🧠:[<think>Okay, let's see. The problem says that f(x) is an odd function defined on all real numbers. When x is positive, f(x) is equal to -x + 2. They want to know what f(x) is when x is less than 0, and the options are A to D. Hmm, right. So, since it's an odd function, I remember that the definition of an odd function is that f(-x) = -f(x) for all x in the domain. That's the key property here. So, when x is greater than 0, they've given f(x) = -x + 2. Now, if x is less than 0, then -x would be positive, right? Because if x is negative, multiplying by -1 makes it positive. So maybe I can use the property of odd functions here. Let me try to work through it step by step.Let's take a value of x that is less than 0. Let's call this x = -a, where a is a positive number. So, if x is negative, then x = -a for some a > 0. Then, according to the definition of an odd function, f(-a) should be equal to -f(a). Because f(-x) = -f(x), so substituting x with a, we get f(-a) = -f(a). Since a is positive, f(a) is given by the expression -a + 2. Therefore, f(-a) = -f(a) = -(-a + 2) = a - 2. But wait, x is -a, so if we want to express this in terms of x, we can substitute back. Since a = -x (because x = -a), then f(x) = a - 2 = (-x) - 2 = -x - 2. So, putting that together, when x is negative, f(x) should be -x - 2. Let me check that against the options. Option D is -x - 2, so that seems like the answer. But let me verify this to make sure I didn't make a mistake.Another way to think about it is to use the definition directly. For any x < 0, f(x) = -f(-x). Because f is odd. Since x is negative, -x is positive, so we can use the given expression for positive x. So f(-x) would be -(-x) + 2 = x + 2. Then, taking the negative of that gives us -x - 2. Yep, that matches what I had before. So that's consistent.Wait, let me make sure I didn't mix up the steps here. Let me break it down again. If x < 0, then -x > 0. Therefore, f(-x) is defined as -(-x) + 2, which simplifies to x + 2. Since the function is odd, f(x) = -f(-x) = -(x + 2) = -x - 2. So that gives f(x) = -x - 2 when x < 0, which is option D. Just to double-check, maybe plug in a value. Let's take x = 1. Then f(1) = -1 + 2 = 1. Now, if x = -1, since the function is odd, f(-1) should be -f(1) = -1. Let's see what the options give for x = -1. Option D: -(-1) - 2 = 1 - 2 = -1. That's correct. Let's try another one. x = 2: f(2) = -2 + 2 = 0. Then f(-2) should be -0 = 0. Applying option D: -(-2) - 2 = 2 - 2 = 0. Perfect, that works. Wait, but just to check if another option might also work. Let's take option B: x - 2. For x = -1, that would be -1 - 2 = -3. But f(-1) should be -1, so that's not right. Option C: x + 2, so for x = -1, that's 1. Which is not equal to -f(1) = -1. Option A: -x + 2. For x = -1, that's 1 + 2 = 3, which is not -1. So only option D gives the correct value. Therefore, the answer should be D: -x - 2.Final Answerboxed{D}</think>The function ( f(x) ) is an odd function defined on (mathbb{R}). When ( x > 0 ), ( f(x) = -x + 2 ). We need to find the expression for ( f(x) ) when ( x < 0 ).An odd function satisfies the property ( f(-x) = -f(x) ) for all ( x ) in the domain. For ( x < 0 ), let ( x = -a ) where ( a > 0 ). Then, using the property of odd functions:[f(-a) = -f(a)]Since ( a > 0 ), we use the given expression ( f(a) = -a + 2 ). Therefore,[f(-a) = -(-a + 2) = a - 2]Substituting back ( a = -x ) (since ( x = -a )), we get:[f(x) = (-x) - 2 = -x - 2]Verifying with specific values:- For ( x = 1 ), ( f(1) = -1 + 2 = 1 ). Then ( f(-1) ) should be (-f(1) = -1). Using option D: ( -(-1) - 2 = 1 - 2 = -1 ), which is correct.- For ( x = 2 ), ( f(2) = -2 + 2 = 0 ). Then ( f(-2) ) should be (-f(2) = 0). Using option D: ( -(-2) - 2 = 2 - 2 = 0 ), which is correct.Thus, the expression for ( f(x) ) when ( x < 0 ) is (boxed{D}).,<think>Okay, let's see. The problem is about an odd function f(x) defined on all real numbers. When x is positive, f(x) is given as -x + 2. The question is asking what the expression for f(x) is when x is less than 0, and the options are A to D. First, I remember that an odd function has the property that f(-x) = -f(x) for all x in its domain. So, for any x, if I replace x with -x, the function value should be the negative of the original function value. That seems important here because the function is defined differently for positive and negative x. Given that when x > 0, f(x) = -x + 2. So, for x < 0, we need to find an expression such that f(x) satisfies the odd function condition. Let me write that down step by step.Let’s consider x < 0. Then, -x is positive. Since -x is positive, we can use the given expression for positive x. According to the definition of an odd function, f(x) = -f(-x). Because x is negative, -x is positive, so f(-x) is equal to -(-x) + 2. Let me compute that.If x is negative, then -x is positive. Therefore, f(-x) = -(-x) + 2. Simplifying that: -(-x) is x, so f(-x) = x + 2. Then, since f(x) is odd, f(x) = -f(-x). Therefore, substituting f(-x) in, f(x) = -(x + 2) = -x - 2.So, for x < 0, the expression should be -x - 2. Looking at the options, D is -x - 2. So, the answer should be D.Wait, let me check again. Maybe I made a mistake in the signs. Let's verify with an example. Let’s take x = -1. Then, since x is negative, we need to compute f(-1). According to the odd function property, f(-1) = -f(1). Because f(1) when x=1 is -1 + 2 = 1. Therefore, f(-1) should be -1. Now, let's plug x = -1 into the options to see which one gives -1.Option A: -(-1) + 2 = 1 + 2 = 3. Not -1. Option B: (-1) - 2 = -3. Not -1. Option C: (-1) + 