💡Alright, so I've got this function f(x) = x - 1/x - ln(x), and I need to do two things with it. First, figure out if it's increasing or decreasing, and second, prove that if two different points have the same derivative, then the sum of the function values at those points is greater than 3 - 2 ln(2). Hmm, okay, let's start with part (1).For part (1), I remember that to determine if a function is increasing or decreasing, I need to look at its derivative. If the derivative is positive everywhere, the function is increasing; if it's negative everywhere, it's decreasing. So, let's find f'(x).f(x) = x - 1/x - ln(x)So, f'(x) would be the derivative of x, which is 1, minus the derivative of 1/x, which is -1/x², minus the derivative of ln(x), which is 1/x. So putting that together:f'(x) = 1 + 1/x² - 1/xHmm, let me write that as:f'(x) = 1 - 1/x + 1/x²Wait, is that right? Let me double-check. The derivative of x is 1, the derivative of -1/x is 1/x², and the derivative of -ln(x) is -1/x. So yes, f'(x) = 1 + 1/x² - 1/x.Maybe I can combine these terms over a common denominator to simplify it. Let's see, the common denominator would be x²:f'(x) = (x²)/x² + 1/x² - (x)/x²So that's:f'(x) = (x² - x + 1)/x²Hmm, okay. So f'(x) is equal to (x² - x + 1)/x². Now, to determine if this is always positive or not, I can look at the numerator and the denominator.The denominator is x², which is always positive for x ≠ 0. So the sign of f'(x) depends on the numerator, x² - x + 1.Is x² - x + 1 always positive? Let's see, it's a quadratic in x. The discriminant is b² - 4ac = (-1)² - 4*1*1 = 1 - 4 = -3, which is negative. That means the quadratic never crosses the x-axis and since the coefficient of x² is positive, it's always positive. So x² - x + 1 > 0 for all real x.Therefore, f'(x) is always positive because both the numerator and denominator are positive. So f(x) is monotonically increasing on its domain, which is (0, ∞) since ln(x) is only defined for x > 0.Okay, that seems straightforward. So part (1) is done.Now, moving on to part (2). It says that if f'(x₁) = f'(x₂) with x₁ ≠ x₂, then prove that f(x₁) + f(x₂) > 3 - 2 ln(2).Alright, so first, let's understand what this means. We have two different points x₁ and x₂ where the derivative of f is the same. Since f is monotonically increasing, its derivative is always positive, so f'(x) is always positive, but it's not necessarily injective. That is, different x's can have the same derivative.So, f'(x₁) = f'(x₂) = m, say. So, both x₁ and x₂ satisfy the equation f'(x) = m.From part (1), we have f'(x) = (x² - x + 1)/x². So setting that equal to m:(x² - x + 1)/x² = mMultiply both sides by x²:x² - x + 1 = m x²Bring all terms to one side:x² - x + 1 - m x² = 0Simplify:(1 - m) x² - x + 1 = 0So, this is a quadratic equation in x. For x₁ and x₂ to be distinct real roots, the discriminant must be positive.The discriminant D is:D = b² - 4ac = (-1)² - 4*(1 - m)*1 = 1 - 4(1 - m) = 1 - 4 + 4m = 4m - 3For two distinct real roots, D > 0, so 4m - 3 > 0 => m > 3/4.So, m must be greater than 3/4 for there to be two distinct points x₁ and x₂ where f'(x) = m.Now, since x₁ and x₂ are roots of the quadratic equation (1 - m)x² - x + 1 = 0, we can use Vieta's formulas to find relationships between x₁ and x₂.Vieta's formulas say that for a quadratic ax² + bx + c = 0, the sum of roots is -b/a and the product is c/a.So, in our case:Sum of roots: x₁ + x₂ = -(-1)/(1 - m) = 1/(1 - m)Product of roots: x₁ x₂ = 1/(1 - m)Wait, that seems a bit odd. Let me double-check.Quadratic equation is (1 - m)x² - x + 1 = 0.So, a = (1 - m), b = -1, c = 1.Thus, sum of roots: x₁ + x₂ = -b/a = -(-1)/(1 - m) = 1/(1 - m)Product of roots: x₁ x₂ = c/a = 1/(1 - m)So, both sum and product are equal to 1/(1 - m). Interesting.So, x₁ + x₂ = x₁ x₂.Hmm, that's an important relationship.So, x₁ + x₂ = x₁ x₂.Let me denote S = x₁ + x₂ and P = x₁ x₂. So, S = P.Also, from the discriminant, we know that m > 3/4.So, m > 3/4.But m is equal to f'(x₁) = f'(x₂). From part (1), f'(x) = (x² - x + 1)/x².So, m = (x₁² - x₁ + 1)/x₁² = (x₂² - x₂ + 1)/x₂².But since x₁ and x₂ are roots of the quadratic, we can also express m in terms of S and P.Wait, maybe it's better to express m in terms of S and P.From the quadratic equation, m = (x² - x + 1)/x².But since x₁ and x₂ satisfy (1 - m)x² - x + 1 = 0, we can write m = (x² - x + 1)/x².Alternatively, maybe we can express m in terms of S and P.Wait, let's see.We have S = x₁ + x₂ = P = x₁ x₂.So, S = P.Also, from the quadratic equation, we have:(1 - m)x² - x + 1 = 0So, for x = x₁ and x = x₂.But maybe instead of that, let's think about f(x₁) + f(x₂).We need to compute f(x₁) + f(x₂) and show that it's greater than 3 - 2 ln(2).Given that f(x) = x - 1/x - ln(x).So, f(x₁) + f(x₂) = (x₁ - 1/x₁ - ln(x₁)) + (x₂ - 1/x₂ - ln(x₂)).Simplify:= (x₁ + x₂) - (1/x₁ + 1/x₂) - (ln(x₁) + ln(x₂))We know that x₁ + x₂ = S and x₁ x₂ = P = S.Also, 1/x₁ + 1/x₂ = (x₁ + x₂)/(x₁ x₂) = S/P = S/S = 1.Similarly, ln(x₁) + ln(x₂) = ln(x₁ x₂) = ln(P) = ln(S).So, f(x₁) + f(x₂) = S - 1 - ln(S)So, f(x₁) + f(x₂) = S - 1 - ln(S)Now, we need to find a lower bound for this expression.Given that m > 3/4, and m is related to S.Wait, let's recall that m = (x² - x + 1)/x².But since x₁ and x₂ are roots, we can relate m to S.Alternatively, let's think about the relationship between m and S.From the quadratic equation, we have:(1 - m)x² - x + 1 = 0So, for x = x₁ and x = x₂.But since x₁ and x₂ are roots, we can express m in terms of S.Wait, let's see.We have S = x₁ + x₂ = P = x₁ x₂.So, S = P.Also, from the quadratic equation, the sum of roots is 1/(1 - m) and the product is 1/(1 - m). So, S = 1/(1 - m).Therefore, 1 - m = 1/S => m = 1 - 1/S.But we also know that m > 3/4.So, 1 - 1/S > 3/4 => 1/S < 1 - 3/4 = 1/4 => S > 4.So, S > 4.Therefore, S > 4.So, f(x₁) + f(x₂) = S - 1 - ln(S), and S > 4.So, we need to find the minimum value of the function g(S) = S - 1 - ln(S) for S > 4.If we can show that g(S) > 3 - 2 ln(2), then we are done.So, let's analyze g(S) = S - 1 - ln(S).Compute its derivative:g'(S) = 1 - 1/SSet derivative to zero to find critical points:1 - 1/S = 0 => S = 1But S > 4, so the critical point at S = 1 is not in our domain.Therefore, g(S) is increasing for S > 1 because g'(S) = 1 - 1/S > 0 when S > 1.Since S > 4, g(S) is increasing on (4, ∞).Therefore, the minimum value of g(S) on S > 4 is approached as S approaches 4 from the right.So, lim_{S→4+} g(S) = 4 - 1 - ln(4) = 3 - ln(4) = 3 - 2 ln(2)But since S > 4, g(S) > 3 - 2 ln(2).Therefore, f(x₁) + f(x₂) = g(S) > 3 - 2 ln(2).Hence, we've proven that f(x₁) + f(x₂) > 3 - 2 ln(2).Wait, let me double-check the steps.1. Found f'(x) correctly, simplified to (x² - x + 1)/x², which is always positive, so f is increasing.2. For part (2), set f'(x₁) = f'(x₂) = m, leading to quadratic equation (1 - m)x² - x + 1 = 0.3. Used Vieta's formulas to find that x₁ + x₂ = x₁ x₂ = 1/(1 - m).4. Expressed f(x₁) + f(x₂) in terms of S = x₁ + x₂, leading to S - 1 - ln(S).5. Noted that S > 4 because m > 3/4, leading to 1 - m < 1/4, so S = 1/(1 - m) > 4.6. Analyzed g(S) = S - 1 - ln(S), found it's increasing for S > 1, so its minimum on S > 4 is at S = 4, giving 3 - 2 ln(2).7. Concluded that since S > 4, g(S) > 3 - 2 ln(2).Yes, that seems correct. I think I covered all the necessary steps and didn't make any errors in the reasoning.

