🧠:<think>Alright, let me try to tackle this problem step by step. So, we have a country with 2001 towns connected by roads, each town has at least one road, but no town is connected to every other town. That means the graph isn't complete, right? We are given the concept of a dominant set ( D ), which is a set of towns such that every town not in ( D ) is connected by a road to some town in ( D ). And we know that every dominant set has at least ( k ) towns. Our goal is to prove that the country can be partitioned into ( 2001 - k ) republics where no two towns in the same republic are connected by a road.Hmm, okay. So, in graph theory terms, we're dealing with a graph ( G ) with 2001 vertices, each with at least degree 1, and no vertex is connected to all others. A dominant set is like a dominating set, which is a well-known concept in graph theory. The size of the smallest dominating set is at least ( k ). We need to show that the chromatic number of the graph, which is the minimum number of colors needed to color the graph such that no two adjacent vertices share the same color, is at most ( 2001 - k ).Wait, is that right? Because if we can color the graph with ( 2001 - k ) colors, then each color class is an independent set, which would correspond to a republic where no two towns are connected by a road. So, partitioning into ( 2001 - k ) republics is equivalent to coloring the graph with ( 2001 - k ) colors.So, maybe I can use some known relationship between the size of the dominating set and the chromatic number? Let me think.I remember that in graph theory, there's a relation between the chromatic number and the size of the dominating set. Specifically, I think that the chromatic number plus the size of the minimum dominating set is less than or equal to the number of vertices in the graph. So, if ( chi(G) ) is the chromatic number and ( gamma(G) ) is the minimum dominating set size, then ( chi(G) + gamma(G) leq |V(G)| ). Is that true? Let me try to recall or derive it. Suppose ( G ) is a graph with ( n ) vertices, and ( gamma(G) ) is the minimum size of a dominating set. If we have a dominating set ( D ) of size ( gamma(G) ), then the remaining ( n - gamma(G) ) vertices are all adjacent to at least one vertex in ( D ). So, we can color the vertices in ( D ) with one color each, and then color the remaining vertices with the same colors, ensuring that no two adjacent vertices share the same color.Wait, that might not directly give the relation. Maybe I need to think differently. Perhaps consider that if we have a dominating set ( D ), then the subgraph induced by ( V(G) setminus D ) has a certain structure that can be colored with fewer colors.Alternatively, maybe we can use induction or some other method. Let me try to think of a proof for the lemma that ( chi(G) + gamma(G) leq n ), where ( n = |V(G)| ).Suppose ( G ) is a connected graph (if it's disconnected, we can consider each component separately). Let ( chi(G) = c ). Then, there exists a coloring of ( G ) with ( c ) colors. Now, let's partition the vertex set into color classes ( S_1, S_2, ldots, S_c ), where each ( S_i ) is an independent set.Now, let's look at the sizes of these color classes. The idea is that the smallest color class can be used as a dominating set. Suppose that ( S_1 ) is the smallest color class. Then, every vertex not in ( S_1 ) must be adjacent to a vertex in ( S_1 ), otherwise, we could recolor those vertices to reduce the number of colors, which would contradict the minimality of ( c ).Wait, is that right? If a vertex not in ( S_1 ) is not adjacent to any vertex in ( S_1 ), then we could assign it the color of ( S_1 ) without any conflict, which would mean we don't need a separate color for it, reducing the chromatic number. So, indeed, every vertex not in ( S_1 ) must be adjacent to some vertex in ( S_1 ). Therefore, ( S_1 ) is a dominating set.So, the size of ( S_1 ) is at least ( gamma(G) ). But ( |S_1| leq frac{n}{c} ) because it's the smallest color class. Wait, that doesn't directly give us the relation. Maybe I need to consider the sum of the sizes or something else.Alternatively, let's consider that ( |S_1| + 2|S_2| + cdots + c|S_c| ) is minimized. Then, similar to the argument above, ( S_1 ) must be a dominating set. So, ( |S_1| geq gamma(G) ). But how does this relate to the chromatic number?Perhaps another approach. If ( D ) is a dominating set of size ( gamma(G) ), then the graph ( G ) can be partitioned into ( D ) and its neighborhood. The neighborhood can be colored with ( chi(G[D]) ) colors, where ( G[D] ) is the subgraph induced by ( D ). But I'm not sure if that helps directly.Wait, maybe considering that the chromatic number of ( G ) is at most ( |V(G)| - gamma(G) ). If we can show that, then we are done because ( gamma(G) geq k ), so ( chi(G) leq 2001 - k ).But how to show that ( chi(G) leq n - gamma(G) )? Let me think again about the relationship between the chromatic number and the dominating set.Suppose we have a dominating set ( D ) of size ( gamma(G) ). Then, every vertex not in ( D ) is adjacent to at least one vertex in ( D ). So, if we color all vertices in ( D ) with distinct colors, and then color the remaining vertices with colors not used by their neighbors in ( D ), we might be able to reduce the number of colors needed.But I need to make sure that this process doesn't require too many colors. Let's try to construct such a coloring.First, color all vertices in ( D ) with ( gamma(G) ) distinct colors. Now, for each vertex ( v ) not in ( D ), it is adjacent to at least one vertex in ( D ). So, ( v ) cannot be colored with any of the colors used by its neighbors in ( D ). Since ( D ) is a dominating set, each ( v ) outside ( D ) has at