2 = 1. Not -1. Option D: -(-1) - 2 = 1 - 2 = -1. Yes, that works. So, D is correct.Therefore, the answer is D. Let me make sure there isn't any other way this could be interpreted. Since the function is odd, it's symmetric about the origin. The given part for x > 0 is a linear function with slope -1 and y-intercept 2. For x < 0, the function should be the reflection over the origin, which would flip both the sign and the x-coordinate. So, reflecting the line -x + 2 over the origin would invert the slope and the intercept. The original slope is -1, so the reflected slope should be 1 (since -(-1) = 1), but wait, hold on. Wait, the reflection over the origin of a function f(x) is -f(-x). Let me compute that.Original function for x > 0: f(x) = -x + 2. Therefore, for x < 0, f(x) = -f(-x) = -[ -(-x) + 2 ] = -[x + 2] = -x - 2. Yes, that's exactly what we had before. So, the slope becomes -1 when reflected? Wait, the original slope for x > 0 is -1. For x < 0, the slope would be the derivative of f(x) = -x - 2, which is -1. Wait, but that seems like the same slope. Hmm. Wait, maybe I need to graph this mentally. If for x > 0, it's a line going downward with slope -1 starting at (0,2). But wait, hold on, if x > 0, then f(x) = -x + 2. At x = 0, that would be 2, but since the function is odd, f(0) must be 0. Hmm, there's a contradiction here. Wait, hold on. The problem says it's an odd function defined on R. But for x > 0, f(x) = -x + 2. But at x = 0, f(0) must be 0 because it's odd. However, if we plug x = 0 into the expression for x > 0, we get f(0) = -0 + 2 = 2, but f(0) must be 0. That's a problem. Wait, so the function is defined piecewise? So for x > 0, f(x) = -x + 2, and for x < 0, it's something else, but at x = 0, f(0) = 0. But the given definition for x > 0 would give f(0) = 2, which contradicts the requirement for an odd function. Therefore, maybe there's a mistake in the problem statement? Or perhaps the function is defined as x > 0, f(x) = -x + 2, and x ≤ 0, f(x) is determined by the oddness? But in that case, at x = 0, f(0) = 0, which would not match the x > 0 formula. Therefore, maybe the problem is intended to have the function defined as x > 0: f(x) = -x + 2, and x < 0: determined by oddness, with f(0) = 0. So, the problem is okay, but we have to be careful about x = 0. But since the question is about x < 0, maybe we can ignore x = 0. Let's proceed. So, when x < 0, we use the oddness property. For x < 0, f(x) = -f(-x). Since -x is positive, f(-x) = -(-x) + 2 = x + 2. Therefore, f(x) = -(x + 2) = -x - 2. So that gives us option D. But let's check again with another example. Take x = -2. Then, f(-2) should be -f(2). Since x = 2 is positive, f(2) = -2 + 2 = 0. Therefore, f(-2) = -0 = 0. Let's plug x = -2 into the options. Option D: -(-2) - 2 = 2 - 2 = 0. That works. If we use option C, x + 2: -2 + 2 = 0, which also gives 0. Wait, that's confusing. Wait, if we use option C for x = -2, it would be 0, but according to the odd function property, it should also be 0. Wait, so maybe both C and D can give 0? Wait, but let's check x = -1 again. When x = -1, according to odd property, f(-1) = -f(1). f(1) = -1 + 2 = 1. Therefore, f(-1) should be -1. Using option D: -(-1) - 2 = 1 - 2 = -1. Correct. Using option C: -1 + 2 = 1. Not correct. So, only option D works for x = -1, but x = -2 gives 0 for both C and D. Wait, but the function is linear for x > 0, so it should also be linear for x < 0. Let me check. If the function for x < 0 is option C: x + 2. Let's see if that's consistent with being odd. If x < 0, then f(x) = x + 2. Then, for x > 0, f(x) should be -f(-x). Let's take x = 1. Then, f(1) should be -f(-1) = -[ (-1) + 2 ] = -[1] = -1. But according to the problem, f(1) = -1 + 2 = 1. So that's a contradiction. Therefore, option C cannot be correct. Similarly, if we take x = 1, f(1) = 1, but if we use option C, then f(-1) = 1, so f(1) should be -1, which contradicts. Therefore, option D is correct. Another check: For x = 3, f(3) = -3 + 2 = -1. Therefore, f(-3) should be -f(3) = 1. Using option D: -(-3) - 2 = 3 - 2 = 1. Correct. Using option C: -3 + 2 = -1. Incorrect. So, yes, D works. So, the conclusion is that when x < 0, f(x) = -x - 2, which is option D. The key idea here is applying the property of odd functions, f(-x) = -f(x), and substituting -x into the expression for positive x, then taking the negative of that result to find the expression for negative x. It's important to carefully handle the signs and verify with specific examples to avoid mistakes.Final Answerboxed{D}</think>Given that the function ( f(x) ) is an odd function defined on (mathbb{R}), and ( f(x) = -x + 2 ) when ( x > 0 ), we need to determine the expression for ( f(x) ) when ( x < 0 ).An odd function satisfies the property ( f(-x) = -f(x) ) for all ( x ). For ( x < 0 ), let ( x = -y ) where ( y > 0 ). Then, we can use the given expression for ( f(y) ) when ( y > 0 ):1. For ( x < 0 ), let ( y = -x ) (so ( y > 0 )).2. By the definition of an odd function, ( f(x) = -f(-x) ).3. Substitute ( -x ) into the expression for positive ( x ): ( f(-x) = -(-x) + 2 = x + 2 ).4. Therefore, ( f(x) = -f(-x) = -(x + 2) = -x - 2 ).To verify, consider specific values:- For ( x = -1 ), ( f(-1) = -f(1) ). Since ( f(1) = -1 + 2 = 1 ), ( f(-1) = -1 ). Using the derived expression, ( f(-1) = -(-1) - 2 = 1 - 2 = -1 ), which is correct.- For ( x = -2 ), ( f(-2) = -f(2) ). Since ( f(2) = -2 + 2 = 0 ), ( f(-2) = 0 ). Using the derived expression, ( f(-2) = -(-2) - 2 = 2 - 2 = 0 ), which is correct.Thus, the expression for ( f(x) ) when ( x < 0 ) is (boxed{D}).]

