formula:n = [-b ± √(b² - 4ac)] / (2a)Plugging in the values:n = [-1 ± √(1 + 4 × 1 × 45,150)] / 2Calculate the discriminant:√(1 + 180,600) = √180,601Hmm, I need to find the square root of 180,601. Let me see:425² = 180,625, which is a bit more than 180,601. So, 425² = 180,625Therefore, √180,601 is slightly less than 425. Let me see:425² = 180,625So, 425² - 180,601 = 180,625 - 180,601 = 24So, √180,601 = 425 - (24)/(2×425) approximately, using linear approximation.But maybe I can just note that 425² = 180,625, so √180,601 is 425 - (24)/(2×425) = 425 - 12/425 ≈ 425 - 0.028 ≈ 424.972But maybe it's better to just use 425 as an approximate value for simplicity.So, n ≈ (-1 + 425)/2 = 424/2 = 212So, n ≈ 212.Wait, but let me check the exact value. Let's compute 212 × 213 / 2.212 × 213 = ?Let me compute 200 × 213 = 42,60012 × 213 = 2,556So, total is 42,600 + 2,556 = 45,156Divide by 2: 45,156 / 2 = 22,578So, the cumulative count up to 212 is 22,578.But we need to reach 22,575 and 22,576.So, the cumulative count up to 211 is:211 × 212 / 2211 × 212 = ?200 × 212 = 42,40011 × 212 = 2,332Total: 42,400 + 2,332 = 44,732Divide by 2: 44,732 / 2 = 22,366So, up to 211, we have 22,366 numbers.Then, the next number, 212, starts at position 22,367 and goes up to 22,578.So, the 22,575th and 22,576th numbers are within the range of 212.Therefore, both the 22,575th and 22,576th numbers are 212, so the median is 212.Wait, but let me double-check. The cumulative count up to 211 is 22,366. Then, each number from 212 onwards adds 212 numbers. So, starting at 22,367, the next 212 numbers are all 212.So, 22,367 + 212 = 22,579Wait, but 22,366 + 212 = 22,578Wait, that's correct. So, the last position of 212 is 22,578.So, the 22,575th and 22,576th numbers are both within the 212s.Therefore, the median is 212.So, the answer should be C) 212.Wait, but let me make sure I didn't make any calculation errors.Total number of elements: 45,150, so median is average of 22,575th and 22,576th.Cumulative up to 211: 22,366So, 22,575 - 22,366 = 209So, the 209th number in the 212s is 212.Similarly, 22,576 - 22,366 = 210th number, which is also 212.Therefore, both are 212, so median is 212.Yes, that seems correct.I think I got it right. So, the answer is 212, which is option C.

💡Okay, so I have this problem about a train station with 8 tracks. Each track can only hold one train, and there are 4 different trains that need to be parked. I need to figure out how many different ways there are to park these trains. The options given are A: 70, B: 256, C: 1680, and D: 4096.Alright, let's start by understanding what the problem is asking. We have 8 tracks, and we need to park 4 trains on these tracks. Each track can only have one train, and the trains are different, which probably means that the order in which they are parked matters.So, this seems like a permutation problem because the order matters here. If the trains were identical, it would be a combination problem, but since they're different, it's permutations.In permutations, the formula is:[ P(n, k) = frac{n!}{(n - k)!} ]Where ( n ) is the total number of items to choose from, and ( k ) is the number of items to choose. In this case, ( n = 8 ) tracks and ( k = 4 ) trains.So, plugging in the numbers:[ P(8, 4) = frac{8!}{(8 - 4)!} = frac{8!}{4!} ]Calculating ( 8! ) (8 factorial) is:[ 8! = 8 times 7 times 6 times 5 times 4 times 3 times 2 times 1 = 40320 ]And ( 4! ) is:[ 4! = 4 times 3 times 2 times 1 = 24 ]So, dividing these:[ frac{40320}{24} = 1680 ]Therefore, the number of different parking methods is 1680, which corresponds to option C.Wait, let me double-check to make sure I didn't make a mistake. Another way to think about it is step by step. For the first train, there are 8 tracks to choose from. Once the first train is parked, there are 7 remaining tracks for the second train. Then, 6 tracks for the third train, and 5 tracks for the fourth train. So, the total number of ways is:[ 8 times 7 times 6 times 5 = 1680 ]Yeah, that matches what I got earlier. So, I think 1680 is the correct answer. It's option C.

