💡Okay, so I have this problem about an odd function f(x) defined on all real numbers. When x is less than 0, f(x) is given as e^x times (x + 1). There are four statements about this function, and I need to figure out how many of them are correct. The options are A:1, B:2, C:3, D:4. Let me take it step by step.First, let me recall what an odd function is. An odd function satisfies the property f(-x) = -f(x) for all x in its domain. That means if I know the function's values for negative x, I can find the values for positive x by reflecting over both the x-axis and y-axis. So, since f(x) is defined for x < 0, I can find f(x) for x > 0 by using this property.Statement ① says that when x > 0, f(x) = e^x(1 - x). Hmm, let me check if this is true. Since f is odd, f(x) = -f(-x). So, for x > 0, f(x) = -f(-x). But f(-x) when x > 0 is f(-x) = e^{-x}(-x + 1). So, f(x) = -e^{-x}(-x + 1). Let me simplify that: -e^{-x}(-x + 1) = e^{-x}(x - 1). So, f(x) = e^{-x}(x - 1) for x > 0. But statement ① says f(x) = e^x(1 - x). That's different. So, ① is incorrect because it's e^{-x}(x - 1), not e^x(1 - x). So, ① is wrong.Moving on to statement ②: The function f(x) has two zeros. A zero of the function is where f(x) = 0. Let's find the zeros. For x < 0, f(x) = e^x(x + 1). Since e^x is never zero, f(x) = 0 when x + 1 = 0, so x = -1. That's one zero. For x > 0, f(x) = e^{-x}(x - 1). Again, e^{-x} is never zero, so f(x) = 0 when x - 1 = 0, so x = 1. That's another zero. What about x = 0? Since f is odd, f(0) = 0. So, actually, there are three zeros: x = -1, x = 0, and x = 1. So statement ② is incorrect because it says two zeros, but there are three. So, ② is wrong.Statement ③: The solution set of f(x) > 0 is (-1, 0) ∪ (1, +∞). Let me analyze where f(x) is positive. For x < 0, f(x) = e^x(x + 1). e^x is always positive, so the sign of f(x) depends on (x + 1). So, x + 1 > 0 when x > -1. But since x < 0, the interval where f(x) > 0 is (-1, 0). For x > 0, f(x) = e^{-x}(x - 1). e^{-x} is positive, so the sign depends on (x - 1). x - 1 > 0 when x > 1. So, f(x) > 0 for x > 1. Therefore, the solution set is indeed (-1, 0) ∪ (1, +∞). So, statement ③ is correct.Statement ④: For all x1, x2 ∈ ℝ, |f(x1) - f(x2)| < 2. This is saying that the function is bounded in the sense that the difference between any two function values is less than 2. Alternatively, this would mean the function is Lipschitz continuous with constant less than 2, but more directly, it's about the maximum difference between any two outputs. Let me think about the maximum and minimum values of f(x).First, let's analyze f(x) for x < 0: f(x) = e^x(x + 1). Let's find its maximum or minimum. Take the derivative: f’(x) = e^x(x + 1) + e^x(1) = e^x(x + 2). Setting f’(x) = 0: e^x(x + 2) = 0. Since e^x is never zero, x + 2 = 0 => x = -2. So, at x = -2, f(x) has a critical point. Let's compute f(-2): e^{-2}(-2 + 1) = e^{-2}(-1) = -1/e² ≈ -0.135. So, it's a local minimum at x = -2.As x approaches -∞, e^x approaches 0, and (x + 1) approaches -∞, but e^x decays faster, so f(x) approaches 0 from below. At x = -1, f(x) = 0. So, for x < 0, f(x) ranges from approximately -0.135 to 0.For x > 0, f(x) = e^{-x}(x - 1). Let's find its extrema. Take the derivative: f’(x) = -e^{-x}(x - 1) + e^{-x}(1) = e^{-x}(-x + 1 + 1) = e^{-x}(-x + 2). Setting f’(x) = 0: e^{-x}(-x + 2) = 0. Again, e^{-x} is never zero, so -x + 2 = 0 => x = 2. So, at x = 2, f(x) has a critical point. Compute f(2): e^{-2}(2 - 1) = e^{-2}(1) ≈ 0.135.As x approaches +∞, e^{-x} approaches 0, and (x - 1) approaches +∞, but e^{-x} decays faster, so f(x) approaches 0 from above. At x = 1, f(x) = 0. So, for x > 0, f(x) ranges from 0 to approximately 0.135.So, overall, the function f(x) ranges from approximately -0.135 to +0.135. Therefore, the maximum difference between any two function values is less than 0.27, which is certainly less than 2. So, statement ④ is correct.Wait, but let me double-check. The maximum value of f(x) is about 0.135, and the minimum is about -0.135. So, the maximum difference |f(x1) - f(x2)| would be when one is at the maximum and the other at the minimum: |0.135 - (-0.135)| = 0.27, which is less than 2. So, yes, statement ④ is correct.So, summarizing:①: Incorrect②: Incorrect③: Correct④: CorrectTherefore, two statements are correct: ③ and ④. So, the number of correct statements is 2, which is option B.**Final Answer**boxed{B}

💡Okay, so I need to find the distance from a point ( M_0 (x_0, y_0, z_0) ) to a plane given by the equation ( Ax + By + Cz + D = 0 ). Hmm, I remember that the distance from a point to a plane can be found using some formula, but I'm not exactly sure what it is. Maybe I can derive it?First, I think about what a plane is. A plane in three-dimensional space can be defined by its normal vector, which is perpendicular to the plane. The coefficients ( A, B, C ) in the plane equation ( Ax + By + Cz + D = 0 ) represent the components of this normal vector. So, the normal vector ( vec{n} ) is ( (A, B, C) ).Now, to find the distance from the point ( M_0 ) to the plane, I need to find the length of the perpendicular segment from ( M_0 ) to the plane. This makes sense because the shortest distance from a point to a plane is along the line perpendicular to the plane.I recall that the formula for the distance from a point to a plane involves the dot product of the normal vector and the vector from a point on the plane to the given point ( M_0 ). But I'm not entirely sure how to set this up.Let me think. If I take a point ( P ) on the plane, then the vector ( vec{PM_0} ) would be from ( P ) to ( M_0 ). The distance we're looking for is the length of the projection of ( vec{PM_0} ) onto the normal vector ( vec{n} ).Wait, actually, since the normal vector is perpendicular to the plane, the distance should be the absolute value of the scalar projection of ( vec{PM_0} ) onto ( vec{n} ) divided by the magnitude of ( vec{n} ).But I don't know the coordinates of point ( P ) on the plane. Maybe there's another way. I think the formula can be derived using the plane equation itself.Let me consider the plane equation ( Ax + By + Cz + D = 0 ). If I plug in the coordinates of ( M_0 ) into this equation, I get ( Ax_0 + By_0 + Cz_0 + D ). This value represents how far ( M_0 ) is from the plane, but it's not normalized by the magnitude of the normal vector.So, to get the actual distance, I need to divide this value by the magnitude of the normal vector ( sqrt{A^2 + B^2 + C^2} ). Also, since distance can't be negative, I should take the absolute value of the numerator.Putting it all together, the distance ( rho ) from ( M_0 ) to the plane is:[rho = frac{|Ax_0 + By_0 + Cz_0 + D|}{sqrt{A^2 + B^2 + C^2}}]Let me double-check this. If I have a plane ( Ax + By + Cz + D = 0 ) and a point ( M_0 ), then the numerator is the signed distance scaled by the normal vector's magnitude, and the denominator normalizes it to get the actual distance. Yes, that makes sense.I think this formula is correct. It aligns with what I remember about the distance from a point to a plane in three-dimensional space. The absolute value ensures the distance is positive, and the denominator accounts for the scaling factor due to the normal vector's length.So, to summarize, the distance from ( M_0 (x_0, y_0, z_0) ) to the plane ( Ax + By + Cz + D = 0 ) is given by:[rho = frac{|Ax_0 + By_0 + Cz_0 + D|}{sqrt{A^2 + B^2 + C^2}}]This formula should work for any point and any plane in three-dimensional space.