🔑:Let's assume x 0 Therefore, f(-x) = -(-x) + 2 = x + 2 Since the function f(x) is an odd function defined on the domain mathbb{R}, we have f(-x) = -f(x), Therefore, f(x) = -f(-x) = -x - 2. Hence, the correct choice is: boxed{text{D}}. Analysis: Assuming x 0, substituting into the given expression to find the expression for f(-x), and then using the definition of an odd function, we obtain the expression for the function f(x) when x < 0.

🧠:[<think>Okay, so there's this problem about six students consulting their teacher about grades. The teacher says two things: A is not the best among the six, and B is not the worst among the six. Also, all six students have different grades. The question is asking for the number of different possible rankings of their grades, with the options given. Let's try to figure this out step by step.First, without any restrictions, the number of possible rankings for six students with distinct grades is 6 factorial, which is 6! = 720. That's the total number of permutations. But here, we have two restrictions: A cannot be the best (i.e., first), and B cannot be the worst (i.e., sixth). So we need to subtract the cases where A is first or B is sixth. But wait, we have to be careful here because there might be overlap between those two cases. If we just subtract them separately, we might be double-counting the overlap where A is first AND B is sixth. So we need to use the principle of inclusion-exclusion here.Let me recall the inclusion-exclusion principle. For two events, the number of elements in their union is equal to the sum of the numbers in each event minus the number in their intersection. So, the number of rankings where A is first OR B is sixth is equal to the number where A is first plus the number where B is sixth minus the number where both A is first and B is sixth.Therefore, the total number of invalid rankings (those that violate either or both conditions) is:Number of invalid = (Number with A first) + (Number with B sixth) - (Number with A first AND B sixth)Then, the valid rankings would be the total permutations minus this invalid number.Let's compute each part.Number with A first: If A is fixed in first place, the remaining 5 students can be arranged in any order. So that's 5! = 120.Number with B sixth: Similarly, if B is fixed in sixth place, the remaining 5 can be arranged in 5! = 120 ways.Number with A first AND B sixth: Here, both A is fixed in first and B is fixed in sixth. The remaining 4 students can be arranged in 4! = 24 ways.So, applying inclusion-exclusion:Invalid = 120 + 120 - 24 = 216.Therefore, the valid rankings would be 720 - 216 = 504.Hmm, 504 is one of the answer choices, option D. So maybe the answer is D) 504. But let me verify this again to make sure I didn't make any mistakes.Wait, let me double-check the logic. The total number of permutations is 720. We subtract the cases where A is first or B is sixth. The inclusion-exclusion here is correct because if we just subtract 120 + 120, we would be subtracting the overlap twice, so we need to add it back once. Therefore, the invalid is indeed 120 + 120 - 24 = 216, and 720 - 216 = 504. That seems right.Alternatively, we can calculate the valid rankings directly by considering the possible positions for A and B.For A: Since A cannot be first, there are 5 possible positions (2nd to 6th).For B: Since B cannot be sixth, there are 5 possible positions (1st to 5th).However, the positions of A and B might affect each other. Let's see. If we fix A's position and B's position, then arrange the rest. But this might get complicated because A and B's positions could overlap. Wait, maybe using multiplication principle here is tricky because the available positions for B depend on where A is placed, and vice versa.Alternatively, maybe use the principle of inclusion-exclusion again, but phrased differently.The number of permutations where A is not first and B is not last is equal to:Total permutations - permutations where A is first - permutations where B is last + permutations where A is first AND B is last.Which is exactly what we did before: 720 - 120 - 120 + 24 = 504. So that's the same result.Alternatively, another approach: For each student, except for the restrictions on A and B, the others can be arranged freely.But perhaps another way is to consider the possible positions of A and B:Case 1: A is not first and B is not sixth. So we need to count the number of permutations where A is in positions 2-6 and B is in positions 1-5, with A and B in different positions.Wait, let's think of it as two separate restrictions. Let me consider the possible positions for A and B.First, A cannot be in position 1. So A has 5 choices (positions 2-6). Then, B cannot be in position 6. So B has 5 choices (positions 1-5). However, depending on where A is placed, some positions might be taken. If A is in position 6, then B can't be in position 6, which is already taken by A. Wait, but if A is in position 6, then B can still choose from positions 1-5. Similarly, if A is not in position 6, then B still can't choose position 6. So perhaps we can split into two cases:Case 1: A is in position 6.Case 2: A is not in position 6.In Case 1: A is in position 6. Then B cannot be in position 6, so B has 5 choices (positions 1-5). The remaining 4 students can be arranged in 4! ways. So total for this case: 1 (position for A) * 5 (positions for B) * 4! = 5 * 24 = 120.Case 2: A is not in position 6. So A can be in positions 2-5 (since A can't be in 1 or 6). Wait, no: A is not in position 1, so positions 2-6. But in Case 2, we are considering A not in 6, so positions 2-5. Wait, positions 2-5? Wait, if A is not in 1 or 6, then A can be in 2-5? Wait, 6 is possible if we don't have the case 1. Wait, no, in Case 2, A is not in 6, so A is in positions 2-5. Wait, no, positions 2-6 except 6, so positions 2-5. Wait, position 6 is handled in Case 1.Wait, no, positions 2-6, but Case 1 is when A is in position 6, and Case 2 is when A is in positions 2-5. Therefore, in Case 2, A has 4 possible positions (positions 2-5). Then, B cannot be in position 6. But since A is not in position 6, position 6 is still available for others, but B can't be there. So B can be in positions 1-5, but excluding the position where A is. Since A is in positions 2-5, then B can be in positions 1-5 except where A is. So for B, there are 4 positions (since one position is taken by A). Then, the remaining 4 students can be arranged in 4! ways. So for Case 2:Number of possibilities = 4 (positions for A) * 4 (positions for B) * 4! = 4*4*24 = 16*24= 384.Then, total number of valid permutations is Case 1 + Case 2 = 120 + 384 = 504. So same result. Hmm, so this approach also gives 504.Therefore, the answer is D) 504.But let me just make sure that there isn't any mistake in this case analysis. In Case 1, A is in position 6. Then B cannot be in position 6, so B has 5 positions (1-5). The rest can be arranged in 4! ways, so 5*24=120. That seems okay.In Case 2, A is not in position 6, so A is in positions 2-5. So 4 positions for A. Then B can't be in position 6, but since A is not in position 6, B can be in positions 1-5 