least one neighbor in ( D ), so we need to assign a color to ( v ) that is different from the color of its neighbor in ( D ).If we have ( gamma(G) ) colors for ( D ), then for each ( v ) not in ( D ), we need to choose a color different from the color of its neighbor in ( D ). Since each ( v ) has only one neighbor in ( D ), we can assign a color to ( v ) from the remaining ( n - gamma(G) ) colors. Wait, but that might not be efficient because we can reuse colors across different parts of the graph.Alternatively, maybe instead of using ( gamma(G) ) colors for ( D ), we can color ( D ) with ( gamma(G) - 1 ) colors, and then use the remaining colors for the other vertices. But I'm not sure if that's possible because ( D ) itself might require ( gamma(G) ) colors if it's a complete graph.Wait, but in our problem, no town is connected to every other town, so ( D ) cannot be a complete graph, right? Because if ( D ) were complete, then each vertex in ( D ) would be connected to all others in ( D ), but since no town is connected to every other town, ( D ) cannot be complete if ( |D| geq 2 ). So, maybe ( D ) is not a complete graph, which means its chromatic number is less than ( |D| ).Hmm, this is getting a bit complicated. Maybe I should look for an existing theorem or lemma that directly relates the chromatic number and the minimum dominating set size.After a quick search in my mind, I recall that there is a lemma which states that for any graph ( G ), ( chi(G) + gamma(G) leq n + 1 ), but that might not be tight enough for our purposes. Wait, in our case, since the graph isn't complete, maybe we can get a better bound.Alternatively, perhaps using the concept of independent sets. If we can find an independent set of size at least ( k ), then we can subtract that from the total and proceed recursively. But I'm not sure if that's the right approach.Wait, going back to the earlier idea. If we have a dominating set ( D ) of size ( gamma(G) ), then ( D ) is a dominating set, so every vertex not in ( D ) is adjacent to at least one vertex in ( D ). So, if we color ( D ) with ( gamma(G) ) colors, and then color the remaining vertices with colors different from their neighbors in ( D ), perhaps we can bound the total number of colors.But I need to ensure that this doesn't require more than ( n - k ) colors. Since ( gamma(G) geq k ), and ( n = 2001 ), if I can show that the chromatic number is at most ( n - gamma(G) ), which would be ( 2001 - k ), then we're done.So, maybe the key idea is that the chromatic number plus the minimum dominating set size is at most ( n ). If that's the case, then ( chi(G) leq n - gamma(G) ), which would give us the desired result.But I need to verify this lemma. Let's consider a graph ( G ) with ( n ) vertices, and suppose ( gamma(G) = k ). Then, if ( chi(G) + k leq n ), we have ( chi(G) leq n - k ).Is this always true? Let's consider some examples. Take a complete graph ( K_n ). The minimum dominating set is 1, and the chromatic number is ( n ). So, ( chi(G) + gamma(G) = n + 1 ), which is ( n + 1 ). But in our problem, the graph isn't complete, so maybe the bound is tighter.In our problem, since no town is connected to every other town, the graph isn't complete, so perhaps for such graphs, ( chi(G) + gamma(G) leq n ).Let me try to think of a proof for this. Suppose ( G ) is a graph where no vertex is adjacent to all others, and ( gamma(G) geq k ). We need to show ( chi(G) leq n - k ).Consider the complement graph ( overline{G} ). The independence number of ( G ) is the clique number of ( overline{G} ). Maybe that's not helpful.Alternatively, think about the fact that since no vertex is adjacent to all others, the maximum degree of ( G ) is at most ( n - 2 ). So, the graph is not a star graph, which might have a high degree.Wait, but the maximum degree being ( n - 2 ) doesn't necessarily help with the chromatic number. For example, a complete graph missing one edge still has chromatic number ( n - 1 ).Hmm, maybe I need to use the concept of a kernel in graph theory. A kernel is an independent dominating set. If the graph has a kernel, then the chromatic number is at least 2 because the kernel is an independent set, and the rest can be colored accordingly.But I don't think all graphs have kernels. It's a well-known result that every directed graph has a kernel, but for undirected graphs, it's not necessarily true.Wait, perhaps another approach. If we can find a large independent set, then we can partition the graph into independent sets.But how large can an independent set be? In general, the size of the largest independent set is at least ( n / (Delta + 1) ), where ( Delta ) is the maximum degree. But since ( Delta leq n - 2 ), this gives a lower bound of ( n / (n - 1) ), which is less than 2, so not helpful.Alternatively, maybe use the fact that the graph is not complete to deduce something about its chromatic number. If the graph is not complete, then its chromatic number is at most ( n - 1 ). But we need a better bound.Wait, going back to the lemma idea. If we can show that ( chi(G) + gamma(G) leq n ), then we're done. Let's try to sketch a proof.Assume ( G ) is connected (if it's disconnected, consider each component). Let ( chi(G) = c ). Consider a coloring of ( G ) with ( c ) colors. Let ( S_1, S_2, ldots, S_c ) be the color classes. Suppose ( |S_1| leq |S_2| leq ldots leq |S_c| ). Then, ( S_1 ) is a dominating set because if there's a vertex not in ( S_1 ) not adjacent to any vertex in ( S_1 ), we could recolor it with color 1, reducing the number of colors, which contradicts minimality.Thus, ( S_1 ) is a dominating set, so ( |S_1| geq gamma(G) ). But the sum of the sizes of all color classes is ( n ). So, ( |S_1| + |S_2| + ldots + |S_c| = n ). Since ( |S_1| geq gamma(G) ), we have ( |S_2| + ldots + |S_c| leq n - gamma(G) ). But ( |S_2| + ldots + |S_c| geq c - 1 ) because each ( |S_i| geq 1 ).Wait, maybe that's not directly helpful. Alternatively, since ( |S_1| geq gamma(G) ), and ( c leq n - gamma(G) ), because ( |S_1| + (c - 1) leq n ) (since each ( |S_i| geq 1 )), so ( gamma(G) + (c - 1) leq n ), which simplifies to ( c leq n - gamma(G) ). So, ( chi(G) leq n - gamma(G) ).Yes, that seems to work. So, if ( G ) is connected, then ( chi(G) + gamma(G) leq n ). For disconnected graphs, we can apply this to each component, and the total chromatic number would be the maximum chromatic number among the components, and the total minimum dominating set size would be the sum of the minimum dominating sets of each component. Hmm, but in our problem, the graph isn't necessarily connected, but the condition on the dominating set is global, so maybe it still holds.Wait, actually, the problem states that each dominant set consists of at least ( k ) towns. So, the minimum dominating set size ( gamma(G) geq k ). Therefore, ( chi(G) leq n - gamma(G) leq n - k ).Thus, the chromatic number is at most ( 2001 - k ), which means we can partition the graph into ( 2001 - k ) independent sets, corresponding to ( 2001 - k ) republics where no two towns in the same republic are connected by a road.So, the key idea was to use the relationship between the chromatic number and the minimum dominating set size, leveraging the fact that the graph isn't complete and each dominating set has a certain size. This allows us to bound the chromatic number and thus prove the desired partition.Final Answerboxed{2001 - k}</think>

🧠:<think>Alright, so I have this inequality to prove: for every positive (a) and (b), (frac{(a+b)^2}{2} + frac{a+b}{4} ge asqrt{b} + bsqrt{a}). Hmm, okay. Let me try to figure out how to approach this. First, I should probably understand what each part of the inequality represents. The left side has two terms: (frac{(a+b)^2}{2}) and (frac{a+b}{4}). The right side is (asqrt{b} + bsqrt{a}). I need to show that the left side is always greater than or equal to the right side for any positive (a) and (b).Maybe I can start by expanding the left side to see if it simplifies into something more manageable. Let's see:[frac{(a+b)^2}{2} = frac{a^2 + 2ab + b^2}{2}]So, the entire left side becomes:[frac{a^2 + 2ab + b^2}{2} + frac{a + b}{4}]To combine these, I can find a common denominator, which is 4:[frac{2a^2 + 4ab + 2b^2 + a + b}{4}]Hmm, that's a bit messy. Maybe I can rearrange the terms:[frac{2a^2 + a + 4ab + 2b^2 + b}{4}]But I'm not sure if that helps. Maybe I should think about the right side, (asqrt{b} + bsqrt{a}). That looks a bit like the AM-GM inequality, which states that for non-negative numbers, the arithmetic mean is greater than or equal to the geometric mean. Wait, if I consider (a) and (b), the AM-GM inequality would say:[frac{a + b}{2} ge sqrt{ab}]But I have (asqrt{b} + bsqrt{a}) on the right side, which is a bit different. Maybe I can manipulate that expression. Let's see:[asqrt{b} + bsqrt{a} = sqrt{a}cdot a + sqrt{b}cdot b = a^{3/2} + b^{3/2}]Hmm, not sure if that helps. Maybe I can think of (asqrt{b}) as (sqrt{a}cdot a + sqrt{b}cdot b), but I'm not sure.Alternatively, perhaps I can use the Cauchy-Schwarz inequality. The Cauchy-Schwarz inequality states that for any real numbers, ((x_1y_1 + x_2y_2)^2 le (x_1^2 + x_2^2)(y_1^2 + y_2^2)). Maybe I can apply that here.Let me set (x_1 = sqrt{a}), (x_2 = sqrt{b}), and (y_1 = sqrt{b}), (y_2 = sqrt{a}). Then, the left side of Cauchy-Schwarz would be:[(sqrt{a}cdot sqrt{b} + sqrt{b}cdot sqrt{a})^2 = (2sqrt{ab})^2 = 4ab]The right side would be:[((sqrt{a})^2 + (sqrt{b})^2)((sqrt{b})^2 + (sqrt{a})^2) = (a + b)^2]So, Cauchy-Schwarz gives us:[4ab le (a + b)^2]Which simplifies to:[(a + b)^2 ge 4ab]That's a known inequality, so that's good. But how does that relate to our original inequality? Well, the left side of our original inequality has (frac{(a+b)^2}{2}), which is half of ((a + b)^2), and we know that ((a + b)^2 ge 4ab), so:[frac{(a + b)^2}{2} ge 2ab]Okay, that's something. But the right side of our original inequality is (asqrt{b} + bsqrt{a}), which we need to compare to (2ab). Hmm, not sure if that's directly helpful.Maybe I can think of (asqrt{b} + bsqrt{a}) in terms of AM-GM. Let's see:Consider the terms (asqrt{b}) and (bsqrt{a}). Their arithmetic mean is:[frac{asqrt{b} + bsqrt{a}}{2}]And their geometric mean is:[sqrt{asqrt{b} cdot bsqrt{a}} = sqrt{ab cdot sqrt{ab}} = (ab)^{3/4}]But I'm not sure how that helps either.Wait, maybe I can bound (asqrt{b} + bsqrt{a}) from above using the terms in the left side of the inequality. Let's think about that.Looking back at the left side:[frac{(a+b)^2}{2} + frac{a+b}{4}]I can write this as:[frac{a^2 + 2ab + b^2}{2} + frac{a + b}{4}]Let me see if I can find a way to relate this to (asqrt{b} + bsqrt{a}). Maybe I can use the AM-GM inequality on some parts of this expression.For example, consider (a^2) and (b^2). By AM-GM:[frac{a^2 + b^2}{2} ge ab]Which is true, but I'm not sure if that directly helps. Alternatively, maybe I can consider the terms (a^2) and (ab). Let's see:[frac{a^2 + ab}{2} ge sqrt{a^2 cdot ab} = a sqrt{ab}]Similarly, for (b^2) and (ab):[frac{b^2 + ab}{2} ge sqrt{b^2 cdot ab} = b sqrt{ab}]Adding these two inequalities together:[frac{a^2 + ab}{2} + frac{b^2 + ab}{2} ge a sqrt{ab} + b sqrt{ab}]Simplifying the left side:[frac{a^2 + 2ab + b^2}{2} ge sqrt{ab}(a + b)]So,[frac{(a + b)^2}{2} ge sqrt{ab}(a + b)]Dividing both sides by (a + b) (since (a + b > 0)):[frac{a + b}{2} ge sqrt{ab}]Which is just the AM-GM inequality again. Okay, but how does this help with our original inequality?Well, we