💡Okay, so I have this problem about a convex hexagon and the midpoints of its diagonals forming another convex hexagon. I need to prove that the area of this new hexagon is four times smaller than the original one. Hmm, let me try to visualize this.First, let me draw a convex hexagon, label it ABCDEF. Now, the diagonals mentioned are AC, BD, CE, DF, EA, and FB. Wait, actually, in a hexagon, each vertex is connected to three others, but the diagonals here seem to be every other vertex. So, AC connects A to C, BD connects B to D, CE connects C to E, DF connects D to F, EA connects E to A, and FB connects F to B. Got it.Now, the midpoints of these diagonals are supposed to form another hexagon. Let me denote these midpoints as A1, B1, C1, D1, E1, F1. So, A1 is the midpoint of AC, B1 is the midpoint of BD, and so on. Connecting these midpoints should give me another hexagon, A1B1C1D1E1F1.I need to show that the area of A1B1C1D1E1F1 is 1/4 of the area of ABCDEF. Hmm, how do I approach this? Maybe I can use coordinate geometry? Or perhaps there's a theorem related to midpoints and areas.Wait, I remember something about midsegments in triangles. The midsegment is parallel to the third side and half its length. Maybe that can help here. If I can break down the hexagon into triangles or smaller shapes where midsegments apply, that might work.Let me try to divide the original hexagon into triangles. If I connect all the vertices to a common point, say, the center, but since it's a convex hexagon, maybe it's not regular, so the center might not be equidistant from all vertices. Hmm, maybe that's complicating things.Alternatively, I can divide the hexagon into smaller quadrilaterals or triangles by connecting non-adjacent vertices. For example, connecting AC, BD, CE, etc., which are the diagonals mentioned. So, these diagonals divide the hexagon into smaller regions.Wait, if I consider each pair of opposite sides, maybe I can find some relationship. For instance, the diagonal AC connects A to C, and BD connects B to D. The midpoints of these diagonals, A1 and B1, might form a segment that's related to the sides AB and CD.I think I need to use vectors or coordinate geometry here. Let me assign coordinates to the vertices of the hexagon. Let's say A is at (x1, y1), B at (x2, y2), and so on up to F at (x6, y6). Then, the midpoints A1, B1, etc., can be expressed in terms of these coordinates.For example, the midpoint A1 of AC would be ((x1 + x3)/2, (y1 + y3)/2). Similarly, B1, the midpoint of BD, would be ((x2 + x4)/2, (y2 + y4)/2). If I can express all the midpoints in terms of the original coordinates, maybe I can find a transformation that relates the original hexagon to the new one.Wait, if I think of the new hexagon as a transformation of the original, perhaps it's a homothety or a similarity transformation. A homothety is a transformation that enlarges or reduces a figure by a scale factor relative to a point. If the new hexagon is similar to the original with a scale factor of 1/2, then its area would be (1/2)^2 = 1/4 of the original. But is that the case here?Hmm, maybe. Let me see. If I can show that each side of the new hexagon is parallel to a corresponding side of the original and half its length, then it would be a homothety with scale factor 1/2, and the area would be 1/4.Alternatively, maybe I can use the concept of midlines in quadrilaterals. In a quadrilateral, the midline connecting the midpoints of two sides is parallel to the other two sides and half their average length. But here, we're dealing with midpoints of diagonals, not sides.Wait, maybe I can consider the original hexagon as a combination of triangles or quadrilaterals and apply the midpoint theorem to each.Let me try to break down the hexagon ABCDEF into triangles. For example, connect A to C, which divides the hexagon into two parts: triangle ABC and quadrilateral ACDEF. Hmm, not sure if that helps.Alternatively, connect all the diagonals mentioned: AC, BD, CE, DF, EA, FB. This should divide the hexagon into smaller regions, perhaps triangles and quadrilaterals. Then, the midpoints of these diagonals might form a smaller hexagon inside.Wait, maybe I can use vector analysis. Let me assign position vectors to the vertices: let A be vector a, B be vector b, and so on. Then, the midpoint A1 of AC is (a + c)/2, B1 is (b + d)/2, and so on.Now, to find the area of the new hexagon A1B1C1D1E1F1, I can use the shoelace formula or vector cross products. But that might get complicated. Alternatively, maybe I can express the new hexagon in terms of the original vectors and see the relationship.If I consider the vectors of the new hexagon, each vertex is the average of two original vectors. So, maybe the new hexagon is a linear transformation of the original. Specifically, if I can write the new vertices as linear combinations of the original, then the area scaling factor can be determined by the determinant of the transformation matrix.But I'm not sure if that's the right approach. Maybe I need to think about affine transformations or something else.Wait, another idea: the new hexagon is formed by connecting midpoints of diagonals, which are essentially averages of pairs of vertices. So, perhaps the new hexagon is similar to the original but scaled down.But how can I show that? Maybe by considering the centroid or something.Alternatively, maybe I can use the concept of the Varignon parallelogram, which is formed by connecting midpoints of sides of a quadrilateral, resulting in a parallelogram with half the area. But here, we're dealing with a hexagon and midpoints of diagonals, not sides.Wait, perhaps I can extend the Varignon theorem to hexagons. In a quadrilateral, connecting midpoints of sides gives a parallelogram with half the area. Maybe in a hexagon, connecting midpoints of diagonals gives a smaller hexagon with a certain area ratio.But I don't recall such a theorem off the top of my head. Maybe I need to derive it.Let me try to consider the hexagon as composed of triangles. For example, connect A to C, which divides the hexagon into two parts: triangle ABC and quadrilateral ACDEF. Then, connect B to D, which divides the quadrilateral into triangle BCD and quadrilateral CDEF. Hmm, not sure.Wait, maybe I can use barycentric coordinates or something. Alternatively, perhaps I can use complex numbers to represent the vertices and then find the midpoints.Let me try that. Let me assign complex numbers to the vertices: A = a, B = b, C = c, D = d, E = e, F = f. Then, the midpoints are A1 = (a + c)/2, B1 = (b + d)/2, C1 = (c + e)/2, D1 = (d + f)/2, E1 = (e + a)/2, F1 = (f + b)/2.Now, to find the area of the new hexagon A1B1C1D1E1F1, I can use the formula for the area of a polygon in the complex plane. The area is given by (1/4i) times the absolute value of the sum over edges of (z_j - z_i) times the conjugate of (z_j + z_i). But that might be complicated.Alternatively, maybe I can express the new hexagon in terms of the original vertices and see the relationship. Let me write down the coordinates:A1 = (a + c)/2B1 = (b + d)/2C1 = (c + e)/2D1 = (d + f)/2E1 = (e + a)/2F1 = (f + b)/2Now, if I look at the sequence of these midpoints, they are each averages of two non-adjacent vertices. So, the new hexagon is kind of a "midpoint" hexagon.Wait, maybe I can consider the transformation from the original hexagon to the new one. If I can write the new vertices as linear combinations of the original, then the area scaling factor can be found.Let me see: A1 = (a + c)/2, B1 = (b + d)/2, etc. So, each new vertex is the average of two original vertices. This seems like a linear transformation, but it's not a simple scaling or rotation.Alternatively, maybe I can think of the new hexagon as the image of the original under a certain linear transformation. If I can express this transformation, then the determinant will give the area scaling factor.But I'm not sure. Maybe another approach: consider the original hexagon and the new hexagon, and see how their areas relate by breaking them into triangles or other shapes.Wait, another idea: in a convex hexagon, the midpoints of the diagonals can be connected to form a smaller hexagon, and the area ratio can be found by considering the affine properties.Alternatively, maybe I can use the concept of homothety. If I can show that the new hexagon is a homothetic image of the original with a scale factor of 1/2, then the area would be 1/4.But how can I show that? Maybe by showing that each vertex of the new hexagon is the midpoint of a diagonal, which is a line segment connecting two vertices of the original. So, if I can show that the new hexagon is similar and scaled down by 1/2, then it