🧠:Okay, so the question is asking if there's a polygon without a center of symmetry that can be divided into two convex polygons, each of which does have a center of symmetry. Hmm, interesting. Let me try to break this down step by step.First, let me recall some definitions to make sure I'm on the right track. A center of symmetry (or point symmetry) for a polygon means that there's a point such that if you rotate the polygon 180 degrees around that point, it maps onto itself. So, every point P in the polygon has a corresponding point Q such that the center is the midpoint between P and Q.Now, the problem is about taking a polygon that doesn't have such a center and splitting it into two convex polygons, each of which does have their own centers of symmetry. So, the original polygon isn't symmetric around a point, but when you cut it into two parts, each part is symmetric.Let me start by considering simple cases. Maybe start with a polygon that's asymmetric but can be split into two symmetric convex parts. Let's think about shapes. A rectangle has a center of symmetry, but if I modify it... Wait, but the original polygon mustn't have a center. So, if I take two rectangles and join them in an asymmetric way, maybe?Wait, but the two convex polygons resulting from the division must each have their own centers. So, for example, if the original polygon is made by joining two rectangles at an angle, such that the overall shape isn't symmetric, but each rectangle is symmetric. But wait, rectangles are convex. So, maybe that's an example. But let me visualize this.Suppose I have two rectangles connected by a thin strip. If the connection is such that the overall shape is not symmetric, but each rectangle is. However, the problem states that the original polygon must be divided into two convex polygons. Wait, but if you split the original polygon into two convex parts, each of those parts must themselves be convex. So, if the original polygon is already made of two convex parts connected in a way that the whole is not convex, but each part is convex and symmetric.Wait, but the original polygon may not be convex, but when divided into two convex polygons, each has a center of symmetry.Alternatively, maybe take two different convex polygons with their own centers and join them in such a way that the combined polygon loses the overall symmetry.Let me try to think of an explicit example.Take a square (which has a center of symmetry) and a rectangle (which also has a center). If I attach them in a way that the combined shape doesn't have a center. For example, attach the square to the rectangle not at the center but off to one side. Then, the combined shape would be asymmetric. However, if we split it back into the square and rectangle, each has its own center.Wait, but the original polygon here is the combined shape. If we divide it into the square and rectangle, then each has a center. But in this case, the original polygon (the combined shape) is divided into two convex polygons (the square and rectangle). So, this seems like it would answer the question in the affirmative. But wait, does the combined shape have a center of symmetry? If the square is attached to one end of the rectangle, then probably not. Let me check.Suppose the rectangle is longer than the square. Let's say the rectangle is of length 4 units and the square is 2x2. If I attach the square to one end of the rectangle, the combined shape would look like a rectangle with a square extension on one end. The original combined shape's center of symmetry would be somewhere, but does it exist?To check symmetry, imagine rotating 180 degrees around the center. If the square is attached on the right end, after rotation, it should be on the left end. But since the original has it only on the right, unless there's a corresponding part on the left, it won't map onto itself. Therefore, the combined shape does not have a center of symmetry. So, if this combined shape is divided into the original square and rectangle, each of which has a center, then yes, this would satisfy the conditions.But wait, is the combined shape a single polygon? Yes, if they are connected. For example, if the square shares a side with the rectangle. Then, the combined figure is a single polygon. However, is this polygon convex? Wait, the problem doesn't state that the original polygon is convex. It just has to be a polygon. So, the combined shape could be non-convex. However, each of the two parts after division must be convex.Wait, in the example I thought of, the combined shape is a rectangle with a square attached to one end. If they share a full side, the resulting shape would be a non-convex polygon? Wait, no. If you attach a square to the end of a rectangle along one side, the resulting shape is actually still convex? Wait, no. Let me visualize this. If the rectangle is, say, horizontal, and the square is attached to the right end, then the overall shape would have all internal angles less than 180 degrees. Wait, the rectangle has four right angles, and attaching a square would just extend one side. The vertices would still have angles of 90 degrees. Wait, maybe it's still convex? Hmm, no, wait. If you have a rectangle and attach a square to one of its sides, the resulting figure is a larger rectangle. Wait, that can't be. For example, if the rectangle is 4x2 and the square is 2x2, attaching them along the 2-unit side would make a 6x2 rectangle, which is convex and has a center of symmetry. That's not good because the original combined shape would have a center.Wait, so that approach doesn't work. Because attaching a square to a rectangle along a side may just make a larger rectangle. So, in that case, the combined shape is symmetric. But I need the combined shape to be asymmetric.Maybe attach the square not along a full side but shifted. Wait, but polygons are made of straight sides. If I attach the square such that it's only partially overlapping? Wait, no. To form a polygon, edges have to be connected properly.Alternatively, attach the square to the rectangle at a corner. But then, that would create a non-convex polygon. For example, imagine a rectangle with a square sticking out from one corner. The resulting shape would be non-convex because the square would create an indentation. Wait, but actually, if you attach a square to a corner, you might end up with a star-like shape, but with right angles. Wait, perhaps.Wait, let me think again. Suppose we have a rectangle (A) and a square (B). If we place the square so that one of its corners is attached to a corner of the rectangle, but rotated. But that would create a non-convex polygon. However, the problem allows the original polygon to be non-convex. The key is that when you split it into two parts, each part is convex and has a center of symmetry.Wait, but if we attach the square to the rectangle in such a way that the combined shape is non-convex, but each individual part (the original rectangle and square) are convex and symmetric. Then, splitting the combined shape into the rectangle and square would satisfy the conditions.But here's the problem: If you attach the square to the rectangle in a way that the combined shape is a single polygon, then the square and rectangle must share an edge, not just a corner. Otherwise, it's two separate polygons. So, for them to form a single polygon, they need to share a common edge. So, for example, take a rectangle and a square, and attach them along one edge. But as I thought earlier, that just forms a larger rectangle, which is convex and has a center. So that's not helpful.Alternatively, maybe attach them at a vertex? But then you can't form a single polygon. Unless you create a non-convex shape.Wait, here's an idea. Take a rectangle and a square, and attach them such that the square is attached to the midpoint of one of the rectangle's sides. But that would create a T-shaped figure. Wait, but a T-shape is not a polygon because of the overlapping edges. No, actually, to make a polygon, the attachment must be such that the two shapes share a common edge without overlapping. Wait, perhaps if you have a rectangle and attach a square to one of its longer sides, but not along the entire side. Wait, but how? The square has four sides. If you attach part of a square's side to part of the rectangle's side, but that would require vertices in the middle of edges, which isn't allowed in polygons.Alternatively, think of a polygon made by connecting a rectangle and a square in a way that creates a non-convex shape. For example, take a rectangle and extend one of its sides into a square, but offset. Wait, maybe like a "L" shape. But an L-shape made of two rectangles. However, an L-shape is a polygon. Is the L-shape convex? No, it's concave because the internal angle at the joint is 270 degrees. So, the L-shape is a concave polygon. Does it have a center of symmetry? If the two arms of the L are equal in length, then yes, it might have a center. For example, a square missing a quarter would have symmetry. Wait, no. Let's see.An L-shape formed by two squares of equal size. If you rotate it 180 degrees, does it map onto itself? Yes, because each arm of the L would correspond to the other arm after rotation. So, the center of symmetry would be the point where the two arms meet. Wait, no. Wait, suppose the L-shape is made by a 2x2 square missing one 1x1 square, forming an L. Then, the center of the original 2x2 square is the center of symmetry. If you rotate 180 degrees, the missing square would be mapped to the opposite corner, but since it's missing, the L-shape would coincide with itself. Wait, actually, that L-shape does have a center of symmetry.Hmm, so maybe that's not a good example. Let's make an asymmetric L-shape. If the two arms of the L are of different lengths. For instance, one arm is a 3x1 rectangle, and the other is a 1x2 rectangle. Then, the overall L-shape would not have a center of symmetry because the arms are different lengths. However, if you split the L-shape into the two rectangles, each of those rectangles has a center of symmetry. So, in this case, the original L-shaped polygon does not have a center of symmetry, but when divided into the two rectangles (each convex and with their own centers), it satisfies the condition.Wait, is that correct? Let me check. Suppose the original polygon is an L-shape made of a horizontal 3x1 rectangle and a vertical 1x2 rectangle. So, the overall shape looks like an L with a longer horizontal arm and a shorter vertical arm. The center of symmetry for the original polygon would require that every point has a corresponding point 180 degrees rotated around the center. Since the horizontal arm is longer, rotating it 180 degrees would map the horizontal arm to a position where there's no corresponding arm of the same length. Hence, the original L-shape does not have a center of symmetry. However, if you split it along the corner where the two rectangles meet, each rectangle (3x1 and 1x2) has its own center of symmetry. Therefore, this would answer the question affirmatively.But wait, hold on. The problem states that the original polygon is divided into two convex polygons. In this case, the L-shape is divided into two rectangles, which are both convex. So, yes. Therefore, this example satisfies all the conditions: the original polygon (L-shape) has no center of symmetry, and it can be divided into two convex polygons (the two rectangles), each of which has a center of symmetry.But let me verify if the L-shape indeed lacks a center of symmetry. Suppose the center of the original L-shape is somewhere. If we pick a point in the horizontal rectangle, its 180-degree rotation should land in the vertical rectangle. But since the horizontal rectangle is longer, there might not be a corresponding point. For example, take the endpoint of the horizontal arm. Rotating 180 degrees around the supposed center should land on another endpoint. However, if the vertical arm is shorter, this might not be possible. Therefore, such a center doesn't exist.Alternatively, let's compute coordinates to check. Let's place the L-shape on a coordinate system. Suppose the horizontal rectangle goes from (0,0) to (3,1), and the vertical rectangle goes from (0,0) to (1,2). The combined polygon has vertices at (0,0), (3,0), (3,1), (1,1), (1,2), (0,2), (0,0). Wait, no, that's not correct. Let me actually plot this properly.Wait, if the horizontal rectangle is 3 units long (from x=0 to x=3) and 1 unit tall (y=0 to y=1). The vertical rectangle is 1 unit wide (x=0 to x=1) and 2 units tall (y=0 to y=2). The overlapping part is the square from (0,0) to (1,1). So, the combined shape has vertices: (0,0), (3,0), (3,1), (1,1), (1,2), (0,2), back to (0,0). So, this is the L-shape.Now, to check for a center of symmetry. Suppose there is a center point (h,k). For every vertex (x,y), there must be a corresponding vertex (2h - x, 2k - y) also in the polygon.Let's check the vertex (3,0). Its symmetric counterpart would be (2h - 3, 2k - 0). This point must also be a vertex. Looking at the list of vertices: (0,0), (3,0), (3,1), (1,1), (1,2), (0,2). The only candidates are the existing vertices. Let's see if (2h - 3, 2k) is one of the other vertices.Suppose (2h - 3, 2k) = (0,2). Then:2h - 3 = 0 => h = 1.52k = 2 => k = 1So, the center would be (1.5, 1). Let's check another vertex. Take (3,1). Its symmetric counterpart would be (2*1.5 - 3, 2*1 -1) = (0,1). But (0,1) is not a vertex. The vertices at y=1 are (3,1) and (1,1). So, (0,1) is not a vertex, which means that the center (1.5,1) doesn't work.Alternatively, check another vertex. Take (1,2). Its symmetric counterpart would be (2h -1, 2k -2). If the center is (1.5,1), then this would be (3 -1, 2 -2) = (2,0). But (2,0) is not a vertex. The vertices on the lower part are (0,0), (3,0). So, (2,0) isn't a vertex. Therefore, the center (1.5,1) doesn't work.Alternatively, maybe another center? Let's try another approach. Suppose the center is at (0.5,1). Then, the symmetric point of (3,0) would be (2*0.5 -3, 2*1 -0) = (-2, 2). Which is not a vertex. So that's not valid.Alternatively, let's check if any pair of vertices are symmetric. For example, (3,0) and some other vertex. If (3,0) maps to (0,2), then the center is midpoint between them: ((3+0)/2, (0+2)/2) = (1.5,1). As before, but as we saw, that doesn't map other vertices correctly.Another pair: (3,1) and (0,1). Their midpoint is (1.5,1). But (0,1) isn't a vertex.Alternatively, (1,2) and (2,0). Midpoint is (1.5,1). But (2,0) isn't a vertex.So, it seems there's no center of symmetry for this L-shaped polygon. Therefore, this example works. The original polygon (L-shape) has no center of symmetry, but it can be divided into two convex polygons (the 3x1 rectangle and the 1x2 rectangle), each of which has a center of symmetry.Therefore, the answer should be yes, such a polygon exists.But let me think if there's any possible flaw in this reasoning. Is the division into two convex polygons valid? The original L-shape can be divided along the line where the two rectangles meet, which is the rectangle from (0,0) to (1,1). But actually, when you split the L-shape into the two rectangles, you're cutting along a line that separates them. So, the two resulting polygons are the horizontal 3x1 rectangle and the vertical 1x2 rectangle, both of which are convex and have centers of symmetry.Therefore, this seems to satisfy all the conditions. So, the answer is yes, and an example is the L-shaped polygon made by joining a 3x1 rectangle and a 1x2 rectangle, which can be split into those two rectangles.Alternatively, another example could be a polygon composed of two different squares attached in an asymmetric way. But the L-shape example seems to work.Wait, just to be thorough, let me consider another example. Suppose you have a regular hexagon, which has a center of symmetry. But if you modify it... Hmm, maybe not. Alternatively, take a convex polygon and a non-convex polygon. Wait, no, the original can be non-convex, but the two parts after division must be convex.Another example: take a parallelogram and a rectangle. If you attach them in a way that the resulting shape is not symmetric. Wait, a parallelogram has a center of symmetry. If you attach a rectangle to it, the combined shape might lose the symmetry. For instance, attach a rectangle to one side of a parallelogram such that the attachment breaks the symmetry. Then, splitting them back would give the two original shapes, each with their own centers. But the combined shape may not have a center.But let's visualize this. Suppose we have a parallelogram (which has a center) and attach a rectangle to one of its sides. If the attachment is not at the midpoint, then the combined shape might not have a center. For example, take a rhombus (a type of parallelogram) with sides of length 2 and acute angles. Attach a rectangle of length 1 and width 0.5 to one of its sides, not centered. The resulting shape would have a protrusion on one side, making it asymmetric. However, if you split it into the rhombus and the rectangle, each has a center. But wait, the original shape here is formed by attaching a rectangle to a rhombus. If the attachment is along a full edge, then the resulting shape is still a single polygon. But would that polygon have a center?If the rhombus has a center, and the rectangle is attached off-center, the combined shape's center would be somewhere between the rhombus and the rectangle. However, rotating 180 degrees might not map the shape onto itself because the rectangle is attached asymmetrically. For example, if the rectangle is attached to the right side of the rhombus, then after rotation, it should be on the left, but since there's no rectangle on the left, it wouldn't match. Therefore, the combined shape lacks a center of symmetry. Dividing it into the rhombus and rectangle, each with their own centers, would satisfy the problem's conditions.But again, this is similar to the L-shape example but with different component shapes. So, multiple examples exist.Therefore, the answer is yes, such a polygon exists. The key idea is that combining two symmetric convex polygons in an asymmetric way results in an overall asymmetric polygon, which can then be divided back into the two symmetric components.I should check if there's any theorem or existing result that might contradict this. For example, if a polygon is divided into two symmetric convex polygons, does the original necessarily have a symmetry? But in the examples we constructed, it doesn't. The centers of the two parts are different, and their combination doesn't enforce a global symmetry. Therefore, such a configuration is possible.Another angle to consider: the centers of the two convex parts. Let's say the first convex polygon has center C1, and the second has center C2. If C1 and C2 are different points, then the original polygon's center would not necessarily be related to C1 or C2. Unless there's some relationship enforced by the way they're joined, but in the examples above, they are joined asymmetrically, so no global center exists.Therefore, the reasoning seems solid. The answer should be yes, and the L-shaped polygon is a valid example.

✅:Step-by-Step Explanation and Answer:To determine if such a polygon exists, we analyze the problem through construction and verification:1. Definitions Recap: - A center of symmetry (or point symmetry) in a polygon means there exists a point ( O ) such that rotating the polygon 180° around ( O ) maps it onto itself. - A convex polygon has all interior angles ≤ 180°, and every line segment between two vertices lies inside the polygon.2. Objective: - Find a polygon ( P ) without a center of symmetry that can be partitioned into two convex polygons ( P_1 ) and ( P_2 ), each with their own centers of symmetry.3. Constructing an Example: - Consider an L-shaped polygon formed by two rectangles: - A horizontal rectangle ( R_1 ) (e.g., 3×1 units). - A vertical rectangle ( R_2 ) (e.g., 1×2 units), attached to one end of ( R_1 ). - Key Properties: - ( R_1 ) and ( R_2 ) are both convex and have individual centers of symmetry. - The combined L-shape is non-convex and asymmetric (due to unequal arm lengths).4. Verification: - Original Polygon (L-shape): - No Center of Symmetry: Testing rotational symmetry about any point reveals mismatched vertices (e.g., the long horizontal arm vs. short vertical arm). - Subpolygons ( R_1 ) and ( R_2 ): - Each is convex and retains its own center of symmetry (midpoints of ( R_1 ) and ( R_2 )).5. Conclusion: - The L-shaped polygon satisfies all conditions: it lacks a global center of symmetry but can be divided into two convex polygons, each with their own centers.Answer: Yes, such a polygon exists. An example is an L-shaped polygon formed by joining a 3×1 rectangle and a 1×2 rectangle. The L-shape has no center of symmetry, but dividing it into the two rectangles yields convex polygons, each with their own symmetry centers.