except where A is. So if A is in position 2, B can be in 1,3,4,5. Similarly for A in 3,4,5. So for each position of A (4 choices), B has 4 choices. Then, the remaining 4 students can be arranged in 4! ways. So 4*4*24=384. Adding 120 gives 504. So yes, that's correct.Alternatively, another way: Use the principle of multiplication with restrictions. For the first position (position 1), we can't have A. So there are 5 choices (since B is allowed here). Wait, but actually, all students except A can be in position 1. Wait, no. Let's think again.Wait, actually, if we approach this step by step, assigning positions:First, assign position 1: cannot be A. So there are 5 choices (any of the other 5 students).Then, assign position 6: cannot be B. So there are 5 choices (since B is excluded, but maybe the student assigned to position 1 is already placed, so perhaps 5 - 1 = 4? Wait, need to be careful.This approach might get complicated because the assignments are dependent. Let's see.Alternatively, use the multiplication principle considering the two restrictions:Total permutations: 6! = 720.Subtract permutations where A is first: 5! = 120.Subtract permutations where B is last: 5! = 120.Add back permutations where A is first AND B is last: 4! = 24.So total valid permutations: 720 - 120 - 120 + 24 = 504. So same answer.Yes, so regardless of the method, we get 504. So the answer is D) 504.Wait, but just to make sure, let's think of another way. Suppose we use the formula for the number of derangements, but I don't think this is a derangement problem. Because we are not deranging all elements, just two specific elements with specific restrictions.Alternatively, think of the problem as two separate constraints:1. A ≠ 1st position.2. B ≠ 6th position.So, the number of permutations where A is not first and B is not last.We can model this as two forbidden positions for A and B.The formula for the number of permutations of n elements with two forbidden positions for two different elements is:Total permutations - permutations with A in forbidden - permutations with B in forbidden + permutations with both A and B in forbidden.Which is exactly what we did. So yeah, 6! - 5! - 5! + 4! = 720 - 120 -120 +24 = 504.Therefore, all approaches lead to 504, so the answer is D.Final Answerboxed{D}</think>The problem involves six students with different grades, where A is not the best (first) and B is not the worst (sixth). We need to find the number of possible rankings.First, calculate the total number of permutations without any restrictions:[6! = 720]Next, subtract the invalid permutations using the principle of inclusion-exclusion. The invalid permutations are those where A is first, B is sixth, or both.1. Number of permutations where A is first:[5! = 120]2. Number of permutations where B is sixth:[5! = 120]3. Number of permutations where both A is first and B is sixth:[4! = 24]Using inclusion-exclusion, the number of invalid permutations is:[120 + 120 - 24 = 216]Subtracting the invalid permutations from the total permutations gives the valid permutations:[720 - 216 = 504]Thus, the number of different possible rankings of their grades is:[boxed{D}],<think>Okay, so I need to solve this problem about the number of different possible rankings for six students' grades given some conditions. Let me read the problem again and make sure I understand it.The problem says: Six students consulted about their grades. The teacher said: A is not the best among the six, B is not the worst among the six, and the six students have different grades. Then the number of different possible rankings of their grades is ( ). The options are (A) 120, (B) 216, (C) 384, (D) 504.Alright, so the question is about permutations of six students with different grades, but with two constraints: A cannot be the best (i.e., highest grade), and B cannot be the worst (i.e., lowest grade). We need to find how many possible rankings there are under these constraints.First, let's recall that without any constraints, the number of possible rankings for six students with distinct grades is 6! = 720. But here we have two restrictions, so we need to subtract the cases that violate these restrictions. However, we have to be careful about overlapping cases where both restrictions are violated, to avoid double-counting.So, let's model this using the principle of inclusion-exclusion. The total number of permutations without any restrictions is 6! = 720.Now, let's calculate the number of permutations where A is the best (i.e., ranked first). If A is fixed in the first position, then the remaining five students can be arranged in 5! ways. So, that's 5! = 120 permutations.Similarly, the number of permutations where B is the worst (i.e., ranked sixth). If B is fixed in the sixth position, the remaining five students can be arranged in 5! ways. That's also 5! = 120 permutations.However, if we subtract both of these from the total, we might be overcounting the cases where both A is the best and B is the worst. So we need to add back in the number of permutations where both A is first and B is sixth. If both A and B are fixed, then the remaining four students can be arranged in 4! ways. That's 4! = 24 permutations.So, according to inclusion-exclusion, the number of invalid permutations (where A is the best or B is the worst) is 120 + 120 - 24 = 216. Therefore, the number of valid permutations is total permutations minus invalid ones: 720 - 216 = 504.Wait a second, 504 is one of the options, option D. But let me verify this again because sometimes inclusion-exclusion can be tricky.Alternatively, maybe there's another way to approach this problem.Let's think about it step by step. The total number of permutations is 6! = 720.Constraint 1: A is not the best. So, how many permutations have A in the first position? 5! = 120. So, the number of permutations where A is not first is 720 - 120 = 600.Constraint 2: B is not the worst. So, similar logic: the number of permutations where B is in the last position is 5! = 120. So, if we subtract that from 720, we get 720 - 120 = 600. But we have two constraints here. However, we can't just subtract both 120s from 720 because some permutations might have both A first and B last, and we subtracted those twice. So we need to adjust for that overlap.So, the formula is:Valid permutations = Total permutations - permutations_with_A_first - permutations_with_B_last + permutations_with_A_first_and_B_lastWhich is exactly what I did earlier: 720 - 120 - 120 + 24 = 504. So that checks out.But let me think if there's another perspective. For example, maybe constructing the valid permutations directly.First, consider the position of A. A cannot be first. So, A can be in positions 2 to 6. Similarly, B cannot be sixth, so B can be in positions 1 to 5. However, the positions of A and B are not independent because if A is in a certain position, it affects where B can be, and vice versa.Alternatively, perhaps we can compute the number of permutations where A is not first and B is not last by considering different cases for A and B's positions.But that might complicate things more. Let's see.Suppose we first