have:[frac{(a + b)^2}{2} ge sqrt{ab}(a + b)]And our left side of the original inequality is (frac{(a + b)^2}{2} + frac{a + b}{4}), which is greater than or equal to (sqrt{ab}(a + b) + frac{a + b}{4}).So, if I can show that:[sqrt{ab}(a + b) + frac{a + b}{4} ge asqrt{b} + bsqrt{a}]Then, combining this with the previous inequality, we would have:[frac{(a + b)^2}{2} + frac{a + b}{4} ge sqrt{ab}(a + b) + frac{a + b}{4} ge asqrt{b} + bsqrt{a}]So, let's see if we can show that:[sqrt{ab}(a + b) + frac{a + b}{4} ge asqrt{b} + bsqrt{a}]Hmm, this seems a bit circular. Maybe I need a different approach.Let me consider substituting variables to simplify the expression. Let me set (x = sqrt{a}) and (y = sqrt{b}). Then, (a = x^2) and (b = y^2). Substituting into the inequality:Left side:[frac{(x^2 + y^2)^2}{2} + frac{x^2 + y^2}{4}]Right side:[x^2 y + y^2 x = xy(x + y)]So, the inequality becomes:[frac{(x^2 + y^2)^2}{2} + frac{x^2 + y^2}{4} ge xy(x + y)]Hmm, not sure if that helps, but let's see. Maybe I can expand the left side:[frac{x^4 + 2x^2 y^2 + y^4}{2} + frac{x^2 + y^2}{4} = frac{x^4}{2} + x^2 y^2 + frac{y^4}{2} + frac{x^2}{4} + frac{y^2}{4}]Combine like terms:[frac{x^4}{2} + frac{x^2}{4} + x^2 y^2 + frac{y^4}{2} + frac{y^2}{4}]Not sure if that helps. Maybe I can think of this as a function in terms of (x) and (y) and see if it's always non-negative.Alternatively, maybe I can consider the difference between the left side and the right side and show that it's always non-negative.Let me define:[D = frac{(a+b)^2}{2} + frac{a+b}{4} - (asqrt{b} + bsqrt{a})]I need to show that (D ge 0) for all positive (a) and (b).Let me try to manipulate (D):[D = frac{a^2 + 2ab + b^2}{2} + frac{a + b}{4} - asqrt{b} - bsqrt{a}]Combine the terms:[D = frac{a^2 + 2ab + b^2}{2} + frac{a + b}{4} - asqrt{b} - bsqrt{a}]Let me write all terms over a common denominator of 4:[D = frac{2a^2 + 4ab + 2b^2 + a + b}{4} - frac{4asqrt{b} + 4bsqrt{a}}{4}]Combine the fractions:[D = frac{2a^2 + 4ab + 2b^2 + a + b - 4asqrt{b} - 4bsqrt{a}}{4}]So, to show that (D ge 0), it's sufficient to show that the numerator is non-negative:[2a^2 + 4ab + 2b^2 + a + b - 4asqrt{b} - 4bsqrt{a} ge 0]Hmm, this looks complicated. Maybe I can group terms in a way that allows me to complete the square or factor.Let me group the (a^2) and (b^2) terms together, and the (ab) terms, and so on:[2a^2 + 2b^2 + 4ab + a + b - 4asqrt{b} - 4bsqrt{a}]Hmm, perhaps I can write this as:[2(a^2 + 2ab + b^2) + a + b - 4asqrt{b} - 4bsqrt{a}]Which is:[2(a + b)^2 + a + b - 4asqrt{b} - 4bsqrt{a}]Still not obvious. Maybe I can think of (a + b) as a single variable. Let me set (s = a + b) and (p = ab). But I'm not sure if that helps because the terms involve square roots.Alternatively, maybe I can use the AM-GM inequality on some of these terms. Let's see:Consider (2a^2) and (2b^2). By AM-GM:[frac{2a^2 + 2b^2}{2} ge sqrt{2a^2 cdot 2b^2} = 2ab]Which gives:[2a^2 + 2b^2 ge 4ab]Okay, that's another known inequality. So, replacing (2a^2 + 2b^2) with (4ab), the numerator becomes:[4ab + 4ab + a + b - 4asqrt{b} - 4bsqrt{a} = 8ab + a + b - 4asqrt{b} - 4bsqrt{a}]Hmm, not sure if that helps. Maybe I can factor out some terms:[8ab + a + b - 4asqrt{b} - 4bsqrt{a} = a(8b + 1 - 4sqrt{b}) + b(8a + 1 - 4sqrt{a})]Not sure if that's useful. Maybe I can think of (8ab) as (4ab + 4ab). Let's try:[8ab = 4ab + 4ab]So, the numerator becomes:[4ab + 4ab + a + b - 4asqrt{b} - 4bsqrt{a}]Now, let's group terms:[(4ab + a) + (4ab + b) - 4asqrt{b} - 4bsqrt{a}]Factor out (a) from the first group and (b) from the second group:[a(4b + 1) + b(4a + 1) - 4asqrt{b} - 4bsqrt{a}]Hmm, maybe I can write this as:[a(4b + 1 - 4sqrt{b}) + b(4a + 1 - 4sqrt{a})]Still not obvious. Maybe I can consider each term separately. Let's look at (a(4b + 1 - 4sqrt{b})). Let me set (t = sqrt{b}), so (b = t^2). Then, the term becomes:[a(4t^2 + 1 - 4t) = a(4t^2 - 4t + 1) = a(2t - 1)^2]Oh, that's interesting! So, (4b + 1 - 4sqrt{b} = (2sqrt{b} - 1)^2). Similarly, (4a + 1 - 4sqrt{a} = (2sqrt{a} - 1)^2).So, substituting back, the numerator becomes:[a(2sqrt{b} - 1)^2 + b(2sqrt{a} - 1)^2]Since squares are always non-negative, and (a) and (b) are positive, each term is non-negative. Therefore, the entire numerator is non-negative:[a(2sqrt{b} - 1)^2 + b(2sqrt{a} - 1)^2 ge 0]Thus, (D ge 0), which means:[frac{(a+b)^2}{2} + frac{a+b}{4} ge asqrt{b} + bsqrt{a}]And that's what we needed to prove!</think>

🧠:<think>Okay, so I've got this equation: (x² - x - 6)^(x + 4) = 1. Hmm, I need to find how many integer values of x satisfy this. Alright, let me think about this step by step.First, I remember that for something like a^b = 1, there are a few cases where this can happen. One is when the base a is 1 because 1 raised to any power is still 1. Another case is when the exponent b is 0 because any non-zero number raised to the power of 0 is 1. Also, if the base a is -1 and the exponent b is even, then (-1)^b would be 1. So, these are the scenarios I need to consider.Alright, let's break it down.Case 1: Base is 1So, if x² - x - 6 = 1, then we have:x² - x - 6 = 1 => x² - x - 7 = 0I need to solve this quadratic equation. Using the quadratic formula:x = [1 ± √(1 + 28)] / 2 => x = [1 ± √29] / 2Hmm, √29 is approximately 5.385, so the solutions are (1 + 5.385)/2 ≈ 3.192 and (1 - 5.385)/2 ≈ -2.192. These aren't integers, so no solutions here.Case 2: Base is -1 and exponent is evenSo, if x² - x - 6 = -1, then:x² - x - 6 = -1 => x² - x - 5 = 0Again, applying the quadratic formula:x = [1 ± √(1 + 20)] / 2 => x = [1 ± √21] / 2√21 is approximately 4.583, so the solutions are (1 + 4.583)/2 ≈ 2.791 and (1 - 4.583)/2 ≈ -1.791. Again, not integers. So, no solutions here either.But wait, even if the base is -1, we need the exponent to be even. So, let's