🔑:1. Statement of the Problem: We are to determine if there exists a polygon without a center of symmetry that can be divided into two convex polygons, each having a center of symmetry.2. Existence of Such a Polygon: We will provide an example to show that such a polygon exists.3. Example Construction: Consider the following polygon, which is constructed by combining two rectangles as shown in the reference image. 4. Explanation: - Each individual rectangle has a center of symmetry. The center of symmetry of a rectangle is the point where its diagonals intersect. - If we combine the rectangles along a shared side, the resulting polygon does not possess a center of symmetry. - This is because if we place a point of symmetry anywhere outside the blue or red segments passing through the midpoints of the shared side, the transformation would map vertices in one portion of the polygon to points outside the polygon.5. Conclusion: - By this example, we can observe that it is indeed possible to have a polygon with no center of symmetry that can be divided into two convex polygons, each possessing a center of symmetry. Conclusion:Therefore, the answer to the problem is that such a polygon exists.[boxed{text{Exists}}]

🧠:Okay, so I need to solve this inequality: (sqrt(2 - x) + 4x - 3)/x ≥ 2. Hmm, let me start by understanding what steps I need to take here. First, maybe I should simplify the inequality as much as possible. But before that, I should note the domain of the function because there's a square root and a denominator involved. Alright, starting with the domain. The expression under the square root, 2 - x, must be non-negative. So, 2 - x ≥ 0 ⇒ x ≤ 2. Then, the denominator x can't be zero, so x ≠ 0. So combining these, the domain is x ≤ 2 and x ≠ 0. But also, since x is in the denominator, x has to be a real number except 0. So the domain is x ∈ (-∞, 0) ∪ (0, 2]. I need to keep this in mind because any solution I find must be within this domain.Now, moving on to the inequality itself: (sqrt(2 - x) + 4x - 3)/x ≥ 2. Maybe I can subtract 2 from both sides to get everything on one side. Let me try that:(sqrt(2 - x) + 4x - 3)/x - 2 ≥ 0.Simplify the left-hand side. Combine the terms over a common denominator:[sqrt(2 - x) + 4x - 3 - 2x]/x ≥ 0.Wait, because 2 is equivalent to 2x/x. So subtracting 2 gives (sqrt(2 - x) + 4x - 3 - 2x)/x ≥ 0. Let me compute the numerator:sqrt(2 - x) + 4x - 3 - 2x = sqrt(2 - x) + 2x - 3.So the inequality becomes (sqrt(2 - x) + 2x - 3)/x ≥ 0.Hmm, that's a bit simpler. Now, the inequality is of the form (A)/x ≥ 0, where A = sqrt(2 - x) + 2x - 3. So the sign of this expression depends on the signs of A and x. Since the denominator is x, and the domain excludes x = 0, the expression is positive when A and x have the same sign, and negative when they have opposite signs. So the inequality (A)/x ≥ 0 is equivalent to (A ≥ 0 and x > 0) or (A ≤ 0 and x < 0). But since our domain is x ≤ 2 and x ≠ 0, we have two intervals to consider: (0, 2] and (-∞, 0). Let me split the problem into these two intervals.First, let's consider x > 0 (since x is in (0, 2]). In this case, the inequality (sqrt(2 - x) + 2x - 3)/x ≥ 0 simplifies to sqrt(2 - x) + 2x - 3 ≥ 0 because x is positive. So we need to solve sqrt(2 - x) + 2x - 3 ≥ 0 for x ∈ (0, 2].Second, for x < 0 (since x ∈ (-∞, 0)), the denominator x is negative, so the inequality (sqrt(2 - x) + 2x - 3)/x ≥ 0 would require the numerator A = sqrt(2 - x) + 2x - 3 to be ≤ 0. So sqrt(2 - x) + 2x - 3 ≤ 0 for x < 0.Therefore, the solution set will be the union of the solutions in these two intervals.Let me tackle each interval separately.Starting with x ∈ (0, 2]. Here, we need sqrt(2 - x) + 2x - 3 ≥ 0.Let me denote f(x) = sqrt(2 - x) + 2x - 3. We need to find where f(x) ≥ 0 on (0, 2].First, let's check the endpoints. At x = 2: sqrt(2 - 2) + 2*2 - 3 = 0 + 4 - 3 = 1 ≥ 0. So x = 2 is a solution. At x approaching 0 from the right: sqrt(2 - 0) + 0 - 3 = sqrt(2) - 3 ≈ 1.414 - 3 ≈ -1.586 < 0. So f(x) starts negative near 0 and becomes positive at x = 2. So there must be some point c in (0, 2) where f(c) = 0. The function f(x) is continuous on (0, 2] because sqrt(2 - x) is continuous there, and 2x - 3 is a polynomial. So by the Intermediate Value Theorem, there is at least one root in (0, 2). Let's find that root.Set f(x) = sqrt(2 - x) + 2x - 3 = 0.Let me solve sqrt(2 - x) = 3 - 2x.Before squaring both sides, note that sqrt(2 - x) is non-negative, so 3 - 2x must also be non-negative. Therefore, 3 - 2x ≥ 0 ⇒ x ≤ 3/2. But since x ∈ (0, 2], the equation is only valid when x ≤ 3/2. So when we square both sides, we have to check solutions against this condition.Squaring both sides: (sqrt(2 - x))² = (3 - 2x)² ⇒ 2 - x = 9 - 12x + 4x².Bring all terms to one side: 4x² - 12x + 9 - (2 - x) = 0 ⇒ 4x² - 12x + 9 - 2 + x = 0 ⇒ 4x² - 11x + 7 = 0.Now, solving 4x² - 11x + 7 = 0.Using quadratic formula: x = [11 ± sqrt(121 - 112)] / 8 = [11 ± sqrt(9)] / 8 = [11 ± 3]/8.So x = (11 + 3)/8 = 14/8 = 7/4 = 1.75, or x = (11 - 3)/8 = 8/8 = 1.Now, check if these solutions satisfy the original equation and the condition x ≤ 3/2 (which is 1.5). So x = 1.75 is 1.75 ≤ 1.5? No, 1.75 is greater than 1.5, so this solution is extraneous. So only x = 1 is a possible solution.Check x = 1 in the original equation: sqrt(2 - 1) + 2*1 - 3 = sqrt(1) + 2 - 3 = 1 + 2 - 3 = 0. So yes, x = 1 is a solution.Therefore, the equation f(x) = 0 has solution x = 1. Now, since f(x) is increasing or decreasing? Let's check the derivative to see the behavior.Wait, maybe not necessary. Since f(x) is continuous on (0, 2], f(0+) ≈ -1.586, f(1) = 0, f(2) = 1. So the function increases from x approaching 0 to x = 2. Wait, but between 1 and 2, does it keep increasing? Let me check f(1.5): sqrt(2 - 1.5) + 2*(1.5) - 3 = sqrt(0.5) + 3 - 3 ≈ 0.707 + 0 ≈ 0.707 ≥ 0. So at x = 1.5, it's positive. So between x = 1 and x = 2, f(x) increases from 0 to 1. But what about between x = 0 and x = 1? Let's check at x = 0.5: sqrt(2 - 0.5) + 2*(0.5) - 3 = sqrt(1.5) + 1 - 3 ≈ 1.225 - 2 ≈ -0.775 < 0. So the function crosses zero at x = 1, and before that, it's negative, after that, positive. Therefore, in the interval (0, 2], f(x) ≥ 0 when x ≥ 1. So the solution for x ∈ (0, 2] is [1, 2].Now, moving to the other interval x < 0. Here, the inequality (sqrt(2 - x) + 2x - 3)/x ≥ 0 requires that the numerator sqrt(2 - x) + 2x - 3 ≤ 0, because x is negative. So we need sqrt(2 - x) + 2x - 3 ≤ 0 for x < 0.Let me analyze this. Let me denote g(x) = sqrt(2 - x) + 2x - 3. We need g(x) ≤ 0 for x < 0.First, check the behavior of g(x) as x approaches -infty. Wait, x is approaching negative infinity, but the domain is x ≤ 2. However, x < 0 here. So as x approaches negative infinity, sqrt(2 - x) becomes sqrt(2 - x) where x is very negative, so 2 - x approaches positive infinity, so sqrt(2 - x) approaches positive infinity. Meanwhile, 2x - 3 approaches negative infinity. So the question is, which term dominates? sqrt(2 - x) grows like sqrt(|x|), and 2x - 3 grows linearly negative. For very large |x| (negative x), the linear term will dominate, so sqrt(2 - x) + 2x - 3 ≈ sqrt(|x|) + 2x. As x approaches negative infinity, this tends to negative infinity. So g(x) approaches -infty as x approaches -infty. At x = 0 (approaching from the left), x = 0-: sqrt(2 - 0) + 0 - 3 = sqrt(2) - 3 ≈ -1.586 < 0. So g(x) is negative at x approaching 0 from the left, and tends to -infty as x approaches -infty. So is g(x) always negative for x < 0?Wait, but maybe there's a point where g(x) = 0 in x < 0? Let's check. Suppose there's a solution where sqrt(2 - x) + 2x - 3 = 0 in x < 0. Let's try solving this equation.sqrt(2 - x) = 3 - 2x.Again, sqrt(2 - x) is non-negative, so 3 - 2x must be non-negative. So 3 - 2x ≥ 0 ⇒ x ≤ 3/2. Which is always true here since x < 0. So square both sides:2 - x = (3 - 2x)^2 ⇒ 2 - x = 9 - 12x + 4x² ⇒ 4x² - 11x + 7 = 0.Wait, this is the same quadratic equation as before. So the solutions are x = 1 and x = 7/4. But these are both positive, so x = 1 is in (0, 2], which we already considered, and x = 7/4 is also in (0, 2]. So there are no solutions for x < 0. Therefore, in x < 0, the equation sqrt(2 - x) + 2x - 3 = 0 has no solutions, so g(x) is always negative? Let's verify with a test point in x < 0. Let's take x = -1.Compute g(-1) = sqrt(2 - (-1)) + 2*(-1) - 3 = sqrt(3) - 2 - 3 ≈ 1.732 - 5 ≈ -3.268 < 0. Another test point, x = -2: sqrt(2 - (-2)) + 2*(-2) - 3 = sqrt(4) - 4 - 3 = 2 - 7 = -5 < 0. So yes, for all x < 0, g(x) is negative. Therefore, sqrt(2 - x) + 2x - 3 ≤ 0 is always true for x < 0. Therefore, in the interval x < 0, the inequality (sqrt(2 - x) + 2x - 3)/x ≥ 0 holds because numerator is negative and denominator is negative, so their quotient is positive. Therefore, all x < 0 satisfy the inequality.Wait, hold on. The inequality in x < 0 is (sqrt(2 - x) + 2x - 3)/x ≥ 0. Since numerator is sqrt(2 - x) + 2x - 3 = g(x) which is ≤ 0 (as we just saw for x < 0) and denominator x is < 0. Therefore, the quotient is ≥ 0 because negative divided by negative is positive. Therefore, all x < 0 satisfy the inequality. So the solution set for x < 0 is (-∞, 0).But wait, the original domain for the inequality is x ≤ 2 and x ≠ 0, so x < 0 is already within the domain. Therefore, combining both intervals, the solutions are (-∞, 0) ∪ [1, 2].But wait, we need to check if there are any restrictions in x < 0. The square root sqrt(2 - x) is defined for x ≤ 2, which is already satisfied here. So all x < 0 are valid. Therefore, the entire interval x < 0 is part of the solution.Therefore, the solution set is x ∈ (-∞, 0) ∪ [1, 2].But wait, let's confirm this with test points in each interval.First, test x = -1: plug into original inequality (sqrt(2 - (-1)) + 4*(-1) - 3)/(-1) ≥ 2.Compute numerator: sqrt(3) - 4 - 3 = sqrt(3) - 7 ≈ 1.732 - 7 ≈ -5.268. Denominator: -1. So the left-hand side is (-5.268)/(-1) = 5.268 ≥ 2. Which is true. So x = -1 satisfies the inequality.Another test point in (0,1), say x = 0.5: original inequality (sqrt(2 - 0.5) + 4*0.5 - 3)/0.5 ≥ 2.Numerator: sqrt(1.5) + 2 - 3 ≈ 1.225 - 1 ≈ 0.225. Denominator: 0.5. So left-hand side ≈ 0.225 / 0.5 ≈ 0.45 ≥ 2? No, 0.45 < 2. So x = 0.5 does not satisfy, which aligns with our solution.Test x = 1: numerator sqrt(1) + 4*1 - 3 = 1 + 4 - 3 = 2. Denominator 1. So 2/1 = 2 ≥ 2. True.Test x = 1.5: numerator sqrt(2 - 1.5) + 4*1.5 - 3 = sqrt(0.5) + 6 - 3 ≈ 0.707 + 3 ≈ 3.707. Denominator 1.5. So 3.707 / 1.5 ≈ 2.47 ≥ 2. True.Test x = 2: numerator sqrt(0) + 8 - 3 = 0 + 5 = 5. Denominator 2. 5/2 = 2.5 ≥ 2. True.So all test points check out. Therefore, the solution is indeed x ∈ (-∞, 0) ∪ [1, 2].Wait, but I should check if x approaches 0 from the left, does the inequality hold? As x approaches 0 from the left, numerator sqrt(2 - x) + 2x - 3 approaches sqrt(2) - 3 ≈ -1.586, which is negative, and denominator approaches 0 from the negative side. So (-1.586)/x where x approaches 0- is (-1.586)/(approaching 0-) which tends to positive infinity. So the left-hand side becomes very large positive, which is certainly ≥ 2. Therefore, near x = 0 from the left, the inequality holds. So x approaching 0 from the left is included in the solution, but x = 0 is excluded.Similarly, approaching x = 1 from below (x approaching 1 from the left), the numerator approaches 0 from negative values (since at x =1, it's zero, and before that it's negative). But for x approaching 1 from the left, since we are in the interval (0,2], the denominator is positive approaching 1. So the left-hand side approaches 0 from below (negative values), but wait, no. Wait, at x just below 1, the numerator is negative (since f(x) = sqrt(2 - x) + 2x - 3, which is negative before x =1), and denominator is positive, so the entire expression is negative, which would not satisfy the inequality. Wait, but according to our previous analysis, in (0, 2], the solution is [1, 2]. So at x approaching 1 from below, the expression is negative, hence does not satisfy the inequality, which matches the solution [1, 2]. Therefore, everything seems consistent.Therefore, the final solution is x ∈ (-∞, 0) ∪ [1, 2].But let me just write this again to confirm:Original inequality: (sqrt(2 - x) + 4x - 3)/x ≥ 2.After simplifying, we found that the solution is x < 0 or x ≥ 1, considering the domain x ≤ 2. Therefore, combining these, x ∈ (-∞, 0) ∪ [1, 2].Yes, that seems correct. I think this is the answer.

✅:To solve the inequality frac{sqrt{2-x} + 4x - 3}{x} geqslant 2,we first determine the domain, which requires (2 - x geq 0) (so (x leq 2)) and (x neq 0). Thus, the domain is (x in (-infty, 0) cup (0, 2]).Next, we simplify the inequality:1. Subtract 2 from both sides: frac{sqrt{2-x} + 4x - 3}{x} - 2 geq 0. 2. Combine terms over a common denominator: frac{sqrt{2-x} + 2x - 3}{x} geq 0. This inequality holds when the numerator and denominator have the same sign. We analyze two cases:Case 1: (x > 0) (i.e., (x in (0, 2]))- The inequality simplifies to (sqrt{2-x} + 2x - 3 geq 0).- Solving (sqrt{2-x} + 2x - 3 = 0), we find (x = 1) (after verifying roots and behavior).- Testing intervals shows the solution here is (x in [1, 2]).Case 2: (x < 0) (i.e., (x in (-infty, 0)))- The inequality simplifies to (sqrt{2-x} + 2x - 3 leq 0).- Analysis shows the left-hand side is always negative for (x < 0), so the inequality holds for all (x < 0).Final Solution:Combining both cases, the solution to the inequality is:x in (-infty, 0) cup [1, 2].Answer: (boxed{(-infty, 0) cup [1, 2]})

🔑:1. Find the Domain of the Inequality: Given the inequality: [ frac{sqrt{2-x} + 4x - 3}{x} geq 2 ] We need to determine the domain of the function. - The term (sqrt{2-x}) implies that (2-x geq 0) or (x leq 2). - The denominator (x) implies that (x neq 0). Therefore, the domain of inequality (11) is (x < 0) or (0 < x leq 2).2. Case 1: ( x < 0 ): For (x < 0), the inequality: [ frac{sqrt{2-x} + 4x - 3}{x} geq 2 ] can be rewritten as: [ sqrt{2-x} + 4x - 3 leq 2x ] Which simplifies to: [ sqrt{2-x} leq 3 - 2x ] We need to verify if this is always true for (x < 0). Since (sqrt{2-x} > 1) for any negative (x) (as (sqrt{2 - x} > sqrt{2})), therefore: [ sqrt{2-x} < (sqrt{2-x})^2 = 2-x ] And for any negative (x): [ 2-x < (2-x) + (1-x) = 3 - 2x ] Hence, (sqrt{2-x} leq 3 - 2x) holds for all (x < 0). Therefore, all (x < 0) are solutions for this case.3. Case 2: ( 0 < x leq 2 ): For (0 < x leq 2), the inequality: [ frac{sqrt{2-x} + 4x - 3}{x} geq 2 ] can be rewritten as: [ sqrt{2-x} + 4x - 3 geq 2x ] Which simplifies to: [ sqrt{2-x} geq 3 - 2x ] Now we analyze it by splitting the interval (0 < x leq 2) into two subintervals: (0 < x leq frac{3}{2}) and ( frac{3}{2} < x leq 2 ). - For (0 < x leq frac{3}{2}): [ sqrt{2-x} geq 3 - 2x ] Both sides are non-negative, and squaring both sides gives: [ 2 - x geq (3 - 2x)^2 ] Simplifying, [ 2 - x geq 9 - 12x + 4x^2 ] [ 4x^2 -11x + 7 leq 0 ] Solving this quadratic inequality: [ x = frac{11 pm sqrt{(-11)^2 - 4 cdot 4 cdot 7}}{2 cdot 4} = frac{11 pm sqrt{121 - 112}}{8} = frac{11 pm 3}{8} ] So, the roots are: [ x = 1 quad text{and} quad x = frac{7}{4} ] Testing the intervals (0 < x leq frac{3}{2}) reveals that the solution lies in (1 leq x leq frac{3}{2}). - For ( frac{3}{2} < x leq 2 ): Here, the left side (sqrt{2-x}) is non-negative while (3 - 2x) is negative because (x > frac{3}{2}). Hence: [ sqrt{2-x} geq 3 - 2x ] is always true for this interval.4. Conclusion: Combining the solutions from both cases, the complete solution to the inequality is the union of: [ (-infty, 0) cup [1, 2] ] Therefore, the solutions to the inequality are: [ boxed{(-infty, 0) cup [1, 2]} ]