place A. Since A can't be first, there are 5 possible positions for A (positions 2-6). Then, for each of these positions, we have to place B such that B is not in position 6. But depending on where A is placed, the available positions for B might be different.Wait, if A is placed in position 2-5, then position 6 is still available, but B can't be in position 6. So if A is in positions 2-5, then B can be in positions 1-5, excluding wherever A is. If A is in position 6, then B can't be in position 6 either, so B has to be in positions 1-5. But if A is in position 6, then position 6 is already taken by A, so B can't be there, so B is in positions 1-5, same as before. Wait, but A cannot be first, but can A be sixth? The problem says A is not the best, so being sixth would mean A is the worst. But there's no restriction on A being the worst. The only restriction is A is not the best (first) and B is not the worst (sixth). So A can be in positions 2-6, and B can be in positions 1-5.So, when placing A and B:Case 1: A is in position 6 (i.e., A is the worst). Then, B cannot be in position 6, but since A is already in position 6, B can be in any of positions 1-5. So, for this case:- Position A: 1 way (position 6)- Position B: 5 choices (positions 1-5)- The remaining 4 students can be arranged in 4! ways- Total for this case: 1 * 5 * 4! = 5 * 24 = 120Case 2: A is not in position 6 (i.e., A is in positions 2-5). Then, there are 4 positions for A. Then, B cannot be in position 6, but also needs to be placed in the remaining positions (since A is in positions 2-5, which are different from position 6). So:- Position A: 4 choices (positions 2-5)- Position B: Since A is in one of positions 2-5, and B can't be in position 6, B can be in positions 1-5 except where A is. So that's 5 - 1 = 4 choices- The remaining 4 students can be arranged in 4! ways- Total for this case: 4 * 4 * 4! = 16 * 24 = 384Wait, hold on, 120 + 384 = 504. So that's the same answer as before. So this method also gives 504. Therefore, the answer is D) 504.But let me check this again. So in Case 1, where A is in position 6 (the worst), then B can be in any of the first 5 positions. So 5 choices for B, then the remaining 4 students can be arranged in 4! ways. So 5 * 24 = 120. That seems correct.In Case 2, A is not in position 6, so A can be in positions 2-5 (4 positions). For each of these, B cannot be in position 6. But since A is in positions 2-5, which are different from position 6, B has to be in the remaining 5 positions (1-5), but excluding the position where A is. Wait, actually, is that correct? Wait, if A is in positions 2-5, then B cannot be in position 6. So the available positions for B are positions 1-5, but one of those positions is already occupied by A. So actually, the available positions for B are positions 1-5 excluding where A is. Since A is in positions 2-5, then B can be in positions 1-5 except for A's position, so 4 choices. Then, the remaining 4 students can be arranged in 4! ways. So yes, 4 * 4 * 24 = 384. Then total is 120 + 384 = 504. So that seems correct.Alternatively, another way: total permutations where A is not first and B is not last. So first, total permutations where A is not first: 6! - 5! = 720 - 120 = 600. Then, from these 600 permutations, subtract the ones where B is last. But wait, in the 600 permutations where A is not first, how many have B last?In the total permutations where A is not first, how many have B last? Let's calculate that. If B is last, then A is not first. So fix B in last position and A is not first. The remaining 4 students can be arranged freely, but A can't be first. Wait, so:If B is fixed in last position, then we have 5 positions left (1-5). A can't be first, so position 1 can be any of the remaining 4 students (excluding A and B). Then, positions 2-5 can be filled with the remaining 4 students, but A is included. Wait, no:Wait, if B is fixed in position 6, then the remaining 5 positions (1-5) need to be filled with A, C, D, E, F. But A cannot be in position 1. So, position 1 can be filled by C, D, E, F (4 choices). Then, positions 2-5 can be filled with the remaining 4 students, which include A. So that's 4! ways. Therefore, the number of permutations where B is last and A is not first is 4 * 4! = 4 * 24 = 96.Therefore, the number of permutations where A is not first and B is not last is 600 (A not first) - 96 (B last in those) = 504. So same answer again.So regardless of the method, the answer seems to be 504, which is option D.But let me check with another approach for thoroughness. Let's use the principle of inclusion-exclusion again.The formula is:|A' ∩ B'| = |U| - |A ∪ B| = |U| - (|A| + |B| - |A ∩ B|)Where:- U is the universal set (all permutations): 6! = 720- A is the set of permutations where A is first: |A| = 5! = 120- B is the set of permutations where B is last: |B| = 5! = 120- A ∩ B is the set of permutations where A is first and B is last: |A ∩ B| = 4! = 24Therefore,|A' ∩ B'| = 720 - (120 + 120 - 24) = 720 - 216 = 504. Yep, same result.So all methods lead to 504, which is option D.I think this is solid. All different approaches confirm that the answer is 504. Therefore, the correct answer is D) 504.Final Answerboxed{D}</think>The problem requires determining the number of different possible rankings of six students' grades given that A is not the best (i.e., not first) and B is not the worst (i.e., not sixth), with all grades being distinct.First, we calculate the total number of permutations without any constraints, which is (6! = 720).Next, we use the principle of inclusion-exclusion to account for the constraints:1. The number of permutations where A is first (best) is (5! = 120).2. The number of permutations where B is sixth (worst) is (5! = 120).3. The number of permutations where both A is first and B is sixth is (4! = 24).Using inclusion-exclusion, the number of invalid permutations (where A is first or B is sixth) is:[120 + 120 - 24 = 216]Subtracting the invalid permutations from the total permutations gives:[720 - 216 = 504]Thus, the number of different possible rankings of their grades is (boxed{D}).,<think>Okay, so we've got this problem here where six students, A to F presumably, consulted their teacher about their grades. The teacher made two statements: A is not the best (so not the top rank), and B is not the worst (so not the bottom rank). Also, all six students have different grades. The question is asking for the number of different possible rankings of their grades, with the options given.First, I need to figure out the total number of possible rankings without any restrictions. Since there are six students and all have different grades, the total permutations would be 6 factorial, which is 6! = 720. So normally, there are 720 possible rankings.But now we have two restrictions: A cannot be first, and B cannot be sixth. So we need to subtract the cases where A is first or B is sixth. However, we have to be careful not to double subtract the cases where both A is first and B is sixth. This is a classic inclusion-exclusion problem.Let me recall the inclusion-exclusion principle here. The formula