check if x + 4 is even for these x values. But since these x aren't integers, it doesn't matter. So, no solutions in this case.Case 3: Exponent is 0If the exponent x + 4 = 0, then:x + 4 = 0 => x = -4Now, let's check if the base is non-zero because 0^0 is undefined. So, plugging x = -4 into the base:(-4)² - (-4) - 6 = 16 + 4 - 6 = 1414 is not zero, so 14^0 = 1, which works. So, x = -4 is a valid solution.Case 4: Base is any number but exponent is 0Wait, I think I already covered that in Case 3. So, that's one solution.Case 5: Base is 0 and exponent is positiveWait, if the base is 0, then 0 raised to any positive exponent is 0, which isn't 1. If the exponent is 0, it's undefined. So, if x² - x - 6 = 0, let's solve that:x² - x - 6 = 0 => (x - 3)(x + 2) = 0 => x = 3 or x = -2So, x = 3 and x = -2.Now, let's check the exponent for these x values.For x = 3:Exponent = 3 + 4 = 7, which is positive. So, 0^7 = 0 ≠ 1. So, x = 3 is not a solution.For x = -2:Exponent = -2 + 4 = 2, which is positive. So, 0^2 = 0 ≠ 1. So, x = -2 is not a solution either.Hmm, so even though the base is 0, since the exponent is positive, it doesn't give us 1. So, these aren't solutions.Wait, but what if the exponent is 0? For x = 3, exponent is 7, which isn't 0. For x = -2, exponent is 2, also not 0. So, in this case, x = 3 and x = -2 don't work because the exponent isn't 0, and 0 raised to a positive exponent is 0, not 1.So, maybe these aren't valid solutions after all.Wait, let me double-check. If x² - x - 6 = 0, then (x² - x - 6)^(x + 4) becomes 0^(x + 4). For this to be 1, we need 0^(x + 4) = 1. But 0 raised to any positive exponent is 0, and 0^0 is undefined. So, unless x + 4 = 0, but then x = -4. But x = -4 makes the base 14, not 0. So, no solution from this case.So, x = 3 and x = -2 don't work because the exponent is positive, making the result 0, not 1.Case 6: Maybe base is -1 and exponent is even?Wait, I think I covered that in Case 2, but let me check again.If the base is -1, then (-1)^b = 1 if b is even.So, if x² - x - 6 = -1, then we have x² - x - 5 = 0, which we already solved and found non-integer roots. So, no solutions here.But just to make sure, let me plug in some integer values near the roots I found.For x = 2: x² - x - 6 = 4 - 2 - 6 = -4, which is not -1.For x = 3: 9 - 3 - 6 = 0, which is not -1.For x = -1: 1 + 1 - 6 = -4, not -1.For x = -2: 4 + 2 - 6 = 0, not -1.So, no integer x makes the base -1.Case 7: Maybe base is some other number that when raised to an exponent gives 1?Hmm, besides 1, -1 with even exponent, 0 with positive exponent is 0, which doesn't help. What about other numbers?Wait, any number raised to the power of 0 is 1, so if the exponent is 0, regardless of the base (as long as base is not 0, because 0^0 is undefined), then the expression is 1.So, in this case, exponent x + 4 = 0 => x = -4. We already checked that x = -4 gives base 14, so 14^0 = 1, which works. So, x = -4 is a solution.Case 8: Maybe base is -1 and exponent is even?Wait, I think I covered that in Case 2, but let me check again.If the base is -1, then (-1)^b = 1 if b is even.So, if x² - x - 6 = -1, then we have x² - x - 5 = 0, which we already solved and found non-integer roots. So, no solutions here.But just to make sure, let me plug in some integer values near the roots I found.For x = 2: x² - x - 6 = 4 - 2 - 6 = -4, which is not -1.For x = 3: 9 - 3 - 6 = 0, which is not -1.For x = -1: 1 + 1 - 6 = -4, not -1.For x = -2: 4 + 2 - 6 = 0, not -1.So, no integer x makes the base -1.Case 9: Maybe base is another root of unity?Wait, but since we're dealing with integers, the base has to be an integer, and the only integers that are roots of unity are 1 and -1. So, no other possibilities there.Case 10: Maybe the base is a fraction that equals 1 when raised to some power?But since we're looking for integer x, the base x² - x - 6 must be an integer, and the exponent x + 4 is also an integer. So, if the base is a fraction, it would have to be 1 or -1 to give 1 when raised to an integer power. But we've already covered those cases.So, summarizing:- From Case 1: No integer solutions.- From Case 2: No integer solutions.- From Case 3: x = -4 is a solution.- From Case 4: No solutions.- From Case 5: No solutions.- From Case 6: No solutions.- From Case 7: x = -4 is a solution.- From Case 8: No solutions.- From Case 9: No solutions.- From Case 10: No solutions.Wait, but earlier I thought x = 3 and x = -2 might be solutions, but then realized they aren't because the exponent is positive, making the expression 0, not 1.But hold on, maybe I missed something. Let me check x = 3 and x = -2 again.For x = 3:(x² - x - 6)^(x + 4) = (9 - 3 - 6)^(7) = (0)^7 = 0 ≠ 1For x = -2:(x² - x - 6)^(x + 4) = (4 + 2 - 6)^(2) = (0)^2 = 0 ≠ 1So, indeed, they don't work.Wait, but what about x = -3?Let me check x = -3:Base: (-3)^2 - (-3) - 6 = 9 + 3 - 6 = 6Exponent: -3 + 4 = 1So, 6^1 = 6 ≠ 1x = -1:Base: 1 + 1 - 6 = -4Exponent: -1 + 4 = 3(-4)^3 = -64 ≠ 1x = 0:Base: 0 - 0 - 6 = -6Exponent: 0 + 4 = 4(-6)^4 = 1296 ≠ 1x = 1:Base: 1 - 1 - 6 = -6Exponent: 1 + 4 = 5(-6)^5 = -7776 ≠ 1x = 2:Base: 4 - 2 - 6 = -4Exponent: 2 + 4 = 6(-4)^6 = 4096 ≠ 1x = 4:Base: 16 - 4 - 6 = 6Exponent: 4 + 4 = 86^8 = a big number, definitely not 1.x = 5:Base: 25 - 5 - 6 = 14Exponent: 5 + 4 = 914^9 is way bigger than 1.x = -5:Base: 25 + 5 - 6 = 24Exponent: -5 + 4 = -124^(-1) = 1/24 ≠ 1x = -6:Base: 36 + 6 - 6 = 36Exponent: -6 + 4 = -236^(-2) = 1/1296 ≠ 1So, none of these x values give us 1 except x = -4.Wait, but earlier I thought x = 3 and x = -2 might be solutions because the base is 0, but the exponent is positive, so 0 raised to positive exponent is 0, not 1. So, they don't work.But wait, what if the exponent is 0? For x = -4, exponent is 0, which works