🧠:Okay, so I need to prove the opposite statement of problem 60668. Let me first make sure I understand what the original problem was. The original statement says that if the binomial coefficient C(n, k) is divisible by n for all 1 ≤ k ≤ n-1, then n is a prime number. So the opposite statement would be: If n is not a prime number (i.e., composite), then there exists some k between 1 and n-1 such that C(n, k) is not divisible by n. Alternatively, maybe the contrapositive? Wait, the contrapositive of "If P, then Q" is "If not Q, then not P". So the original statement is P implies Q, where P is "C(n, k) divisible by n for all 1 ≤ k ≤ n-1", and Q is "n is prime". So the contrapositive would be "If n is not prime, then there exists some k in 1 ≤ k ≤ n-1 such that C(n, k) is not divisible by n". Therefore, proving the contrapositive would be equivalent to proving the original statement. But the user is asking to prove the opposite of the original statement. Wait, no. Wait, the original statement is "If C(n,k) divisible by n for all k=1,...,n-1, then n is prime". The opposite statement would be "There exists a composite number n such that C(n,k) is divisible by n for all k=1,...,n-1". But maybe the user wants the negation of the original statement. Let me check.The original problem's statement: If C(n,k) divisible by n for all 1 ≤k≤n−1, then n is prime. The opposite would be: It is not the case that if C(n,k) divisible by n for all 1≤k≤n−1, then n is prime. Which in logic is: There exists an n where C(n,k) is divisible by n for all 1≤k≤n−1, but n is not prime. So, to disprove the original statement, we need to find a composite number n such that all binomial coefficients C(n,k) for 1 ≤k≤n−1 are divisible by n. Hence, the task is to prove that such composite numbers exist, thereby showing that the original statement is false.Wait, but the problem says "Prove the opposite statement to what was in problem 60668". So perhaps more precisely, the original statement is "If for all k in 1≤k≤n−1, n divides C(n,k), then n is prime". The opposite would be "There exists a composite n such that for all k in 1≤k≤n−1, n divides C(n,k)". Therefore, the user wants to show that the original statement is false by providing a counterexample. So the solution would be to find a composite number n where all the binomial coefficients C(n,k) for 1 ≤k≤n−1 are divisible by n. If such a number exists, then the original implication is not always true.But are there composite numbers with this property? Wait, primes satisfy that all C(p,k) for 1≤k≤p−1 are divisible by p. So primes have this property. The original problem was saying that if a number has this property, then it's prime. So the converse: if n has the property, then it's prime. To disprove this, we need a composite n with the same property. Therefore, such numbers are called "anti-primes" or perhaps "Carmichael numbers" but in the context of binomial coefficients. Wait, actually, I recall that numbers such as primes have the property that all C(n,k) for 1≤k≤n−1 are divisible by n. But composite numbers normally do not. However, there might be exceptions. For example, n=4: C(4,1)=4, which is divisible by 4. C(4,2)=6, which is not divisible by 4. So 4 is composite and fails. n=6: C(6,1)=6 divisible by 6. C(6,2)=15, not divisible by 6. So 6 fails. n=8: C(8,1)=8 divisible by 8. C(8,2)=28, not divisible by 8 (28 ÷8=3.5). So no. n=9: C(9,1)=9 divisible by 9. C(9,2)=36, which is 36 ÷9=4, so divisible. C(9,3)=84 ÷9=9.333… which is not integer. So 84 is not divisible by 9. So 9 fails. n=10: C(10,1)=10 divisible by 10. C(10,2)=45, not divisible by 10. So 10 fails. Hmm, n=105? Wait, perhaps a square of a prime? Let's try n=25. C(25,1)=25 divisible by 25. C(25,2)=300, 300 ÷25=12, divisible. C(25,3)=2300, 2300 ÷25=92, divisible. Wait, maybe higher k? Wait, C(25,5)=53130. 53130 ÷25=2125.2, which is not integer. So 53130 is not divisible by 25. Hence, n=25 is composite and C(25,5) is not divisible by 25. Therefore, 25 fails.Wait, but maybe there are composite numbers where all binomial coefficients are divisible by them. Alternatively, perhaps n=4 is the only composite number where at least one binomial coefficient is not divisible. But no, as shown above, for composite numbers, there's often at least one binomial coefficient that is not divisible by n. But maybe for some composite numbers, all binomial coefficients are divisible? Is that possible?Alternatively, perhaps if n is a prime power. Let me try n= p² where p is prime. Let's take p=2, n=4: C(4,2)=6, not divisible by 4. p=3, n=9: C(9,3)=84, which is 84 ÷9=9.333… not divisible. So n=p² is composite and fails.Wait, how about n=1? But 1 is neither prime nor composite. So not relevant. n=0? Not a positive integer. So maybe n=341? 341 is composite (11*31). Let's check C(341,1)=341, divisible. C(341,2)= (341*340)/2=341*170=57970. 57970 ÷341=170. So divisible. Wait, but 341 is composite. Wait, hold on, 341*170 = 57970. So 57970 ÷341=170. So C(341,2) is divisible by 341. Then, what about C(341,3)= (341*340*339)/6. Let's compute. 341 divided by 6 is 341/6=56.833… So (341*340*339)/6= (341/6)*340*339. But 341 is 11*31. So 6 divides into 340? Wait, 340 is divisible by 2 and 5 and 17. 339 is 3*113. So 341=11*31. Let's see if the entire binomial coefficient is divisible by 341. So C(341,3)= (341*340*339)/6. Since 341 is 11*31, we need to check if 11 and 31 divide the numerator. The numerator is 341*340*339. Divided by 6. 341 is already in the numerator. So the denominator is 6=2*3. Let's check divisibility by 11 and 31. Since 341 is part of the numerator, the numerator is 341*(340*339). Therefore, the numerator is divisible by 341, and then divided by 6. Therefore, C(341,3)= (341*(340*339))/6. Since 341 is already factored out, and divided by 6. So the question is, is (340*339)/6 an integer? 340 is even, 339 is divisible by 3. So 340=2*170, 339=3*113. So 340*339=2*170*3*113=6*170*113. Divided by 6 gives 170*113=19210. So C(341,3)=341*19210. Therefore, C(341,3)=341*19210, which is divisible by 341. Therefore, C(341,3) is divisible by 341. Wait, so for n=341, C(n,1), C(n,2), C(n,3) are divisible by n. Is this the case for all k?Wait, let's check C(n, k) for k=4. C(341,4)= (341*340*339*338)/24. Similarly, since 341 is in the numerator, the rest of the terms would be (340*339*338)/24. Let's see if this is an integer. 340=2*2*5*17, 339=3*113, 338=2*13*13. So the product is (2^3 *5*17*3*113*13^2). Divided by 24=2^3*3. So cancel out 2^3 and 3, leaving 5*17*113*13^2, which is an integer. Therefore, C(341,4) is divisible by 341. Wait a second, so does this hold for all k? If so, then n=341 would be a composite number where all C(n,k) are divisible by n, which would contradict the original statement.But wait, but n=341 is composite (11*31). So if all C(341,k) are divisible by 341, then that's a counterexample. But does this hold?Alternatively, perhaps n=341 satisfies this. Wait, is there a theorem about such numbers? Maybe similar to Wieferich primes or something else. Alternatively, maybe numbers n such that for all 1≤k≤n−1, n divides C(n,k). Primes have this property due to the fact that C(p,k)=p*(p-1)*...*(p-k+1)/k! and since p is prime and k < p, p does not divide k!, so the entire binomial coefficient is divisible by p. For composite numbers, normally, this fails. But perhaps some composites have this property. Such numbers are called "absolute pseudoprimes" or similar. Wait, in the literature, Carmichael numbers are composite numbers n such that b^{n-1} ≡1 mod n for all integers b coprime to n. But binomial coefficients... Maybe there's a different term.Alternatively, let's consider Kummer's theorem, which relates the highest power of a prime dividing a binomial coefficient. But n is composite here. Alternatively, Lucas' theorem. Wait, Lucas' theorem gives the binomial coefficients modulo a prime. But n is composite here.Alternatively, considering that for a prime p, C(p,k) ≡0 mod p for 1≤k≤p-1. So for composite n, if n divides all C(n,k), then n is called a "Wieferich-like" number for binomial coefficients. But I'm not sure if such numbers exist. However, in the case of n=341, let's check for k=11. Wait, 341=11*31. Let me check C(341,11). Is this divisible by 341? That seems more complicated, but maybe there's a pattern.Alternatively, consider n= p^m where p is prime. For example, n= p^2. Then C(p², p) = (p²)! / (p! * (p² - p)! ). Let's check if this is divisible by p². For p=2: C(4,2)=6, which is not divisible by 4. For p=3: C(9,3)=84, which is not divisible by 9. So p^2 does not work. For n= p*q where p and q are distinct primes. Let's take p=2, q=3, n=6: C(6,2)=15 not divisible by 6. p=2, q=5, n=10: C(10,2)=45 not divisible by 10. p=3, q=5, n=15: C(15,5)=3003. 3003 ÷15=200.2, not integer. So 3003 is not divisible by 15. Hence, composite numbers of the form p*q don't seem to work.Wait, but what about n=341=11*31. Let's check C(341,11). Let's try to compute modulo 11 and modulo 31. If C(341,11) is divisible by both 11 and 31, then it's divisible by 341. Since 341=11*31 and 11 and 31 are primes.Compute C(341,11) mod 11. By Lucas' theorem, since 341 in base 11 is 3*11 + 1*11^0= 3*11 +1=34 (wait, 11*31=341, so 341 divided by 11 is 31, which is 2*11 +9, so in base 11, 341 is 2*11^2 +9*11 +0. Wait, maybe I need to write 341 in base 11. Let's compute:11^3=1331, which is larger than 341. 11^2=121.341 ÷121=2 with remainder 341 - 2*121=341-242=99.99 ÷11=9 with remainder 0.So 341 in base 11 is 2*11^2 +9*11 +0*1, i.e., (2,9,0)_11.Similarly, 11 in base 11 is (1,0)_11.According to Lucas' theorem, C(341,11) ≡ C(2,1)*C(9,0)*C(0,0) mod 11.C(2,1)=2, C(9,0)=1, C(0,0)=1. So total is 2*1*1=2 mod 11. Hence, C(341,11) ≡2 mod 11. Therefore, C(341,11) is congruent to 2 modulo 11, which is not zero. Therefore, 11 does not divide C(341,11), which implies that 341 does not divide C(341,11). Therefore, n=341 is composite, but C(n,11) is not divisible by n. Thus, n=341 is not a counterexample.Hmm, so maybe my earlier reasoning was flawed. So, how to find such a composite number?Alternatively, maybe there are no such composite numbers, meaning the original statement is actually true. But I need to verify. Wait, the original statement is: If C(n,k) is divisible by n for all 1 ≤k ≤n−1, then n is prime. So if this is true, then there are no composite numbers with this property. Therefore, the opposite statement would be false. But the user is asking to prove the opposite statement, implying that the original is false.But if the original statement is true, then the opposite is false, and the user's request is invalid. So, first, I need to verify whether the original statement is true or not.Let me recall that for prime numbers p, C(p,k) is divisible by p for all 1 ≤k ≤p−1. This is a well-known result. For composite numbers, it's generally not the case. For example, take n=4, composite, C(4,2)=6, not divisible by 4. n=6, composite, C(6,2)=15, not divisible by 6. So for composite n, there exists some k where C(n,k) is not divisible by n.But is this always the case? Suppose n is composite. Is there always some k such that C(n,k) is not divisible by n?If yes, then the original statement is true. If no, then there exists a composite n where all C(n,k) are divisible by n, hence the original statement is false.So, is there a composite number n where all C(n,k) for 1 ≤k ≤n−1 are divisible by n?To answer this, perhaps using a theorem. I remember that if n is composite, then there exists a prime p dividing n such that p^k does not divide C(n,p) for some exponent k. Wait, perhaps not. Alternatively, consider that if n is composite, then it has a prime divisor p. Then, consider C(n,p). Is this divisible by n? Let's check with n=9, p=3. C(9,3)=84, 84 ÷9=9.333… Not divisible. For n=15, p=3. C(15,3)=455, 455 ÷15≈30.333… Not divisible. For n=21, p=3. C(21,3)=1330, 1330 ÷21≈63.333… Not divisible. So seems like for composite n, choosing k as a prime divisor of n might result in C(n,k) not divisible by n.Alternatively, let's take n=pq where p and q are distinct primes. Let's take k=p. Then C(n,p) = (pq)! / (p! (pq - p)! ). Let's see if this is divisible by pq. Let's factor out p and q.C(pq, p) = [pq * (pq-1) * ... * (pq - p +1)] / [p!]. Let's count the number of times p divides numerator and denominator.In the numerator, pq is divisible by p, and the term (pq - p +1) is pq - p +1 = p(q -1) +1. So among the p consecutive terms in the numerator, there's exactly one multiple of p (pq) and the rest are not divisible by p. Therefore, the numerator has exactly one factor of p. The denominator p! has exactly one factor of p (from the term p). Therefore, the total number of p's in numerator and denominator cancels out, leaving no factors of p. Similarly for q: in the numerator, pq is divisible by q, but the other terms (pq-1), ..., (pq -p +1) are consecutive integers. If q > p, then none of these terms except pq would be divisible by q. If q < p, then similarly. Let's take n=15=3*5, k=3. C(15,3)=455. 455 is divisible by 5 (455=5*91), but 455 ÷15=30.333… So not divisible by 15=3*5. Because while 455 is divisible by 5, it's not divisible by 3. 455 ÷3=151.666… Hence, 15 does not divide 455. So even if the binomial coefficient is divisible by one prime factor, it may not be divisible by the other.Therefore, for n=pq with p and q distinct primes, C(n,p) is divisible by p (since numerator has pq which is divisible by p, and denominator p! has p, so cancels one p), but the remaining part may not be divisible by q. Similarly, the remaining part is [ (pq-1)*...*(pq - p +1) ] / ( (p-1)! ). Let's see for n=15, p=3: numerator terms are 15,14,13. So 15*14*13. Denominator is 3! =6. So (15*14*13)/6 = (15/3)*(14*13)/2=5*(91)=455. So 455 is divisible by 5 (from 15/3=5), but not by 3. Hence, 455 is divisible by 5 but not by 3, so not divisible by 15.Hence, for n=pq, C(n,p) is divisible by p but not necessarily by q. Hence, n=pq does not divide C(n,p). Therefore, there exists a k=p such that C(n,k) is not divisible by n. Hence, n composite implies there exists k such that C(n,k) is not divisible by n.This suggests that the original statement is true. Therefore, the opposite statement is false, and there are no composite numbers n where all C(n,k) are divisible by n.But the user is asking to prove the opposite statement. Hmm, perhaps there was a misunderstanding. Maybe the user wants the converse proved? Wait, the original problem 60668 is proving that if all C(n,k) are divisible by n, then n is prime. The user is asking to prove the opposite, which would be: if n is prime, then all C(n,k) are divisible by n. But that's not the opposite, that's the converse. Wait, no. The original implication is P→Q, where P is "all C(n,k) divisible by n" and Q is "n is prime". The opposite could be ~P→~Q, but that's the inverse. The converse is Q→P. The contrapositive is ~Q→~P. The contrapositive is logically equivalent, but the opposite (negation) of P→Q is P ∧ ~Q.So the original statement is "If P, then Q". The opposite (negation) is "P and not Q". Therefore, to prove the opposite statement would be to show that there exists an n where P holds (all C(n,k) divisible by n) and Q does not hold (n is not prime). But if no such n exists, then the original statement is true.Given the previous analysis, such n does not exist. Hence, the opposite statement cannot be proven because it's false. Therefore, the original statement is true. Hence, the user's request to prove the opposite is based on a misunderstanding, or perhaps the user wants the contrapositive proved.Wait, perhaps the user wants us to prove the contrapositive. The contrapositive of "If P, then Q" is "If not Q, then not P". So the contrapositive is "If n is not prime (composite), then there exists some k with 1≤k≤n−1 such that C(n,k) is not divisible by n". Which is actually a true statement. Therefore, maybe the user refers to this contrapositive as the "opposite statement". But in logical terms, the contrapositive is equivalent, not the opposite. The opposite would be the negation.But given the user's instruction, perhaps they meant the contrapositive. Because proving the contrapositive would indeed prove the original statement. Since the original statement and the contrapositive are logically equivalent.If that's the case, then we need to prove that if n is composite, then there exists some k such that C(n,k) is not divisible by n. Which we have already discussed.So, summarizing, the steps would be:1. Assume n is composite.2. Then n has a prime divisor p.3. Consider k=p.4. Show that C(n,p) is not divisible by n.But earlier examples show that this might not always hold. Wait, for n=9, p=3, C(9,3)=84, not divisible by 9. For n=25, p=5, C(25,5)=53130, not divisible by 25. For n=12, p=3, C(12,3)=220, 220 ÷12≈18.333… Not divisible. For n=8, p=2, C(8,2)=28, not divisible by 8. So seems like taking k=p where p is a prime divisor of n, C(n,p) is not divisible by n. Hence, this would imply that for composite n, there exists k (specifically, k being a prime divisor of n) such that C(n,k) is not divisible by n. Hence, proving the contrapositive.Therefore, to formalize this:Let n be composite, so n=ab where a,b>1.Let p be a prime divisor of n. Let k=p.Compute C(n,p) = n!/(p!(n-p)!).We need to show that C(n,p) is not divisible by n.Using Legendre's formula, which gives the exponent of a prime in the factorial of a number. The exponent of p in n! is floor(n/p) + floor(n/p²) + ..., similarly for the exponent in p! and (n-p)!.The exponent of p in C(n,p) = [exponent of p in n!] - [exponent of p in p!] - [exponent of p in (n-p)!].Let’s compute this:For n = p*m (since p divides n), so n = p*m.Exponent of p in n! = floor(p*m/p) + floor(p*m/p²) + ... = m + floor(m/p) + floor(m/p²) + ...Exponent of p in p! = 1 + 0 + 0 + ... =1.Exponent of p in (n - p)! = exponent of p in (p*m - p)! = exponent of p in p*(m -1)! = exponent of p in p*(m-1)! =1 + exponent of p in (m-1)!.Therefore, exponent of p in C(n,p) = [m + floor(m/p) + floor(m/p²) + ...] -1 - [1 + exponent of p in (m-1)!] = m + floor(m/p) + floor(m/p²) + ... -1 -1 - exponent of p in (m-1)!.But exponent of p in (m-1)! = floor((m-1)/p) + floor((m-1)/p²) + ... So,Exponent of p in C(n,p) = m -2 + [floor(m/p) - floor((m-1)/p)] + [floor(m/p²) - floor((m-1)/p²)] + ... Now, floor(m/p) - floor((m-1)/p) is 1 if p divides m, otherwise 0. Similarly for higher powers. But if m < p, then floor(m/p) =0, floor((m-1)/p)=0, so difference is 0. Similarly for higher powers.If m < p, then exponent of p in C(n,p) = m -2 +0 +0 +... = m -2.Since n=p*m and m >=2 (since n is composite and p is a prime divisor), m >=2. Therefore, exponent of p in C(n,p) is m -2 + ... (possible contributions from higher terms if m >=p).But if m < p, then exponent of p in C(n,p) is m -2. For this exponent to be at least 1 (so that p divides C(n,p)), m -2 >=1 => m >=3. But if m >=3 and p > m, then n=p*m >=p*3. However, p is a prime divisor of n, so p <=m. Wait, this is confusing.Alternatively, take a specific example where n=p^2. Then m=p.Exponent of p in n! = floor(p²/p) + floor(p²/p²) + ... = p +1 +0 +...=p+1.Exponent of p in p! =1.Exponent of p in (n-p)! = exponent of p in (p² -p)! = exponent of p in p(p-1)! =1 +0=1.Therefore, exponent of p in C(n,p) = (p+1) -1 -1= p -1.Thus, exponent of p in C(p²,p) is p-1. Since C(p²,p) = p² * (p² -1)*...*(p² -p +1)/p!.The exponent of p in the numerator is p +1 (from p²!) and in the denominator is1 (from p!). So total exponent is p +1 -1 =p. Wait, but according to the previous calculation, it's p-1. There's a discrepancy here.Wait, let's recalculate for n=p², k=p:Exponent of p in numerator (p²!):floor(p²/p) + floor(p²/p²) + floor(p²/p³) + ...= p +1 +0+...=p+1.Exponent of p in denominator p!*(p² -p)!:Exponent in p! = floor(p/p) + floor(p/p²)+...=1+0+...=1.Exponent in (p² -p)! = floor((p² -p)/p) + floor((p² -p)/p²)+...= floor(p-1) +0+...=p-1.Therefore, total exponent in denominator is1 + (p-1)=p.Thus, exponent of p in C(p²,p)= (p+1) -p=1.Therefore, exponent of p in C(p²,p) is1, meaning p divides C(p²,p) but p² does not. Hence, C(p²,p) is divisible by p but not by p². Therefore, C(p²,p) is not divisible by n=p². Therefore, for n=p², k=p, C(n,k) is not divisible by n.Therefore, for n=p² (a composite number), there exists k=p where C(n,k) is not divisible by n. Hence, this supports the original statement.Similarly, for n=p*q with p and q distinct primes, take k=p. Then, as in the previous example with n=15, the exponent of p in C(n,p) is 1 -1 -0=0? Wait, let's compute.For n=p*q, k=p.Exponent of p in numerator n! = floor(p*q/p) + floor(p*q/p²) +...=q +0+...=q.Exponent of p in denominator p!*(n-p)!:Exponent in p! =1.Exponent in (n-p)! = exponent in (p*q -p)! = exponent in p(q-1)! =1 + exponent of p in (q-1)!.If q-1 < p, then exponent of p in (q-1)! is0. So total exponent in denominator:1 +0=1.Therefore, exponent of p in C(n,p)= q -1 -0= q-1.Since q is a prime different from p, and q >=2, then q-1 >=1. Therefore, exponent of p in C(n,p) is q-1 >=1, so p divides C(n,p). But does n=p*q divide C(n,p)?Well, C(n,p) = (p*q)! / (p! * (p*q -p)! ). The exponent of p in C(n,p) is q-1. The exponent of p in n=p*q is1. Therefore, unless q-1 >=1 and 1 <= q-1, which is true, but the exponent of p in C(n,p) is q-1, which is at least1, so p divides C(n,p), but does q divide C(n,p)?To check divisibility by q, compute exponent of q in C(n,p).Exponent of q in numerator n! = floor(p*q/q) + floor(p*q/q²) +...=p +0+...=p.Exponent of q in denominator p!*(n-p)!:Exponent in p! =0 (since q >p, as primes, assuming p < q).Exponent in (n-p)! = exponent in (p*q -p)! = exponent in p(q-1)!.Since q is a prime, and q-1 < q (assuming q>2), then exponent of q in (q-1)! is0. Therefore, exponent of q in denominator is0.Therefore, exponent of q in C(n,p)=p -0 -0=p. Therefore, q divides C(n,p) exactly p times. Hence, since n=p*q, to have n divide C(n,p), we need both p and q to divide C(n,p), and since they are coprime, it's equivalent to p*q divides C(n,p). But the exponent of p in C(n,p) is q-1, and the exponent of q is p. Therefore, for n=p*q to divide C(n,p), we need q-1 >=1 and p >=1, which is true, but how many times does n divide into C(n,p)?The multiplicity of p in C(n,p) is q-1, and the multiplicity of q is p. Therefore, the multiplicity of p*q in C(n,p) is the minimum of q-1 and p. Unless both q-1 >=1 and p >=1, which they are, but for n=p*q to divide C(n,p), we need that the multiplicity of p is at least1 (which it is) and the multiplicity of q is at least1 (which it is). However, the exponent of p in n is1, and in C(n,p) it's q-1. To have n divide C(n,p), we need q-1 >=1 and p >=1. Both hold, but it's not about exponents. Wait, perhaps this approach is confusing.Alternatively, consider specific example: n=15=3*5, k=3. C(15,3)=455. 455 divided by 15=30.333… Not an integer. So even though p=3 divides 455 (since 455=5*7*13), 15=3*5 does not divide 455. Hence, C(15,3) is divisible by 3 but not by 5. Wait, 455 ÷5=91, which is integer. Wait, 455 is 5*91, so it is divisible by 5. Wait, then why is 455 not divisible by 15? Because 91=13*7, which is 7*13. So 455=5*7*13. Therefore, 455 is divisible by 5 and 7 and 13, but not by 3. Hence, 455 ÷15=455/(3*5)= (455/5)/3=91/3≈30.333… Not integer. Therefore, even though C(n,p) is divisible by p and q, it's not divisible by n=p*q because the factor of p and q are not present simultaneously in the binomial coefficient. Wait, but 455 is divisible by 5 and 7 and 13, but not by 3. Wait, but n=15=3*5. So C(n,3) must be divisible by 3 and 5. But 455 is divisible by 5 but not by 3. Hence, even though the exponent of q=5 in C(n,p) is p=3, which is 3, and exponent of p=3 is q-1=2, which is 2. Therefore, 3² divides C(n,p) and 5³ divides C(n,p). Therefore, C(n,p) is divisible by 3²*5³= 9*125=1125. But 455 is divisible by 5 once and by 7 and 13. So this contradicts the previous calculation. Therefore, there must be an error in the exponent calculation.Wait, let's compute the exponent of p=3 in C(15,3). C(15,3)=455.Exponent of 3 in 15!:floor(15/3) + floor(15/9) + floor(15/27) +...=5 +1 +0=6.Exponent of 3 in 3! = floor(3/3) +...=1.Exponent of 3 in 12! = floor(12/3) + floor(12/9) +...=4 +1=5.Therefore, exponent of 3 in C(15,3)=6 -1 -5=0. Hence, 3 does not divide C(15,3). Which matches the fact that 455 is not divisible by 3. But according to the previous general formula, exponent of p in C(n,p)=q-1=5-1=4. Which contradicts. Therefore, my general formula was incorrect. Therefore, the earlier approach has a mistake. Hence, need to correct the exponent calculation.Let’s re-express the exponent of p in C(n,p) when n=p*q.n=p*q, p and q distinct primes.Exponent of p in n! = floor(p*q/p) + floor(p*q/p²)+...=q + floor(q/p)+... Since q <p (assuming p >q), floor(q/p)=0. So exponent of p in n! =q.Exponent of p in denominator: p! and (n-p)!.Exponent of p in p! =1.Exponent of p in (n-p)! = exponent of p in (p*q -p)! = exponent of p in p(q-1)! =1 + exponent of p in (q-1)!.Since q <p (assuming p >q), (q-1)! does not contain p, so exponent of p in (q-1)! =0.Therefore, exponent of p in (n-p)! =1 +0=1.Hence, total exponent of p in denominator=1+1=2.Therefore, exponent of p in C(n,p)=q -2.Similarly, exponent of q in C(n,p):Exponent of q in numerator n! =floor(p*q/q) + floor(p*q/q²)+...=p +0= p.Exponent of q in denominator p!*(n-p)!:Exponent of q in p! =0 (since q >p).Exponent of q in (n-p)! = exponent of q in (p*q -p)! = exponent of q in p(q-1)!.Since q is prime and q-1 <q, exponent of q in (q-1)! =0. Hence, exponent of q in (n-p)! =0.Therefore, exponent of q in denominator=0+0=0.Exponent of q in C(n,p)=p -0= p.Therefore, C(n,p) has exponents of p as q-2 and q as p.But in the specific case of n=15=3*5, p=3, q=5:Exponent of 3 in C(15,3)=5 -2=3. But actual exponent of 3 in 455 is0. Contradiction.Therefore, there's a mistake in the calculation. Wait, in the specific case of n=15, p=3, q=5:Exponent of 3 in 15! is floor(15/3)+floor(15/9)+floor(15/27)=5+1+0=6.Exponent of 3 in denominator: 3! and 12!.Exponent of 3 in 3! =1.Exponent of 3 in 12! =floor(12/3)+floor(12/9)=4+1=5.Total exponent in denominator=1+5=6.Therefore, exponent of 3 in C(15,3)=6 -6=0.But according to the general formula, exponent of p=3 should be q-2=5-2=3. Which is wrong. Therefore, the general formula is incorrect.The mistake in the general approach is that when n=p*q, and p and q are primes with p < q, then (n-p)! = (p*q -p)! =p(q-1)!. However, when factoring this, if q-1 >=p, then there may be other multiples of p in (q-1)!.But in the example, q=5, so q-1=4. Therefore, (q-1)! =4!=24. Which has exponent of p=3 as floor(4/3)=1. Hence, exponent of p in (q-1)! =1. Therefore, exponent of p in (n-p)! = exponent of p in p*(q-1)! =1 +1=2. Hence, total exponent of p in denominator=1 (from p!) +2 (from (n-p)!)=3. Therefore, exponent of p in C(n,p)=q -3=5-3=2. But in reality, exponent of p in C(n,p)=6 -1 -5=0. Hence, there is a mistake.Therefore, the previous general approach is flawed. Therefore, need a different approach.Alternative approach:If n is composite, then it has a prime factor p such that p <= sqrt(n).Consider k=p.Compute C(n,p). We need to show that n does not divide C(n,p).Using Legendre's formula, the exponent of p in n! is floor(n/p) + floor(n/p²) + ... The exponent of p in p! is1 + floor(p/p²) + ...=1.The exponent of p in (n-p)! is floor((n-p)/p) + floor((n-p)/p²) + ... Let’s denote n = p*m. Then, (n-p) = p*(m-1).Therefore, exponent of p in (n-p)! = floor(p*(m-1)/p) + floor(p*(m-1)/p²) + ...= (m-1) + floor((m-1)/p) + floor((m-1)/p²) + ... Therefore, exponent of p in C(n,p)= [floor(n/p) + floor(n/p²)+...] - [1] - [ (m-1) + floor((m-1)/p) + floor((m-1)/p²) + ... ]Since n=p*m, floor(n/p)=m, floor(n/p²)=floor(m/p), etc.Therefore, exponent of p in numerator n! =m + floor(m/p) + floor(m/p²)+...Exponent in denominator p!*(n-p)! =1 + (m-1) + floor((m-1)/p) + floor((m-1)/p²)+...= m + floor((m-1)/p) + floor((m-1)/p²)+... -1.Therefore, exponent of p in C(n,p)= [m + floor(m/p) + floor(m/p²)+...] - [m + floor((m-1)/p) + floor((m-1)/p²)+... -1 ]=1 + [floor(m/p) - floor((m-1)/p)] + [floor(m/p²) - floor((m-1)/p²)] + ... Now, floor(m/p) - floor((m-1)/p) is1 if p divides m, otherwise 0.Similarly for higher terms.Therefore, exponent of p in C(n,p) =1 + (1 if p divides m else 0) + (1 if p² divides m else 0) + ... Therefore, exponent of p in C(n,p) is at least1. Hence, p divides C(n,p).But n=p*m. So for n to divide C(n,p), m must also divide C(n,p)/p.But m and p may not be coprime. For example, if n=p², then m=p. exponent of p in C(n,p)=1 +1=2 (since p divides m=p). Hence, exponent of p in C(n,p)=2. But n=p², so exponent of p in n is2. Therefore, p² divides C(n,p) only if exponent of p in C(n,p)>=2. Which it is. Therefore, for n=p², C(n,p) is divisible by p². Wait, but earlier calculation for n=9=3², C(9,3)=84, which is 84 ÷9=9.333… Not divisible by 9. Contradiction.Wait, for n=9=3², k=3. C(9,3)=84. Exponent of 3 in 84 is 84=3*28, so exponent of 3 is1. But according to the formula above, exponent of p=3 in C(9,3)=1 +1=2. But actual exponent is1. Hence, discrepancy.Therefore, the formula must be wrong. Let’s recalculate for n=9=3², k=3.Exponent of 3 in 9!:floor(9/3)+floor(9/9)+floor(9/27)=3+1+0=4.Exponent of 3 in denominator: 3! and 6!.Exponent of 3 in 3! =1.Exponent of 3 in 6! =floor(6/3)+floor(6/9)=2+0=2.Total exponent in denominator=1+2=3.Therefore, exponent of 3 in C(9,3)=4 -3=1. Hence, exponent is1, not2. Therefore, the general formula is incorrect. Therefore, our previous step missed something.Re-examizing the exponent of p in C(n,p) when n=p²:Exponent in numerator p²! = floor(p²/p) + floor(p²/p²) +...=p +1.Exponent in denominator p!*(p² -p)!:Exponent in p! =1.Exponent in (p² -p)! = exponent in p(p-1)! =1 + exponent of p in (p-1)! =1 +0=1.Therefore, total exponent in denominator=1 +1=2.Hence, exponent of p in C(p²,p)= (p +1) -2= p -1.For p=3, exponent=3 -1=2. But actual exponent in C(9,3)=1. Therefore, contradiction. Hence, this formula is also wrong. Therefore, there must be an error in the calculation.The correct exponent of p in C(n,p) when n=p²:Numerator: p²! has exponent floor(p²/p) + floor(p²/p²) =p +1.Denominator: p!*(p² -p)!.Exponent in p! =1.Exponent in (p² -p)! = floor((p² -p)/p) + floor((p² -p)/p²)=floor(p-1) +0= p-1.Therefore, exponent in (p² -p)! =p-1.Therefore, total exponent in denominator=1 + (p-1)=p.Exponent in numerator= p +1.Thus, exponent in C(n,p)= (p +1) -p=1.Therefore, for n=p², exponent of p in C(n,p)=1. Hence, p² does not divide C(n,p). Therefore, C(n,p) is divisible by p but not by p². Hence, n=p² does not divide C(n,p).This matches our example with n=9: C(9,3)=84, divisible by3 but not by9.Therefore, general formula for exponent of p in C(n,p) when n=p*m: exponent = [floor(n/p) + floor(n/p²) + ...] - [floor(p/p) + floor(p/p²) + ...] - [floor((n-p)/p) + floor((n-p)/p²) + ...]Which for n=p*m, and assuming m <p (to avoid multiple factors), then floor(n/p)=m, floor(n/p²)=0.floor(p/p)=1, floor(p/p²)=0.floor((n-p)/p)=floor((p*m -p)/p)=floor(p(m-1)/p)=m-1.floor((n-p)/p²)=0.Therefore, exponent of p in C(n,p)=m -1 - (m-1)=0. Which contradicts previous calculations. Wait, for example, n=15=3*5, m=5, p=3.Exponent of p=3 in numerator=5.Exponent in denominator=1 (from p!) + floor((15-3)/3)+floor((15-3)/9)+...=1 + floor(12/3)+floor(12/9)=1 +4 +1=6.Thus, exponent in C(n,p)=5 -6= -1, which is impossible. Hence, this approach is flawed.Therefore, the correct way is to realize that Legendre's formula applies to prime exponents in factorials. However, when calculating the exponent in a binomial coefficient, it's the exponent in the numerator minus the sum of exponents in the denominators.But when n is composite and we take k as a prime divisor of n, the exponent of p in C(n,p) can be calculated as:Exponent in numerator (n!)= floor(n/p) + floor(n/p²) + ...Exponent in denominator (p!*(n-p)!)= [floor(p/p) + floor(p/p²) + ...] + [floor((n-p)/p) + floor((n-p)/p²) + ...]For n=p*m:Exponent in numerator= m + floor(m/p) + floor(m/p²) + ...Exponent in denominator=1 + [floor(p*m -p)/p + floor((p*m -p)/p²) + ...] =1 + [floor(p(m-1)/p) + floor(p(m-1)/p²) + ...] =1 + (m-1) + floor((m-1)/p) + floor((m-1)/p²) + ...Therefore, exponent of p in C(n,p)= [m + floor(m/p) + floor(m/p²) + ...] -1 - (m-1) - [floor((m-1)/p) + floor((m-1)/p²) + ...]Simplify:= m -1 - (m-1) + [floor(m/p) - floor((m-1)/p)] + [floor(m/p²) - floor((m-1)/p²)] + ...=0 + [floor(m/p) - floor((m-1)/p)] + [floor(m/p²) - floor((m-1)/p²)] + ...Each term [floor(m/p^i) - floor((m-1)/p^i)] is1 if p^i divides m, else0.Therefore, exponent of p in C(n,p) equals the number of exponents i such that p^i divides m.If m is not divisible by p, then all terms are0, so exponent of p in C(n,p)=0.If m is divisible by p, say m=p^k*t where t is not divisible by p, then the number of exponents i where p^i divides m is k. Therefore, exponent of p in C(n,p)=k.Therefore, exponent of p in C(n,p)= number of times p divides m.But n=p*m, so m=n/p. Therefore, exponent of p in m= exponent of p in n -1.If n=p*m where m is divisible by p, then exponent of p in n is at least2. Therefore, exponent of p in C(n,p)= exponent of p in m= exponent of p in n -1.But for n=p*m where m is not divisible by p, exponent of p in C(n,p)=0.Hence, if n is composite and has a prime divisor p such that p divides n exactly once (i.e., p^2 does not divide n), then exponent of p in C(n,p)=0 (if m is not divisible by p) or exponent of p in C(n,p)=k where k=exponent of p in m= exponent of p in n -1.Wait, this is getting too convoluted. Let's take specific examples.Case 1: n=p*q where p and q are distinct primes.Then m=q (assuming n=p*q).Exponent of p in C(n,p)= exponent of p in numerator (n!)= floor(p*q/p) + floor(p*q/p²)+...=q +0=q.Exponent of p in denominator= exponent in p! + exponent in (n-p)!.Exponent in p!=1.Exponent in (n-p)! = exponent in (p*q -p)! = exponent in p(q-1)! =1 + exponent of p in (q-1)!.Since q is a prime different from p and q-1 <p (assuming p <q), exponent of p in (q-1)!=0.Therefore, exponent of p in denominator=1 +1=2.Therefore, exponent of p in C(n,p)=q -2.If q >=2, then exponent of p in C(n,p)>=0. But for C(n,p) to be divisible by p, exponent must be >=1. So q -2 >=1 => q >=3. If q=2 and p=2, but n=4, which is p². For n=6=2*3, q=3. Then exponent of p=2 in C(6,2)=3 -2=1. So C(6,2)=15. 15 is divisible by3 but not by2. Wait, exponent of2 in15 is0. Contradiction.Therefore, the formula must be incorrect. Therefore, our general approach to compute exponents is leading to inconsistencies with actual examples. Hence, we need a different strategy.Alternative approach:If n is composite, then there exists a prime p dividing n. Consider the smallest prime p dividing n.If p² divides n, then consider k=p. C(n,p)=n*(n-1)*...*(n-p+1)/p!.Since p² divides n, n= p²*m.C(n,p)= (p²*m)*(p²*m -1)*...*(p²*m -p +1)/p!.In the numerator, there are p terms. The term p²*m is divisible by p². The other terms p²*m -1, ..., p²*m -p +1 are consecutive integers. Exactly one of these terms is divisible by p (since they are p consecutive numbers). Let's say it's p²*m -t, where t is between0 and p-1. Therefore, this term is divisible by p but not by p². Therefore, the numerator has exactly two factors of p: one from p²*m and one from p²*m -t. However, the denominator p! has two factors of p if p=2 (since 2! =2, which has one factor), wait p=2: denominator 2! has one factor of2. For p=3: denominator3! has one factor of3.Therefore, for p >=2, the denominator p! has one factor of p. Therefore, the numerator has two factors of p (from p²*m and from another term), and the denominator has one factor of p. Therefore, total factors of p in C(n,p)=2 -1=1. Therefore, C(n,p) is divisible by p but not by p². Therefore, since n=p²*m, which is divisible by p², C(n,p) is not divisible by n. Hence, n does not divide C(n,p).If n is composite and square-free (i.e., product of distinct primes), then take p as the smallest prime dividing n, and consider k=p. Then, C(n,p)=n*(n-1)*...*(n-p+1)/p!.Since n is square-free, p divides n exactly once. In the numerator, n is divisible by p, and among the p-1 other terms, none are divisible by p (since they are consecutive numbers from n-1 to n-p+1, and n ≡0 mod p, so n-1 ≡-1 mod p, n-2 ≡-2 mod p, ..., n-p+1 ≡-p+1 mod p). None of these terms are divisible by p. Hence, the numerator has one factor of p, and the denominator p! has one factor of p. Therefore, C(n,p) is divisible by n/p (since n/p is an integer), but not necessarily by p. Wait, no. C(n,p) has (n/p)*(n-1)*...*(n-p+1)/ (p-1)!.But since n is composite and square-free, n=p*q where q>p (since p is the smallest prime divisor).Then C(n,p)= (p*q)!/(p! (p*q -p)!).We need to check if this is divisible by p*q.The exponent of p in C(n,p) is floor(p*q/p) + floor(p*q/p²) +... - [floor(p/p) + floor(p/p²)+...] - [floor((p*q -p)/p) + floor((p*q -p)/p²)+...].= q +0 -1 - (q-1 +0)=q -1 - (q-1)=0.Therefore, exponent of p in C(n,p)=0. Hence, p does not divide C(n,p). Therefore, C(n,p) is not divisible by p, hence not by n=p*q. Therefore, for square-free composite n, taking k=p (smallest prime divisor) gives a binomial coefficient not divisible by n.Therefore, in both cases (whether n is square-free or not), there exists a k such that C(n,k) is not divisible by n. Hence, the original statement holds: if all C(n,k) are divisible by n, then n must be prime. Therefore, the opposite statement (that there exists a composite n with this property) is false.Hence, the conclusion is that the original statement is true, and thus its opposite is false. However, the user asked to prove the opposite statement. This suggests there might be confusion. However, based on the analysis, there are no composite numbers n for which all C(n,k) are divisible by n. Hence, the original statement is true, and the opposite cannot be proven.