💡Okay, so I have this problem here about finding the slope of the tangent line to the curve y = f(x) at the point (1, f(1)). The function f(x) is given as f(x) = g(x) + x². They also mention that the equation of the tangent line to the curve y = g(x) at the point (1, g(1)) is y = 2x + 1. Hmm, I need to figure out what the slope of the tangent line to f(x) at that point is.Alright, let's break this down. First, I know that the slope of the tangent line to a curve at a particular point is given by the derivative of the function at that point. So, for f(x), the slope of the tangent line at x = 1 would be f'(1). Similarly, for g(x), the slope of the tangent line at x = 1 is g'(1).They've given me the equation of the tangent line to g(x) at (1, g(1)) as y = 2x + 1. Since the slope of this tangent line is 2, that means g'(1) must be 2. That's straightforward because the slope of the tangent line is the derivative at that point.Now, since f(x) is defined as g(x) + x², I can find f'(x) by differentiating both terms. The derivative of g(x) is g'(x), and the derivative of x² is 2x. So, f'(x) = g'(x) + 2x.To find the slope of the tangent line to f(x) at x = 1, I need to evaluate f'(1). Plugging in x = 1 into the derivative, we get f'(1) = g'(1) + 2*1. We already know that g'(1) is 2, so substituting that in, f'(1) = 2 + 2 = 4.Wait, let me make sure I didn't skip any steps or make a mistake. So, f(x) = g(x) + x², so f'(x) = g'(x) + 2x. At x = 1, f'(1) = g'(1) + 2*1. The tangent line to g(x) at x = 1 is y = 2x + 1, so the slope there is 2, meaning g'(1) = 2. Therefore, f'(1) = 2 + 2 = 4. Yeah, that seems right.I don't think I made any errors here. The key was recognizing that the slope of the tangent line to g(x) at x = 1 gives us g'(1), and then using that to find f'(1) by differentiating f(x). It all adds up to 4. So, the slope of the tangent line to f(x) at (1, f(1)) is 4.**Final Answer**The slope of the tangent line to the curve ( y = f(x) ) at the point ( (1, f(1)) ) is boxed{A}.

💡Okay, so I'm trying to solve this problem about frogs jumping on stones arranged in a circle. There are 23 stones and 22 frogs, each numbered from 1 to 22. Each frog jumps a number of stones equal to its number every minute. Frog 22 jumps 1 stone counter-clockwise, which is equivalent to jumping 22 stones clockwise because there are 23 stones in total. The goal is to prove that at some point, at least 6 stones will be empty.Hmm, let me start by understanding the setup. We have 23 stones arranged in a circle, so each stone can be labeled from 1 to 23. The frogs are initially placed randomly on these stones, and each minute, every frog jumps a certain number of stones forward in the clockwise direction. Specifically, frog number i jumps i stones. Since there are 23 stones, jumping i stones forward is equivalent to moving to the position (current position + i) mod 23.Wait, but frog 22 jumps 22 stones forward, which is the same as jumping 1 stone backward because 22 mod 23 is -1. So, frog 22 is effectively moving counter-clockwise by 1 stone each minute.Now, the problem is to show that at some point in time, at least 6 stones will be empty. That means, at some minute k, there will be at least 6 stones with no frogs on them. Since there are 23 stones and 22 frogs, if 6 stones are empty, that leaves 17 stones occupied by the 22 frogs. So, on average, each occupied stone would have about 22/17 ≈ 1.29 frogs. But since frogs can't split, some stones will have 1 frog, and some will have 2 or more.I think this might relate to the pigeonhole principle. The idea being that if we can show that the frogs must cluster on fewer stones at some point, then the remaining stones will be empty. But how do we formalize this?Let me think about the positions of the frogs over time. Each frog's position after k minutes can be described by a linear function of k. Specifically, for frog i, its position after k minutes is (a_i + i*k) mod 23, where a_i is the initial position of frog i.So, the key here is that each frog's movement is linear in k, with a different coefficient for each frog. Since the frogs have different jump distances, their movements are different modulo 23.I recall that in modular arithmetic, if you have different coefficients, the sequences generated by these linear functions will eventually cover all residues modulo 23, but since we have 22 frogs, each with a different step size, maybe their positions will overlap in some way.Wait, actually, each frog's movement is a linear congruence. For each frog i, the position after k minutes is (a_i + i*k) mod 23. So, if we fix k, we can think of each frog's position as a_i + i*k mod 23.Now, if we consider all frogs at time k, their positions are determined by these linear functions. The question is whether, for some k, these positions will overlap sufficiently to leave at least 6 stones empty.I think another way to look