💡Okay, so I have this probability problem here, and I need to figure it out step by step. Let me read it again: If 2 cards are randomly drawn without replacement from 6 cards labeled 1, 2, 3, 4, 5, and 6, then the probability that the product of the numbers on the 2 cards is a multiple of 4 is ( ). The options are A: 1/5, B: 1/3, C: 2/5, D: 2/3.Alright, so first, I need to understand what the problem is asking. We have six cards, each labeled from 1 to 6. We're going to draw two cards without replacement, meaning once we draw a card, we don't put it back before drawing the second one. We need to find the probability that when we multiply the numbers on these two cards, the result is a multiple of 4.Probability is generally calculated as the number of successful outcomes divided by the total number of possible outcomes. So, I need to figure out two things:1. The total number of ways to draw two cards from six.2. The number of those pairs where the product is a multiple of 4.Let me start with the total number of possible outcomes. Since we're drawing two cards from six without replacement, this is a combination problem because the order doesn't matter. That is, drawing card 1 and then card 2 is the same as drawing card 2 and then card 1 in terms of the pair we end up with.The formula for combinations is C(n, k) = n! / (k!(n - k)!), where n is the total number of items, and k is the number of items to choose.So, plugging in the numbers, C(6, 2) = 6! / (2!(6 - 2)!) = (6 × 5 × 4!) / (2 × 1 × 4!) = (6 × 5) / 2 = 30 / 2 = 15.So, there are 15 possible pairs of cards we could draw.Now, onto the successful outcomes: the number of pairs where the product is a multiple of 4.To find this, I need to think about what makes a product a multiple of 4. A number is a multiple of 4 if it has at least two factors of 2, meaning it's divisible by 4. So, for the product of two numbers to be a multiple of 4, at least one of the following must be true:1. At least one of the numbers is a multiple of 4.2. Both numbers are even (since the product of two even numbers is divisible by 4).Wait, hold on. Let me think about that. If both numbers are even, does that always make the product a multiple of 4? Let's test it with some examples.Take 2 and 4: 2 × 4 = 8, which is a multiple of 4. Good.Take 2 and 6: 2 × 6 = 12, which is a multiple of 4. Hmm, 12 divided by 4 is 3, so yes, it's a multiple.Take 4 and 6: 4 × 6 = 24, which is a multiple of 4.Wait, but what about 2 and 2? Well, in this case, we don't have duplicate cards, so that's not an issue here.But wait, in our set, the even numbers are 2, 4, and 6. So, if I pair any two of these, their product will be a multiple of 4.But let me think again. If I have two even numbers, say 2 and 6, their product is 12, which is 4 × 3, so yes, it's a multiple of 4. Similarly, 2 and 4 is 8, which is 4 × 2, and 4 and 6 is 24, which is 4 × 6.So, yes, pairing any two even numbers will give a product that's a multiple of 4.Additionally, if one of the numbers is a multiple of 4, even if the other is odd, the product will still be a multiple of 4.For example, 4 and 1: 4 × 1 = 4, which is a multiple of 4.4 and 3: 4 × 3 = 12, which is a multiple of 4.4 and 5: 4 × 5 = 20, which is a multiple of 4.So, in this case, the pairs that result in a product that's a multiple of 4 are:1. Any pair that includes the number 4.2. Any pair that includes two even numbers (which are 2, 4, and 6).But wait, if I include both conditions, I might be double-counting some pairs. Specifically, the pair (4, 2) and (4, 6) are already included in both categories. So, I need to be careful not to count them twice.Let me list all possible pairs and then identify which ones meet the criteria.The total pairs are:1. (1, 2)2. (1, 3)3. (1, 4)4. (1, 5)5. (1, 6)6. (2, 3)7. (2, 4)8. (2, 5)9. (2, 6)10. (3, 4)11. (3, 5)12. (3, 6)13. (4, 5)14. (4, 6)15. (5, 6)Now, let's go through each pair and see if their product is a multiple of 4.1. (1, 2): 1 × 2 = 2 → Not a multiple of 4.2. (1, 3): 1 × 3 = 3 → Not a multiple of 4.3. (1, 4): 1 × 4 = 4 → Multiple of 4.4. (1, 5): 1 × 5 = 5 → Not a multiple of 4.5. (1, 6): 1 × 6 = 6 → Not a multiple of 4.6. (2, 3): 2 × 3 = 6 → Not a multiple of 4.7. (2, 4): 2 × 4 = 8 → Multiple of 4.8. (2, 5): 2 × 5 = 10 → Not a multiple of 4.9. (2, 6): 2 × 6 = 12 → Multiple of 4.10. (3, 4): 3 × 4 = 12 → Multiple of 4.11. (3, 5): 3 × 5 = 15 → Not a multiple of 4.12. (3, 6): 3 × 6 = 18 → Not a multiple of 4.13. (4, 5): 4 × 5 = 20 → Multiple of 4.14. (4, 6): 4 × 6 = 24 → Multiple of 4.15. (5, 6): 5 × 6 = 30 → Not a multiple of 4.Now, let's count how many of these are multiples of 4:- (1, 4)- (2, 4)- (2, 6)- (3, 4)- (4, 5)- (4, 6)That's 6 pairs.Wait, but earlier I thought that (2, 6) might not be a multiple of 4, but 2 × 6 is 12, which is 4 × 3, so it is a multiple of 4. So, that pair is valid.So, total successful pairs are 6.But wait, earlier I thought that the successful pairs were 5, but now I'm getting 6. Let me check again.Looking back at the list:1. (1, 4) – yes2. (2, 4) – yes3. (2, 6) – yes4. (3, 4) – yes5. (4, 5) – yes6. (4, 6) – yesYes, that's 6 pairs.But wait, in my initial thought process, I thought that (2, 6) might not be a multiple of 4, but it is. So, actually, there are 6 successful pairs.But wait, let me check the product of (2,6): 2×6=12, which is divisible by 4 (12/4=3). So, yes, it is a multiple of 4.Therefore, the number of successful pairs is 6.So, the probability is 6/15, which simplifies to 2/5.Wait, but hold on. Let me double-check because I might have made a mistake earlier.Wait, 6 successful pairs out of 15 total pairs is 6/15, which simplifies to 2/5. So, that would be option C.But in the initial thought process, I thought it was 5 pairs, but upon recounting, it's 6.Wait, let me recount the successful pairs:1. (1,4)2. (2,4)3. (2,6)4. (3,4)5. (4,5)6. (4,6)Yes, that's 6 pairs.So, 6/15 simplifies to 2/5.Therefore, the probability is 2/5, which is option C.But wait, let me think again. Is there another way to approach this problem to confirm?Another approach could be to calculate the probability of the complementary event, i.e., the probability that the product is not a multiple of 4, and then subtract that from 1.So, total pairs: 15.Number of pairs where product is not a multiple of 4: 15 - 6 = 9.So, probability of not being a multiple of 4 is 9/15 = 3/5.Therefore, probability of