✅:The original statement asserts that if all binomial coefficients ( C(n, k) ) for ( 1 leq k leq n-1 ) are divisible by ( n ), then ( n ) must be prime. To prove the opposite statement, we need to demonstrate that there exists a composite ( n ) where all ( C(n, k) ) are divisible by ( n ). However, through rigorous analysis, it is shown that no such composite numbers exist. Here's the step-by-step reasoning:1. Prime Property: For a prime ( p ), all ( C(p, k) ) (where ( 1 leq k leq p-1 )) are divisible by ( p ). This follows because ( p ) divides the numerator ( p! ) but not the denominator ( k!(p-k)! ).2. Composite Numbers: For composite ( n ): - Let ( p ) be the smallest prime divisor of ( n ). - Consider ( k = p ). Analyze ( C(n, p) ).3. Case 1 – ( n ) is a prime power (( n = p^m )): - Compute the exponent of ( p ) in ( C(p^m, p) ). The numerator ( p^m! ) contains ( leftlfloor frac{p^m}{p} rightrfloor + leftlfloor frac{p^m}{p^2} rightrfloor + dots = p^{m-1} + p^{m-2} + dots + 1 ). - The denominator ( p!(p^m - p)! ) reduces the exponent by ( 1 + leftlfloor frac{p^m - p}{p} rightrfloor + dots = 1 + (p^{m-1} - 1) ). - The resulting exponent of ( p ) in ( C(p^m, p) ) is 1, insufficient for ( p^m ) to divide ( C(p^m, p) ).4. Case 2 – ( n ) is square-free (( n = pq ), distinct primes ( p, q )): - ( C(pq, p) )’s numerator contains ( pq ), but the denominator ( p!(pq - p)! ) removes one factor of ( p ). The remaining terms lack sufficient factors of ( q ), making ( pq ) unable to divide ( C(pq, p) ).5. General Composite ( n ): - For any composite ( n ), choosing ( k ) as the smallest prime divisor ( p ) leads to ( C(n, p) ) not divisible by ( n ), as shown in both cases.Conclusion: No composite number ( n ) exists where all ( C(n, k) ) are divisible by ( n ). Hence, the original statement is true, and its opposite cannot hold. Therefore, the final answer is:(boxed{text{The original statement is true; there are no composite } n text{ with all } C(n, k) text{ divisible by } n.})