at this is by considering the number of distinct positions occupied by the frogs at time k. If we can show that for some k, the number of distinct positions is at most 17, then the number of empty stones would be at least 6.So, how can we show that the number of distinct positions is at most 17? Maybe by considering collisions between frogs. If two frogs end up on the same stone, that reduces the number of distinct positions.Each collision between two frogs reduces the count of distinct positions by 1. So, if we can show that there are enough collisions, then the number of distinct positions will be small enough.But how do we count the number of possible collisions? For each pair of frogs, there might be a time k when they collide, i.e., when (a_i + i*k) ≡ (a_j + j*k) mod 23. Rearranging, this gives (a_i - a_j) ≡ (j - i)*k mod 23.So, for each pair of frogs i and j, there is a unique k modulo 23 that satisfies this equation. That means, for each pair, there is exactly one k in {1, 2, ..., 23} where they collide.Since there are 22 frogs, the number of pairs is C(22, 2) = 231. Each pair collides at exactly one k in 1 to 23. So, over the 23 possible k values, there are 231 collisions spread out.By the pigeonhole principle, the average number of collisions per k is 231 / 23 ≈ 10.04. So, on average, each k has about 10 collisions. But since we can't have a fraction of a collision, some k must have at least 11 collisions.Wait, so if for some k, there are 11 collisions, that means 11 pairs of frogs are on the same stone. Each collision reduces the number of distinct positions by 1. So, starting with 22 frogs, if there are 11 collisions, the number of distinct positions would be 22 - 11 = 11. But 11 is less than 17, so that would mean more than 6 stones are empty.But the problem asks to prove that at least 6 stones are empty, so maybe I'm overcomplicating it. Let me think again.If we have 22 frogs and 23 stones, the maximum number of stones that can be occupied without any empty stones is 22. But we need to show that at some point, at least 6 stones are empty, meaning that only 17 stones are occupied.So, if we can show that for some k, the number of distinct positions is at most 17, then 23 - 17 = 6 stones are empty.To do this, let's consider the number of collisions. If the number of collisions is at least 22 - 17 = 5, then the number of distinct positions is at most 17.But wait, each collision reduces the number of distinct positions by 1. So, starting with 22 frogs, if we have c collisions, the number of distinct positions is 22 - c.We need 22 - c ≤ 17 ⇒ c ≥ 5.But earlier, we saw that the average number of collisions per k is about 10, so there must be some k with at least 11 collisions, which would give c ≥ 11, leading to 22 - 11 = 11 distinct positions, which is way more than needed.But maybe I'm miscalculating. Let me think differently.Each collision corresponds to two frogs being on the same stone. So, if we have c collisions, that means c pairs of frogs are overlapping, but the number of distinct positions is 22 - c.Wait, no, that's not quite right. Because each collision reduces the count by 1, but multiple collisions can overlap on the same stone. For example, three frogs on the same stone would result in 3 choose 2 = 3 collisions, but only reduces the distinct positions by 2.So, the relationship between the number of collisions and the number of distinct positions isn't linear. Therefore, my previous approach might not be accurate.Perhaps a better way is to consider the concept of residues. Each frog's position after k minutes is a linear function of k. Since the frogs have different step sizes, their movements are different modulo 23.I remember that in modular arithmetic, if you have functions like a_i + i*k mod 23, the number of distinct residues can be analyzed using properties of linear congruences.Specifically, for each k, the positions of the frogs are given by a_i + i*k mod 23. Since the step sizes i are all different modulo 23 (except for i=22, which is equivalent to -1), each frog's movement is a distinct linear function.Now, if we consider the set of positions {a_i + i*k mod 23 | i=1,2,...,22}, we want to show that for some k, the size of this set is at most 17.To do this, let's consider the total number of possible triples (i, j, k) such that a_i + i*k ≡ a_j + j*k mod 23. As before, this simplifies to (a_i - a_j) ≡ (j - i)*k mod 23.For each pair (i, j), there is exactly one k in {1, 2, ..., 23} that satisfies this equation. Therefore, there are C(22, 2) = 231 such triples.By the pigeonhole principle, since there are 23 possible k values, the average number of collisions per k is 231 / 23 ≈ 10.04. Therefore, there must exist some k where the number of collisions is at least 11.Now, if for a particular k, there are 11 collisions, that means 11 pairs of frogs are on the same stone. However, as I thought earlier, this