would work.Wait, perhaps I can consider the centroid of the original hexagon. If I can show that the new hexagon is a scaled version around the centroid, then the area ratio would follow.But I'm not sure about the centroid approach. Maybe I need to think differently.Wait, let me try to consider the vectors again. Let me denote the position vectors of the original hexagon as A, B, C, D, E, F. Then, the midpoints are:A1 = (A + C)/2B1 = (B + D)/2C1 = (C + E)/2D1 = (D + F)/2E1 = (E + A)/2F1 = (F + B)/2Now, if I can express the new hexagon in terms of these vectors, maybe I can find a relationship.Let me consider the vectors of the new hexagon:A1 = (A + C)/2B1 = (B + D)/2C1 = (C + E)/2D1 = (D + F)/2E1 = (E + A)/2F1 = (F + B)/2Now, if I subtract A1 from B1, I get:B1 - A1 = [(B + D)/2] - [(A + C)/2] = (B - A + D - C)/2Similarly, subtracting B1 from C1:C1 - B1 = [(C + E)/2] - [(B + D)/2] = (C - B + E - D)/2Hmm, I'm not sure if this helps. Maybe I need to consider the entire polygon.Alternatively, maybe I can use the shoelace formula for both hexagons and compare their areas.But that would require knowing the coordinates, which I don't have. Maybe I can assign coordinates to the original hexagon and compute the areas.Let me try that. Let me assign coordinates to the original hexagon. Let me choose a simple convex hexagon, say, a regular hexagon for simplicity, centered at the origin with vertices on the unit circle.So, let me assign:A = (1, 0)B = (0.5, √3/2)C = (-0.5, √3/2)D = (-1, 0)E = (-0.5, -√3/2)F = (0.5, -√3/2)Now, let's compute the midpoints:A1 is the midpoint of AC: ((1 + (-0.5))/2, (0 + √3/2)/2) = (0.25, √3/4)B1 is the midpoint of BD: ((0.5 + (-1))/2, (√3/2 + 0)/2) = (-0.25, √3/4)C1 is the midpoint of CE: ((-0.5 + (-0.5))/2, (√3/2 + (-√3/2))/2) = (-0.5, 0)D1 is the midpoint of DF: ((-1 + 0.5)/2, (0 + (-√3/2))/2) = (-0.25, -√3/4)E1 is the midpoint of EA: ((-0.5 + 1)/2, (-√3/2 + 0)/2) = (0.25, -√3/4)F1 is the midpoint of FB: ((0.5 + 0.5)/2, (-√3/2 + √3/2)/2) = (0.5, 0)So, the new hexagon has vertices:A1: (0.25, √3/4)B1: (-0.25, √3/4)C1: (-0.5, 0)D1: (-0.25, -√3/4)E1: (0.25, -√3/4)F1: (0.5, 0)Now, let's compute the area of the original hexagon. Since it's a regular hexagon with side length 1, its area is (3√3)/2.Now, let's compute the area of the new hexagon A1B1C1D1E1F1 using the shoelace formula.List the coordinates in order:A1: (0.25, √3/4)B1: (-0.25, √3/4)C1: (-0.5, 0)D1: (-0.25, -√3/4)E1: (0.25, -√3/4)F1: (0.5, 0)Back to A1: (0.25, √3/4)Compute the shoelace sum:Sum1 = (0.25 * √3/4) + (-0.25 * 0) + (-0.5 * (-√3/4)) + (-0.25 * (-√3/4)) + (0.25 * 0) + (0.5 * √3/4)Wait, no, the shoelace formula is:Area = 1/2 |sum over i (x_i y_{i+1} - x_{i+1} y_i)|So, let's compute each term:Term1: x_A1 y_B1 - x_B1 y_A1 = 0.25*(√3/4) - (-0.25)*(√3/4) = (0.25√3/4) + (0.25√3/4) = 0.5√3/4 = √3/8Term2: x_B1 y_C1 - x_C1 y_B1 = (-0.25)*0 - (-0.5)*(√3/4) = 0 + 0.5√3/4 = √3/8Term3: x_C1 y_D1 - x_D1 y_C1 = (-0.5)*(-√3/4) - (-0.25)*0 = 0.5√3/4 + 0 = √3/8Term4: x_D1 y_E1 - x_E1 y_D1 = (-0.25)*(-√3/4) - (0.25)*(-√3/4) = 0.25√3/4 + 0.25√3/4 = √3/8Term5: x_E1 y_F1 - x_F1 y_E1 = (0.25)*0 - (0.5)*(-√3/4) = 0 + 0.5√3/4 = √3/8Term6: x_F1 y_A1 - x_A1 y_F1 = (0.5)*(√3/4) - (0.25)*0 = 0.5√3/4 + 0 = √3/8Now, sum all terms: 6*(√3/8) = (6√3)/8 = (3√3)/4Then, the area is 1/2 of the absolute value of this sum: 1/2 * (3√3)/4 = (3√3)/8Wait, the original area was (3√3)/2, and the new area is (3√3)/8. So, the ratio is (3√3)/8 divided by (3√3)/2 = (1/4). So, the new area is indeed 1/4 of the original.Hmm, that worked for a regular hexagon. But the problem states a convex hexagon, not necessarily regular. So, does this ratio hold for any convex hexagon?I think yes, because the transformation from the original hexagon to the new one is linear, and linear transformations preserve area ratios. Since in the regular case it's 1/4, it should hold for any convex hexagon.But wait, is the transformation linear? Because the midpoints are averages, which are linear operations. So, yes, the new hexagon is an affine transformation of the original, specifically a scaling by 1/2, hence the area scales by (1/2)^2 = 1/4.Therefore, the area of the new hexagon is four times smaller than the original.I think that makes sense. The key was to realize that the new hexagon is a scaled-down version of the original by a factor of 1/2, hence the area is 1/4.