🔑:1. Assume the contrary: Suppose ( n ) is a composite number, meaning that it has at least one prime divisor ( p ).2. Consider the binomial coefficient: We focus on the binomial coefficient ( C_{n}^{p} ), where ( 1 leq p leq n-1 ).3. Expression for the binomial coefficient: [ C_{n}^{p} = frac{n!}{p!(n-p)!} ]4. Simplification using factorial properties: [ C_{n}^{p} = frac{n (n-1) (n-2) cdots (n-p+1)}{1 cdot 2 cdot 3 cdots p} ]5. Factor analysis: - Since ( n ) is composite, it means ( n ) has prime divisors, including possibly ( p ). - The numerator ( n (n-1) (n-2) cdots (n-p+1) ) consists of ( p ) consecutive integers from ( n ) down to ( n-p+1 ). - Because ( p ) is one of the prime divisors, and ( p leq n-1 ), none of the factors in the numerator are divisible by ( p ) except possibly ( n ) itself.6. Cancellation and divisibility: - In the denominator ( 1 cdot 2 cdot 3 cdots p ), the integer ( p ) appears. - Therefore, the fraction has ( p ) in the denominator but not in the numerator (except for ( n ), the only factor possible that’s exactly ( n ), which is contradicted by it being a combination and integer multiple within). 7. Conclusion from divisibility: - Thus, the binomial coefficient ( C_{n}^{p} ) (if ( n ) is composite) fails to be an integer because there exists a prime ( p ) in the denominator which does not have a matching factor in the numerator (except the trivial contradiction with ( n )). 8. Final contradiction: - Therefore, ( C_{n}^{p} ) does not divide evenly by ( n ) if ( n ) is a composite number, contradicting our initial composite assumption.9. Conclude that: - If ( C_{n}^{k} ) is divisible by ( n ) for all ( 1 leq k leq n-1 ), then ( n ) must be a prime number.[boxed{text{Hence, } n text{ is a prime number.}}]

🧠:Alright, let me try to figure out this problem. So, the question is: someone picked 677 distinct natural numbers from 1 to 2022, and they claim that the sum of any two of these numbers isn't divisible by 6. We need to check if this claim is correct. Hmm, okay. Let me break this down step by step.First, I need to recall some number theory concepts. Since the problem is about sums not being divisible by 6, modular arithmetic might be useful here, especially modulo 6. If two numbers add up to a multiple of 6, then their residues modulo 6 must add up to 0 modulo 6. So, possible residues are 0, 1, 2, 3, 4, 5. Let's note which pairs of residues add up to 0 mod 6:- 0 and 0: 0 + 0 = 0 mod 6- 1 and 5: 1 + 5 = 6 ≡ 0 mod 6- 2 and 4: 2 + 4 = 6 ≡ 0 mod 6- 3 and 3: 3 + 3 = 6 ≡ 0 mod 6So, these are the pairs that would sum to a multiple of 6. Therefore, to ensure that no two numbers in the selected set have a sum divisible by 6, we must avoid selecting numbers in such pairs.Let me categorize the numbers from 1 to 2022 based on their residues modulo 6. There are six residue classes: 0, 1, 2, 3, 4, 5. For each residue class, I can calculate how many numbers fall into each category.Let's compute the count for each residue:- For residue 0: Numbers divisible by 6. The numbers are 6, 12, 18, ..., up to 2022. To find how many there are, divide 2022 by 6. 2022 ÷ 6 = 337. So, there are 337 numbers congruent to 0 mod 6.- Similarly, residues 1, 2, 3, 4, 5: Each of these will have either 337 or 338 numbers. Since 6*337 = 2022, but 2022 ÷ 6 is exactly 337, so each residue class from 1 to 5 will have 337 numbers. Wait, hold on. Wait, 2022 divided by 6 is exactly 337. So 6*337 = 2022. Therefore, each residue class from 0 to 5 has exactly 337 numbers. Wait, but hold on, 6*337 = 2022, so starting from 1 to 2022, each residue class has exactly 337 numbers. Because 2022 is divisible by 6. Wait, 2022 ÷ 6 = 337. So yes, each residue class 0,1,2,3,4,5 has exactly 337 numbers. Therefore, each residue class has 337 numbers.Wait, but let me verify this. For example, numbers congruent to 1 mod 6 would be 1, 7, 13, ..., up to the largest number less than or equal to 2022. The last term would be 6*337 +1? Wait, 6*337 = 2022, so 6*337 +1 = 2023, which is beyond. So the last term is 2022 -5 = 2017. Wait, 2017 mod 6 is 2017 ÷6 is 336*6=2016, so 2017 is 1 mod 6. So 2017 is the last number congruent to 1 mod 6. Then, how many terms? Let's compute (2017 -1)/6 +1 = (2016)/6 +1 = 336 +1 = 337. So yes, 337 numbers in each residue class. Similarly for others. So each residue class has exactly 337 numbers. So 0,1,2,3,4,5 each have 337 numbers. Got it.Now, the problem is to select numbers such that no two sum to a multiple of 6. As we saw earlier, pairs that sum to 0 mod 6 are (0,0), (1,5), (2,4), (3,3). Therefore, to avoid such pairs, the selection must ensure:1. At most one number from residue 0 mod 6. Because if you pick two numbers that are 0 mod 6, their sum is 0+0=0 mod 6, which is divisible by 6. So we can have at most one number from residue 0.2. For residues 1 and 5: We can choose numbers from residue 1 or residue 5, but not both. Because if we pick one from 1 and one from 5, their sum is 1+5=6 ≡0 mod6. Similarly, if we pick two from residue 1, their sum is 2 mod6, which is okay. Similarly, two from residue 5 would sum to 10 ≡4 mod6, which is also okay. Wait, but actually, the problem is only about the sum being divisible by 6. So residues 1 and 5 can't coexist. Similarly, residues 2 and 4 can't coexist. Residue 3 can have at most one number because two 3s sum to 6, which is 0 mod6.Wait, hold on. Let me clarify:- If we choose numbers from residue 0: Can have at most one number. Otherwise, two 0s sum to 0 mod6.- For residues 1 and 5: These are complementary. So if we include any numbers from residue 1, we cannot include any from residue 5, and vice versa. Because 1 +5 ≡0 mod6. But if we include only residue 1, then two residues 1 would sum to 2 mod6, which is okay. Similarly, two residues 5 would sum to 10 ≡4 mod6, okay. So the restriction is between residues 1 and 5.Similarly, residues 2 and 4 are complementary. So if we include numbers from residue 2, we can't include any from residue 4, and vice versa. Because 2 +4 ≡0 mod6. But within residue 2, two numbers sum to 4 mod6, which is okay. Similarly, two residues 4 sum to 8 ≡2 mod6.Residue 3: If we include two numbers from residue 3, their sum is 3 +3=6 ≡0 mod6. So we can have at most one number from residue 3.Therefore, the strategy is to select numbers from:- At most one from residue 0.- Either all from residue 1 or all from residue 5.- Either all from residue 2 or all from residue 4.- At most one from residue 3.So to maximize the number of numbers selected, we need to choose the larger between residue 1 and 5, between 2 and 4, and decide whether to include residue 0 and residue 3.But first, let's note that residue classes 1,5; 2,4; 3; and 0 are separate. So we need to make choices for each pair.Since residues 1 and 5 have the same size (337 each), same with 2 and 4 (337 each). So choosing either 1 or 5 gives 337 numbers. Similarly, choosing either 2 or 4 gives 337 numbers. Then, we can take at most 1 from residue 0 and at most 1 from residue 3.So the maximum possible size is 337 (from 1 or 5) + 337 (from 2 or 4) +1 (from 0) +1 (from 3) = 337 +337 +1 +1= 676.Wait, that gives 676. But the person selected 677 numbers. Hmm. So that would be a problem because according to this calculation, the maximum possible is 676. Therefore, the person's claim might not be correct if they selected 677 numbers. Therefore, the answer would be that he is not correct; such a selection of 677 numbers is impossible without having at least one pair summing to a multiple of 6.But let me verify again.Wait, is 337 +337 +1 +1= 676? Yes. 337*2=674, plus 1+1=2, so total 676. So the maximum number of numbers we can select without having any two sum to 6 is 676. Therefore, selecting 677 numbers would necessarily violate this condition, so the person must be incorrect.But wait, let me check again whether my calculation is correct. Let me think through again.The logic is:- Residues 1 and 5 are complementary; you can pick at most 337 from one of them.- Similarly, residues 2 and 4; you can pick at most 337 from one of them.- Residues 0 and 3: at most 1 each.So total maximum is 337 + 337 +1 +1 = 676.Therefore, 677 numbers cannot be selected without having two numbers whose sum is divisible by 6.But let me make sure that there isn't a different combination that allows more numbers.Wait, for example, what if you take residues 1, 2, 3, 4, 5, but avoid the conflicting pairs? But no, because 1 and 5 conflict, 2 and 4 conflict, 3 conflicts with itself. So you can't take both 1 and 5, or both 2 and 4. So the maximum is indeed the sum of the largest in each complementary pair plus at most 1 from each singleton.Wait, but what if you don't take the maximum from both pairs? For example, suppose you take 337 from residue 1, 336 from residue 2, then you could take 1 from residue 4? Wait, but if you take even one from residue 4, then you can't take any from residue 2. So if you take 337 from residue 1, you can take 337 from residue 2, but not take anything from 4 or 5. Alternatively, take 337 from residue 1, 337 from residue 4, but then 1 and 4 don't conflict? Wait, 1 +4=5 mod6, which is not 0, so that's okay. Wait, hold on, 1 +4=5 mod6, which is allowed. Similarly, 1 +2=3 mod6, which is allowed. Wait, maybe there's a different way to combine residues?Wait a minute, maybe my initial approach is too restrictive. Let me re-examine the pairs that sum to 0 mod6.The forbidden pairs are:- Two numbers ≡0 mod6.- A number ≡1 mod6 and a number ≡5 mod6.- A number ≡2 mod6 and a number ≡4 mod6.- Two numbers ≡3 mod6.Therefore, to avoid these pairs, the set must satisfy:- At most one number ≡0 mod6.- At most one number ≡3 mod6.- If you include any numbers ≡1 mod6, you can't include any ≡5 mod6, and vice versa.- If you include any numbers ≡2 mod6, you can't include any ≡4 mod6, and vice versa.But perhaps there's another way to include numbers from different residues, as long as their pairwise sums aren't 0 mod6.Wait, for example, if I take numbers from residues 1, 2, 3. Let's check the sums:- 1 +1=2 mod6 (okay)- 1 +2=3 mod6 (okay)- 1 +3=4 mod6 (okay)- 2 +2=4 mod6 (okay)- 2 +3=5 mod6 (okay)- 3 +3=0 mod6 (not allowed, so only one 3 allowed)Similarly, if I take residues 1,2, and 3 (with at most one 3), then the sums are all safe.But in this case, how many numbers can we take? If we take residue 1 (337 numbers), residue 2 (337 numbers), and residue 3 (1 number). Then total is 337 +337 +1=675. Then add residue 0 (1 number), total 676. So same as before.Alternatively, if we take residues 1, 4, and 3. Let's check:- 1 +4=5 mod6 (okay)- 1 +1=2 mod6 (okay)- 4 +4=8≡2 mod6 (okay)- 3 +1=4 mod6 (okay)- 3 +4=7≡1 mod6 (okay)But in this case, taking residues 1 and 4: but 1 and 4 don't conflict. Wait, but 2 and 4 do conflict? Wait, no. The conflicting pairs are 1 and 5, 2 and 4, 0 and 0, 3 and 3.Wait, so if we take residue 1 and residue 4, that's okay. Because 1 +4=5 mod6, which is not 0. Similarly, 1 and 2 is okay. Wait, maybe I made a mistake earlier. Let me clarify the forbidden pairs again:The problem states that the sum of any two numbers is not divisible by 6. So any two numbers in the set should not add up to 0 mod6. So pairs that sum to 0 mod6 are:- 0 and 0 (sum 0)- 1 and 5 (sum 6)- 2 and 4 (sum 6)- 3 and 3 (sum 6)Therefore, to avoid these, the rules are:- At most one 0.- At most one 3.- If you have any 1s, you can't have any 5s.- If you have any 2s, you can't have any 4s.But if you choose, for example, 1s and 4s, that's allowed, because 1 +4=5 mod6, which is not 0. Similarly, 2s and 5s would be allowed. So maybe there's a way to combine residues from different pairs?Wait, this changes things. So previously, I thought that you have to choose between 1 or 5, and between 2 or 4. But actually, you can choose residues from different non-conflicting pairs as long as their individual sums don't hit 0 mod6.For example, choosing residues 1 and 2: 1 +2=3 mod6, which is okay. So if you take residues 1, 2, and maybe 0 and 3, that's allowed. But how many numbers can you take?But in that case, you can take 337 (from 1) + 337 (from 2) +1 (from 0) +1 (from 3) = 676 again. Similarly, taking 1 and 4: 337 (from 1) +337 (from 4) +1 +1=676.Wait, but maybe you can take some combination of residues that are not in conflicting pairs but allow more numbers?Wait, but let's think about the total numbers. Let's consider that residues 1,2,3,4,5,0 each have 337 numbers. To maximize the number of selected numbers, we need to include as many as possible without having forbidden pairs.The forbidden pairs are:- (0,0)- (1,5)- (2,4)- (3,3)So the maximum set would be the union of:- At most one from residue 0.- The larger of residue 1 or 5.- The larger of residue 2 or 4.- At most one from residue 3.But since residues 1 and 5 are the same size, as are 2 and 4. So choosing either gives 337 each. Then adding 1 from 0 and 1 from 3 gives total 337 +337 +1 +1=676.Alternatively, is there a way to include more numbers by overlapping residues?Wait, suppose we take residue 1 and residue 2. Then we can take all 337 from residue 1 and all 337 from residue 2. Then, since 1 and 2 don't conflict, their sums are 3 mod6, which is okay. Additionally, we can take one number from residue 0 and one from residue 3. That's 337 +337 +1 +1=676.Alternatively, if we take residue 1, residue 2, residue 3 (but only one from residue 3), and residue 0 (one). But even then, you can't take more than one from residue 3. So same total.Alternatively, if we skip residue 0 and residue 3, and take residues 1, 2, 4, 5. But wait, 1 and 5 conflict, 2 and 4 conflict, so we can't take all of them. So no, that's worse.Wait, another approach: What if we partition the numbers into classes where each class is a group of numbers that can coexist without their sums being divisible by 6. Then, the maximum size is the sum of the maximums of each class.But I think the previous approach is correct. Let me think of another way. Maybe using graph theory. Imagine each number as a vertex, and connect two numbers with an edge if their sum is divisible by 6. Then, the problem reduces to finding the maximum independent set in this graph. The question is whether the maximum independent set is at least 677. If it's exactly 676, then the person's claim is wrong.But calculating the maximum independent set for such a graph is complicated. However, given the modular structure, we can exploit the residue classes. The graph would have connections between:- All pairs in residue 0.- All pairs between residue 1 and residue 5.- All pairs between residue 2 and residue 4.- All pairs in residue 3.Therefore, the graph is composed of a clique for residue 0 (since every two numbers in residue 0 sum to 0 mod6), a clique for residue 3, and bicliques between 1-5 and 2-4.In such a case, the maximum independent set would be the largest possible selection of vertices with no edges between them. Which translates to:- At most one vertex from residue 0 (since it's a clique).- At most one vertex from residue 3 (since it's a clique).- Choosing either all vertices from residue 1 or all from residue 5 (since they form a biclique; selecting both would create edges).- Similarly, choosing either all from residue 2 or all from residue 4.Thus, the maximum independent set is indeed 337 (from 1 or 5) + 337 (from 2 or 4) +1 (from 0) +1 (from 3) = 676.Therefore, it's impossible to have an independent set of size 677. Hence, the person must have at least one pair summing to a multiple of 6, so his assertion is incorrect.But wait, let me check if there's another way to structure the set. Suppose instead of taking maximum from both 1/5 and 2/4, we take some from one and some from another? For example, take x from residue 1 and y from residue 5 such that x + y <= 337 (since if you take any from residue 1, you can't take any from residue 5). No, actually, if you take any from residue 1, you can't take any from residue 5. Similarly for 2 and 4. So it's either all from 1 or all from 5, and similarly for 2 and 4. Therefore, you can't take a mix.Therefore, the maximum is indeed 337 +337 +1 +1=676.Thus, the conclusion is that it's impossible to select 677 numbers without having at least one pair summing to a multiple of 6. Therefore, the person's claim is incorrect.Wait, but let me double-check the counts again. Each residue class has 337 numbers. So residue 0: 337, residues 1-5: 337 each.If you take one from residue 0, one from residue 3, then 337 from residue 1 and 337 from residue 2. Total is 1 +1 +337 +337=676. Correct.Alternatively, if you take none from residue 0 and none from residue 3, then you can take 337 +337 +337=1011, but this is invalid because residues 1,2,4,5 would have overlaps in forbidden pairs. Wait, no, if you don't take from 0 or 3, then you can take residues 1,2,4,5. But residues 1 and 5 can't coexist, residues 2 and 4 can't coexist. So actually, the maximum in that case would be 337 (from 1 or 5) + 337 (from 2 or 4). So 674, which is less than 676. Therefore, including residue 0 and 3 gives a better count.Another check: if you take two from residue 0, but that's forbidden. Similarly, two from residue 3 is forbidden. So, yes, the maximum is 676.Therefore, the answer should be that the person is not correct; selecting 677 numbers is impossible without having at least one pair summing to a multiple of 6.But wait, the problem states that the person chose 677 numbers and claims that no two sum to a multiple of 6. The question is, is he correct? So our conclusion is that he cannot be correct, because the maximum possible is 676. Therefore, the answer is no, he is not correct.Another way to see this is through the pigeonhole principle. If you have 677 numbers, considering the residues mod6, by the pigeonhole principle, at least one residue class must contain at least ceil(677/6)=113 numbers. But this line of reasoning might not directly apply because the restrictions are between different residue classes.Alternatively, perhaps considering that in order to avoid the forbidden pairs, the selection is limited to certain residues, and exceeding the maximum size of such a selection necessarily causes a conflict.Therefore, the final answer is that the person is incorrect; it's impossible to choose 677 numbers under the given conditions.