being a multiple of 4 is 1 - 3/5 = 2/5.Yes, that confirms it.Alternatively, let's think about the conditions for the product not being a multiple of 4.For the product not to be a multiple of 4, the product must be either odd or exactly divisible by 2 but not by 4.An odd product occurs when both numbers are odd. The even product that's not a multiple of 4 occurs when one number is even but not divisible by 4, and the other is odd.Wait, let's break it down.First, the product is odd only if both numbers are odd. The odd numbers in the set are 1, 3, 5. So, the number of pairs with both odd numbers is C(3,2) = 3.These pairs are:1. (1,3)2. (1,5)3. (3,5)So, 3 pairs.Next, the product is even but not a multiple of 4. This happens when one number is even but not divisible by 4, and the other is odd.The even numbers not divisible by 4 in the set are 2 and 6.So, the pairs would be:- (2,1), (2,3), (2,5)- (6,1), (6,3), (6,5)That's 6 pairs.So, total pairs where the product is not a multiple of 4 is 3 (odd products) + 6 (even but not multiple of 4) = 9.Therefore, the number of pairs where the product is a multiple of 4 is 15 - 9 = 6.So, probability is 6/15 = 2/5.Yes, that matches my earlier conclusion.Therefore, the correct answer is C: 2/5.Wait, but hold on a second. Let me think about the even numbers again.We have even numbers: 2, 4, 6.Numbers divisible by 4: only 4.So, if we pair 4 with any other number, the product is a multiple of 4.Number of such pairs: 5 (since 4 can pair with 1,2,3,5,6).Additionally, pairing two even numbers that are not 4: which are 2 and 6.So, pairing 2 and 6: 2×6=12, which is a multiple of 4.So, that's one more pair.Therefore, total successful pairs: 5 +1 =6.Yes, that's consistent.So, 6 successful pairs out of 15.Therefore, probability is 6/15=2/5.So, the answer is C: 2/5.I think that's solid. I've approached it in multiple ways: listing all pairs, calculating complementary probability, and considering the conditions for the product to be a multiple of 4.Therefore, I'm confident that the correct answer is C: 2/5.

💡Okay, so I have this problem where I need to prove that if I take n distinct fractions from the interval (0,1), the sum of their denominators is at least (1/3) * n^(3/2). Hmm, that seems a bit abstract, but let me try to break it down.First, let me understand what the problem is asking. I have n fractions, each between 0 and 1, and they're all different. Each fraction can be written as a/b, where a and b are positive integers with a < b. The denominators b are what we're interested in, and we need to show that when we add up all these denominators, the total is at least (1/3) times n raised to the power of 3/2.Alright, so maybe I should start by considering how fractions are distributed in the interval (0,1). Fractions with smaller denominators are more "spread out," while fractions with larger denominators are closer together. For example, with denominator 2, we have 1/2. With denominator 3, we have 1/3 and 2/3. With denominator 4, we have 1/4, 2/4 (which simplifies to 1/2, but we already have that), 3/4. So, each denominator b gives us (b-1) fractions, but some of them might simplify to fractions with smaller denominators.Wait, so maybe it's better to think about fractions in their reduced form. That is, fractions where the numerator and denominator have no common factors other than 1. So, for each denominator b, the number of fractions in (0,1) is φ(b), where φ is Euler's totient function. But I'm not sure if that's directly useful here.Alternatively, maybe I can think about the Farey sequence. The Farey sequence of order n is the set of reduced fractions between 0 and 1 with denominators less than or equal to n, arranged in increasing order. But again, I'm not sure if that's the right approach.Let me try a different angle. Suppose I have n fractions, each with denominator b_i. I need to show that the sum of all b_i is at least (1/3) * n^(3/2). Maybe I can use some inequality or estimation here.Perhaps I can use the Cauchy-Schwarz inequality. That might be a good tool for sums involving denominators. Let me recall that the Cauchy-Schwarz inequality states that for any real numbers a_i and b_i,(Σ a_i b_i)^2 ≤ (Σ a_i^2)(Σ b_i^2)But I'm not sure how to apply that directly here. Maybe I need to relate the denominators to something else.Wait, another thought: the number of distinct fractions with denominator b is b-1, but considering reduced fractions, it's φ(b). But since we're dealing with n fractions, perhaps the denominators can't be too small, otherwise, we wouldn't have enough fractions.So, if we have n fractions, each with a denominator b_i, maybe the denominators can't all be too small, otherwise, we wouldn't have n distinct fractions. So, perhaps the denominators have to be spread out enough to allow for n distinct fractions.Let me think about the number of fractions with denominators up to some number t. If I fix t, then the number of fractions with denominators ≤ t is roughly proportional to t^2, since for each denominator b, there are about b fractions. But actually, it's the sum of φ(b) from b=1 to t, which is approximately (3/π²) t^2. But I'm not sure if that's necessary here.Alternatively, maybe I can use an averaging argument. If the sum of the denominators is S, then the average denominator is S/n. If I can show that the average denominator is at least (1/3) n^(1/2), then S would be at least (1/3) n^(3/2).So, how can I show that the average denominator is at least (1/3) n^(1/2)? Maybe by considering that if the denominators were smaller on average, we wouldn't have enough distinct fractions.Let me formalize this idea. Suppose that the average denominator is less than (1/3) n^(1/2). Then, the total number of possible fractions with denominators up to (1/3) n^(1/2) is roughly (1/3)^2 n, which is (1/9) n. But we have n fractions, which is much larger than (1/9) n. Therefore, this is a contradiction, meaning that the average denominator must be at least (1/3) n^(1/2).Wait, that seems too hand-wavy. Let me try to make it more precise.Suppose that the sum of the denominators S < (1/3) n^(3/2). Then, the average denominator is S/n < (1/3) n^(1/2). So, if the average denominator is less than (1/3) n^(1/2), then most denominators are less than, say, (2/3) n^(1/2). But how does that relate to the number of distinct fractions?Each denominator b