because 14^0 = 1.Wait, is there any other x that makes the exponent 0? Only x = -4.So, so far, I have only x = -4 as a solution.But wait, earlier I thought x = 3 and x = -2 might be solutions, but they aren't.Wait, but let me think again. If the base is 0, then 0^exponent is 0 only if exponent is positive. If exponent is 0, it's undefined. So, in this case, if x = 3 or x = -2, the base is 0, but the exponent is 7 and 2 respectively, both positive. So, 0^7 = 0 and 0^2 = 0, neither of which is 1. So, they don't work.So, only x = -4 works.But wait, let me think again. Maybe I missed something.Wait, what if the base is 1? We checked that, but the solutions weren't integers.What about the base being -1 with even exponent? We checked that, but no integer solutions.What about the exponent being 0? Only x = -4 works.So, only one solution: x = -4.But wait, the answer options are A)1, B)2, C)3, D)4, E)5.Wait, but the assistant initially thought there were three solutions: x = 3, -2, -4. But after checking, only x = -4 works.Wait, maybe I made a mistake in the initial thought process. Let me check.Wait, in the initial thought process, the assistant considered x = 3 and x = -2 as solutions because the base is 0, but in reality, 0 raised to any positive exponent is 0, not 1. So, those shouldn't be considered solutions.So, only x = -4 works, which would make the answer A)1.But wait, let me double-check.Wait, in the initial thought process, the assistant considered x = 3 and x = -2 as solutions because the base is 0, but in reality, 0 raised to any positive exponent is 0, not 1. So, those shouldn't be considered solutions.So, only x = -4 works, making the answer A)1.But wait, in the initial thought process, the assistant found three solutions: x = 3, -2, -4, but upon closer inspection, only x = -4 works.So, maybe the correct answer is 1, which is option A.But wait, let me check x = -4 again.x = -4:Base: (-4)^2 - (-4) - 6 = 16 + 4 - 6 = 14Exponent: -4 + 4 = 0So, 14^0 = 1, which is correct.x = 3:Base: 9 - 3 - 6 = 0Exponent: 70^7 = 0 ≠ 1x = -2:Base: 4 + 2 - 6 = 0Exponent: 20^2 = 0 ≠ 1So, indeed, only x = -4 works.Wait, but in the initial thought process, the assistant concluded three solutions, but upon re-evaluation, only one works. So, maybe the assistant made a mistake.Alternatively, maybe I missed something.Wait, maybe there's another case where the base is a non-integer that somehow equals 1 when raised to an integer exponent. But since x is integer, the base must be integer, so that's not possible.Alternatively, maybe the base is -1 and exponent is even, but we already saw that x² - x - 6 = -1 leads to non-integer x.Wait, unless x is such that x² - x - 6 = -1 and x + 4 is even, but since x isn't integer, that doesn't help.So, perhaps the only solution is x = -4.But wait, let me check x = 0:Base: 0 - 0 - 6 = -6Exponent: 4(-6)^4 = 1296 ≠ 1x = 1:Base: 1 - 1 - 6 = -6Exponent: 5(-6)^5 = -7776 ≠ 1x = 2:Base: 4 - 2 - 6 = -4Exponent: 6(-4)^6 = 4096 ≠ 1x = -1:Base: 1 + 1 - 6 = -4Exponent: 3(-4)^3 = -64 ≠ 1x = -3:Base: 9 + 3 - 6 = 6Exponent: 16^1 = 6 ≠ 1x = -4:As before, works.x = -5:Base: 25 + 5 - 6 = 24Exponent: -124^-1 = 1/24 ≠ 1x = -6:Base: 36 + 6 - 6 = 36Exponent: -236^-2 = 1/1296 ≠ 1x = 4:Base: 16 - 4 - 6 = 6Exponent: 86^8 = large number ≠ 1x = 5:Base: 25 - 5 - 6 = 14Exponent: 914^9 ≠ 1x = 6:Base: 36 - 6 - 6 = 24Exponent: 1024^10 ≠ 1So, indeed, only x = -4 works.Therefore, the number of integer solutions is 1, which is option A.</think>

🧠:<think>Okay, so I have this problem about guests sitting around a round table with n seats. Each guest has an invitation number from 1 to n. The waiter is serving them in a specific order: first, he picks any guest and places them anywhere. Then, for each subsequent guest, he moves around the table a number of seats equal to the previous guest's invitation number and serves the next guest there. The goal is to figure out for which values of n this is possible.Let me try to break this down.First, I think it's helpful to visualize the table as a circle with n seats, each labeled from 0 to n-1 for simplicity, maybe. The waiter starts by placing one guest anywhere, say seat 0. Then, the next guest is placed by moving a number of seats equal to the previous guest's invitation number. So if the first guest was number k, the next guest is placed at seat (0 + k) mod n. Then, the next guest is placed by moving k seats from there, and so on.Wait, but the problem says the waiter selects the guests in an order, not necessarily the order of their invitation numbers. So maybe the sequence of guests is chosen such that each subsequent guest is placed by moving the number of seats equal to the previous guest's invitation number.So, starting with guest A, then from A's seat, move A's number of seats to place guest B, then from B's seat, move B's number of seats to place guest C, etc.The key is that each move is determined by the previous guest's invitation number. So the challenge is to arrange the guests in such an order that all seats get filled without repetition before all guests are seated.This sounds a bit like a permutation problem, where we need a permutation of the numbers 1 to n such that starting from any point, moving the specified number of seats each time, we visit every seat exactly once.Wait, actually, since the table is circular, this is similar to constructing a permutation where each step is determined by the current element's value. This is reminiscent of a permutation generated by a specific kind of mapping.I think this is similar to a concept in mathematics called a "complete mapping" or perhaps a "derangement," but I'm not entirely sure. Maybe it's more related to cyclic permutations or something.Let me think about small cases.Case