💡Alright, so I have this problem here about vectors, and I need to find the unit vector in the same direction as vector AB, which is given as (-1, 2). Hmm, okay, let's break this down step by step. I remember that a unit vector is a vector with a magnitude of 1, and it points in the same direction as the original vector. So, essentially, I need to scale down the vector AB so that its length becomes 1, but keep its direction the same.First, I think I need to find the magnitude of vector AB. The magnitude of a vector (x, y) is calculated using the formula sqrt(x² + y²). So, applying that to vector AB, which is (-1, 2), the magnitude should be sqrt((-1)² + 2²). Let me compute that: (-1) squared is 1, and 2 squared is 4. Adding those together gives 1 + 4 = 5. So, the magnitude is sqrt(5). Okay, that seems straightforward.Now, to find the unit vector, I need to divide each component of the original vector by its magnitude. That means I'll take the x-component, which is -1, and divide it by sqrt(5), and do the same for the y-component, which is 2. So, the unit vector should be (-1/sqrt(5), 2/sqrt(5)). Hmm, but I remember that it's often preferred to rationalize the denominator when expressing vectors. So, let me rationalize those components.For the x-component: -1/sqrt(5) can be rationalized by multiplying numerator and denominator by sqrt(5), which gives (-1*sqrt(5))/(sqrt(5)*sqrt(5)) = (-sqrt(5))/5. Similarly, for the y-component: 2/sqrt(5) becomes (2*sqrt(5))/5. So, putting it all together, the unit vector is (-sqrt(5)/5, 2*sqrt(5)/5).Let me double-check my calculations to make sure I didn't make any mistakes. The magnitude was sqrt(5), that's correct. Dividing each component by sqrt(5) gives (-1/sqrt(5), 2/sqrt(5)), and rationalizing gives (-sqrt(5)/5, 2*sqrt(5)/5). Yeah, that seems right.Looking at the options provided:A: (sqrt(5)/5, -2*sqrt(5)/5)B: (2*sqrt(5)/5, -sqrt(5)/5)C: (-sqrt(5)/5, 2*sqrt(5)/5)D: (-1/2, 1)So, my calculated unit vector matches option C exactly. Just to be thorough, let me check the other options to ensure none of them could also be correct or if I made a mistake in my calculations.Option A has the x-component positive and the y-component negative, which is the opposite direction of our original vector AB. So, that's not correct.Option B has both components scaled by sqrt(5)/5, but the x-component is 2*sqrt(5)/5 and the y-component is -sqrt(5)/5. That doesn't match our original vector's direction either.Option D is (-1/2, 1). Let me check the magnitude of this vector to see if it's a unit vector. The magnitude would be sqrt((-1/2)² + 1²) = sqrt(1/4 + 1) = sqrt(5/4) = sqrt(5)/2, which is not 1. So, this isn't a unit vector.Therefore, option C is the correct answer. It has the same direction as vector AB and is scaled appropriately to have a magnitude of 1.

💡Okay, so I have this problem where I need to find the minimum possible value of the sum of all numbers in an 80x80 table. Each entry in the table is given by the logarithm expression (log_{x_k} frac{x_i}{16}), where (x_1, x_2, ldots, x_{100}) are natural numbers greater than 1. First, I need to understand the structure of the problem. There are 100 natural numbers (x_1) to (x_{100}), each greater than 1. These numbers are used to fill an 80x80 table. Each cell in the table is determined by the row index (i) and the column index (k), and the value at that cell is (log_{x_k} frac{x_i}{16}).My goal is to find the minimum possible sum of all these logarithmic expressions. Since the table is 80x80, there are 6400 entries. Each entry is a logarithm, so I need to figure out how to choose the (x_i) and (x_k) such that the sum of all these logarithms is as small as possible.Let me start by breaking down the logarithmic expression. The expression (log_{x_k} frac{x_i}{16}) can be rewritten using logarithmic properties. Specifically, (log_b frac{a}{c} = log_b a - log_b c). So, applying this property, I can rewrite each entry as:[log_{x_k} frac{x_i}{16} = log_{x_k} x_i - log_{x_k} 16]This means each entry is the difference of two logarithms. So, the sum of all entries in the table can be expressed as the sum of all (log_{x_k} x_i) minus the sum of all (log_{x_k} 16).Let me write this out:[text{Total Sum} = sum_{i=1}^{80} sum_{k=1}^{80} left( log_{x_k} x_i - log_{x_k} 16 right)]This can be separated into two sums:[text{Total Sum} = sum_{i=1}^{80} sum_{k=1}^{80} log_{x_k} x_i - sum_{i=1}^{80} sum_{k=1}^{80} log_{x_k} 16]Let me analyze each part separately.First, the sum (sum_{i=1}^{80} sum_{k=1}^{80} log_{x_k} x_i). This is the sum over all rows and columns of (log_{x_k} x_i). Since (x_i) and (x_k) are both variables, this seems a bit complex. However, I notice that for each pair ((i, k)), we have a term (log_{x_k} x_i). I recall that (log_b a = frac{1}{log_a b}). So, (log_{x_k} x_i = frac{1}{log_{x_i} x_k}). This might be useful, but I'm not sure yet. Maybe I can think about the relationship between these terms.Wait, perhaps I can consider the sum over all (i) and (k) of (log_{x_k} x_i). If I fix (i), then for each row (i), I'm summing (log_{x_k} x_i) over all columns (k). Similarly, if I fix (k), for each column (k), I'm summing (log_{x_k} x_i) over all rows (i).But I'm not sure if that helps. Maybe I can think about the sum in terms of variables. Let me denote (y_{i,k} = log_{x_k} x_i). Then, the first sum is just