✅:The person's assertion is incorrect. Here's the step-by-step reasoning:1. Residue Classes Modulo 6: Numbers from 1 to 2022 are categorized by their residues modulo 6 (0, 1, 2, 3, 4, 5). Each residue class contains exactly 337 numbers since 2022 ÷ 6 = 337.2. Forbidden Pairs: To avoid sums divisible by 6, the following pairs of residues are prohibited: - (0, 0), (1, 5), (2, 4), (3, 3).3. Selection Constraints: - Residue 0: At most 1 number (since two 0s sum to 0 mod 6). - Residues 1 and 5: Choose entirely from one group (337 numbers) but not both. - Residues 2 and 4: Choose entirely from one group (337 numbers) but not both. - Residue 3: At most 1 number (since two 3s sum to 0 mod 6).4. Maximum Possible Selection: - 337 (from 1 or 5) + 337 (from 2 or 4) + 1 (from 0) + 1 (from 3) = 676 numbers.5. Conclusion: Since 677 exceeds the maximum allowable selection (676), at least one forbidden pair must exist. Therefore, the person’s claim is false.Answer: boxed{text{No}}

🔑:1. Understand the Problem Statement: - We have to determine whether it is possible to choose 677 distinct natural numbers from the list (1, 2, 3, ldots, 2022) such that the sum of any two chosen numbers is not divisible by 6.2. Analyze Remainders when Divided by 6: - When a number is divided by 6, the remainders possible are 0, 1, 2, 3, 4, and 5.3. Count the Remainders: - Calculate how many numbers in the list (1, 2, 3, ldots, 2022) leave each remainder when divided by 6: - Remainder 0: (6, 12, 18, ldots, 2022) -- (337) numbers. - Remainder 1: (1, 7, 13, ldots, 2021) -- (337) numbers. - Remainder 2: (2, 8, 14, ldots, 2022) -- (337) numbers. - Remainder 3: (3, 9, 15, ldots, 2022) -- (337) numbers. - Remainder 4: (4, 10, 16, ldots, 2020) -- (336) numbers. - Remainder 5: (5, 11, 17, ldots, 2021) -- (336) numbers.4. Condition for Sum of Two Numbers: - The sum of two numbers should not be divisible by 6: - (0 + 0 = 0) (divisible by 6) - (1 + 5 = 6) (divisible by 6) - (2 + 4 = 6) (divisible by 6) - (3 + 3 = 6) (divisible by 6)5. Determine Limits on Selections: - We can only select at most one number with remainder 0. - We can only select at most one number with remainder 3. - We cannot select both 1 and 5 together, so maximum from 1 and 5 is 337. - We cannot select both 2 and 4 together, so maximum from 2 and 4 is 337.6. Calculate Maximum Possible Selections: - Sum the allowed choices: [ 1 (text{{one number with remainder 0}}) + 1 (text{{one number with remainder 3}}) + 337 (text{{remainder 1 or 5}}) + 337 (text{{remainder 2 or 4}}) ] [ = 1 + 1 + 337 + 337 = 676 ]7. Conclusion: - The maximum number of distinct natural numbers that can be selected under the given condition is 676. - Since 677 is more than the maximum 676, the Chudak's assertion is false.[boxed{text{No, he is wrong.}}]

🧠:Okay, let me try to work through this problem step by step. The problem is about a binomial expansion of (left(sqrt{x} + frac{1}{2 sqrt[4]{x}}right)^n), arranged in descending powers of (x). It says that the coefficients of the first three terms form an arithmetic sequence. We need to find how many terms in the expansion have an integer exponent of (x).First, let me recall the binomial expansion formula. The expansion of ((a + b)^n) is given by the sum from (k=0) to (n) of (binom{n}{k} a^{n - k} b^k). Each term in the expansion is (binom{n}{k} a^{n - k} b^k).In this case, (a = sqrt{x}) and (b = frac{1}{2 sqrt[4]{x}}). So, each term in the expansion will be (binom{n}{k} (sqrt{x})^{n - k} left(frac{1}{2 sqrt[4]{x}}right)^k).Let me simplify the exponents first. The exponent of (x) in each term comes from ((sqrt{x})^{n - k}) and (left(frac{1}{sqrt[4]{x}}right)^k).Breaking this down:- ((sqrt{x})^{n - k}) is (x^{frac{n - k}{2}}).- (left(frac{1}{sqrt[4]{x}}right)^k) is (x^{-frac{k}{4}}).So, combining these, the exponent of (x) in the term is (frac{n - k}{2} - frac{k}{4} = frac{n}{2} - frac{k}{2} - frac{k}{4} = frac{n}{2} - frac{3k}{4}).Therefore, each term has an exponent of (x) equal to (frac{n}{2} - frac{3k}{4}). The coefficients of the terms are (binom{n}{k} times left(frac{1}{2}right)^k), since the (b) term has a coefficient of (frac{1}{2}).Now, the expansion is arranged in descending powers of (x). Since the exponent is (frac{n}{2} - frac{3k}{4}), the term with the highest power of (x) occurs when (k = 0), which is just ((sqrt{x})^n = x^{n/2}). The next term, with (k = 1), will have an exponent of (frac{n}{2} - frac{3}{4}), and so on. So, as (k) increases, the exponent of (x) decreases by (frac{3}{4}) each time. Therefore, the terms are ordered from (k = 0) to (k = n) in descending exponents.The problem states that the coefficients of the first three terms form an arithmetic sequence. Let's denote the coefficients of the first three terms as (C_0), (C_1), and (C_2), corresponding to (k = 0), (k = 1), and (k = 2).First, let's compute these coefficients:- (C_0 = binom{n}{0} times left(frac{1}{2}right)^0 = 1 times 1 = 1)- (C_1 = binom{n}{1} times left(frac{1}{2}right)^1 = n times frac{1}{2} = frac{n}{2})- (C_2 = binom{n}{2} times left(frac{1}{2}right)^2 = frac{n(n - 1)}{2} times frac{1}{4} = frac{n(n - 1)}{8})An arithmetic sequence requires that the difference between consecutive terms is constant. Therefore, the difference between (C_1) and (C_0) should be equal to the difference between (C_2) and (C_1). So, we have:(C_1 - C_0 = C_2 - C_1)Substituting the values:(frac{n}{2} - 1 = frac{n(n - 1)}{8} - frac{n}{2})Let me compute each side:Left side: (frac{n}{2} - 1)Right side: (frac{n(n - 1)}{8} - frac{n}{2} = frac{n(n - 1) - 4n}{8} = frac{n^2 - n - 4n}{8} = frac{n^2 - 5n}{8})So, the equation becomes:(frac{n}{2} - 1 = frac{n^2 - 5n}{8})To solve for (n), multiply both sides by 8 to eliminate denominators:Left side: (8 times left(frac{n}{2} - 1right) = 4n - 8)Right side: (n^2 - 5n)So, the equation is:(4n - 8 = n^2 - 5n)Bring all terms to one side:(n^2 - 5n - 4n + 8 = 0)Simplify:(n^2 - 9n + 8 = 0)Factor the quadratic equation:Looking for two numbers that multiply to 8 and add to -9. The numbers -1 and -8 do that.So, (n^2 - 9n + 8 = (n - 1)(n - 8) = 0)Therefore, the solutions are (n = 1) and (n = 8).But we need to check if these solutions are valid in the original problem context.First, if (n = 1): The expansion would have only two terms, but the problem mentions the first three terms. So, (n = 1) is invalid because there are not three terms. Thus, we discard (n = 1).Therefore, the valid solution is (n = 8).So, the expansion is (left(sqrt{x} + frac{1}{2 sqrt[4]{x}}right)^8).Now, the next part of the problem asks: How many terms in the expansion have an integer exponent of (x)?We need to determine for each term (k = 0, 1, ..., 8), whether the exponent of (x) is an integer.Earlier, we found that the exponent of (x) in each term is (frac{n}{2} - frac{3k}{4}). Since (n = 8), substitute that in:Exponent (= frac{8}{2} - frac{3k}{4} = 4 - frac{3k}{4}).We need (4 - frac{3k}{4}) to be an integer. Let's denote the exponent as (E = 4 - frac{3k}{4}).So, (E) must be an integer. Let's express this as:(4 - frac{3k}{4} in mathbb{Z})Multiply both sides by 4 to eliminate denominator:(16 - 3k in 4mathbb{Z}), which means (16 - 3k) must be divisible by 4. Alternatively, (3k equiv 16 mod 4). Since 16 mod 4 is 0, so:(3k equiv 0 mod 4)Which simplifies to:(3k equiv 0 mod 4)Since 3 and 4 are coprime, this implies that (k equiv 0 mod 4). Therefore, (k) must be a multiple of 4.Given that (k) is an integer between 0 and 8 inclusive, the possible values of (k) that are multiples of 4 are 0, 4, 8.Therefore, there are 3 terms where the exponent of (x) is an integer.Wait, but let me check this again. Let's compute the exponent for each possible (k):For (k = 0): (E = 4 - 0 = 4) (integer)For (k = 1): (4 - 3/4 = 3.25) (not integer)For (k = 2): 4 - 6/4 = 4 - 1.5 = 2.5 (not integer)For (k = 3): 4 - 9/4 = 4 - 2.25 = 1.75 (not integer)For (k = 4): 4 - 12/4 = 4 - 3 = 1 (integer)For (k = 5): 4 - 15/4 = 4 - 3.75 = 0.25 (not integer)For (k = 6): 4 - 18/4 = 4 - 4.5 = -0.5 (not integer)For (k = 7): 4 - 21/4 = 4 - 5.25 = -1.25 (not integer)For (k = 8): 4 - 24/4 = 4 - 6 = -2 (integer)So indeed, only k=0, 4, 8 give integer exponents. Therefore, there are 3 terms.But let me make sure that my reasoning about the congruence is correct. The exponent is 4 - (3k)/4. For this to be integer, (3k)/4 must be an integer subtracted from 4. Therefore, (3k)/4 must be an integer. Wait, no, 4 - (3k)/4 must be integer. So, (3k)/4 must be such that when subtracted from 4, it's an integer. So, (3k)/4 must be a number that is an integer when subtracted from 4. Therefore, (3k)/4 must be a fractional number that, when subtracted from 4, gives an integer. But 4 is an integer, so (3k)/4 must also be a fractional number such that 4 - (3k)/4 is integer. Alternatively, (3k)/4 must be an integer or a half-integer or something?Wait, perhaps a better way: Let me denote E = 4 - (3k)/4. For E to be integer, (3k)/4 must be an integer because 4 is integer, so 4 - integer = integer. Wait, but (3k)/4 can be a fractional number such that when subtracted from 4, the result is integer. For example, if (3k)/4 is 0.25, then 4 - 0.25 = 3.75, which is not integer. But if (3k)/4 is integer, then 4 - integer = integer. Therefore, to have E integer, (3k)/4 must be integer. Therefore, 3k must be divisible by 4. So 3k ≡ 0 mod 4. As 3 and 4 are coprime, this implies that k ≡ 0 mod 4. Therefore, k must be multiples of 4. So k=0,4,8. Therefore, 3 terms. Which matches with our earlier enumeration.Therefore, the answer is 3 terms. So, the number of terms with integer exponents is 3.But wait, let me check again with k=4:k=4: exponent = 4 - (3*4)/4 = 4 - 3 = 1 (integer)k=8: exponent = 4 - (3*8)/4 = 4 - 6 = -2 (integer)Yes, those are integers. And k=0 gives exponent 4, which is also integer.Therefore, there are 3 terms. But wait, the problem states "the exponent of x is an integer". So, each term's exponent must be integer. So, yes, 3 terms.But hold on, let me check if there's any other possible k where exponent is integer. For example, if n was different, but here n=8. Let's check for k=0,1,...,8:k=0: exponent 4 (integer)k=1: 4 - 3/4 = 3.25 (non-integer)k=2: 4 - 6/4 = 2.5 (non-integer)k=3: 4 - 9/4 = 1.75 (non-integer)k=4: 1 (integer)k=5: 4 - 15/4 = 0.25 (non-integer)k=6: 4 - 18/4 = -0.5 (non-integer)k=7: 4 - 21/4 = -1.25 (non-integer)k=8: -2 (integer)So, only 0,4,8. So, three terms. Therefore, the answer is 3.But let me confirm once again. Maybe I made a mistake in the arithmetic progression part.Wait, the problem says that the coefficients of the first three terms form an arithmetic sequence. So, we found that n=8 is the solution. Then, in the expansion, the first three terms correspond to k=0,1,2. Their coefficients are 1, n/2=4, and n(n-1)/8=8*7/8=7. So, 1,4,7. The differences are 3 and 3, which is indeed an arithmetic sequence. So, that checks out. So n=8 is correct.Therefore, with n=8, the number of terms with integer exponents is 3. So the answer is 3.But wait, let me check if the term with k=4 is present. Since n=8, the expansion has 9 terms (from k=0 to k=8). So, all k from 0 to 8 are valid. Therefore, k=0,4,8 are within this range, so three terms.Alternatively, is there a term with k=4? Yes, when k=4, that's the fifth term. The exponents are 4, 3.25, 2.5, 1.75, 1, 0.25, -0.5, -1.25, -2. So, at k=4, exponent 1, which is integer. So, yes, three terms.Hence, the answer is 3.Final Answerboxed{3}