n=1: Trivial, only one guest. So n=1 works.Case n=2: Two guests. Let's say guest 1 and guest 2. If we start with guest 1, then the next guest must be placed 1 seat away from guest 1. Since it's a round table, moving 1 seat from guest 1 would take us to the only other seat. So we place guest 2 there. Then, starting from guest 2, moving 2 seats would bring us back to guest 2's seat, which is already occupied. So we can't place anyone else, but n=2 only has two guests, so it's done. Wait, but does that work? So starting with guest 1, then moving 1 seat to place guest 2, and that's all. So n=2 works.Wait, actually, if you start with guest 2, then moving 2 seats would bring you back to the starting seat, so you can't place guest 1 because it's already taken. So starting with guest 2 doesn't work, but starting with guest 1 does. So n=2 is possible.Case n=3: Three guests, 1,2,3.Let's try starting with guest 1. Then next seat is (current +1) mod 3, so seat 1. Place guest 2 there. Next, from seat 1, move 2 seats: (1+2)=3 mod 3 is 0. So seat 0 is already taken by guest 1. So we can't place guest 3. That's a problem.Alternatively, starting with guest 2. From seat 0, move 2 seats: seat 2. Place guest 3 there. Then from seat 2, move 3 seats: (2+3)=5 mod 3=2. Already taken. So stuck.Starting with guest 3. From seat 0, move 3 seats: 0+3=3 mod 3=0. Already taken. So stuck.Wait, is there another way? Maybe not starting at seat 0? Wait, no, the first guest can be placed anywhere, but since the table is circular, it's equivalent to starting at seat 0.Hmm, seems like n=3 doesn't work.Wait, maybe I made a mistake. Let me try again.Starting with guest 1 at seat 0. Next, move 1 seat to seat 1, place guest 2. Then, from seat 1, move 2 seats to seat 0, which is already taken. So no good.Starting with guest 2 at seat 0. Next, move 2 seats to seat 2, place guest 3. Then from seat 2, move 3 seats: 2+3=5 mod 3=2, which is already taken. So stuck.Starting with guest 3 at seat 0. Next, move 3 seats: 0+3=3 mod 3=0, already taken. So no.So n=3 doesn't work.Case n=4: Four guests, 1,2,3,4.Let me try starting with guest 1 at seat 0. Next, move 1 seat to seat 1, place guest 2. From seat 1, move 2 seats to seat 3, place guest 3. From seat 3, move 3 seats: 3+3=6 mod 4=2, place guest 4. From seat 2, move 4 seats: 2+4=6 mod 4=2, already taken. So stuck, but all guests are seated except for seat 2. Wait, no, guests 1,2,3,4 are placed at 0,1,3,2, which covers all seats. Wait, so that works? Let me check:Start at seat 0: guest 1.Move 1 seat to seat 1: guest 2.Move 2 seats from seat 1: seat 3: guest 3.Move 3 seats from seat 3: 3+3=6 mod 4=2: guest 4.Now, all seats are filled. So n=4 works.Wait, but in my initial attempt, I thought it didn't work, but maybe I messed up. So n=4 works.Case n=5: Let's see.Starting with guest 1 at seat 0.Move 1 seat: seat 1: guest 2.Move 2 seats: seat 3: guest 3.Move 3 seats: seat 6 mod 5=1: already taken. So stuck.Alternatively, starting with guest 2 at seat 0.Move 2 seats: seat 2: guest 3.Move 3 seats: seat 5 mod 5=0: already taken. So stuck.Starting with guest 3 at seat 0.Move 3 seats: seat 3: guest 4.Move 4 seats: seat 7 mod 5=2: guest 5.Move 5 seats: seat 7 mod 5=2: already taken. So stuck.Wait, perhaps another order.Alternatively, starting with guest 1, then 3, then 4, etc.Wait, maybe I need to try a different permutation.But this might take too long. Maybe there's a pattern.From n=1: works.n=2: works.n=3: doesn't work.n=4: works.n=5: doesn't work? Maybe.Wait, is there a pattern? n even works, n odd doesn't?Wait, n=1 is odd and works. Hmm, but n=1 is trivial.Wait, n=3 and n=5 don't work, n=4 works.Wait, maybe n is a power of 2?Wait, n=4 is a power of 2. Let's check n=6.n=6: Let's try.Starting with guest 1 at seat 0.Move 1: seat 1: guest 2.Move 2: seat 3: guest 3.Move 3: seat 6 mod 6=0: already taken. So stuck.Alternatively, starting with guest 2 at seat 0.Move 2: seat 2: guest 3.Move 3: seat 5: guest 4.Move 4: seat 9 mod 6=3: guest 5.Move 5: seat 8 mod 6=2: already taken. So stuck.Alternatively, starting with guest 3 at seat 0.Move 3: seat 3: guest 4.Move 4: seat 7 mod 6=1: guest 5.Move 5: seat 6 mod 6=0: already taken. So stuck.Alternatively, starting with guest 4 at seat 0.Move 4: seat 4: guest 5.Move 5: seat 9 mod 6=3: guest 6.Move 6: seat 9 mod 6=3: already taken. So stuck.Wait, maybe starting with a different guest.Alternatively, starting with guest 5 at seat 0.Move 5: seat 5: guest 6.Move 6: seat 11 mod 6=5: already taken. So stuck.Hmm, n=6 doesn't seem to work either.Wait, maybe n=4 is special because it's a power of 2. Let's try n=8.But before that, maybe I should think about the general case.Perhaps the problem is related to whether n is a power of 2. Because in n=4, it worked, and n=8 might work.Alternatively, maybe it's related to whether n is even or odd.Wait, n=2 works, n=4 works, n=6 doesn't. So maybe it's not just even numbers.Wait, n=6 didn't work, so it's not all even numbers. Hmm.Alternatively, maybe n must be a power of 2.Wait, n=8: Let's try.Starting with guest 1 at seat 0.Move 1: seat 1: guest 2.Move 2: seat 3: guest 3.Move 3: seat 6: guest 4.Move 4: seat 10 mod 8=2: guest 5.Move 5: seat 7: guest 6.Move 6: seat 13 mod 8=5: guest 7.Move 7: seat 12 mod 8=4: guest 8.Now, all seats are filled: 0,1,3,6,2,7,5,4.Wait, that seems to work. So n=8 works.Wait, but n=6 didn't work. So maybe n must be a power of 2.Wait, n=1 is 2^0, n=2 is 2^1, n=4 is 2^2, n=8 is 2^3. So maybe n must be a power of 2.But wait, let me check n=5 again.n=5:If I start with guest 3.Seat 0: guest 3.Move 3: seat 3: guest 4.Move 4: seat 7 mod 5=2: guest 5.Move 5: seat 7 mod 5=2: already taken.Wait, stuck.Alternatively, starting with guest 5.Seat 0: guest 5.Move 5: seat 5 mod 