doesn't directly translate to 11 fewer distinct positions because multiple collisions can overlap on the same stone.But perhaps we can bound the number of distinct positions. Let's denote the number of distinct positions as D. Each collision reduces D by 1, but if multiple collisions occur on the same stone, the reduction is less.However, even if all 11 collisions were on the same stone, the number of distinct positions would be reduced by 10 (since 11 frogs on one stone instead of 11 separate stones). But in reality, the collisions are spread out, so the reduction is somewhere between 1 and 10.But regardless, having 11 collisions implies that the number of distinct positions is significantly reduced. Let's see:If we have 11 collisions, that means 11 pairs of frogs are overlapping. The minimum number of distinct positions occurs when all these collisions are on the same stone, which would mean 11 frogs on one stone and the remaining 11 frogs on separate stones, giving D = 1 + 11 = 12. But this is just the minimum; in reality, the collisions are spread out, so D would be higher.But we need D ≤ 17. So, if we can show that for some k, D ≤ 17, then we're done.Alternatively, maybe we can use the concept of residues and the fact that the step sizes are all different. Since each frog's step size is unique modulo 23, except for frog 22 which is equivalent to -1, the positions after k minutes are all distinct unless there's a collision.Wait, no, that's not necessarily true. Even with different step sizes, two frogs can still end up on the same stone for some k.But perhaps we can use the fact that the step sizes are co-prime to 23. Since 23 is prime, all step sizes except 23 itself (which is equivalent to 0) are co-prime to 23. Therefore, each frog's movement is a generator of the additive group modulo 23.This means that for each frog, its positions after k minutes will cycle through all 23 stones as k increases. However, since all frogs have different step sizes, their cycles are different.But how does this help us? Maybe by considering that the positions of the frogs are spread out in a certain way.Alternatively, perhaps we can use the concept of the Chinese Remainder Theorem or something related to the distribution of residues.Wait, another idea: consider the function f(k) = number of distinct positions at time k. We need to show that f(k) ≤ 17 for some k.Since f(k) can vary between 1 and 22 (since there are 22 frogs), but we need to show it goes down to 17.But how? Maybe by considering the average value of f(k) over all k.The average number of distinct positions over all k is equal to the sum over k of f(k) divided by 23.But calculating this average might be complicated. Alternatively, perhaps we can use the fact that the total number of frog positions over all k is 22*23, since each frog visits 23 stones over 23 minutes.But the total number of distinct positions across all k is 23*23 = 529, but that's not directly helpful.Wait, maybe another approach: consider the expected number of distinct positions at a random k.But this might not be straightforward either.Wait, going back to the collision idea. We have 231 collisions spread over 23 k values, so average 10.04 collisions per k. Therefore, there must be some k with at least 11 collisions.Each collision reduces the number of distinct positions by 1, but as I thought earlier, multiple collisions can overlap. However, even if all 11 collisions were on separate stones, that would reduce the number of distinct positions by 11, leading to D = 22 - 11 = 11, which is way below 17.But in reality, the collisions are spread out, so the reduction is less. However, even if only 5 collisions are on separate stones, reducing D by 5, we get D = 17, which is exactly what we need.Wait, so if for some k, there are at least 5 collisions on separate stones, then D ≤ 17.But how do we ensure that? Because collisions could overlap on the same stone, reducing the number of distinct positions less than the number of collisions.But perhaps we can argue that since the average number of collisions is about 10, there must be some k where the number of distinct positions is low enough.Alternatively, maybe we can use the probabilistic method. Consider the expected number of distinct positions at a random k.But I'm not sure. Maybe another angle: consider that each stone can be occupied by multiple frogs. The total number of frogs is 22, so if we have D stones occupied, the average number of frogs per stone is 22/D.We need to show that D ≤ 17, which would mean the average is at least 22/17 ≈ 1.29. But this doesn't directly help.Wait, perhaps using the concept of the inclusion-exclusion principle or something related to the distribution of the frogs.Alternatively, maybe considering that the frogs' movements are permutations of the stones. Since each frog's movement is a permutation (because the step size is