the sum of all (y_{i,k}) for (i, k = 1) to (80).Now, the second sum is (sum_{i=1}^{80} sum_{k=1}^{80} log_{x_k} 16). Since 16 is a constant, this is the same as (sum_{i=1}^{80} sum_{k=1}^{80} log_{x_k} 16). But since the inner sum over (i) doesn't affect the logarithm, which only depends on (x_k), this simplifies to (80 times sum_{k=1}^{80} log_{x_k} 16). Wait, actually, for each column (k), we have 80 terms of (log_{x_k} 16), so the entire second sum is (80 times sum_{k=1}^{80} log_{x_k} 16).So, putting it all together, the total sum is:[text{Total Sum} = sum_{i=1}^{80} sum_{k=1}^{80} log_{x_k} x_i - 80 times sum_{k=1}^{80} log_{x_k} 16]Now, I need to find the minimum of this expression. Since (x_1, x_2, ldots, x_{100}) are natural numbers greater than 1, they can be 2, 3, 4, etc. However, since the table only uses the first 80 (x_k) and (x_i), perhaps the other 20 (x)s are irrelevant? Wait, no, the problem says (x_1, x_2, ldots, x_{100}), but the table only uses 80 of them. So, actually, the table uses (x_1) to (x_{80}) for both rows and columns. Wait, no, the problem says "at the intersection of the (i)-th row and the (k)-th column, the number (log_{x_k} frac{x_i}{16}) is written." So, (x_i) is used for the rows, and (x_k) is used for the columns. So, the table uses (x_1) to (x_{80}) for the rows and (x_1) to (x_{80}) for the columns. So, actually, the (x)s used are the first 80, and the remaining 20 are not used. So, perhaps I can ignore the last 20 (x)s since they don't affect the table.Therefore, I can assume that (x_1, x_2, ldots, x_{80}) are the variables we need to choose, each greater than 1, and the rest (x_{81}) to (x_{100}) can be arbitrary since they don't affect the table.So, now, focusing on (x_1) to (x_{80}), each greater than 1, natural numbers.So, the total sum is:[text{Total Sum} = sum_{i=1}^{80} sum_{k=1}^{80} log_{x_k} x_i - 80 times sum_{k=1}^{80} log_{x_k} 16]I need to minimize this expression.Let me think about how to approach this. Maybe I can consider the first sum and the second sum separately.First, let's consider the first sum: (sum_{i=1}^{80} sum_{k=1}^{80} log_{x_k} x_i).I can rewrite this as:[sum_{k=1}^{80} sum_{i=1}^{80} log_{x_k} x_i]Which is:[sum_{k=1}^{80} left( sum_{i=1}^{80} log_{x_k} x_i right)]So, for each column (k), we have the sum over all rows (i) of (log_{x_k} x_i).Similarly, the second sum is:[80 times sum_{k=1}^{80} log_{x_k} 16]Which is:[80 times sum_{k=1}^{80} log_{x_k} 16]So, the total sum is:[sum_{k=1}^{80} left( sum_{i=1}^{80} log_{x_k} x_i right) - 80 times sum_{k=1}^{80} log_{x_k} 16]Now, perhaps I can combine these terms for each (k). Let me denote for each (k):[S_k = sum_{i=1}^{80} log_{x_k} x_i - 80 times log_{x_k} 16]Then, the total sum is:[sum_{k=1}^{80} S_k]So, if I can find the minimum of each (S_k), then summing them up will give me the total minimum.So, let's focus on (S_k):[S_k = sum_{i=1}^{80} log_{x_k} x_i - 80 times log_{x_k} 16]I can factor out (log_{x_k}) since it's a common base:[S_k = sum_{i=1}^{80} log_{x_k} x_i - 80 times log_{x_k} 16 = sum_{i=1}^{80} log_{x_k} x_i - sum_{i=1}^{80} log_{x_k} 16]Wait, that's not correct. The second term is 80 times (log_{x_k} 16), so it's equivalent to (sum_{i=1}^{80} log_{x_k} 16). So, actually, (S_k) can be written as:[S_k = sum_{i=1}^{80} left( log_{x_k} x_i - log_{x_k} 16 right) = sum_{i=1}^{80} log_{x_k} left( frac{x_i}{16} right)]But that's just the original expression for each column (k). Hmm, maybe another approach.Alternatively, let's consider that for each (k), (S_k) is:[S_k = sum_{i=1}^{80} log_{x_k} x_i - 80 times log_{x_k} 16]Let me factor out (log_{x_k}):[S_k = log_{x_k} left( prod_{i=1}^{80} x_i right) - 80 times log_{x_k} 16]Because the sum of logs is the log of the product.So,[S_k = log_{x_k} left( prod_{i=1}^{80} x_i right) - log_{x_k} (16^{80})]Which simplifies to:[S_k = log_{x_k} left( frac{prod_{i=1}^{80} x_i}{16^{80}} right)]So, (S_k) is the logarithm base (x_k) of the product of all (x_i) divided by (16^{80}).So, the total sum is:[sum_{k=1}^{80} log_{x_k} left( frac{prod_{i=1}^{80} x_i}{16^{80}} right)]Hmm, this seems a bit abstract. Maybe I can think about the relationship between the (x_i) and (x_k).Wait, perhaps I can consider that for each (k), (x_k) is a variable, and the other (x_i)s are variables as well. So, maybe I can choose all (x_i)s and (x_k)s to be the same value to simplify things.Let me assume that all (x_i = x) for some natural number (x > 1). Then, each (x_k = x) as well.Then, the expression becomes:[log_{x} left( frac{x}{16} right)]So, each entry in the table is (log_{x} left( frac{x}{16} right)). Then, the total sum is 80x80 times this value.So, total sum is:[6400 times log_{x} left( frac{x}{16} right)]Simplify this:[6400 times left( log_{x} x - log_{x} 16 right) = 6400 times (1 - log_{x} 16)]Since (log_{x} x = 1).Now, (log_{x} 16) can be written as (frac{log 16}{log x}), but perhaps it's better to express it in terms of base 2.Since 16 is (2^4), so:[log_{x} 16 = log_{x} 2^4 = 4 log_{x} 2]So, the total sum becomes:[6400 times (1 - 4 log_{x} 2)]Now, I need to minimize this expression with respect to (x), where (x) is a natural number greater than 1.Let me denote (y = log_{x} 2). Then, (y = frac{ln 2}{ln x}), so as (x) increases, (y) decreases.Our expression becomes:[6400 times (1 - 4y)]To minimize this, since it's linear in (y), and (y) is positive, the expression is minimized when (y) is maximized. Because as (y) increases, the term (1 - 4y) decreases.But (y = log_{x} 2), which is maximized when (x) is minimized. Since (x) is a natural number greater than 1, the smallest possible (x) is 2.So, if (x = 2), then (y = log_{2} 2 = 1). Plugging this back in:[6400 times (1 - 4 times 1) = 6400 times (-3) = -19200]So, if all (x_i = x_k = 2), the total sum is -19200.Is this the minimum possible? Let me check if choosing different (x_i) and (x_k) can lead to a smaller sum.Suppose I choose some (x_i) to be 2 and others to be larger. For example, if I set some (x_i = 2) and others (x_i = 4), how does that affect the sum?Wait, but in the expression for (S_k), we have:[S_k = log_{x_k} left( frac{prod_{i=1}^{80} x_i}{16^{80}} right)]If I set some (x_i) to be larger, say 4, then the product (prod x_i) increases, which would make the argument of the logarithm larger, but since the logarithm is base (x_k), which could also be 2 or 4, it's not straightforward.Alternatively, perhaps if I set some (x_k) to be larger, the logarithm base increases, which would make the logarithm smaller (since for a larger base, the logarithm of a fixed number is smaller). But then, the term (log_{x_k} x_i) would also be affected.This seems complicated. Maybe the minimal sum is achieved when all (x_i = x_k = 2), as this makes each term (log_{2} frac{2}{16} = log_{2} frac{1}{8} = -3), and with 6400 terms, the total sum is -19200.But let me verify this.Suppose I set (x_k = 2) for all (k), and (x_i = 2) for all (i). Then each entry is (log_{2} frac{2}{16} = log_{2} frac{1}{8} = -3). So, each of the 6400 entries is -3, so the total sum is indeed -19200.If I set some (x_i) to be larger than 2, say 4, then for those rows, the term (log_{x_k} x_i) would be (log_{2} 4 = 2), but the term (log_{x_k} 16) would be (log_{2} 16 = 4). So, each entry in those rows would be (2 - 4 = -2), which is larger than -3. So, the total sum would increase, which is not desirable since we are looking for the minimum.Similarly, if I set some (x_k) to be larger than 2, say 4, then for those columns, the term (log_{x_k} x_i) would be (log_{4} x_i), which is smaller than (log_{2} x_i), and the term (log_{x_k} 16) would be (log_{4} 16 = 2). So, each entry in those columns would be (log_{4} x_i - 2). If (x_i = 2), then (log_{4} 2 = 0.5), so the entry would be (0.5 - 2 = -1.5), which is larger than -3. If (x_i = 4), then (log_{4} 4 = 1), so the entry would be (1 - 2 = -1), which is still larger than -3. If (x_i = 8), then (log_{4} 8 = 1.5), so the entry would be (1.5 - 2 = -0.5), which is even worse.Therefore, setting any (x_k) larger than 2 would result in some entries being larger than -3, thus increasing the total sum. Similarly, setting any (x_i) larger than 2 would also result in some entries being larger than -3.What if I set some (x_k) to be smaller than 2? But (x_k) must be a natural number greater than 1, so the smallest possible is 2. So, I can't set any (x_k) smaller than 2.Therefore, it seems that setting all (x_i = x_k = 2) gives the minimal possible sum of -19200.Wait, but let me think again. Suppose I set some (x_k) to be 2 and others to be larger. For example, suppose I set half of the (x_k) to be 2 and the other half to be 4. Then, for the columns with (x_k = 2), the entries would be (log_{2} frac{x_i}{16}). If (x_i = 2), then each entry is (-3). If (x_i = 4), then each entry is (log_{2} frac{4}{16} = log_{2} frac{1}{4} = -2). So, for those columns, the entries would be -2 or -3 depending on (x_i).For the columns with (x_k = 4), the entries would be (log_{4} frac{x_i}{16}). If (x_i = 2), then (log_{4} frac{2}{16} = log_{4} frac{1}{8} = log_{4} 2^{-3} = -3 times log_{4} 2 = -3 times 0.5 = -1.5). If (x_i = 4), then (log_{4} frac{4}{16} = log_{4} frac{1}{4} = -1). If (x_i = 8), then (log_{4} frac{8}{16} = log_{4} frac{1}{2} = -0.5).So, in this case, for the columns with (x_k = 4), the entries are less negative or even positive if (x_i) is large enough. Therefore, the total sum would be a mix of -3, -2, -1.5, -1, -0.5, etc., which would result in a total sum larger than -19200.Therefore, setting any (x_k) larger than 2 would result in some entries being less negative, thus increasing the total sum. Therefore, the minimal sum is achieved when all (x_k = 2) and all (x_i = 2), resulting in each entry being -3 and the total sum being -19200.I think this is the minimal possible value. Let me just check if there's any other configuration that could lead to a smaller sum.Suppose I set some (x_i) to be 16. Then, for those rows, the term (log_{x_k} x_i) would be (log_{x_k} 16). If (x_k = 2), then (log_{2} 16 = 4), so the entry would be (4 - log_{2} 16 = 4 - 4 = 0). If (x_k = 4), then (log_{4} 16 = 2), so the entry would be (2 - log_{4} 16 = 2 - 2 = 0). So, setting (x_i = 16) would result in some entries being 0, which is worse than -3.Alternatively, if I set some (x_i) to be larger than 16, the entries would be positive, which is even worse.What if I set some (x_i) to be less than 2? But (x_i) must be a natural number greater than 1, so the smallest possible is 2. So, I can't set any (x_i) smaller than 2.Therefore, the minimal sum is indeed achieved when all (x_i = x_k = 2), resulting in each entry being -3 and the total sum being -19200.I think this is the correct answer.