5=0: already taken. Stuck.Hmm.Alternatively, maybe starting with guest 2.Seat 0: guest 2.Move 2: seat 2: guest 3.Move 3: seat 5 mod 5=0: already taken. Stuck.Alternatively, starting with guest 4.Seat 0: guest 4.Move 4: seat 4: guest 5.Move 5: seat 9 mod 5=4: already taken. Stuck.Hmm.Alternatively, maybe the problem is that for odd n, the permutation doesn't cover all seats because of some parity issue.Wait, in the case of n=4, which is even, the permutation worked. For n=8, which is also even, it worked. But n=2 worked, n=6 didn't. Hmm.Wait, maybe it's not just even numbers, but numbers where n is a power of 2.Because n=6 is not a power of 2, and it didn't work, while n=4 and n=8, being powers of 2, worked.But n=2 is also a power of 2, and it worked.Wait, but n=1 is 2^0, which worked, but it's trivial.So maybe the answer is that n must be a power of 2.But I need to verify this.Alternatively, maybe the problem is related to the concept of a "complete mapping" in combinatorics, where a permutation f of the elements of a group satisfies that the mapping g(x) = f(x) + x is also a permutation.In this case, the group is the additive group modulo n, and the permutation is determined by the steps.Wait, if we think of the permutation as a sequence where each step is determined by the previous element, then it's similar to a permutation where the differences are the elements themselves.Wait, more formally, if we have a permutation π of {0,1,2,...,n-1}, then π_{i+1} ≡ π_i + a_{π_i} mod n, where a_{π_i} is the value of the guest at position π_i.But in our problem, the permutation is determined by the guest's numbers, so perhaps it's more about the permutation of guests rather than seats.Wait, maybe it's better to model this as a graph where each node is a seat, and each node has an edge to the node obtained by moving a_i seats, where a_i is the guest's number.Then, the problem reduces to finding a Hamiltonian cycle in this graph.But the graph is directed, and each node has out-degree 1, since from each seat, moving a_i seats leads to exactly one other seat.Wait, but in our case, the guest's number determines the step, so the edges are determined by the permutation of guests.Wait, perhaps not. Maybe the guests are assigned to seats, and each guest's number determines the step to the next guest's seat.So, if we have a permutation of guests, then the movement is determined by their numbers.Wait, I'm getting a bit confused.Alternatively, maybe the problem is similar to the concept of a "permutation cycle." If the permutation of guests can be arranged such that the movement steps form a single cycle covering all seats, then it's possible.So, for the permutation to form a single cycle, the step sizes must satisfy certain conditions.In particular, if the step sizes are such that the greatest common divisor (gcd) of the step sizes and n is 1, then the cycle can cover all seats.Wait, but in our case, the step sizes are the guest's numbers, which are 1 to n. So, if n is a power of 2, then the step sizes include 1, which has gcd 1 with n, so it can generate all seats.Wait, but in n=4, we saw that starting with guest 1, we can traverse all seats. Similarly, in n=8, starting with guest 1, we can traverse all seats.But in n=6, starting with guest 1, we get stuck.Wait, maybe the key is that if n is a power of 2, then the permutation can be arranged such that the steps generate a single cycle.But I'm not entirely sure.Alternatively, maybe the problem is related to the concept of a "complete mapping" where the permutation and its differences are both permutations.In this case, the permutation of guests would have to satisfy that the differences (steps) are also a permutation, which is our case.But I'm not sure about the exact conditions for the existence of such permutations.Wait, I recall that complete mappings exist for abelian groups of odd order, but not necessarily for all groups. But I'm not sure.Alternatively, maybe the problem is related to the concept of a "Langford sequence" or something similar, but I'm not sure.Wait, perhaps I should look for a pattern.From the cases I tried:n=1: works.n=2: works.n=3: doesn't work.n=4: works.n=5: doesn't work.n=6: doesn't work.n=8: works.So it seems that n=1,2,4,8 work, which are powers of 2.Therefore, maybe the answer is that n must be a power of 2.But I need to confirm this.Wait, let me try n=7.n=7: Starting with guest 1.Seat 0: guest 1.Move 1: seat 1: guest 2.Move 2: seat 3: guest 3.Move 3: seat 6: guest 4.Move 4: seat 10 mod 7=3: already taken. So stuck.Alternatively, starting with guest 2.Seat 0: guest 2.Move 2: seat 2: guest 3.Move 3: seat 5: guest 4.Move 4: seat 9 mod 7=2: already taken. Stuck.Starting with guest 3.Seat 0: guest 3.Move 3: seat 3: guest 4.Move 4: seat 7 mod 7=0: already taken. Stuck.Starting with guest 4.Seat 0: guest 4.Move 4: seat 4: guest 5.Move 5: seat 9 mod 7=2: guest 6.Move 6: seat 8 mod 7=1: guest 7.Move 7: seat 8 mod 7=1: already taken. So stuck.Hmm.Alternatively, starting with guest 5.Seat 0: guest 5.Move 5: seat 5: guest 6.Move 6: seat 11 mod 7=4: guest 7.Move 7: seat 11 mod 7=4: already taken. Stuck.Starting with guest 6.Seat 0: guest 6.Move 6: seat 6: guest 7.Move 7: seat 13 mod 7=6: already taken. Stuck.Starting with guest 7.Seat 0: guest 7.Move 7: seat 7 mod 7=0: already taken. Stuck.So n=7 doesn't work.Similarly, n=9.Wait, maybe it's too time-consuming, but from the pattern, it seems that only powers of 2 work.Therefore, the answer is that n must be a power of 2.But wait, n=6 is even but not a power of 2, and it didn't work. So the answer is that n must be a power of 2.But wait, n=1 is a power of 2, n=2,4,8, etc.So, to conclude, the waiter can serve all guests if and only if n is a power of 2.</think>