✅:To solve the problem, we start by considering the binomial expansion of (left(sqrt{x} + frac{1}{2 sqrt[4]{x}}right)^n). The exponents of (x) in each term of the expansion are given by (frac{n}{2} - frac{3k}{4}). The coefficients of the first three terms form an arithmetic sequence, leading us to solve for (n).The coefficients of the first three terms (for (k = 0, 1, 2)) are:- (C_0 = 1)- (C_1 = frac{n}{2})- (C_2 = frac{n(n - 1)}{8})Setting up the arithmetic sequence condition (C_1 - C_0 = C_2 - C_1), we solve:[frac{n}{2} - 1 = frac{n(n - 1)}{8} - frac{n}{2}]This simplifies to the quadratic equation (n^2 - 9n + 8 = 0), giving solutions (n = 1) and (n = 8). Since (n = 1) does not provide three terms, we use (n = 8).Next, we determine the number of terms with integer exponents. The exponent of (x) for each term is:[4 - frac{3k}{4}]For this to be an integer, (frac{3k}{4}) must be an integer, implying (k) must be a multiple of 4. The valid values of (k) (0, 4, 8) yield three terms with integer exponents.Thus, the number of terms in the expansion with integer exponents is (boxed{3}).

🔑:1. Consider the binomial expansion expression given by: [ left(sqrt{x} + frac{1}{2 sqrt[4]{x}}right)^n ]2. We need to find the general term (the ( (r+1) )-th term) of this binomial expansion. The general term in the expansion can be written as: [ T_{r+1} = binom{n}{r} (sqrt{x})^{n-r} left(frac{1}{2 sqrt[4]{x}}right)^r ]3. Simplify the general term ( T_{r+1} ): [ T_{r+1} = binom{n}{r} (sqrt{x})^{n-r} left(frac{1}{2 sqrt[4]{x}}right)^r = binom{n}{r} (sqrt{x})^{n-r} left(frac{1}{2}right)^r left(sqrt[4]{x}right)^{-r} ]4. Express ( sqrt{x} ) and ( sqrt[4]{x} ) in terms of powers of ( x ): [ sqrt{x} = x^{1/2} quad text{and} quad sqrt[4]{x} = x^{1/4} ]5. Substitute these into the general term: [ T_{r+1} = binom{n}{r} (x^{1/2})^{n-r} left(frac{1}{2}right)^r (x^{1/4})^{-r} ]6. Combine the exponents of ( x ): [ T_{r+1} = binom{n}{r} left(frac{1}{2}right)^r x^{frac{n-r}{2}} x^{-frac{r}{4}} ]7. Simplify the exponent of ( x ): [ T_{r+1} = binom{n}{r} left(frac{1}{2}right)^r x^{frac{n-r}{2} - frac{r}{4}} = binom{n}{r} left(frac{1}{2}right)^r x^{frac{2n-3r}{4}} ]8. Based on the given information, the coefficients of the first three terms form an arithmetic sequence; hence, we set the first three terms of the expansion: - First term: [ a_1 = binom{n}{0} left(frac{1}{2}right)^0 = 1 ] - Second term: [ a_2 = binom{n}{1} left(frac{1}{2}right)^1 = frac{n}{2} ] - Third term: [ a_3 = binom{n}{2} left(frac{1}{4}right) = frac{n(n-1)}{8} ]9. Since these coefficients ( 1, frac{n}{2}, frac{n(n-1)}{8} ) form an arithmetic sequence, we use the common difference ( d ) to establish the condition: Using ( frac{n}{2} - 1 = frac{n(n-1)}{8} - frac{n}{2} ): [ frac{n}{2} - 1 = frac{n(n-1) - 4n}{8} ] Simplify the equation: [ frac{n}{2} - 1 = frac{n^2 - 5n}{8} ]10. Multiply through by 8 to clear the denominator: [ 4n - 8 = n^2 - 5n ]11. Rearrange to form a quadratic equation: [ n^2 - 9n + 8 = 0 ]12. Factorize the quadratic equation: [ (n-8)(n-1) = 0 ]13. Solve for ( n ): [ n = 8 quad text{or} quad n = 1 ] Since ( n = 1 ) does not provide a meaningful expansion, we use ( n = 8 ).14. For ( n = 8 ), we substitute into the exponent formula to get integer exponent terms: [ frac{2n-3r}{4} = frac{16-3r}{4} ] The result must be an integer: [ frac{16-3r}{4} in mathbb{Z} ]15. Solving the integer condition for ( r ): [ 16 - 3r quad text{must be divisible by 4} ] Thus, [ 3r = 16 implies r = 0, 4, 8 ] These values of ( r ) correspond to the integer-valued exponents of ( x ). Conclusion:[boxed{3}]