co-prime to 23), the composition of all frogs' movements might have certain properties.But I'm not sure. Maybe I need to think differently.Let me consider the problem in terms of residues. For each k, the positions of the frogs are a_i + i*k mod 23. We can think of this as a linear transformation on the initial positions.Since the step sizes are all different, the transformations are all different. Therefore, the set of positions after k minutes is a shifted version of the initial positions, but shifted by different amounts for each frog.Now, if we consider the sum of all positions modulo 23, maybe we can find some invariant or property that helps us.But I'm not sure. Alternatively, maybe considering that the frogs' movements are such that their positions after k minutes form an arithmetic progression with different differences.Wait, another idea: since each frog's movement is linear, the difference between any two frogs' positions after k minutes is (i - j)*k mod 23. So, for any two frogs i and j, their relative position after k minutes is determined by (i - j)*k mod 23.This means that for any pair of frogs, their relative position cycles through all possible differences as k increases.But how does this help? Maybe by considering that for some k, many pairs of frogs will have the same relative position, leading to collisions.But I'm not sure. Maybe I need to think about the number of distinct differences.Wait, since the step sizes are all different, the differences (i - j) are also all different modulo 23, except for pairs where i - j ≡ -(j - i).But I'm not sure. Maybe this is too vague.Wait, going back to the collision idea. We have 231 collisions spread over 23 k values, so some k must have at least 11 collisions. Each collision reduces the number of distinct positions by 1, but as I thought earlier, the reduction could be less if multiple collisions overlap.However, even if all 11 collisions were on the same stone, that would mean 11 frogs on one stone, reducing the number of distinct positions by 10 (from 22 to 12). But we only need to reduce it by 5 to get to 17.Therefore, even if the collisions are spread out, with 11 collisions, it's likely that the number of distinct positions is significantly reduced.But to make this rigorous, maybe we can use the following argument:Suppose that for some k, there are c collisions. Then, the number of distinct positions D satisfies D = 22 - c + t, where t is the number of stones with more than one frog. Wait, no, that's not quite right.Actually, each collision corresponds to two frogs being on the same stone. So, if we have c collisions, that means c pairs of frogs are overlapping. However, if multiple frogs are on the same stone, the number of collisions is C(m, 2) for m frogs on that stone.Therefore, the total number of collisions c is equal to the sum over all stones of C(m_s, 2), where m_s is the number of frogs on stone s.So, c = Σ C(m_s, 2).We need to relate c to D, the number of distinct positions. Since D is the number of stones with m_s ≥ 1, and c is the sum of C(m_s, 2) over all stones.We know that c ≥ C(22, 2) / 23 ≈ 10.04, so c ≥ 11 for some k.But how does c relate to D? We can use the inequality that Σ C(m_s, 2) ≥ C(Σ m_s, 2) / D, which is a form of Jensen's inequality because C(m, 2) is convex.So, Σ C(m_s, 2) ≥ C(22, 2) / D.But we have Σ C(m_s, 2) = c ≥ 11.Therefore, 11 ≥ C(22, 2) / D ⇒ 11 ≥ 231 / D ⇒ D ≥ 231 / 11 ⇒ D ≥ 21.Wait, that can't be right because D cannot be greater than 22. Wait, maybe I messed up the inequality.Actually, Jensen's inequality for convex functions states that Σ C(m_s, 2) ≥ C(Σ m_s, 2) / D.So, Σ C(m_s, 2) ≥ C(22, 2) / D.Given that Σ C(m_s, 2) = c ≥ 11,11 ≥ 231 / D ⇒ D ≥ 231 / 11 ⇒ D ≥ 21.But D cannot be greater than 22, so this suggests that D ≥ 21, which is not helpful because we need D ≤ 17.Wait, maybe I applied Jensen's inequality incorrectly. Let me double-check.Jensen's inequality for convex functions says that the average of the function is at least the function of the average. So, for convex f, E[f(X)] ≥ f(E[X]).In our case, f(m) = C(m, 2), which is convex. So, the average of f(m_s) over stones is at least f of the average m_s.But the average m_s is 22 / D, since there are D stones with m_s frogs each.Therefore, Σ C(m_s, 2) / D ≥ C(22 / D, 2).Multiplying both sides by D,Σ C(m_s, 2) ≥ D * C(22 / D, 2).But C(22 / D, 2) = (22/D)(22/D - 1)/2.So,c ≥ D * (22/D)(22/D - 1)/2 = (22(22 - D))/ (2D).But we know that c ≥ 11,So,11 ≥ (22(22 - D)) / (2D).Simplify:11 ≥ (22*22 - 22D) / (2D)Multiply both sides by 2D:22D ≥ 22*22 - 22D22D + 22D ≥ 22*2244D ≥ 484D ≥ 484 / 44D ≥ 11.So, this tells us that D ≥ 11, which is not helpful because we need D ≤ 17.Hmm, maybe this approach isn't working. Let me think differently.Perhaps instead of trying to bound D directly, I can consider the total number of times stones are occupied over all k.Each stone is occupied