contributes at most b-1 fractions, but in reduced form, it's φ(b). But maybe I can just consider the maximum number of fractions with denominators up to t. The maximum number is roughly t^2, as each denominator b can contribute up to b fractions. So, if t is about (1/3) n^(1/2), then t^2 is about (1/9) n, which is much less than n. Therefore, to have n fractions, we need denominators up to at least some t where t^2 is proportional to n, which would mean t is proportional to n^(1/2). Therefore, the sum S should be at least proportional to n * n^(1/2) = n^(3/2). But I need to make this precise. Maybe I can use the Cauchy-Schwarz inequality in a clever way.Let me consider the sum S = Σ b_i. I want to relate this to the number of fractions n. Let me think of each fraction as a point (a_i, b_i) in the grid, where a_i < b_i and gcd(a_i, b_i) = 1. The number of such points with b_i ≤ t is roughly proportional to t^2, as before.Alternatively, maybe I can use the following approach: for each denominator b, count how many fractions it contributes. Let m_b be the number of fractions with denominator b. Then, n = Σ m_b, and S = Σ b m_b.I need to minimize S given that n = Σ m_b, where m_b ≤ φ(b). But since φ(b) ≤ b-1, we have m_b ≤ b-1.But this seems complicated. Maybe instead, I can use the Cauchy-Schwarz inequality on the sum S.Let me write S = Σ b_i. Let me consider the sum over i=1 to n of 1, which is n. By Cauchy-Schwarz,(Σ 1 * b_i) ≤ sqrt( (Σ 1^2) (Σ b_i^2) )But that gives S ≤ sqrt(n Σ b_i^2), which is not helpful because I need a lower bound on S.Alternatively, maybe I can use the AM ≥ GM inequality. The arithmetic mean of the denominators is S/n, and the geometric mean is (Π b_i)^(1/n). But I don't know how to relate the geometric mean to n.Wait, another idea: consider the number of fractions with denominator b. For each b, the number of fractions is at most b-1, but in reduced form, it's φ(b). However, to maximize the number of fractions, we need to include as many as possible, so perhaps the minimal sum S occurs when we include the smallest possible denominators.But if we include the smallest denominators, we can get more fractions with smaller denominators, but the sum S would be smaller. However, we need to have n distinct fractions, so maybe the minimal sum S is achieved when we include the smallest possible denominators, but arranged in such a way that we have n fractions.Wait, let me think about it differently. Suppose we have n fractions, each with denominator b_i. The number of fractions with denominator b is at most φ(b). So, to have n fractions, we need to cover enough denominators such that Σ φ(b_i) ≥ n.But φ(b) is roughly b for large b, but actually, it's less. The average value of φ(b) is about (3/π²) b. So, if we have denominators up to t, the total number of fractions is roughly (3/π²) t^2. So, to have n fractions, we need t^2 ≈ (π²/3) n, so t ≈ sqrt( (π²/3) n ). Therefore, the sum S would be roughly Σ b from b=1 to t, which is about t(t+1)/2 ≈ t^2 / 2 ≈ (π²/6) n. But this is a constant multiple of n, which is much less than n^(3/2). So, this approach might not be giving me the right bound.Wait, maybe I'm missing something. The problem states that the sum of denominators is at least (1/3) n^(3/2). So, it's a lower bound, not an upper bound. So, I need to show that the sum can't be too small.Perhaps I can use the fact that the number of fractions with denominator ≤ t is at most t(t+1)/2, which is about t^2 / 2. So, if we have n fractions, then t^2 / 2 ≥ n, so t ≥ sqrt(2n). Therefore, the largest denominator must be at least sqrt(2n). But that's just the maximum denominator, not the sum.Alternatively, maybe I can use the convexity of the function f(b) = b. Since the sum S is the sum of b_i, and we have n terms, perhaps by Jensen's inequality, the minimal sum occurs when the denominators are as small as possible. But I need to ensure that the fractions are distinct.Wait, another approach: consider arranging the fractions in order. Each fraction a/b must be greater than the previous one. So, the next fraction must have a denominator that allows it to be larger than the previous one. Maybe this imposes some structure on the denominators.Alternatively, think about the mediant property. The mediant of two fractions a/b and c/d is (a+c)/(b+d). It lies between the two fractions. But I'm not sure if that helps here.Wait, maybe I can use the concept of Farey sequences. In a Farey sequence of order n, the fractions are arranged in order, and each fraction has a denominator ≤ n. But again, I'm not sure.Let me try to think about the minimal sum S. To minimize S, we need to choose fractions with the smallest possible denominators. So, the first few fractions would be 1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 2/5, 3/5, 4/5, etc. Each time, we include the next smallest denominator.If I list them out, the denominators would be 2, 3, 3, 4, 4, 5, 5, 5, 5, etc. So, the sum S would be 2 + 3 + 3 + 4 + 4 + 5 + 5 + 5 + 5 + ... Now, if I can estimate the sum S for the first n fractions in this sequence, maybe I can find a lower bound.Let me see. The number of fractions with denominator b is φ(b). So, the total number of fractions up to denominator t is Σ_{b=2}^t φ(b). We know that Σ_{b=1}^t φ(b) ≈ (3/π²) t². So, Σ_{b=2}^t φ(b) ≈ (3/π²) t² - 1 (since φ(1)=1). So, if we have n fractions, then approximately, (3/π²) t² ≈ n, so t ≈ sqrt( (π²/3) n ). Therefore, the largest denominator t is roughly proportional to sqrt(n). But the sum S is Σ_{b=2}^t b * m_b, where m_b is the number of fractions with denominator b. Since m_b = φ(b), which is roughly b on average, the sum S would be roughly Σ_{b=2}^t b * φ(b). But φ(b) is about (3/π²) b on average, so S ≈ Σ_{b=2}^t (3/π²) b². The sum of b² from 1 to t is about t³ / 3. So, S ≈ (3/π²) * (t³ / 3) = t³ / π². But t ≈ sqrt( (π²/3) n ), so t³ ≈ ( (π²/3) n )^(3/2) = (π²/3)^(3/2) n^(3/2). Therefore, S ≈ (π²/3)^(3/2) / π² * n^(3/2) = ( (π²/3)^(3/2) ) / π² * n^(3/2). Simplifying, (π²/3)^(3/2) = (π²)^(3/2) / (3)^(3/2) = π³ / (3√3). So, S ≈ (π³ / (3√3)) / π² * n^(3/2) = (π / (3√3)) n^(3/2). Since π / (3√3) is approximately 1.047 / 5.196 ≈ 0.201, which is less than 1/3. Hmm, that's a problem because the problem states that S should be at least (1/3) n^(3/2), but my estimate gives a smaller coefficient. Maybe my approximation is too rough.Alternatively, perhaps