🧠:Okay, so the question is: Can we select 7 points on a plane such that in any 3 of them, there must be two points with a distance of 1 between them? Hmm, interesting. Let me try to figure this out step by step.First, I need to understand what exactly is being asked. We need a set of 7 points in the plane. The condition is that no matter which 3 points we pick from this set, at least two of them must be exactly 1 unit apart. So, in other words, there are no three points in the set that are all mutually more than 1 unit apart from each other. That means every triple must contain at least one pair at distance 1.I remember something about graphs where edges represent distances. Maybe if I model this problem using graph theory. Let me think. If we consider each point as a vertex and connect two vertices with an edge if they are exactly 1 unit apart, then the condition is that every 3-vertex subset contains at least one edge. In graph terms, the complement of this graph should not contain a triangle (a set of three vertices with no edges between them). So, the complement graph should be triangle-free.Wait, so the problem reduces to finding a graph on 7 vertices such that its complement is triangle-free. Because if the complement has a triangle, that would correspond to three points in the original set where all distances are not 1, violating the condition. Therefore, the complement graph must be triangle-free. So, we need a graph G on 7 vertices where the complement graph G' is triangle-free. Then, by Turán's theorem, the maximum number of edges in a triangle-free graph on n vertices is given by the Turán number T(n,2), which is floor(n²/4). For n=7, that would be floor(49/4) = 12. So, the complement graph can have at most 12 edges. Therefore, the original graph G must have at least C(7,2) - 12 = 21 - 12 = 9 edges.So, G must have at least 9 edges, and G' must be triangle-free with 12 edges. Hmm. But does such a graph G exist? Turán's theorem gives the maximum number of edges without containing a complete graph of a certain size, but here we are talking about triangle-free graphs. So, the maximum triangle-free graph on 7 vertices has 12 edges, according to Turán. So, if G' is such a graph, then G would have 9 edges.But how does this relate to the original problem? If we can find such a graph G where every triple has at least one edge, then placing the points in the plane such that edges correspond to distance 1. However, the problem is whether such a configuration of points is possible in the plane. Because even if the graph exists combinatorially, it might not be realizable in the plane with the given distances.So, this is a question about unit distance graphs. A unit distance graph is a graph that can be embedded in the plane such that two vertices are adjacent if and only if their distance is exactly 1. So, the problem is asking if there's a unit distance graph on 7 vertices whose complement is triangle-free, or equivalently, that every 3-vertex set contains at least one edge.But I need to check if such a graph can be embedded in the plane. There might be constraints from geometry that prevent certain graphs from being realized. For example, the famous Moser spindle is a unit distance graph with 7 vertices and 11 edges, but it's 4-chromatic. Wait, but the Moser spindle has triangles, right? Let me recall. The Moser spindle is two tetrahedrons sharing a common edge, which does form some triangles. So, if the complement of the Moser spindle's graph is triangle-free, then maybe that would work? Wait, but the Moser spindle has 7 vertices and 11 edges. Then, its complement would have 21 - 11 = 10 edges. But we need the complement to have 12 edges. So, that's not enough.Alternatively, perhaps another configuration. Let me think about known unit distance graphs. The regular hexagon with its center is a unit distance graph with 7 vertices. Each vertex on the hexagon is 1 unit apart from its neighbors, and the center is 1 unit from all six perimeter vertices. So, in this case, the graph would have 6 + 6 = 12 edges. Wait, the center connected to 6 vertices, and the hexagon has 6 edges. So total 12 edges. Then, the complement would have 21 - 12 = 9 edges. But the complement graph here would have edges between the non-adjacent hexagon vertices. For example, in the hexagon, each vertex is connected to its two neighbors, so in the complement, each vertex is connected to the three vertices opposite to it (distance more than 1). But in a regular hexagon, the distance between opposite vertices is 2 units, which is more than 1, so those edges would not be in the original unit distance graph. So, in the complement, we have edges between these opposite vertices, forming triangles? Let's see.In the complement of the hexagon plus center graph, the hexagon's complement would include edges between every pair of vertices that are not adjacent in the hexagon. So, each vertex in the hexagon is connected to three others in the complement: the two that are two steps away (distance sqrt(3)) and the opposite vertex (distance 2). The center is connected to none in the complement, since it was connected to all in the original graph. Wait, no. The original graph has the center connected to all six hexagon vertices. So, in the complement graph, the center is not connected to any of the hexagon vertices. The complement edges are only between the hexagon vertices themselves. Specifically, the hexagon's complement is a graph where each vertex is connected to the three vertices not adjacent to it in the hexagon. So, each vertex is connected to three others. But does this complement graph contain triangles?Let me pick three vertices from the hexagon. Suppose we pick three alternate vertices, say every other vertex. In the original hexagon, these are not adjacent, so in the complement, they are connected. Wait, no. Wait, in the original hexagon, edges are adjacent vertices. So, in the complement, edges are between non-adjacent vertices. So, if we pick three alternate vertices in the hexagon, each pair is non-adjacent, so each pair is connected in the complement. Therefore, the complement graph contains a triangle. Therefore, the complement of the hexagon plus center graph does contain triangles, which would mean that in the original graph, there exists three points with all pairwise distances not equal to 1. Which violates the condition. Therefore, the regular hexagon plus center does not satisfy the problem's condition.So that configuration is invalid. Hmm. So, maybe the regular hexagon with center is not a good candidate. Let's think of another configuration. The Moser spindle, as I mentioned earlier, has 7 vertices and 11 edges. Let's analyze its complement. The complement would have 21 - 11 = 10 edges. If the complement has 10 edges, is it triangle-free? If it's not triangle-free, then the Moser spindle doesn't work. But how can I check that?The Moser spindle is constructed by taking two rhombuses with 60-degree angles, sharing a common edge. So, it's two overlapping triangles, each with edges of length 1. Wait, actually, the Moser spindle is a unit distance graph with 7 vertices and 11 edges. Let me recall its structure. It consists of two tetrahedrons (each tetrahedron is a complete graph K4) sharing a common edge. But each tetrahedron would require 6 edges, but sharing an edge would result in 6 + 6 - 1 = 11 edges. However, in reality, the Moser spindle is a specific graph with 7 vertices: it's two diamonds (each diamond is a rhombus) connected by an edge. Wait, maybe my memory is off. Let me try to visualize it.The Moser spindle has two vertices connected by an edge. Each of these two vertices is part of a diamond shape (a rhombus) with two other vertices. Each diamond has two edges of length 1. Then, the two diamonds are connected by another edge. Wait, maybe not. Alternatively, the Moser spindle is constructed by taking two equilateral triangles, rotating one by 30 degrees, and connecting corresponding vertices. Hmm, I might need to look up the exact structure, but since I can't do that, perhaps I should just think of properties.The key point is that the Moser spindle is a unit distance graph with 7 vertices, 11 edges, and is 4-chromatic. But how does this relate to our problem? If the complement of the Moser spindle has 10 edges, and if that complement is triangle-free, then the original graph would satisfy that every 3 vertices have at least one edge. But does the complement of the Moser spindle have a triangle?Alternatively, perhaps it's easier to think in terms of specific examples. Let me try to construct a set of 7 points and see if it satisfies the condition.Alternatively, perhaps consider the Erdos's unit distance problem, but that's more about the maximum number of unit distances. Maybe not directly applicable here.Wait, another thought. The problem is similar to the concept of a graph with no independent set of size 3. Because an independent set in the graph would correspond to three points with no two at distance 1, which is exactly what we need to avoid. So, the graph must have the property that its independence number is at most 2. So, we need a unit distance graph on 7 vertices with independence number 2. Then, such a graph would satisfy that every set of three points contains at least one edge (i.e., no independent set of size 3).So, the question reduces to: does there exist a unit distance graph on 7 vertices with independence number 2?In graph theory, a graph with independence number 2 is a graph where every three vertices contain at least one edge, which is exactly our condition.So, does such a graph exist in the plane as a unit distance graph?Alternatively, perhaps known graphs with independence number 2. For example, the complement of a triangle-free graph (as we discussed earlier) must be such that the original graph has no independent set of size 3. But whether such a graph can be represented as a unit distance graph is another question.Alternatively, perhaps consider specific configurations. For example, consider placing points in such a way that they form a lot of unit distances. Maybe start with a regular polygon and add points.Wait, the Moser spindle is supposed to have a high number of edges for a unit distance graph. It has 7 vertices and 11 edges, which is quite dense. If its complement has 10 edges, which is fewer than Turán's 12 edges for triangle-free, so maybe the complement could contain triangles. Wait, Turán's theorem says that the maximum number of edges in a triangle-free graph on 7 vertices is 12. So, if a graph has 10 edges, it's possible that it is triangle-free, but it's also possible that it contains triangles. So, the complement of the Moser spindle might have triangles. Hence, the Moser spindle itself might have independent sets of size 3, which would violate our condition. Therefore, the Moser spindle is not suitable.Alternatively, if I can find a unit distance graph with 7 vertices that is triangle-free complement, i.e., the complement has no triangles, which would mean the original graph has no independent sets of size 3.Wait, but Turán's theorem says that the maximum edges in a triangle-free graph is 12, so if the complement has 12 edges, then the original graph has 9 edges. So, maybe such a graph with 9 edges where the complement is triangle-free. But can such a graph be embedded as a unit distance graph?Alternatively, perhaps look for a specific example. Let me try to construct such a graph.Suppose we have 7 points arranged in some configuration. Let's start with two equilateral triangles sharing a common edge. That gives us 5 points. Then, add two more points somewhere such that they form unit distances with several others.Wait, let me draw this mentally. Two equilateral triangles, each with side length 1, sharing a common edge. So, points A, B, C form an equilateral triangle, and points B, C, D form another equilateral triangle. So, points A, B, C, D with B and C shared. Then, we have edges AB, BC, CA, BD, CD. Then, adding two more points, E and F. Suppose E is placed such that it's 1 unit away from A and D. Then, F is placed 1 unit away from B and E. Then, G is placed 1 unit away from C and D. Hmm, but this is getting complicated. Maybe this could lead to a graph with 9 edges, but I need to ensure that every three points have at least one edge. It's getting too vague.Alternatively, think of known graphs. The Wagner graph is a 3-regular graph on 8 vertices, but we need 7 vertices. Maybe the complement of the Wagner graph? Not sure.Alternatively, perhaps use graph properties. If the complement is triangle-free, then according to Turán's theorem, it's maximized at 12 edges. So, the original graph must have 9 edges. So, we need a 7-vertex unit distance graph with 9 edges, whose complement is triangle-free. Let's see if such a graph can exist.But how do we check if such a graph can be embedded in the plane? For example, the complete graph on 7 vertices is impossible as a unit distance graph because you can't have all pairwise distances equal to 1. But we don't need a complete graph, just 9 edges.Alternatively, perhaps arrange points in a hexagonal lattice with some points at unit distance. Wait, in a hexagonal lattice, each point has six neighbors at unit distance. But arranging 7 points in such a way might lead to multiple unit distances, but again, ensuring that any three points have at least one unit distance.Alternatively, perhaps start with a known unit distance graph and modify it. For example, the Moser spindle has 7 vertices and 11 edges. If we remove two edges from it, we get a 9-edge graph. Then, check if the complement of this modified graph is triangle-free. But removing edges could introduce independent sets. Hmm, not sure.Alternatively, consider that if a graph has independence number 2, it is the complement of a triangle-free graph. So, according to Turán's theorem, the complement has at most 12 edges. Therefore, the original graph must have at least 9 edges. So, the minimal such graph would have 9 edges. If such a graph is a unit distance graph, then the answer is yes. Otherwise, no.So, the question now becomes: is there a unit distance graph on 7 vertices with 9 edges, whose complement is triangle-free.Alternatively, perhaps think of specific coordinates. Let's try to place 7 points in the plane and check distances.Let me attempt a construction. Let's start with a regular hexagon with side length 1, and place a point at the center. As before, the regular hexagon with center has 7 points. The center is 1 unit away from each vertex. The edges of the hexagon are 1 unit. However, as we saw earlier, the complement graph contains triangles. For example, three alternate vertices on the hexagon are all distance 2 apart from each other (since in a regular hexagon, opposite vertices are 2 units apart). Therefore, in the complement graph, these three form a triangle. Therefore, in the original graph, these three points have no edges between them, meaning they are an independent set of size 3. Therefore, this configuration does not satisfy the condition.Therefore, the regular hexagon plus center is invalid.Another idea: use multiple overlapping unit triangles. For example, arrange three equilateral triangles in a star-like pattern. But I need to ensure that any three points have at least one unit distance.Alternatively, place points in two overlapping unit equilateral triangles. For example, two triangles sharing a common vertex. Suppose we have triangle ABC and triangle ADE, both equilateral with side length 1, sharing vertex A. Then, points B, C, D, E are each 1 unit from A. But distances between B and D, B and E, C and D, C and E, D and E may vary. Let's compute some distances.Assuming all triangles are in the same plane. If ABC and ADE are both equilateral triangles with side 1, and all points are in the same plane, then the distance between B and D depends on the angle between the two triangles. If the two triangles are placed such that angle between them is 60 degrees, then maybe BD is 1 unit as well. Wait, but that might form a tetrahedron, which is impossible in the plane. Hmm.Alternatively, place the two triangles in such a way that points B, A, D are colinear, but that would mean the triangles are back-to-back, forming a rhombus. Wait, if ABC and ADC are two equilateral triangles sharing a common side AC, then ABCD is a rhombus with sides of length 1 and angles 60 and 120 degrees. Then, the distance BD would be sqrt(3). So, BD is sqrt(3), which is more than 1. Then, adding more points. Let's say we have points A, B, C, D as the rhombus, then add points E, F, G.But this is getting too vague. Maybe a different approach.Suppose we have a set of points where every pair is either 1 unit apart or some other distance. To ensure that in any three points, at least one pair is 1 unit apart. So, the goal is to prevent having three points all pairwise distances greater than 1.One way to ensure this is to have the graph be such that its complement is triangle-free. So, if the complement graph has no triangles, then there are no three points all pairwise non-adjacent (i.e., all pairwise distances not 1). Therefore, such a graph would satisfy the condition.But as we saw, Turán's theorem says that the maximum number of edges in a triangle-free graph on 7 vertices is 12. Therefore, the complement graph can have up to 12 edges, so the original graph must have at least 9 edges. So, the question is whether such a graph (with 9 edges and complement triangle-free) can be realized as a unit distance graph.Alternatively, perhaps such a graph exists. For example, the Wagner graph is a triangle-free graph on 8 vertices, but we need 7. Maybe the complement of the 7-vertex cycle? Wait, a 7-cycle has a complement which is another 7-cycle. But the complement of a 7-cycle would be a 7-vertex graph where each vertex is connected to the vertices two steps away in the original cycle. The complement would have edges forming a 7-cycle as well. Wait, no. The complement of a cycle graph C7 is a graph where two vertices are adjacent if they are not adjacent in C7. So, in C7, each vertex is connected to its two neighbors. In the complement, each vertex is connected to the other four vertices. However, the complement graph of C7 is a 4-regular graph on 7 vertices, and it's actually the union of two 7-cycles? Not sure. But in any case, the complement of C7 is not triangle-free. For example, in the complement, pick three vertices that are spaced two apart in the original cycle. Then, they might form a triangle.Alternatively, let's take specific graphs. For example, the Turán graph T(7,2) is a complete bipartite graph that maximizes the number of edges without containing a triangle. The Turán graph T(n,r) is a complete r-partite graph. For r=2, it's a complete bipartite graph with partitions as equal as possible. For n=7, partitions would be 3 and 4. So, T(7,2) is K_{3,4}, which has 3*4=12 edges. So, K_{3,4} is triangle-free. Therefore, the complement of K_{3,4} is a graph on 7 vertices consisting of two cliques, one of size 3 and one of size 4, with no edges between them. Wait, no. The complement of a complete bipartite graph K_{m,n} is the union of a complete graph on m vertices and a complete graph on n vertices, plus all edges between the two partitions. Wait, no. Wait, the original graph has edges between the two partitions. The complement would have edges within each partition. So, the complement of K_{3,4} is a graph where the 3-vertex partition forms a complete graph K3, the 4-vertex partition forms a complete graph K4, and there are no edges between them. Therefore, the complement has 3 + 6 + 0 = 9 edges, which matches the earlier calculation (21 - 12 = 9). So, the complement of Turán's graph T(7,2) is a graph composed of a K3 and a K4 with no edges between them.So, if the original graph is the complement of K_{3,4}, which is two cliques K3 and K4 with no edges between them, then can such a graph be a unit distance graph? That is, can we place 7 points in the plane such that all edges within the K3 and K4 are unit distances, and there are no edges between K3 and K4. But in the plane, arranging two cliques (K3 and K4) such that all intra-clique edges are length 1, and all inter-clique edges are not 1.But arranging a K3 (triangle) with all sides 1 is straightforward. Similarly, arranging a K4 (complete graph on 4 vertices) with all edges 1 is impossible in the plane because a complete graph on 4 vertices requires some edges to be longer than others. Wait, no. In the plane, you can't have a K4 as a unit distance graph. The complete graph K4 cannot be embedded in the plane with all edges equal to 1. The only complete graphs that can be embedded as unit distance graphs in the plane are up to K3 (equilateral triangle) and K2 (two points). K4 requires 4 points with all pairwise distances 1, which is impossible in the plane. Therefore, the complement of Turán's graph K_{3,4} cannot be realized as a unit distance graph, because it contains a K4 component, which is impossible.Therefore, this approach doesn't work. So, the Turán graph's complement is not realizable as a unit distance graph.So, perhaps another way. Maybe instead of two cliques, arrange points in such a way that you have multiple overlapping unit distance triangles, but without forming a K4.Alternatively, consider constructing the graph incrementally. Start with a triangle ABC, all sides 1. Then, add a point D connected to A and B. Then, add a point E connected to B and C. Then, a point F connected to C and A. Then, a point G connected to D and E. Wait, but this might not cover all required edges. Let's see:- Triangle ABC: AB=BC=CA=1.- Point D connected to A and B: AD=BD=1. So, D is somewhere such that it's 1 unit from A and B. The only such point is the reflection of C over the line AB, but that might be 2 units away from C. Wait, if ABC is equilateral, then D would be the other intersection point of the two circles around A and B with radius 1. In that case, the distance from D to C would be 2 units. So, DC=2, which is more than 1. So, in the graph, D is connected to A and B, but not to C.Similarly, add point E connected to B and C. Similarly, E is 1 unit from B and C, but distance from E to A would be something. Similarly, F connected to C and A. Then, G connected to D and E. Each time, the new point is connected to two existing points at distance 1.But in this case, the graph is being built with multiple edges, but the problem is ensuring that every three points have at least one edge. If we follow this construction, there might still be triples without any edges. For example, points D, E, F. If D is connected only to A and B, E to B and C, F to C and A, and G to D and E, then the triple D, E, F might not have any edges among them. Since D is connected to A, B; E to B, C; F to C, A; G to D, E. So, D, E, F: D and E are connected through G? No, G is connected to D and E, but in the graph, D and E are not directly connected unless specified. Wait, in the graph as described, D is connected to A and B, E to B and C, F to C and A, and G to D and E. So, the edges are AB, BC, CA, AD, BD, BE, CE, CF, AF, DG, EG. That's 11 edges. Wait, but we started with triangle ABC (3 edges), then added D connected to A and B (2 more edges), E connected to B and C (2 edges), F connected to C and A (2 edges), and G connected to D and E (2 edges). Total 3+2+2+2+2=11 edges. So, this graph has 7 vertices and 11 edges. But we need a graph with 9 edges. Hmm.But even so, in this graph, are there triples without any edges? Let's check. Take points D, E, F. D is connected to A and B, E to B and C, F to C and A. Among D, E, F themselves, are any connected? If not, then they form an independent set of size 3. So, in the graph, D-E-F would have no edges between them, hence violating the condition. Therefore, this construction doesn't work.So, perhaps this approach of adding points connected to two existing points is insufficient. Maybe need to connect each new point to more existing points.Alternatively, arrange points in a configuration where each new point is connected to three existing points. For example, start with a triangle, then each new point is part of a new triangle with two existing points. But this might lead to overlapping triangles and more edges.Alternatively, the graph known as the "Hoffman graph" or other named graphs, but I'm not sure.Alternatively, think of the problem in terms of the chromatic number. If a unit distance graph has a high chromatic number, it might have certain properties. The Moser spindle is 4-chromatic, which means it contains an induced subgraph that requires four colors. But high chromatic number doesn't directly imply anything about the independence number.Wait, but if a graph has chromatic number k, then its independence number is at least n/k. So, if a graph has chromatic number 4, then independence number is at least 7/4 ≈ 1.75. So, at least 2. Which matches our requirement that the independence number is at most 2. So, if a unit distance graph on 7 vertices has chromatic number 4, then its independence number is at least 2. But we need it to be exactly 2. Wait, but actually, the independence number is the size of the largest independent set. So, if it's 2, then the chromatic number is at least ceiling(n/α) = ceiling(7/2) = 4. So, a graph with independence number 2 must have chromatic number at least 4. The Moser spindle is 4-chromatic, which suggests that its independence number is 2. Wait, maybe the Moser spindle has independence number 2. If that's the case, then in the Moser spindle, every three vertices contain at least one edge. Then, if the Moser spindle can be realized as a unit distance graph, then the answer is yes.But does the Moser spindle have independence number 2? Let's check. The Moser spindle is a 4-chromatic graph with 7 vertices. For a graph to be 4-chromatic, it must have an independence number at most floor(7/4) = 1.75, which is not possible. Wait, actually, the relation is that the chromatic number times the independence number is at least the number of vertices. So, χ(G) * α(G) ≥ n. So, if χ(G) = 4, then α(G) ≥ 7/4 ≈ 1.75, which implies α(G) ≥ 2. So, the independence number is at least 2. But we need it to be exactly 2. So, if the Moser spindle has α(G) = 2, then it would satisfy that every three vertices contain an edge. So, does the Moser spindle have an independence number of 2?Let me recall that the Moser spindle is a graph formed by two tetrahedrons (K4) sharing a common edge. Wait, but in reality, the Moser spindle is constructed by connecting two pairs of nodes with edges. Alternatively, let me think of it as follows: start with two diamonds (each a rhombus with two triangles) connected by an edge. This forms a graph with 7 nodes where each node has degree at least 3. To find an independent set of size 3, we need three nodes with no edges between them.Suppose we try to find three nodes with no edges. In the Moser spindle, due to its construction, every pair of nodes is either connected directly or through a short path. However, due to its complexity, there might be an independent set of size 3. For example, take one vertex from each diamond not including the shared edge. But I need to visualize it better.Alternatively, since the Moser spindle is 4-chromatic, any independent set can't be too large. Specifically, since 4 colors are needed, the graph cannot be colored with 3 colors, implying that the independence number is less than 7/3 ≈ 2.33, so it must be 2. Therefore, the independence number is 2. Therefore, the Moser spindle has α(G) = 2. Therefore, in the Moser spindle, every set of three vertices contains at least one edge. Therefore, if the Moser spindle is a unit distance graph, then it would satisfy the problem's condition.But is the Moser spindle a unit distance graph? Yes, by construction. The Moser spindle is specifically constructed as a unit distance graph. Therefore, the answer to the original question is yes: such a set of 7 points exists, and the Moser spindle is an example.Wait, but earlier I thought the complement of the Moser spindle might have triangles, but if the Moser spindle itself has independence number 2, then its complement must be such that it does not have independent sets of size 3, which is a bit confusing. Let me clarify.The Moser spindle has α(G) = 2, which means there are no independent sets of size 3. Therefore, in the original graph, every three vertices have at least one edge. Therefore, the answer is yes, we can select 7 points (the Moser spindle) such that any three have at least one pair at distance 1.But earlier, I thought the complement of the Moser spindle might have triangles, which would correspond to three points in the original graph with no edges, but if the original graph has no independent sets of size 3, then the complement cannot have triangles. Wait, no. Wait, the complement having a triangle would correspond to the original graph having an independent set of size 3. Therefore, if the original graph has no independent sets of size 3, then the complement has no triangles. Therefore, if the Moser spindle has α(G) = 2, then its complement is triangle-free. So, the Moser spindle is an example of such a graph.But earlier, I calculated that the Moser spindle has 11 edges, so its complement has 21 - 11 = 10 edges. According to Turán's theorem, the maximum number of edges in a triangle-free graph on 7 vertices is 12, so 10 edges is under that. Therefore, the complement could be triangle-free. Therefore, if the complement is triangle-free, then the original graph has the desired property.But to confirm, let's check if the Moser spindle indeed has α(G) = 2. If that's the case, then the answer is yes.A quick search in my memory tells me that the Moser spindle is indeed a graph with 7 vertices, 11 edges, chromatic number 4, and independence number 2. Therefore, it has no independent set of size 3, meaning any three vertices must contain at least one edge. Since it can be embedded in the plane as a unit distance graph, the answer to the problem is yes.Therefore, the required configuration is the Moser spindle, which is a unit distance graph with 7 vertices where every three points contain at least one pair at distance 1.

✅:The problem asks whether 7 points can be placed on a plane such that in any subset of 3 points, at least two are 1 unit apart. This is equivalent to constructing a unit distance graph with 7 vertices where every 3-vertex subset contains at least one edge (i.e., the graph has an independence number of 2). Key Steps:1. Graph Representation: Model the problem as a graph ( G ) where vertices represent points, and edges indicate a unit distance. The requirement translates to ( G ) having no independent set of size 3.2. Complement Graph Analysis: The complement graph ( G' ) must be triangle-free (no three vertices with all pairwise non-edges in ( G )). By Turán's theorem, the maximum edges in a triangle-free graph on 7 vertices is 12, so ( G ) must have at least ( binom{7}{2} - 12 = 9 ) edges.3. Moser Spindle: A known unit distance graph with 7 vertices and 11 edges. It is 4-chromatic, implying its independence number ( alpha(G) leq 2 ). This ensures every 3-vertex subset contains an edge.4. Verification: The Moser spindle’s structure prevents independent sets of size 3. Its complement has 10 edges (below Turán’s 12-edge threshold), confirming ( G' ) is triangle-free. Conclusion:Yes, such a configuration exists. The Moser spindle is a valid example of 7 points in the plane where any 3 points include at least two at unit distance.Answer: boxed{Yes}

🔑:To solve the problem, we need to explore if we can place 7 points on a plane such that any selection of 3 points includes a pair with a distance of 1 unit between them.1. Construct a Rhombus: Consider a rhombus with side lengths 1 and an internal angle of (60^circ). Let the vertices of the rhombus be named (A_1), (A_2), (A_3), and (A_4) such that (A_1) and (A_3) are the acute angle vertices (i.e., opposite each other).2. Rotate the Rhombus: Rotate the rhombus around point (A_1) to obtain a new rhombus with vertices (A_1, A_2', A_3'), and (A_4'), ensuring that (A_3 A_3' = 1).3. Set of 7 Points: Consider the set of 7 points: [ {A_1, A_2, A_2', A_3, A_3', A_4, A_4'} ] 4. Possible Pairings and Distances: For any 3 points chosen from this set of 7, we want to show that there always exists at least one pair of points at distance 1. - Any selection of 3 points contains at least two points within either the original set (S) or the rotated set (T): [ S = {A_1, A_2, A_3, A_4} ] [ T = {A_1, A_2', A_3', A_4'} ]5. Distances within Sets (S) and (T): In each of the sets (S) and (T), consider the 6 pairings of points. Out of these pairings, 5 of them will be at a distance of 1, as both are rhombuses with side length 1. - Within set (S), the pairs (A_1 A_2, A_2 A_3, A_3 A_4, A_4 A_1) and the diagonals (other than (A_1 A_3)) are all 1. - Similarly, within set (T), the pairs (A_1 A_2', A_2' A_3', A_3' A_4', A_4' A_1) and the diagonals (other than (A_1 A_3')) are 1.6. Address Mixed Sets: If 3 points chosen include exactly two points from (S) (excluding the specific point pair (A_1) and (A_3)), then one of those points from set (S) or (T) sets must be at a distance of 1 unit from the other. However, if the three points include (A_1) and (A_3) from (S), this forces one point from (T), possibly including (A_1) again which typically would maintain the distance 1 between points from (T). Hence it’s shown: - If the 3 points are {A_1, A_3, A_3'}, then (A_3 A_3' = 1).7. Conclusion: For any selection of 3 points from the set ({A_1, A_2, A_2', A_3, A_3', A_4, A_4'}), there is always a distance of 1 unit between at least one pair of points. Therefore, it is possible to select 7 points on a plane meeting the given requirement.[boxed{text{Yes}}]