for the number of elements in the union of two sets is |A ∪ B| = |A| + |B| - |A ∩ B|. Therefore, the number of invalid rankings (where A is first or B is sixth) is equal to the number of rankings where A is first plus the number where B is sixth minus the number where both A is first and B is sixth.Calculating each part:1. Number of rankings where A is first: If A is fixed in the first position, the remaining 5 students can be arranged in any order. So that's 5! = 120.2. Number of rankings where B is sixth: Similarly, if B is fixed in the sixth position, the remaining 5 can be arranged in 5! = 120 ways.3. Number of rankings where both A is first and B is sixth: Here, both A and B are fixed. The remaining 4 students can be arranged in 4! = 24 ways.Therefore, the number of invalid rankings is 120 + 120 - 24 = 216.Subtracting this from the total number of rankings: 720 - 216 = 504. So the answer should be 504, which is option D.Wait a minute, let me double-check to make sure I didn't make a mistake here. So total permutations 720. Subtract the invalid ones where A is first or B is last. Calculated as 120 + 120 - 24 = 216. 720 - 216 = 504. That seems right.But let me think again. Maybe there's another way to approach this problem. For example, instead of subtracting the invalid cases, maybe calculate the valid ones directly.The valid rankings are those where A is not first and B is not last. So, how many ways can we arrange the students such that A is not in position 1 and B is not in position 6.For the first position, since A cannot be there, there are 5 choices (B, C, D, E, F). For the sixth position, since B cannot be there, there are 5 choices (A, C, D, E, F). However, these two choices are not independent. If we choose someone for the first position, it affects who is available for the sixth position and vice versa.Wait, maybe this direct method is more complicated because the choices for the first and sixth positions might interfere with each other. Let's try to model it.Case 1: The first position is occupied by someone other than A and B.Then, for the first position, there are 4 choices (C, D, E, F). Then, for the sixth position, since B cannot be there, there are 4 choices (A and the remaining three from the first group). Wait, maybe not. Let me think.Wait, if the first position is occupied by, say, C, then the sixth position cannot be B, so available for sixth position are A, D, E, F. So 4 choices. Then the remaining 4 positions can be filled by the remaining 4 students (including B). So permutations would be 4 * 4 * 4! ?Wait, no. Let's break it down:Case 1: First position is neither A nor B.There are 4 choices (C, D, E, F) for the first position.Then, for the sixth position, since B is allowed here? Wait, no. The sixth position cannot be B regardless. So regardless of who is first, the sixth position can't be B. So if the first position is someone else, the sixth position can be any of the remaining 5 except B. Wait, but if first position is, say, C, then the remaining students are A, B, D, E, F. So sixth position can't be B, so choices are A, D, E, F. So 4 choices.Similarly, if first position is D, same thing: sixth position can be A, C, E, F. So 4 choices.Therefore, for Case 1, first position: 4 choices (C, D, E, F). Then, sixth position: 4 choices (excluding B and the first chosen). Then, the remaining 4 positions can be filled by the remaining 4 students (including B). So 4 * 4 * 4! = 4 * 4 * 24 = 384.Case 2: First position is B.Wait, can the first position be B? The restriction is that A cannot be first. There is no restriction on B being first. So yes, B can be first.If first position is B, then the sixth position cannot be B. So now, first position is B, so remaining students are A, C, D, E, F. For the sixth position, which can't be B, so available are A, C, D, E, F. So 5 choices? Wait, but the sixth position can be any except B. Since B is already placed in first position, the sixth position can be any of the remaining 5. Wait, but the original restriction is that B is not the worst (i.e., not sixth). But B is already in first, so sixth can be any of the remaining. Wait, no. The problem says "B is not the worst among the six", so B cannot be in sixth position. But if B is already in first position, then sixth position can be anyone else. So, if first is B, sixth can be A, C, D, E, F (5 choices), and the remaining 4 positions can be filled by the remaining 4 students. So number of arrangements here is 1 (for B in first) * 5 (choices for sixth) * 4! (for the rest). So 1 * 5 * 24 = 120.Therefore, total valid permutations would be Case1 + Case2 = 384 + 120 = 504. Which matches the previous result.So this confirms that the answer is 504, which is option D.Alternatively, maybe another approach. For each position, assign ranks such that A ≠ 1 and B ≠ 6.Total permutations: 6! = 720.Subtract the permutations where A is 1 or B is 6.Number of permutations with A in 1st: 5! = 120.Number of permutations with B in 6th: 5! = 120.But we have to subtract the overlap where both A is 1st and B is 6th. That would be 4! = 24.Thus, invalid permutations: 120 + 120 - 24 = 216.Valid permutations: 720 - 216 = 504. Same result.So all methods lead to 504. Therefore, the correct answer should be D) 504.Final Answerboxed{D}</think>The problem involves determining the number of different possible rankings of six students' grades, given that A is not the best (not first) and B is not the worst (not sixth), with all students having different grades.First, calculate the total number of possible rankings without any restrictions:[6! = 720]Next, use the principle of inclusion-exclusion to subtract the invalid cases where A is first or B is sixth:1. Number of rankings where A is first:[5! = 120]2. Number of rankings where B is sixth:[5! = 120]3. Number of rankings where both A is first and B is sixth:[4! = 24]The number of invalid rankings is:[120 + 120 - 24 = 216]Subtracting the invalid cases from the total permutations gives:[720 - 216 = 504]Thus, the number of different possible rankings of their grades is:[boxed{D}],<think>Okay, so there's this problem about six students who consulted their teacher about grades. The teacher said three things: A is not the best, B is not the worst, and all six students have different grades. The question is asking for the number of different possible rankings of their grades. The options are 120, 216, 384, and 504. Let me try to figure this out step by step.First, let me parse the problem. We have six students, each with distinct grades. So, their rankings are permutations from 1st (best) to 6th (worst). Without any restrictions, the number of possible rankings would be 6! = 720. But there are two restrictions here: A cannot be the best (1st place), and B cannot be the worst (6th place). We need to subtract the cases that violate these restrictions from the total. However, since these are two separate restrictions, we have to be careful about overlapping cases where both restrictions are violated. So, inclusion-exclusion principle might be needed here.Let me recall the inclusion-exclusion principle. The formula for the number