💡Alright, so I have this problem about complex numbers and similar triangles. Let me try to wrap my head around it step by step. First, the problem states that there are five points on a plane: A, B, C, U, V. These points correspond to complex numbers a, b, c, u, v respectively. The key information is that the triangles AUV, VBU, and UVC are directly similar to each other. I need to prove that triangle ABC is also directly similar to these triangles.Okay, so similarity in triangles means that the corresponding angles are equal, and the sides are proportional. In terms of complex numbers, similarity can be expressed using complex ratios. If two triangles are similar, the ratio of their corresponding sides can be represented as a complex number, which includes both scaling and rotation.Let me recall that for two triangles to be directly similar, the ratio of their corresponding sides must be equal. So, if triangle AUV is similar to triangle VBU, then the ratio of the sides AV to BU should be equal to the ratio of AU to BV, and so on for the other sides.But since all three triangles AUV, VBU, and UVC are directly similar, their corresponding side ratios must all be equal. That gives us a chain of equal ratios.Let me try to write this down. If triangle AUV is similar to triangle VBU, then:(v - a)/(u - a) = (u - v)/(b - v)Similarly, since triangle VBU is similar to triangle UVC, we have:(u - v)/(b - v) = (c - u)/(v - u)And since triangle AUV is similar to triangle UVC, we have:(v - a)/(u - a) = (c - u)/(v - u)So, all these ratios are equal. Let me denote this common ratio as k, which is a complex number representing the similarity transformation (rotation and scaling).So, we have:(v - a)/(u - a) = k(u - v)/(b - v) = k(c - u)/(v - u) = kNow, I need to relate these to triangle ABC. To show that triangle ABC is similar to the others, I need to find a ratio between the sides of ABC and the sides of AUV, VBU, or UVC.Let me consider the sides of triangle ABC: AB, BC, and CA. These correspond to the complex numbers b - a, c - b, and a - c.I need to express these in terms of the given ratios. Let me see if I can find expressions for b - a, c - b, and a - c using the given equalities.From the first ratio, (v - a)/(u - a) = k, so v - a = k(u - a).From the second ratio, (u - v)/(b - v) = k, so u - v = k(b - v).From the third ratio, (c - u)/(v - u) = k, so c - u = k(v - u).Let me write these equations:1. v - a = k(u - a)2. u - v = k(b - v)3. c - u = k(v - u)Now, I can try to solve these equations step by step.Starting with equation 1: v - a = k(u - a). Let me rearrange this:v = a + k(u - a)Similarly, equation 2: u - v = k(b - v). Let me substitute v from equation 1 into equation 2.u - [a + k(u - a)] = k(b - [a + k(u - a)])Simplify the left side:u - a - k(u - a) = k(b - a - k(u - a))Let me factor out (u - a):(u - a)(1 - k) = k(b - a - k(u - a))Hmm, this seems a bit complicated. Maybe I can express b in terms of u and v.From equation 2: u - v = k(b - v)So, u - v = k b - k vRearranging:k b = u - v + k vSo, b = (u - v + k v)/kSimilarly, from equation 3: c - u = k(v - u)So, c = u + k(v - u)Now, I have expressions for b and c in terms of u, v, and k.Let me write these:b = (u - v + k v)/kc = u + k(v - u)Now, let me compute b - a and c - a.First, b - a:b - a = [(u - v + k v)/k] - aSimilarly, c - a = [u + k(v - u)] - aLet me see if I can express these in terms of (v - a) and (u - a), which are related to k.From equation 1: v - a = k(u - a)So, u - a = (v - a)/kSimilarly, v - a = k(u - a) implies that u = (v - a)/k + aLet me substitute u into the expression for b - a.First, let's compute b - a:b - a = [(u - v + k v)/k] - a= [u - v + k v - k a]/k= [u - v + k(v - a)]/kBut from equation 1, v - a = k(u - a), so v - a = k(u - a)Thus, k(v - a) = k^2(u - a)So, substituting back:b - a = [u - v + k^2(u - a)]/kNow, let me express u - v in terms of k.From equation 2: u - v = k(b - v)But I already have an expression for b, so maybe it's better to express u - v in terms of u and v.Wait, from equation 2: u - v = k(b - v)But I have b expressed in terms of u and v, so maybe it's circular.Alternatively, from equation 1: v = a + k(u - a)So, u - v = u - [a + k(u - a)] = u - a - k(u - a) = (1 - k)(u - a)Therefore, u - v = (1 - k)(u - a)So, going back to b - a:b - a = [u - v + k^2(u - a)]/k= [(1 - k)(u - a) + k^2(u - a)]/kFactor out (u - a):= [(1 - k + k^2)(u - a)]/kSimilarly, from equation 1, u - a = (v - a)/kSo, substituting:= [(1 - k + k^2)(v - a)/k]/k= (1 - k + k^2)(v - a)/k^2Now, let me compute c - a:c - a = [u + k(v - u)] - a= u - a + k(v - u)= (u - a) + k(v - u)Again, from equation 1: u - a = (v - a)/kAnd v - u = -(u - v) = -(1 - k)(u - a) = -(1 - k)(v - a)/kSo, substituting:c - a = (v - a)/k + k[-(1 - k)(v - a)/k]Simplify:= (v - a)/k - k(1 - k)(v - a)/k= (v - a)/k - (1 - k)(v - a)Factor out (v - a):= (v - a)[1/k - (1 - k)]= (v - a)[(1 - k(1 - k))/k]= (v - a)[(1 - k + k^2)/k]So, c - a = (1 - k + k^2)(v - a)/kNow, let's look at b - a and c - a:b - a = (1 - k + k^2)(v - a)/k^2c - a = (1 - k + k^2)(v - a)/kSo, c - a = k(b - a)That's interesting. So, the ratio (c - a)/(b - a) = kBut from equation 1, (v - a)/(u - a) = kSo, (c - a)/(b - a) = (v - a)/(u - a)This suggests that the ratio of sides CA to AB is equal to the ratio of sides UV to AU.But in triangle ABC, side AB corresponds to b - a, BC corresponds to c - b, and CA corresponds to a - c.Similarly, in triangle AUV, sides are AU = u - a, UV = v - u, and VA = a - v.Wait, but in our case, we have (c - a)/(b - a) = k, which is the same as (v - a)/(u - a).So, this suggests that the ratio of CA to AB is equal to the ratio of VA to AU.But in triangle ABC, the sides are AB, BC, CA, and in triangle AUV, the sides are AU, UV, VA.So, if the ratio of CA to AB is equal to the ratio of VA to AU, and if the angles are preserved, then triangle ABC is similar to triangle AUV.But I need to ensure that all corresponding sides have the same ratio and that the angles are equal.Wait, let me think about the angles. Since all the given triangles are directly similar, their angles are equal. So, if I can show that the angles of triangle ABC correspond to the angles of triangle AUV, then ABC is similar to AUV.Alternatively, since