by some number of frogs at each k. The total number of frog-stone-occupancies over all k is 22*23 = 506.If we consider the average number of frogs per stone per k, it's 506 / (23*23) ≈ 0.94.But this doesn't directly help.Wait, another idea: consider that for each stone, the number of times it's occupied over all k is equal to the number of frogs whose trajectory passes through it.Since each frog's trajectory cycles through all stones, each stone is visited by each frog exactly once every 23 minutes. Therefore, each stone is occupied by 22 frogs over 23 minutes, but spread out.Wait, no, actually, each frog visits each stone exactly once every 23 minutes, so over 23 minutes, each stone is visited by each frog once. Therefore, over 23 minutes, each stone is occupied by 22 frogs, but at different times.But this means that for each stone, the number of times it's occupied is 22, spread over 23 minutes. Therefore, on average, each stone is occupied about 22/23 ≈ 0.956 times per minute.But again, this doesn't directly help.Wait, maybe considering that the total number of frog-stone-occupancies is 22*23 = 506, and there are 23 stones, so the average number of frogs per stone per minute is 506 / (23*23) ≈ 0.94.But we need to find a minute where the number of frogs is ≤ 17, which would mean that the number of empty stones is ≥ 6.But how do we show that such a minute exists?Perhaps by contradiction: assume that for all k, the number of frogs is ≥ 18, meaning that the number of empty stones is ≤ 5.Then, the total number of frog-stone-occupancies would be ≥ 18*23 = 414.But the total is 506, so 506 - 414 = 92 frog-stone-occupancies are "extra".But I'm not sure how to use this.Alternatively, maybe using the probabilistic method: consider the expected number of frogs on a stone at a random k.The expected number is 22/23 ≈ 0.956. But this is the average, not the maximum.Wait, but we need to find a k where the number of frogs is low enough.Alternatively, perhaps using the concept of derangements or something similar.Wait, another approach: consider that the frogs' movements are such that their positions after k minutes are all distinct unless there's a collision.But since we have 22 frogs and 23 stones, if all frogs were on distinct stones, there would be 1 empty stone. But we need to show that at some k, there are at least 6 empty stones.Wait, but how? Maybe by considering that the frogs' movements are such that they can't all stay spread out indefinitely.Wait, perhaps considering that the frogs' movements are permutations, and the composition of these permutations must have certain properties.But I'm not sure. Maybe I need to think about the problem differently.Wait, going back to the initial idea: for each pair of frogs, there's exactly one k where they collide. So, over 23 k values, each pair collides exactly once.Therefore, the total number of collisions is 231, as we have C(22, 2) pairs.By the pigeonhole principle, since there are 23 k values, the average number of collisions per k is 231 / 23 ≈ 10.04. Therefore, there must be some k where the number of collisions is at least 11.Now, each collision corresponds to two frogs being on the same stone. So, if we have 11 collisions, that means 11 pairs of frogs are overlapping.However, the number of distinct positions D is related to the number of collisions c. Specifically, D = 22 - c + t, where t is the number of stones with more than one frog.Wait, no, that's not quite right. Let me think again.If we have c collisions, that means c pairs of frogs are overlapping. Each collision reduces the number of distinct positions by 1, but if multiple frogs are on the same stone, the number of collisions increases.For example, if three frogs are on the same stone, that's C(3, 2) = 3 collisions, but the number of distinct positions is reduced by 2 (from 3 to 1).So, in general, if we have m frogs on a stone, that contributes C(m, 2) collisions and reduces the number of distinct positions by m - 1.Therefore, the total reduction in distinct positions is Σ (m_s - 1) = Σ m_s - D = 22 - D.But the total number of collisions is Σ C(m_s, 2).So, we have:Σ C(m_s, 2) = candΣ (m_s - 1) = 22 - DWe can relate these two using the identity:Σ C(m_s, 2) = (Σ m_s^2 - Σ m_s) / 2But Σ m_s = 22, so:c = (Σ m_s^2 - 22) / 2Therefore,Σ m_s^2 = 2c + 22Now, we can use the Cauchy-Schwarz inequality, which states that (Σ m_s^2) ≥ (Σ m_s)^2 / D = 22^2 / D = 484 / DSo,2c + 22 ≥ 484 / DWe know that c ≥ 11, so:2*11 + 22 ≥ 484 / D ⇒ 44 ≥ 484 / D ⇒ D ≥ 484 / 44 ⇒ D ≥ 11Again, this gives us D ≥ 11, which is not helpful.Wait, but we need to find an upper bound on D, not a lower bound.Perhaps we can rearrange the inequality:From Σ m_s^2 = 2c + 22 ≥ 484 / DWe have 2c + 22 ≥ 484 / D ⇒ D ≥ 484 / (2c + 22)We want to find D ≤ 17, so:484 / (2c + 22) ≤ 17 ⇒ 484 ≤ 17*(2c + 22) ⇒ 484 ≤ 34c + 374 ⇒ 110 ≤ 34c ⇒ c ≥ 110 / 34 ≈ 3.235So, c ≥ 4But we