of elements in the union of two sets is |A ∪ B| = |A| + |B| - |A ∩ B|. Therefore, the number of permutations that either have A as the best or B as the worst (or both) is equal to the number of permutations with A as the best plus the number with B as the worst minus the number with both A as the best and B as the worst. Then, subtracting this from the total permutations gives the valid number.So, breaking it down:Total permutations: 6! = 720.Number of permutations where A is the best (1st place): If A is fixed in 1st place, then the remaining 5 students can be arranged in 5! = 120 ways.Number of permutations where B is the worst (6th place): Similarly, if B is fixed in 6th place, the remaining 5 students can be arranged in 5! = 120 ways.Number of permutations where both A is the best and B is the worst: If A is 1st and B is 6th, the remaining 4 students can be arranged in 4! = 24 ways.Therefore, using inclusion-exclusion:Number of invalid permutations = 120 + 120 - 24 = 216.Therefore, the number of valid permutations = Total - Invalid = 720 - 216 = 504.Looking at the options, D is 504. So, that would be the answer. Wait, but let me double-check my reasoning to make sure.Wait, another way to think about it: For each position, how many choices do we have? Let me consider A and B first.A cannot be in 1st place. So, A has 5 possible positions (2nd to 6th). Similarly, B cannot be in 6th place, so B has 5 possible positions (1st to 5th). However, since the positions of A and B are not independent, we can't just multiply 5*5 and then permute the rest. Hmm, maybe this approach is more complicated.Alternatively, maybe using inclusion-exclusion is the correct way. Let's confirm.Total permutations: 720.Subtract permutations where A is first: 720 - 120 = 600.But then, we also need to subtract permutations where B is last. However, if we subtract another 120, we would be subtracting the cases where both A is first and B is last twice. So, we have to add back those cases where both A is first and B is last. Which is 24. So, total valid permutations = 720 - 120 - 120 + 24 = 504. Yep, same result. So, that seems correct.Alternatively, let's think about constructing the permutation step by step.First, place A. A can't be first, so there are 5 choices for A's position.Then, place B. B can't be sixth, so depending on where A was placed, B's available positions might be affected. Wait, if A is in position 6, then B can't be in 6, but A is already there, so B still has 5 positions. Wait, this seems complicated. Let me see.Suppose we first place A. Since A can't be first, A has 5 possible positions: 2, 3, 4, 5, 6.Then, for each of these placements of A, we need to place B. However, if A is placed in position 6 (the worst), then B can't be in position 6, but since A is already there, B is already excluded. So, in that case, B can be in positions 1-5, which are 5 positions. If A is placed in positions 2-5, then B can't be in position 6. So, regardless of where A is placed (except position 6), B has 5 positions available. Wait, but if A is in position 6, B can't be in 6, but A is already there, so B has 5 positions. If A is not in position 6, then B has 5 positions (since B can't be in 6). Wait, in both cases, B has 5 positions? Let me check.If A is in position 6, then B cannot be in 6. But since A is already in 6, B's available positions are 1-5. If A is not in position 6, say positions 2-5, then B cannot be in 6, so B can be in 1-5, but A is already in 2-5, so B has 5 positions as well. Wait, but if A is in position 1, which is not allowed. Wait, A can't be in position 1. So, A is in positions 2-6. So, when A is in position 6, B can be in 1-5. When A is in positions 2-5, B can be in 1-5 and 6, but wait, B can't be in 6. Wait, no, the original problem says B cannot be the worst, which is 6th. So regardless of where A is placed (except in 6th), B cannot be in 6th. Wait, even if A is in 6th, B still cannot be in 6th? Wait, the problem says "B is not the worst among the six". So, regardless of where A is, B cannot be in 6th. So, even if A is in 6th, B can't be in 6th. Wait, but A is already in 6th. So, in that case, B just can't be in 6th, but since A is already there, it's already impossible. So, in effect, when A is in 6th, B can be anywhere else (positions 1-5). When A is in positions 2-5, then B can't be in 6th, so B has positions 1-5, but position where A is already placed. So, in that case, if A is in positions 2-5, then B has 5 positions (1-5) but excluding the position where A is. Wait, no. Wait, when A is in positions 2-5, then B can't be in 6th, so B must be in 1-5. However, since A is already in one of positions 2-5, B can't be in the same position as A, so B has 4 positions (1-5 excluding A's position). But if A is in position 6, then B can be in 1-5, with 5 positions. Hmm, so this complicates things.Therefore, breaking it down into cases:Case 1: A is in position 6.Then, B cannot be in 6, but A is already there. So, B can be in positions 1-5. Therefore, for B, there are 5 choices. Then, the remaining 4 students can be arranged in the remaining 4 positions in 4! = 24 ways. So, total for this case: 1 (position for A) * 5 (positions for B) * 24 = 1 * 5 * 24 = 120.Case 2: A is in positions 2-5 (positions 2,3,4,5).There are 4 possible positions for A. For each of these, B cannot be in position 6, so B can be in positions 1-5, but excluding the position where A is. So, B has 4 choices (since one position is taken by A). Then, the remaining 4 students can be arranged in the remaining 4 positions in 4! = 24 ways. So, total for this case: 4 (positions for A) * 4 (positions for B) * 24 = 4 * 4 * 24 = 384.Wait, adding up Case 1 and Case 2: 120 + 384 = 504. Which matches the previous answer. So, that's another way to confirm. Therefore, the answer is 504, which is option D.Alternatively, let's use complementary counting again. The total number of permutations is 6! = 720. The number of permutations where A is first is 5! = 120. The number of permutations where B is last is 5! = 120. The number of permutations where both A is first and B is last is 4! = 24. Therefore, by inclusion-exclusion, the number of permutations where A is first or B is last is 120 + 120 - 24 = 216. Therefore, the number of valid permutations is 720 - 216 = 504. Yep, same answer.Therefore, the answer is D) 504.Final Answerboxed{D}</think>To determine the number of different possible rankings of the six students' grades given the restrictions that A is not the best (1st place) and B is not the worst (6th place), we use the inclusion-exclusion principle.1. Total permutations without restrictions: [ 6! = 720 ]2. Permutations where A is the best (1st place): [ 5! = 120 ]3. Permutations where B is the worst (6th place): [ 5! = 120 ]4. Permutations where both A is the best and B is the worst: [ 4! = 24 ]Using the inclusion-exclusion principle, we calculate the number of invalid permutations:[120 + 120 - 24 = 216]Subtracting the invalid permutations from the total permutations gives the valid permutations:[720 - 216 = 504]Thus, the number of different possible rankings of their grades is (boxed{D}).]