we have the ratio of sides CA to AB equal to the ratio of VA to AU, and if we can show that another pair of sides has the same ratio, then by SAS similarity, the triangles are similar.Let me check another pair of sides. Let's compute (c - b)/(a - b) and see if it relates to another ratio.Wait, from the expressions above, we have:c - a = (1 - k + k^2)(v - a)/kSimilarly, b - a = (1 - k + k^2)(v - a)/k^2So, (c - a) = k(b - a)Therefore, (c - a)/(b - a) = kSimilarly, from triangle AUV, (v - a)/(u - a) = kSo, the ratio of CA to AB is equal to the ratio of VA to AU.Now, let's compute another ratio, say (c - b)/(a - b).From the expressions for b and c:b = (u - v + k v)/kc = u + k(v - u)So, c - b = [u + k(v - u)] - [(u - v + k v)/k]Let me compute this:= u + k(v - u) - (u - v + k v)/k= u + k v - k u - (u - v + k v)/kLet me factor out terms:= u(1 - k) + k v - (u - v + k v)/kNow, let me write (u - v + k v)/k as (u - v)/k + vSo,= u(1 - k) + k v - [(u - v)/k + v]= u(1 - k) + k v - (u - v)/k - v= u(1 - k) - (u - v)/k + k v - vLet me combine the terms with u:= u[(1 - k) - 1/k] + v[ (k - 1) + 1/k ]Hmm, this is getting complicated. Maybe there's a better way.Alternatively, since we have expressions for b and c in terms of u, v, and k, perhaps we can express (c - b) in terms of (v - u) or something else.Wait, from equation 3: c - u = k(v - u)So, c = u + k(v - u)From equation 2: u - v = k(b - v)So, b = (u - v)/k + vTherefore, c - b = [u + k(v - u)] - [(u - v)/k + v]Let me compute this:= u + k v - k u - (u - v)/k - v= u(1 - k) + k v - v - (u - v)/k= u(1 - k) + v(k - 1) - (u - v)/kFactor out (1 - k):= (1 - k)(u - v) - (u - v)/k= (u - v)[(1 - k) - 1/k]= (u - v)[(k(1 - k) - 1)/k]= (u - v)[(k - k^2 - 1)/k]Hmm, not sure if this helps.Wait, from equation 1: v - a = k(u - a)From equation 2: u - v = k(b - v)From equation 3: c - u = k(v - u)We have a system of three equations with variables u, v, b, c in terms of a and k.But since a is fixed, maybe we can express everything in terms of a and k.Wait, let me try to express b and c in terms of a and k.From equation 1: v = a + k(u - a)From equation 2: u - v = k(b - v)Substitute v from equation 1 into equation 2:u - [a + k(u - a)] = k(b - [a + k(u - a)])Simplify left side:u - a - k u + k a = k(b - a - k(u - a))Factor u terms:u(1 - k) + a(k - 1) = k(b - a - k u + k a)Simplify right side:k(b - a) - k^2(u - a)So, we have:u(1 - k) + a(k - 1) = k(b - a) - k^2(u - a)Let me collect like terms:Left side: u(1 - k) + a(k - 1)Right side: k(b - a) - k^2 u + k^2 aBring all terms to left side:u(1 - k) + a(k - 1) - k(b - a) + k^2 u - k^2 a = 0Factor u terms:u(1 - k + k^2) + a(k - 1 - k^2) - k b + k a = 0Wait, this seems messy. Maybe I should solve for b in terms of u and v first.From equation 2: u - v = k(b - v)So, b = (u - v)/k + vSimilarly, from equation 3: c = u + k(v - u)So, c = u + k v - k u = u(1 - k) + k vNow, let me express b and c in terms of u and v.So, b = (u - v)/k + v = u/k - v/k + v = u/k + v(1 - 1/k)Similarly, c = u(1 - k) + k vNow, let me compute b - a and c - a.From equation 1: v = a + k(u - a)So, u - a = (v - a)/kTherefore, u = a + (v - a)/kLet me substitute u into the expressions for b and c.First, b:b = u/k + v(1 - 1/k)= [a + (v - a)/k]/k + v(1 - 1/k)= a/k + (v - a)/k^2 + v - v/kSimilarly, c:c = u(1 - k) + k v= [a + (v - a)/k](1 - k) + k v= a(1 - k) + (v - a)(1 - k)/k + k vLet me simplify b first:b = a/k + (v - a)/k^2 + v - v/k= a/k + v/k^2 - a/k^2 + v - v/kCombine like terms:= (a/k - a/k^2) + (v/k^2 + v - v/k)= a(1/k - 1/k^2) + v(1/k^2 + 1 - 1/k)Similarly, c:c = a(1 - k) + (v - a)(1 - k)/k + k v= a(1 - k) + (v - a)(1 - k)/k + k vLet me expand (v - a)(1 - k)/k:= v(1 - k)/k - a(1 - k)/kSo, c becomes:= a(1 - k) + v(1 - k)/k - a(1 - k)/k + k vCombine like terms:= a(1 - k - (1 - k)/k) + v( (1 - k)/k + k )Simplify the coefficients:For a:= a[ (1 - k) - (1 - k)/k ] = a[ (k(1 - k) - (1 - k))/k ] = a[ (k - k^2 - 1 + k)/k ] = a[ (2k - k^2 - 1)/k ]For v:= v[ (1 - k)/k + k ] = v[ (1 - k + k^2)/k ]So, c = a(2k - k^2 - 1)/k + v(1 - k + k^2)/kSimilarly, for b:b = a(1/k - 1/k^2) + v(1/k^2 + 1 - 1/k)Let me simplify the coefficients:For a:= a(1/k - 1/k^2) = a(k - 1)/k^2For v:= v(1/k^2 + 1 - 1/k) = v(1 + (1 - k)/k^2) = v( (k^2 + 1 - k)/k^2 )So, b = a(k - 1)/k^2 + v(k^2 + 1 - k)/k^2Now, let me express b - a and c - a.First, b - a:= [a(k - 1)/k^2 + v(k^2 + 1 - k)/k^2] - a= a(k - 1)/k^2 - a + v(k^2 + 1 - k)/k^2= a[ (k - 1)/k^2 - 1 ] + v(k^2 + 1 - k)/k^2= a[ (k - 1 - k^2)/k^2 ] + v(k^2 + 1 - k)/k^2Similarly, c - a:= [a(2k - k^2 - 1)/k + v(1 - k + k^2)/k ] - a= a(2k - k^2 - 1)/k - a + v(1 - k + k^2)/k= a[ (2k - k^2 - 1)/k - 1 ] + v(1 - k + k^2)/k= a[ (2k - k^2 - 1 - k)/k ] + v(1 - k + k^2)/k= a[ (k - k^2 - 1)/k ] + v(1 - k + k^2)/kNow, let me compare b - a and c - a.From b - a:= a(k - 1 - k^2)/k^2 + v(k^2 + 1 - k)/k^2From c - a:= a(k - k^2 - 1)/k + v(1 - k + k^2)/kNotice that the coefficients of a and v in c - a are k times the coefficients in b - a.Specifically:c - a = k * [a(k - 1 - k^2)/k^2 + v(k^2 + 1 - k)/k^2 ] = k(b - a)So, c - a = k(b - a)This is a crucial relation. It tells us that the vector from a to c is k times the vector from a to b.In complex numbers, this implies that the transformation from b - a to c - a is a multiplication by k, which represents a rotation and scaling. Since k is the same as the ratio we had before, this suggests that triangle ABC is similar to triangle AUV.But let me make sure. Since c - a = k(b - a), this means that the side CA is k times the side AB, and the angle between them is determined by the argument of k.Similarly, in triangle AUV, we have v - a = k(u - a), so side VA is k times side AU, with the same angle.Therefore, triangle ABC has sides AB, BC, CA, and triangle AUV has sides AU, UV, VA. The ratios of corresponding sides are equal (k), and the angles are preserved because the multiplication by k includes the rotation.Hence, by the definition of similarity, triangle ABC is directly similar to triangle AUV, and since AUV is similar to VBU and UVC, triangle ABC is similar to all of them.I think that's the gist of it. I had to go through expressing b and c in terms of u, v, and k, then showing that the sides of ABC relate by the same ratio k, which confirms similarity.