I need to consider that the sum S is actually larger because the denominators are not just up to t, but spread out more. Maybe I need a better way to estimate S.Wait, another idea: use the Cauchy-Schwarz inequality in a different way. Let me consider the sum S = Σ b_i. I can write S = Σ b_i * 1. Then, by Cauchy-Schwarz,(Σ b_i * 1)^2 ≤ (Σ b_i^2)(Σ 1^2) = (Σ b_i^2) * nSo, S² ≤ n Σ b_i²But I need a lower bound on S, so maybe I can find an upper bound on Σ b_i².Wait, but that seems tricky. Alternatively, maybe I can use the fact that the number of fractions with denominator ≤ t is at most t(t+1)/2, as I thought earlier. So, if we have n fractions, then t(t+1)/2 ≥ n, so t ≥ sqrt(2n) - 1/2. Therefore, the largest denominator is at least sqrt(2n) - 1/2.But that's just the maximum denominator, not the sum. However, if the maximum denominator is at least sqrt(2n), then the sum S is at least sqrt(2n). But that's much less than n^(3/2). So, that approach isn't sufficient.Wait, maybe I can use the fact that the sum of the first t integers is t(t+1)/2. So, if I have denominators up to t, the sum S is at least t(t+1)/2. But we know that t(t+1)/2 ≥ n, so t ≥ sqrt(2n). Therefore, S ≥ t(t+1)/2 ≥ (sqrt(2n))^2 / 2 = (2n)/2 = n. But n is much less than n^(3/2), so that's not helpful.Hmm, maybe I need to consider that the denominators are spread out in a way that their sum is large. Perhaps using the concept of harmonic series or something similar.Wait, another approach: consider the reciprocals of the denominators. If I have n fractions, each with denominator b_i, then the reciprocals 1/b_i must sum to something. But I'm not sure.Alternatively, think about the fact that each fraction a/b must be distinct, so for each denominator b, the numerators a must be distinct and less than b. So, for each b, we can have at most b-1 fractions. Therefore, the total number of fractions is at most Σ_{b=2}^t (b-1) = t(t-1)/2. So, if we have n fractions, then t(t-1)/2 ≥ n, so t ≥ sqrt(2n). But again, this gives a lower bound on the maximum denominator, not the sum.Wait, maybe I can use the fact that the sum S is at least the sum of the first t integers, where t is such that t(t-1)/2 ≥ n. So, t ≈ sqrt(2n), and S ≥ t(t+1)/2 ≈ (sqrt(2n))^2 / 2 = n. But again, that's not enough.I think I'm stuck here. Maybe I need to look for a different inequality or a different way to relate the sum of denominators to the number of fractions.Wait, another idea: use the Cauchy-Schwarz inequality in the form:(Σ b_i) * (Σ 1/b_i) ≥ (Σ 1)^2 = n²So, S * (Σ 1/b_i) ≥ n²Therefore, Σ 1/b_i ≥ n² / SBut I need to relate Σ 1/b_i to something else. Maybe I can find an upper bound on Σ 1/b_i.Wait, the sum Σ 1/b_i is the sum of reciprocals of the denominators. Since each denominator b_i is at least 2, the sum Σ 1/b_i is at most n/2. But that would give:n² / S ≤ n/2 ⇒ S ≥ 2nBut 2n is still less than n^(3/2) for n ≥ 4, so that's not helpful.Wait, maybe I can get a better upper bound on Σ 1/b_i. Since the fractions are distinct, each denominator b_i must be unique? No, wait, denominators can repeat as long as the fractions are distinct. For example, 1/2 and 1/3 have different denominators, but 1/3 and 2/3 have the same denominator.So, denominators can repeat, but the fractions must be distinct. Therefore, for each denominator b, the number of fractions with that denominator is at most b-1. So, if I have m_b fractions with denominator b, then m_b ≤ b-1.Therefore, the total number of fractions n = Σ m_b ≤ Σ (b-1) = Σ b - Σ 1 = Σ b - t, where t is the number of denominators. But I'm not sure if that helps.Wait, maybe I can use the inequality between the arithmetic mean and the harmonic mean. The harmonic mean of the denominators is n / (Σ 1/b_i). The arithmetic mean is S/n. We know that the harmonic mean ≤ arithmetic mean. So,n / (Σ 1/b_i) ≤ S/n ⇒ Σ 1/b_i ≥ n² / SBut I already used that. Maybe I need another approach.Wait, perhaps I can consider that for each denominator b, the number of fractions with that denominator is at most φ(b), which is ≤ b. So, n ≤ Σ φ(b) ≤ Σ b. Therefore, Σ b ≥ n. But again, that's just S ≥ n, which is not enough.Wait, but actually, Σ φ(b) is roughly (3/π²) Σ b, so Σ b ≈ (π²/3) Σ φ(b) ≈ (π²/3) n. So, S ≈ (π²/3) n, which is still linear in n, not n^(3/2).Hmm, I'm not getting anywhere with this. Maybe I need to think differently. Let me try to look for a known result or theorem that relates to this.I recall that in the Farey sequence, the number of terms of order n is about (3/π²) n². But that's not directly helpful here.Wait, another idea: consider the sum of denominators as a sum over b multiplied by the number of fractions with that denominator. So, S = Σ b * m_b, where m_b is the number of fractions with denominator b. We know that Σ m_b = n, and m_b ≤ φ(b).But to minimize S, we need to choose the smallest possible b's as much as possible. So, the minimal S occurs when we take the smallest possible denominators, each contributing as many fractions as possible.So, let's try to construct such a set of fractions. Start with denominator 2: we have 1 fraction (1/2). Then denominator 3: 2 fractions (1/3, 2/3). Denominator 4: 2 fractions (1/4, 3/4). Denominator 5: 4 fractions (1/5, 2/5, 3/5, 4/5). Denominator 6: 2 fractions (1/6, 5/6). And so on.So, the number of fractions per denominator is φ(b). So, to get n fractions, we need to sum φ(b) until we reach n. The sum of φ(b) up to b = t is approximately (3/π²) t². So, t ≈ sqrt( (π²/3) n ). Therefore, the largest denominator is about sqrt(n). But the sum S is Σ b * m_b, where m_b = φ(b). So, S ≈ Σ b * φ(b) up to b = t. The sum Σ b * φ(b) is known to be roughly (3/π²) * (t³)/3 = t³ / π². Substituting t ≈ sqrt( (π²/3) n ), we get S ≈ ( (π²/3) n )^(3/2) / π² = ( (π²/3)^(3/2) ) / π² * n^(3/2). Calculating (π²/3)^(3/2) = (π²)^(3/2) / (3)^(3/2) = π³ / (3√3). Therefore, S ≈ (π³ / (3√3)) / π² * n^(3/2) = (π / (3√3)) n^(3/2). Now, π / (3√3) is approximately 3.1416 / (3 * 1.732) ≈ 3.1416 / 5.196 ≈ 0.605. So, S ≈ 0.605 n^(3/2). Since 0.605 > 1/3 ≈ 0.333, this suggests that S ≥ (1/3) n^(3/2). Therefore, the sum of the denominators is at least (1/3) n^(3/2).I think this makes sense. By considering the minimal sum S when choosing the smallest possible denominators, we find that S is proportional to n^(3/2), and the constant factor is greater than 1/3, which gives the desired lower bound.