already know that c ≥ 11, so this is automatically satisfied.Therefore, this approach doesn't help us directly.Wait, maybe another way: since we have c collisions, and c ≥ 11, and we have Σ m_s^2 = 2c + 22, we can use the fact that the maximum number of frogs on a stone is at most 22, but that's trivial.Alternatively, perhaps considering that the number of stones with more than one frog is at least c / C(m, 2), but I'm not sure.Wait, let's think about it differently. Suppose that for some k, the number of distinct positions D is 18. Then, the number of stones with more than one frog is 22 - 18 = 4. Each of these 4 stones can have at least 2 frogs, contributing at least C(2, 2) = 1 collision per stone, so total collisions c ≥ 4.But we have c ≥ 11, which is much higher. Therefore, if D = 18, c ≥ 4, but we have c ≥ 11, which implies that D must be less than 18.Wait, that's a good point. If D = 18, then the number of stones with more than one frog is 4, and the minimum number of collisions is 4. But since we have c ≥ 11, which is much higher, this suggests that D must be significantly less than 18.In fact, let's formalize this. Suppose that D = 17. Then, the number of stones with more than one frog is 22 - 17 = 5. Each of these 5 stones can have at least 2 frogs, contributing at least 5 collisions. But we have c ≥ 11, which is more than 5, so D must be less than 17.Wait, no, that's not necessarily true. Because the number of collisions can be higher if some stones have more than 2 frogs.For example, if one stone has 4 frogs, that contributes C(4, 2) = 6 collisions, and the remaining 4 stones have 2 frogs each, contributing 4 collisions, totaling 10 collisions. But we have c ≥ 11, so we need at least 11 collisions.Therefore, to get c ≥ 11, we need at least 11 collisions, which could be achieved by, say, one stone with 5 frogs (C(5, 2)=10) and one stone with 2 frogs (C(2, 2)=1), totaling 11 collisions. This would correspond to D = 23 - (5 + 2 - 1) = 23 - 6 = 17 stones occupied, which is exactly what we need.Therefore, if for some k, there are 11 collisions, it's possible that D = 17, meaning 6 stones are empty.But how do we ensure that such a k exists?Well, since the average number of collisions per k is about 10.04, there must be some k where c ≥ 11. For that k, as shown above, D ≤ 17, meaning at least 6 stones are empty.Therefore, we've shown that there exists a k where the number of collisions is at least 11, which implies that the number of distinct positions D is at most 17, leaving at least 6 stones empty.So, the conclusion is that at some point, at least 6 stones will be empty.

💡Okay, so I have this problem where the sum of two numbers is 363. When I divide the larger number by the smaller one, I get a quotient of 16 and a remainder of 6. I need to find out what the larger number is. Hmm, let's see how I can approach this.First, I'll try to understand what the problem is saying. There are two numbers, let's call them x and y, and x is larger than y. Their sum is 363, so I can write that as:x + y = 363Now, when I divide x by y, I get a quotient of 16 and a remainder of 6. I remember that division can be expressed as:x = y * quotient + remainderSo in this case, that would be:x = 16y + 6Okay, so now I have two equations:1. x + y = 3632. x = 16y + 6I think I can substitute the second equation into the first one to find the value of y. Let me try that.Substituting x from the second equation into the first equation:(16y + 6) + y = 363Now, let's simplify this equation. Combine like terms:16y + y + 6 = 36317y + 6 = 363Next, I'll subtract 6 from both sides to isolate the term with y:17y = 363 - 617y = 357Now, to find y, I'll divide both sides by 17:y = 357 / 17Let me do this division. 17 times 20 is 340, and 357 minus 340 is 17. So that's 20 + 1 = 21. So y is 21.Now that I have y, I can find x using the second equation:x = 16y + 6x = 16 * 21 + 6Let me calculate 16 times 21. 16 times 20 is 320, and 16 times 1 is 16, so 320 + 16 = 336. Then add 6:x = 336 + 6 = 342So the larger number is 342. Let me just double-check to make sure everything adds up.First, check the sum: 342 + 21 = 363. That's correct.Next, check the division: 342 divided by 21. Let's see, 21 times 16 is 336, and 342 minus 336 is 6. So the quotient is 16 and the remainder is 6. That matches the problem statement.Everything seems to check out. So, the larger number is definitely 342.

💡To find the value of (cos 315^circ), I start by recognizing that (315^circ) is located in the fourth quadrant of the unit circle. In the fourth quadrant, the cosine values are positive.Next, I determine the reference angle for (315^circ) by subtracting it from (360^circ):[360^circ - 315^circ = 45^circ]This means that (cos 315^circ) is equal to (cos 45^circ).Finally, I recall that (cos 45^circ) is (frac{sqrt{2}}{2}). Therefore, the value of (cos 315^circ) is:[cos 315^circ = frac{sqrt{2}}{2}]