💡Okay, so I have this problem where I need to prove that for any natural numbers ( n ) and ( p ), the equation[left(frac{L_{n}+sqrt{5} F_{n}}{2}right)^{p}=frac{L_{n p}+sqrt{5} F_{n p}}{2}]holds true. First, I need to recall what Fibonacci numbers and Lucas numbers are. I know that Fibonacci numbers start with ( F_1 = 1 ), ( F_2 = 1 ), and each subsequent number is the sum of the two preceding ones. So, ( F_{n+1} = F_n + F_{n-1} ). Similarly, Lucas numbers start with ( L_1 = 1 ), ( L_2 = 3 ), and follow the same recurrence relation: ( L_{n+1} = L_n + L_{n-1} ).I remember that both Fibonacci and Lucas numbers can be expressed using the golden ratio, which is ( alpha = frac{1 + sqrt{5}}{2} ) and its conjugate ( beta = frac{1 - sqrt{5}}{2} ). These are the roots of the quadratic equation ( x^2 - x - 1 = 0 ), which is the characteristic equation for both sequences.For Fibonacci numbers, the closed-form expression is known as Binet's formula:[F_n = frac{alpha^n - beta^n}{sqrt{5}}]And for Lucas numbers, the closed-form is:[L_n = alpha^n + beta^n]Okay, so now I can express both ( F_n ) and ( L_n ) in terms of ( alpha ) and ( beta ). Let me try to plug these into the left-hand side of the equation I need to prove.The left-hand side is:[left(frac{L_n + sqrt{5} F_n}{2}right)^p]Substituting the expressions for ( L_n ) and ( F_n ):[left(frac{alpha^n + beta^n + sqrt{5} cdot frac{alpha^n - beta^n}{sqrt{5}}}{2}right)^p]Simplifying inside the brackets:The ( sqrt{5} ) and ( frac{1}{sqrt{5}} ) cancel out, so:[left(frac{alpha^n + beta^n + (alpha^n - beta^n)}{2}right)^p]Combine like terms:[left(frac{2alpha^n}{2}right)^p = (alpha^n)^p = alpha^{np}]So, the left-hand side simplifies to ( alpha^{np} ).Now, let's look at the right-hand side:[frac{L_{np} + sqrt{5} F_{np}}{2}]Again, substituting the closed-form expressions:[frac{alpha^{np} + beta^{np} + sqrt{5} cdot frac{alpha^{np} - beta^{np}}{sqrt{5}}}{2}]Simplify inside the brackets:The ( sqrt{5} ) and ( frac{1}{sqrt{5}} ) cancel out:[frac{alpha^{np} + beta^{np} + (alpha^{np} - beta^{np})}{2}]Combine like terms:[frac{2alpha^{np}}{2} = alpha^{np}]So, the right-hand side also simplifies to ( alpha^{np} ).Since both sides simplify to the same expression, ( alpha^{np} ), the original equation holds true.Wait, but let me double-check if I missed anything. I substituted the closed-form expressions correctly, right? For Fibonacci, it's ( frac{alpha^n - beta^n}{sqrt{5}} ), and for Lucas, it's ( alpha^n + beta^n ). Yes, that seems correct.When I substituted into the left-hand side, I had ( L_n + sqrt{5} F_n ), which became ( alpha^n + beta^n + sqrt{5} cdot frac{alpha^n - beta^n}{sqrt{5}} ). The ( sqrt{5} ) cancels out, leaving ( alpha^n + beta^n + alpha^n - beta^n ), which simplifies to ( 2alpha^n ). Dividing by 2 gives ( alpha^n ), and raising it to the power ( p ) gives ( alpha^{np} ).On the right-hand side, substituting ( L_{np} + sqrt{5} F_{np} ) gives ( alpha^{np} + beta^{np} + sqrt{5} cdot frac{alpha^{np} - beta^{np}}{sqrt{5}} ). Again, the ( sqrt{5} ) cancels, resulting in ( alpha^{np} + beta^{np} + alpha^{np} - beta^{np} ), which simplifies to ( 2alpha^{np} ). Dividing by 2 gives ( alpha^{np} ).So, both sides are equal to ( alpha^{np} ), which means the original equation is indeed true for any natural numbers ( n ) and ( p ).I think that covers it. Maybe I should consider if there are any special cases or if the properties hold for all ( n ) and ( p ). Since both Fibonacci and Lucas numbers are defined for all natural numbers and the closed-form expressions hold for all ( n ), this proof should be general.Another thing to consider is whether the operations I performed are valid. Specifically, raising ( alpha^n ) to the power ( p ) is straightforward because exponents add when you multiply like bases. So, ( (alpha^n)^p = alpha^{np} ), which is correct.Also, the cancellation of ( sqrt{5} ) terms seems valid because multiplying ( sqrt{5} ) by ( frac{1}{sqrt{5}} ) gives 1, effectively removing those terms. I don't see any errors in the steps I took, so I feel confident that this proof is correct.