💡Alright, so I've got this problem here with parametric equations, and I need to figure out what kind of curve they represent. The equations are:x = cos²θy = sin²θAnd the options are: A straight line, a circle, a line segment, or a ray. Hmm, okay. Let me think about how to approach this.First off, parametric equations define x and y in terms of a third variable, which in this case is θ. So, θ is like a parameter that varies, and as it does, x and y change accordingly. My goal is to eliminate the parameter θ and find a relationship directly between x and y. That should help me figure out what kind of curve it is.I remember that cos²θ + sin²θ = 1. That's one of those fundamental trigonometric identities. So, if I add x and y together, I get:x + y = cos²θ + sin²θ = 1So, x + y = 1. That looks like the equation of a straight line. But wait, the options include a line segment and a ray as well. So, I can't just stop here; I need to consider the range of x and y.Let me think about the possible values of cos²θ and sin²θ. Since cosine and sine functions oscillate between -1 and 1, their squares will always be between 0 and 1. So, cos²θ is between 0 and 1, and sin²θ is also between 0 and 1. That means both x and y are between 0 and 1.So, x is in [0,1] and y is in [0,1]. Therefore, the points (x, y) lie on the line x + y = 1, but only in the region where both x and y are between 0 and 1. That would be the line segment connecting the points (1,0) and (0,1).Wait, but let me make sure I'm not missing something. Could it be a ray instead? A ray starts at a point and goes off infinitely in one direction. But since both x and y are bounded between 0 and 1, it can't extend beyond those points. So, it's not a ray. It's a finite line segment between (1,0) and (0,1).Let me double-check. If θ varies from 0 to 2π, what happens to x and y?When θ = 0:x = cos²0 = 1y = sin²0 = 0So, the point is (1,0).When θ = π/2:x = cos²(π/2) = 0y = sin²(π/2) = 1So, the point is (0,1).When θ = π:x = cos²π = 1y = sin²π = 0Back to (1,0).Similarly, θ = 3π/2 brings us back to (0,1). So, as θ increases, the point (x,y) moves back and forth between (1,0) and (0,1) along the line x + y = 1.Therefore, the curve is indeed a line segment between (1,0) and (0,1).Wait, but what if θ is allowed to go beyond 0 to 2π? Like, if θ can be any real number, does that change anything? Well, cosine and sine are periodic with period 2π, so their squares will also be periodic with the same period. So, even if θ goes beyond 2π, the points (x,y) will just keep oscillating between (1,0) and (0,1). So, it's still just the line segment between those two points.So, putting it all together, the parametric equations describe a line segment from (1,0) to (0,1). Therefore, the correct answer should be C: A line segment.I think that makes sense. I don't see any reason to think it's a full straight line or a circle or a ray. It's definitely a finite segment because of the bounds on x and y. Yeah, I'm pretty confident now.

💡Alright, so I'm trying to figure out whether there are infinitely many pairs of natural numbers (a, b) where a is not equal to b, and the number of distinct prime factors of a + b is equal to the sum of the number of distinct prime factors of a and b individually. That is, C(a + b) = C(a) + C(b). First, I need to understand what C(n) means. It's the count of distinct prime divisors of n. For example, C(10) is 2 because 10 has two prime factors: 2 and 5. Similarly, C(11) is 1 because 11 is a prime number itself, and C(12) is 2 because 12 has prime factors 2 and 3.So, the problem is asking if there are infinitely many pairs (a, b) with a ≠ b such that when you add a and b, the number of distinct prime factors of the sum is exactly the sum of the distinct prime factors of a and b individually.I think a good starting point is to look for patterns or specific types of numbers where this condition might hold. Maybe if a and b are powers of the same prime, or perhaps if they are multiples of different primes.Let me try some small numbers to see if I can find any pairs that satisfy the condition.Let's take a = 2 and b = 4.C(2) = 1 (since 2 is prime)C(4) = 1 (since 4 = 2^2, only one prime factor)a + b = 6C(6) = 2 (since 6 = 2 * 3, two prime factors)So, C(a + b) = 2, and C(a) + C(b) = 1 + 1 = 2. This works!Okay, so (2, 4) is a valid pair. Let's try another pair.How about a = 2 and b = 8.C(2) = 1C(8) = 1a + b = 10C(10) = 2Again, C(a + b) = 2, and C(a) + C(b) = 1 + 1 = 2. So, (2, 8) also works.Hmm, seems like when a and b are powers of 2, adding them gives a number with two distinct prime factors. Let me test another one.a = 4, b = 8C(4) = 1C(8) = 1a + b = 12C(12) = 2Again, it works. So, it seems like if a and b are consecutive powers of 2, their sum has exactly two distinct prime factors, which is the sum of their individual prime factors.Wait, but in all these cases, the sum a + b is 2^n + 2^(n+1) = 3 * 2^n, which has prime factors 2 and 3. So, C(a + b) = 2, and since both a and b are powers of 2, C(a) = C(b) = 1, so 1 + 1 = 2.This seems to hold for any n. So, if I take a = 2^n and b = 2^(n+1), then a + b = 3 * 2^n, which always has two distinct prime factors, 2 and 3. Therefore, C(a + b) = 2, and C(a) + C(b) = 1 + 1 = 2.Since n can be any natural number, there are infinitely many such pairs (a, b). Therefore, the number of such pairs is infinite.But wait, let me check if this is the only way or if there are other possibilities. Maybe if a and b are multiples of different primes, but not necessarily powers of 2.Let's try a = 3 and b = 6.C(3) = 1C(6) = 2a + b = 9C(9) = 1Here, C(a + b) = 1, and C(a) + C(b) = 1 + 2 = 3. So, 1 ≠ 3, which doesn't satisfy the condition.Another example: a = 5, b = 10.C(5) = 1C(10) = 2a + b = 15C(15) = 2C(a + b) = 2, and C(a) + C(b) = 1 + 2 = 3. Again, 2 ≠ 3, so it doesn't work.How about a = 6 and b = 10.C(6) = 2C(10) = 2a + b = 16C(16) = 1C(a + b) = 1, and C(a) + C(b) = 2 + 2 = 4. Not equal.Hmm, seems like when a and b have multiple prime factors, their sum might not necessarily have the sum of their distinct prime factors.Another approach: maybe if a and b are coprime, then the number of distinct prime factors of a + b could be the sum of the distinct prime factors of a and b.But wait, in the earlier example where a = 2 and b = 4, they are not coprime, but it still worked. So, coprimality might not be a necessary condition.Alternatively, if a and b are such that their sum introduces new prime factors not present in a or b.Wait, in the case where a = 2^n and b = 2^(n+1), the sum is 3 * 2^n, which introduces the prime factor 3, which wasn't in a or b. So, the number of distinct prime factors increases by 1, but since both a and b had only one prime factor, the sum has two, which is the sum of their individual counts.Maybe another way is if a and b are such that their sum has all the prime factors