💡Okay, so I have this problem here: I need to find the number of positive integers ( n < 3^8 ) such that the number of positive integers ( k ) (where ( 1 leq k leq frac{n}{3} )) for which ( frac{n!}{(n-3k)! cdot k! cdot 3^{k+1}} ) is not an integer is exactly 216. Hmm, that's a mouthful. Let me try to break it down.First, I need to understand what the expression ( frac{n!}{(n-3k)! cdot k! cdot 3^{k+1}} ) represents. It looks like a combination formula multiplied by some power of 3. Maybe it's related to multinomial coefficients or something similar. Let me rewrite it to see if that helps:[frac{n!}{(n-3k)! cdot k! cdot 3^{k+1}} = binom{n}{3k} cdot frac{(3k)!}{k! cdot 3^{k+1}}]Okay, so it's the binomial coefficient ( binom{n}{3k} ) multiplied by ( frac{(3k)!}{k! cdot 3^{k+1}} ). I wonder if this simplifies further or if there's a combinatorial interpretation.The problem states that this expression is not an integer for exactly 216 values of ( k ). So, I need to find all ( n < 3^8 ) such that exactly 216 values of ( k ) make this fraction non-integer.I recall that for a fraction to be an integer, the denominator must divide the numerator. So, ( (n-3k)! cdot k! cdot 3^{k+1} ) must divide ( n! ). Therefore, the expression is not an integer if ( (n-3k)! cdot k! cdot 3^{k+1} ) does not divide ( n! ).Maybe I can use Legendre's formula here, which tells us the exponent of a prime ( p ) in the factorial of a number. Legendre's formula says that the exponent of a prime ( p ) in ( m! ) is ( sum_{i=1}^{infty} leftlfloor frac{m}{p^i} rightrfloor ). Since we're dealing with powers of 3, let's focus on the exponent of 3 in the numerator and denominator.Let me denote ( v_3(m!) ) as the exponent of 3 in ( m! ). Then, the exponent of 3 in the numerator ( n! ) is ( v_3(n!) ), and in the denominator, it's ( v_3((n-3k)!) + v_3(k!) + (k+1) ).So, the expression is not an integer if:[v_3(n!) - v_3((n-3k)!) - v_3(k!) - (k+1) < 0]Which simplifies to:[v_3(n!) - v_3((n-3k)!) - v_3(k!) < k + 1]Hmm, that seems a bit complicated. Maybe there's a better way to think about this. I remember that in combinatorics, the number of carries when adding numbers in base 3 affects the divisibility of binomial coefficients. Specifically, Lucas' theorem tells us that ( binom{n}{m} ) is divisible by 3 if and only if there is a carry when adding ( m ) and ( n - m ) in base 3.But in this case, we're dealing with ( binom{n}{3k} ) and an additional factor. Maybe I should think about the base-3 representation of ( n ) and ( 3k ).Let me write ( n ) in base 3. Suppose ( n = a_0 + a_1 cdot 3 + a_2 cdot 3^2 + dots + a_s cdot 3^s ), where each ( a_i ) is 0, 1, or 2. Similarly, ( 3k ) in base 3 would be ( k ) shifted left by one digit, so if ( k = b_0 + b_1 cdot 3 + dots + b_t cdot 3^t ), then ( 3k = b_0 cdot 3 + b_1 cdot 3^2 + dots + b_t cdot 3^{t+1} ).When we compute ( binom{n}{3k} ), according to Lucas' theorem, this binomial coefficient is not divisible by 3 if and only if each digit of ( 3k ) in base 3 is less than or equal to the corresponding digit of ( n ). Since ( 3k ) has zeros in the units place and the digits of ( k ) shifted left, this condition translates to the digits of ( k ) (in base 3) being less than or equal to the corresponding digits of ( n ) except for the units digit, which is zero.But wait, in our case, we have an additional factor ( frac{(3k)!}{k! cdot 3^{k+1}} ). Let me analyze this part. The term ( frac{(3k)!}{k! cdot 3^{k+1}} ) can be rewritten as ( frac{(3k)!}{k! cdot 3^{k} cdot 3} = frac{(3k)!}{k! cdot 3^{k}} cdot frac{1}{3} ).The term ( frac{(3k)!}{k! cdot 3^{k}} ) is known as the multinomial coefficient for dividing into three groups of size ( k ), and it's always an integer. However, dividing by 3 again might make it non-integer. So, the entire expression ( frac{n!}{(n-3k)! cdot k! cdot 3^{k+1}} ) is not an integer if either ( binom{n}{3k} ) is not divisible by 3 or if the additional division by 3 makes it non-integer.Wait, but ( frac{(3k)!}{k! cdot 3^{k}} ) is an integer, so dividing by 3 again would mean that the entire expression is not an integer only if ( binom{n}{3k} ) is not divisible by 3, right? Because if ( binom{n}{3k} ) is divisible by 3, then the division by 3 would still leave an integer.So, putting it all together, the expression ( frac{n!}{(n-3k)! cdot k! cdot 3^{k+1}} ) is not an integer if and only if ( binom{n}{3k} ) is not divisible by 3. Therefore, the number of ( k ) such that the expression is not an integer is equal to the number of ( k ) such that ( binom{n}{3k} ) is not divisible by 3.Now, according to Lucas' theorem, ( binom{n}{m} ) is not divisible by 3 if and only if in the base-3 representations of ( n ) and ( m ), each digit of ( m ) is less than or equal to the corresponding digit of ( n ). So, in our case, ( m = 3k ), which in base 3 has zeros in the units place and the digits of ( k ) shifted left.Therefore, ( binom{n}{3k} ) is not divisible by 3 if and only if each digit of ( k ) in base 3 is less than or equal to the corresponding digit of ( n ) (excluding the units digit, which is zero). So, the number of such ( k ) is equal to the product of (each digit of ( n ) in base 3 plus 1), except for the units digit, which is fixed at zero.Wait, let me clarify that. If ( n ) in base 3 is ( a_0 a_1 a_2 dots a_s ), then ( 3k ) in base 3 is ( 0 b_0 b_1 dots b_t ), where ( b_i ) are the digits of ( k ). For ( binom{n}{3k} ) to not be divisible by 3, each ( b_i ) must be less than or equal to ( a_{i+1} ). Therefore, the number of valid ( k ) is the product of ( (a_1 + 1)(a_2 + 1) dots (a_s + 1) ).But in our problem, the number of such ( k ) is given as 216. So, we need to find all ( n < 3^8 ) such that the product of ( (a_1 + 1)(a_2 + 1) dots (a_7 + 1) ) equals 216.Wait, ( n < 3^8 ), so ( n ) can be represented in base 3 with up to 8 digits. Let me denote ( n ) as ( a_0 a_1 a_2 dots a_7 ) in base 3, where each ( a_i ) is 0, 1, or 2. Then, the number of ( k ) such that ( binom{n}{3k} ) is not divisible by 3 is ( (a_1 + 1)(a_2 + 1) dots (a_7 + 1) ).Given that this product equals 216, we need to find all possible combinations of ( a_1, a_2, dots, a_7 ) such that their increments multiply to 216. Let's factorize 216 to understand the possible combinations.216 factors into ( 2^3 cdot 3^3 ). So, we need to express 216 as a product of integers greater than or equal to 1, where each integer is ( a_i + 1 ) and ( a_i ) is 0, 1, or 2. Therefore, each ( a_i + 1 ) can be 1, 2, or 3.So, we need to find the number of ways to write 216 as a product of seven factors, each being 1, 2, or 3. Since 216 = 2^3 * 3^3, we need to distribute these exponents among the seven factors.Each factor can contribute to the exponents of 2 and 3. Let's think of each factor as a pair (x, y), where x is the exponent of 2 and y is the exponent of 3 in that factor. Since each factor is 1, 2, or 3, their exponents are:- 1: (0, 0)- 2: (1, 0)- 3: (0, 1)We need the total exponents across all seven factors to be (3, 3). So, we need to distribute 3 exponents of 2 and 3 exponents of 3 among the seven factors, where each factor can contribute at most 1 exponent of 2 and at most 1 exponent of 3.This is equivalent to choosing 3 positions out of 7 to contribute an exponent of 2, and 3 positions out of 7 to contribute an exponent of 3. However, a factor cannot contribute both an exponent of 2 and 3 because 2 and 3 are coprime, and the factors are 1, 2, or 3. So, a factor can be either 2, contributing to the exponent of 2, or 3, contributing to the exponent of 3, or 1, contributing nothing.Therefore, we need to choose 3 positions for the factor 2 and 3 positions for the factor 3, ensuring that these positions are distinct. The remaining positions will be 1.The number of ways to choose 3 positions for 2 and 3 positions for 3 is ( binom{7}{3} times binom{4}{3} ). Let me compute that:( binom{7}{3} = 35 ) and ( binom{4}{3} = 4 ), so total ways are ( 35 times 4 = 140 ).Wait, but hold on. The problem states that ( n < 3^8 ), which means ( n ) can have up to 8 digits in base 3, but in our earlier analysis, we considered only 7 digits (from ( a_1 ) to ( a_7 )). The units digit ( a_0 ) doesn't affect the number of ( k ) because ( 3k ) has a zero in the units place. So, ( a_0 ) can be 0, 1, or 2 independently.Therefore, for each of the 140 combinations of ( a_1 ) to ( a_7 ), ( a_0 ) can be 0, 1, or 2. So, the total number of ( n ) is ( 140 times 3 = 420 ).But wait, the problem says "the number of positive integer ( n < 3^8 )". Since ( 3^8 = 6561 ), and ( n ) is less than that, ( n ) can indeed have up to 8 digits in base 3, including the units digit. However, in our analysis, we considered ( a_0 ) separately because it doesn't affect the number of ( k ). So, our total count is 420.But let me double-check. If we have 7 digits (from ( a_1 ) to ( a_7 )) contributing to the product, and each can be 0, 1, or 2, but their increments multiply to 216, which we found corresponds to 140 combinations. Then, ( a_0 ) can be 0, 1, or 2, so 3 choices. Therefore, total ( n ) is 140 * 3 = 420.Yes, that seems correct. So, the number of positive integers ( n < 3^8 ) satisfying the given condition is 420.

💡Okay, so I have this problem with four propositions about power functions, and I need to figure out which ones are correct. Let me take each one step by step.First, let me recall what a power function is. A power function is generally of the form ( y = x^a ), where ( a ) is a real number. The behavior of these functions can vary a lot depending on the exponent ( a ). So, I need to analyze each proposition based on different values of ( a ).**Proposition ①: Power functions are increasing functions in the first quadrant.**Hmm, the first quadrant is where both ( x ) and ( y ) are positive. So, if I consider ( x > 0 ), is ( y = x^a ) always increasing?Well, let's think about different exponents. If ( a = 2 ), then ( y = x^2 ) is indeed increasing for ( x > 0 ). But what if ( a = -1 )? Then ( y = x^{-1} = frac{1}{x} ), which is actually decreasing in the first quadrant because as ( x ) increases, ( y ) decreases. So, not all power functions are increasing in the first quadrant. Therefore, proposition ① is incorrect.**Proposition ②: The graph of a power function passes through the points (0, 0) and (1, 1).**Let me check this. For ( x = 1 ), ( y = 1^a = 1 ) regardless of ( a ), so it does pass through (1, 1). But what about (0, 0)? If ( a > 0 ), then ( y = 0^a = 0 ), so it passes through (0, 0). However, if ( a leq 0 ), especially if ( a = 0 ), ( y = x^0 = 1 ) for all ( x neq 0 ), which doesn't pass through (0, 0). If ( a < 0 ), like ( a = -1 ), then ( y = x^{-1} ) is undefined at ( x = 0 ), so it doesn't pass through (0, 0) either. Therefore, proposition ② is only true when ( a > 0 ), but since power functions can have any real exponent, it's not always true. So, proposition ② is incorrect.**Proposition ③: If the power function ( y = x^a ) is an odd function, then ( y = x^a ) is an increasing function on its domain.**Okay, an odd function satisfies ( f(-x) = -f(x) ). For ( y = x^a ) to be odd, ( (-x)^a = -x^a ). This implies that ( (-1)^a = -1 ). So, ( (-1)^a = -1 ) when ( a ) is an odd integer. For example, ( a = 1, 3, 5 ), etc. So, if ( a ) is an odd integer, then ( y = x^a ) is an odd function.Now, is such a function always increasing on its domain? Let's take ( a = 1 ): ( y = x ) is indeed increasing everywhere. What about ( a = 3 ): ( y = x^3 ) is also increasing everywhere because its derivative ( 3x^2 ) is always non-negative. But wait, what about ( a = -1 ): ( y = x^{-1} = frac{1}{x} ). Is this function increasing? Let's check its derivative: ( y' = -frac{1}{x^2} ), which is always negative for ( x neq 0 ). So, ( y = x^{-1} ) is actually decreasing on its domain. But wait, is ( y = x^{-1} ) an odd function? Let me check: ( (-x)^{-1} = -frac{1}{x} = -y ), so yes, it is odd. So, here's a case where the power function is odd but not increasing. Therefore, proposition ③ is incorrect.**Proposition ④: The graph of a power function cannot appear in the fourth quadrant.**The fourth quadrant is where ( x > 0 ) and ( y < 0 ). For a power function ( y = x^a ), if ( x > 0 ), then ( y ) will be positive if ( a ) is any real number because any positive number raised to any power is still positive. Wait, is that true?Let me think. If ( a ) is positive, ( x^a ) is positive. If ( a ) is negative, ( x^a = frac{1}{x^{-a}} ), which is still positive. If ( a = 0 ), ( x^0 = 1 ), which is positive. So, for all real numbers ( a ), ( x > 0 ) implies ( y > 0 ). Therefore, the graph of a power function cannot be in the fourth quadrant where ( y ) is negative. So, proposition ④ is correct.**Summary:**- Proposition ①: Incorrect (depends on the exponent).- Proposition ②: Incorrect (only true for ( a > 0 )).- Proposition ③: Incorrect (counterexample with ( a = -1 )).- Proposition ④: Correct.So, only proposition ④ is correct.**Final Answer**The correct proposition number is boxed{④}.