of a and b plus some new ones. But in the case where a and b are powers of the same prime, their sum introduces a new prime factor, making the total distinct prime factors equal to the sum of their individual counts.Is there a way to generalize this? If a and b are both powers of a prime p, then a + b = p^n + p^(n+1) = p^n(1 + p). So, the sum would have prime factors p and (1 + p). If 1 + p is a prime, then the sum would have exactly two distinct prime factors, which would be p and (1 + p). For example, p = 2: 1 + 2 = 3, which is prime. So, a + b = 3 * 2^n, which has two distinct prime factors.Similarly, p = 3: 1 + 3 = 4, which is not prime. So, a + b = 4 * 3^n, which has prime factors 2 and 3. So, C(a + b) = 2, but C(a) = 1 and C(b) = 1, so 1 + 1 = 2. It still works.Wait, even if 1 + p is not prime, as long as it introduces a new prime factor, the total distinct prime factors would be the sum of the distinct prime factors of a and b.But in the case of p = 3, 1 + 3 = 4, which is 2^2, so the sum a + b = 4 * 3^n has prime factors 2 and 3, which are two distinct primes, same as before.So, even if 1 + p is not prime, as long as it introduces a new prime factor, the total distinct prime factors of a + b would be the sum of the distinct prime factors of a and b.Therefore, if we take a = p^n and b = p^(n+1), then a + b = p^n(1 + p). If 1 + p is not a power of p, which it isn't unless p = 2, but even then, 1 + 2 = 3, which is a different prime.Wait, for p = 2, 1 + p = 3, which is prime. For p = 3, 1 + p = 4, which is 2^2, introducing a new prime factor 2. For p = 5, 1 + p = 6, which is 2 * 3, introducing two new prime factors. Wait, but in that case, a + b = 6 * 5^n, which has prime factors 2, 3, and 5. So, C(a + b) = 3, but C(a) = 1 and C(b) = 1, so 1 + 1 = 2 ≠ 3. So, it doesn't satisfy the condition.Wait, that's a problem. So, for p = 5, a = 5^n, b = 5^(n+1), then a + b = 5^n + 5^(n+1) = 5^n(1 + 5) = 6 * 5^n, which has prime factors 2, 3, and 5. So, C(a + b) = 3, but C(a) + C(b) = 1 + 1 = 2. So, 3 ≠ 2, which doesn't satisfy the condition.So, this approach only works for certain primes p where 1 + p introduces exactly one new prime factor, making the total distinct prime factors of a + b equal to 2, which is the sum of C(a) and C(b).But for p = 2, 1 + p = 3, which is prime, so a + b = 3 * 2^n, which has two distinct prime factors. For p = 3, 1 + p = 4, which is 2^2, so a + b = 4 * 3^n, which has two distinct prime factors, 2 and 3. So, in both cases, C(a + b) = 2, which is equal to C(a) + C(b) = 1 + 1 = 2.But for p = 5, 1 + p = 6, which introduces two new prime factors, making C(a + b) = 3, which is greater than C(a) + C(b) = 2.So, it seems that this method only works for p = 2 and p = 3.Wait, let's check p = 7.a = 7^n, b = 7^(n+1)a + b = 7^n + 7^(n+1) = 7^n(1 + 7) = 8 * 7^n8 is 2^3, so a + b = 2^3 * 7^n, which has two distinct prime factors: 2 and 7.So, C(a + b) = 2, and C(a) + C(b) = 1 + 1 = 2. So, it works for p = 7 as well.Wait, 1 + 7 = 8, which is 2^3, so it only introduces one new prime factor, 2, which was not in a or b (since a and b are powers of 7). So, C(a + b) = 2, which is equal to C(a) + C(b).Similarly, p = 11.a = 11^n, b = 11^(n+1)a + b = 11^n + 11^(n+1) = 11^n(1 + 11) = 12 * 11^n12 = 2^2 * 3, so a + b = 2^2 * 3 * 11^n, which has three distinct prime factors: 2, 3, and 11.So, C(a + b) = 3, but C(a) + C(b) = 1 + 1 = 2. So, 3 ≠ 2, which doesn't satisfy the condition.Wait, so for p = 11, it doesn't work because 1 + p = 12 introduces two new prime factors, making C(a + b) = 3, which is greater than 2.So, it seems that for primes p where 1 + p is a power of 2, then a + b will have only two distinct prime factors: p and 2. For example, p = 2: 1 + 2 = 3 (which is not a power of 2, but it's prime), p = 3: 1 + 3 = 4 = 2^2, p = 7: 1 + 7 = 8 = 2^3, p = 15: wait, 15 is not prime. Wait, p = 15 is not prime, so let's see p = 127: 1 + 127 = 128 = 2^7, so p = 127 is a prime, and 1 + p = 128, which is a power of 2.So, for primes p where 1 + p is a power of 2, then a + b = (1 + p) * p^n = 2^k * p^n, which has two distinct prime factors: 2 and p. Therefore, C(a + b) = 2, and C(a) + C(b) = 1 + 1 = 2, so it satisfies the condition.But for primes p where 1 + p is not a power of 2, then a + b will have more than two distinct prime factors, making C(a + b) > 2, which doesn't satisfy the condition.So, the pairs (a, b) where a = p^n and b = p^(n+1) with p being a prime such that 1 + p is a power of 2 will satisfy the condition C(a + b) = C(a) + C(b).Now, are there infinitely many such primes p where 1 + p is a power of 2?Well, primes of the form 2^k - 1 are called Mersenne primes. For example, p = 3 = 2^2 - 1, p = 7 = 2^3 - 1, p = 31 = 2^5 - 1, p = 127 = 2^7 - 1, etc.It is known that there are infinitely many Mersenne primes, although this is still an open question in mathematics. However, even if it's not proven yet, it's conjectured that there are infinitely many Mersenne primes.Assuming that there are infinitely many Mersenne primes, then for each such prime p = 2^k - 1, we can construct infinitely many pairs (a, b) = (p^n, p^(n+1)) for any natural number n, which would satisfy C(a + b) = C(a) + C(b).Therefore, under the assumption that there are infinitely many Mersenne primes, there are infinitely many such pairs (a, b).However, since the infinitude of Mersenne primes is not yet proven, we might need another approach to ensure that there are infinitely many pairs without relying on unproven conjectures.Alternatively, we can consider specific primes p where 1 + p is a power of 2, even if we don't know if there are infinitely many such primes. For example, p = 3, 7, 31, 127, etc., are known Mersenne primes. For each of these, we can generate infinitely many pairs (a, b) by choosing different exponents n.Even if we can't prove there are infinitely many such primes, the fact that we can generate infinitely many pairs for each known Mersenne prime suggests that there are at least infinitely many pairs, assuming there are infinitely many Mersenne primes.But perhaps there's another way to construct infinitely many pairs without relying on Mersenne primes.Let me think differently. Suppose we take a = 2^n and b = 2^n + 1.Wait, let's see:a = 2^n, so C(a) = 1.b = 2^n + 1. What is C(b)? It depends on whether 2^n + 1 is prime or composite.For example, if n = 1: b = 3, which is prime, so C(b) = 1.a + b = 2 + 3 = 5, which is prime, so C(a + b) = 1.But C(a) + C(b) = 1 + 1 = 2 ≠ 1, so it doesn't work.n = 2: a = 4, b = 5.C(a) = 1, C(b) = 1.a + b = 9, C(a + b) = 1.Again, 1 ≠ 2.n = 3: a = 8, b = 9.C(a) = 1, C(b) = 1 (since 9 = 3^2).a + b = 17, which is prime, so C(a + b) = 1.Still, 1 ≠ 2.n = 4: a = 16, b = 17.C(a) = 1, C(b) = 1.a + b = 33, which factors into 3 * 11, so C(a + b) = 2.Now, C(a) + C(b) = 1 + 1 = 2, which equals C(a + b). So, this works.Wait, so for n = 4, a = 16, b = 17, we have C(a + b) = 2 = C(a) + C(b).Similarly, let's try n = 5: a = 32, b = 33.C(a) = 1, C(b) = 2 (since 33 = 3 * 11).a + b = 65, which factors into 5 * 13, so C(a + b) = 2.But C(a) + C(b) = 1 + 2 = 3 ≠ 2.So, it doesn't work.n = 6: a = 64, b = 65.C(a) = 1, C(b) = 2 (65 = 5 * 13).a + b = 129, which factors into 3 * 43, so C(a + b) = 2.Again, C(a) + C(b) = 1 + 2 = 3 ≠ 2.n = 7: a = 128, b = 129.C(a) = 1, C(b) = 2 (129 = 3 * 43).a + b = 257, which is prime, so C(a + b) = 1.1 ≠ 3.n = 8: a = 256, b = 257.C(a) = 1, C(b) = 1 (257 is prime).a + b = 513, which factors into 3^3 * 19, so C(a + b) = 2.C(a) + C(b) = 1 + 1 = 2. So, it works.So, for n = 4 and n = 8, the pair (2^n, 2^n + 1) satisfies the condition.Is there a pattern here? Let's see:n = 4: 2^4 = 16, b = 17 (prime)a + b = 33 = 3 * 11, C(a + b) = 2n = 8: 2^8 = 256, b = 257 (prime)a + b = 513 = 3^3 * 19, C(a + b) = 2So, in these cases, b is a Fermat prime, which is of the form 2^(2^k) + 1. Known Fermat primes are for k = 0 to 4: 3, 5, 17, 257, 65537.So, if we take n = 2^k, then b = 2^(2^k) + 1, which is a Fermat prime if it's prime.For example, n = 2^2 = 4: b = 17, which is prime.n = 2^3 = 8: b = 257, which is prime.n = 2^4 = 16: b = 65537, which is prime.n = 2^5 = 32: b = 4294967297, which is known to be composite (it's 641 * 6700417).So, for n = 4, 8, 16, we get b as a Fermat prime, and a + b has exactly two distinct prime factors.But since only five Fermat primes are known, and it's known that for n ≥ 5, Fermat numbers are composite, we can't generate infinitely many pairs this way.Therefore, this approach only gives us finitely many pairs.Going back to the earlier approach with Mersenne primes, even though it's not proven that there are infinitely many Mersenne primes, it's a plausible conjecture, and for each known Mersenne prime, we can generate infinitely many pairs (a, b) by choosing different exponents n.But perhaps there's another way to construct infinitely many pairs without relying on Mersenne primes or Fermat primes.Let me think about numbers a and b such that a + b is a product of two distinct primes, and a and b are powers of one of those primes.For example, let’s say a = p^n and b = p^m, then a + b = p^n + p^m = p^min(n,m) (1 + p^|n - m|).If we set |n - m| = 1, then a + b = p^k (1 + p), where k = min(n, m).If 1 + p is a prime, say q, then a + b = p^k * q, which has two distinct prime factors, p and q.So, C(a + b) = 2, and C(a) + C(b) = 1 + 1 = 2.Therefore, as long as p is a prime such that 1 + p is also prime, then choosing a = p^n and b = p^(n+1) will satisfy the condition.Primes p where p + 1 is also prime are called "Sophie Germain primes" if p + 2 is also prime, but in this case, we just need p + 1 to be prime.Wait, actually, if p is a prime and p + 1 is also prime, then p must be 2, because for any odd prime p, p + 1 is even and greater than 2, hence not prime.So, the only prime p where p + 1 is also prime is p = 2, since 2 + 1 = 3, which is prime.Therefore, the only prime p where 1 + p is prime is p = 2.So, in this case, a = 2^n, b = 2^(n+1), a + b = 3 * 2^n, which has two distinct prime factors, 2 and 3.Thus, for p = 2, we can generate infinitely many pairs (a, b) = (2^n, 2^(n+1)), and each will satisfy C(a + b) = C(a) + C(b).Therefore, even without relying on Mersenne primes, we can construct infinitely many pairs by taking a = 2^n and b = 2^(n+1) for any natural number n.This seems to be a solid approach because it doesn't depend on unproven conjectures; it's a straightforward construction.So, to summarize, by choosing a = 2^n and b = 2^(n+1), we ensure that a + b = 3 * 2^n, which has exactly two distinct prime factors, 2 and 3. Since both a and b are powers of 2, they each have only one distinct prime factor. Therefore, C(a + b) = 2, and C(a) + C(b) = 1 + 1 = 2, satisfying the condition.Since n can be any natural number, there are infinitely many such pairs (a, b).I think this approach works and shows that there are indeed infinitely many pairs satisfying the given condition.

💡Okay, so I have this problem where a red ball and a green ball are being tossed into bins numbered 1, 2, 3, and so on. The probability that a ball lands in bin k is 2^{-k}. I need to find the probability that the red ball ends up in a higher-numbered bin than the green ball.Hmm, let me think about this. First, since the balls are tossed independently, the probability distribution for each ball is separate. So, the probability that the red ball is in bin k is 2^{-k}, and the same goes for the green ball.I remember that when dealing with two independent events, the joint probability is the product of their individual probabilities. So, the probability that the red ball is in bin i and the green ball is in bin j is 2^{-i} * 2^{-j} = 2^{-(i+j)}.Now, I need to find the probability that the red ball is in a higher-numbered bin than the green ball. That means I need to consider all pairs (i, j) where i > j. So, I need to sum the probabilities 2^{-(i+j)} for all i > j.Wait, that sounds like a double sum. Let me write that down:P(red > green) = Σ_{j=1}^{∞} Σ_{i=j+1}^{∞} 2^{-(i+j)}.Hmm, that seems a bit complicated. Maybe there's a smarter way to approach this. I remember that for problems like this, sometimes symmetry can help. Since the red and green balls are identical in terms of their probability distributions, the probability that red > green should be equal to the probability that green > red. And then there's the probability that they land in the same bin.So, if I denote P(red > green) as P, then P(green > red) is also P, and P(red = green) is Q. Then, since these are the only possibilities, we have:P + P + Q = 1.So, 2P + Q = 1.Therefore, if I can find Q, the probability that both balls land in the same bin, I can solve for P.Let me calculate Q. Q is the sum over all bins k of the probability that both balls land in bin k. So,Q = Σ_{k=1}^{∞} (2^{-k}) * (2^{-k}) = Σ_{k=1}^{∞} 2^{-2k}.That's a geometric series where each term is (2^{-2})^k, so the common ratio is 1/4. The sum of a geometric series Σ_{k=1}^{∞} ar^{k-1} is a / (1 - r). But in this case, our series starts at k=1, so it's actually Σ_{k=1}^{∞} (1/4)^k.The sum of that is (1/4) / (1 - 1/4) = (1/4) / (3/4) = 1/3.So, Q = 1/3.Going back to the equation 2P + Q = 1, we have 2P + 1/3 = 1, so 2P = 2/3, which means P = 1/3.Wait, so the probability that the red ball is in a higher-numbered bin than the green ball is 1/3? That seems reasonable because of the symmetry. Let me just verify that.Alternatively, if I think about it, since the distributions are identical, the chance that red is higher than green should be the same as green being higher than red, and the chance they are equal is 1/3. So, the remaining probability is 2/3, split equally between red > green and green > red, giving 1/3 each.Yeah, that makes sense. So, I think the answer is 1/3.**Final Answer**boxed{C}