🔑:7. D.Let A be the set of all possible rankings where A ranks first, and B be the set of all possible rankings where B ranks last. Then|A|=|B|=5!,|A cap B|=4!.The number of all possible rankings where A ranks first or B ranks last is|A cup B|=|A|+|B|-|A cap B|=216.According to the teacher, the number of all possible rankings of the six students' scores is6!-216=504

🧠:[<think>Okay, let's try to solve this problem step by step. So, we have two people, Xiao Cheng and Xiao Chen, starting from points A and B respectively, heading towards each other. They leave at 8:00 AM and meet at 9:40 AM, which means the meeting time is 1 hour and 40 minutes after they start. Converting that to hours, that's 1 + 40/60 = 1 + 2/3 = 5/3 hours, right?So, let's denote Xiao Cheng's speed as v1 and Xiao Chen's speed as v2. The distance between A and B would then be the sum of the distances each traveled before meeting. So, distance = v1 * (5/3) + v2 * (5/3) = (v1 + v2) * (5/3). Let's call this equation (1).Now, Xiao Cheng says that if he had traveled 10 km/h faster, they would have met 10 minutes earlier. Let's parse that. If his speed was v1 + 10, then their meeting time would have been 9:40 AM minus 10 minutes, which is 9:30 AM. So, the time taken would be 1 hour and 30 minutes, which is 1.5 hours or 3/2 hours.In this scenario, the distance between A and B would still be the same, so:(v1 + 10 + v2) * (3/2) = distance.But we already have the distance from equation (1), so we can set them equal:(v1 + v2) * (5/3) = (v1 + 10 + v2) * (3/2)Let me write this as equation (2).Then, Xiao Chen says that if he had set off half an hour earlier, they would have met 20 minutes earlier. Let's think about that. If Xiao Chen started at 7:30 AM instead of 8:00 AM, then the meeting time would have been 20 minutes earlier than 9:40 AM, which is 9:20 AM. So, the time from Xiao Chen's start would be from 7:30 AM to 9:20 AM, which is 1 hour and 50 minutes, or 1 + 50/60 = 1 + 5/6 = 11/6 hours. Meanwhile, Xiao Cheng would have started at 8:00 AM, so his travel time would be from 8:00 AM to 9:20 AM, which is 1 hour and 20 minutes, or 1 + 20/60 = 1 + 1/3 = 4/3 hours.In this case, the distance between A and B is the sum of Xiao Chen's distance (starting earlier) and Xiao Cheng's distance. So:v2 * (11/6) + v1 * (4/3) = distance.But again, distance is the same as equation (1), so:(v1 + v2) * (5/3) = v2 * (11/6) + v1 * (4/3)Let me call this equation (3).Now, we have two equations (2) and (3) with two variables v1 and v2. Let's solve them.First, equation (2):(v1 + v2) * (5/3) = (v1 + v2 + 10) * (3/2)Multiply both sides by 6 to eliminate denominators:(v1 + v2) * 10 = (v1 + v2 + 10) * 910(v1 + v2) = 9(v1 + v2) + 9010(v1 + v2) - 9(v1 + v2) = 90(v1 + v2) = 90So, that's nice, we found that the sum of their speeds is 90 km/h. Let's note this as equation (4).Then, equation (3):(v1 + v2) * (5/3) = v2 * (11/6) + v1 * (4/3)But from equation (4), (v1 + v2) = 90, so substitute:90 * (5/3) = (11/6)v2 + (4/3)v1Calculate left side: 90 * (5/3) = 150So:150 = (11/6)v2 + (4/3)v1Multiply both sides by 6 to eliminate denominators:900 = 11v2 + 8v1Now, since we know v1 + v2 = 90, we can express v1 = 90 - v2 and substitute into the equation:900 = 11v2 + 8(90 - v2)Expand the right side:900 = 11v2 + 720 - 8v2Combine like terms:900 - 720 = (11v2 - 8v2)180 = 3v2Thus, v2 = 60 km/hThen, v1 = 90 - v2 = 90 - 60 = 30 km/hSo, Xiao Cheng's speed is 30 km/h, and Xiao Chen's speed is 60 km/h.Now, the distance between A and B is (v1 + v2) * (5/3) = 90 * (5/3) = 150 km.Wait, let's check if this fits with the other scenarios.First, if Xiao Cheng's speed is 30 + 10 = 40 km/h, then their combined speed is 40 + 60 = 100 km/h. The time taken to meet would be distance / combined speed = 150 / 100 = 1.5 hours, which is 1 hour 30 minutes, so meeting at 8:00 AM + 1:30 = 9:30 AM, which is 10 minutes earlier. That checks out.For Xiao Chen starting half an hour earlier: he starts at 7:30 AM, so from 7:30 AM to meeting time, which would be 9:20 AM, that's 1 hour 50 minutes or 11/6 hours. His distance is 60 * (11/6) = 110 km. Xiao Cheng starts at 8:00 AM, so from 8:00 AM to 9:20 AM is 1 hour 20 minutes or 4/3 hours. His distance is 30 * (4/3) = 40 km. Total distance is 110 + 40 = 150 km, which matches. So the meeting time is 20 minutes earlier than original 9:40 AM, which is correct.Therefore, the distance between A and B is 150 kilometers.Final AnswerThe distance between A and B is boxed{150} kilometers.</think>Xiao Cheng and Xiao Chen set off from locations A and B respectively at 8:00 AM and met at 9:40 AM. The time taken to meet is 1 hour and 40 minutes, which is (frac{5}{3}) hours. Let their speeds be (v_1) and (v_2). The distance between A and B is:[text{Distance} = (v_1 + v_2) times frac{5}{3}]Xiao Cheng mentioned that if he traveled 10 km/h faster, they would have met 10 minutes earlier (9:30 AM), which is 1.5 hours. The distance equation becomes:[(v_1 + 10 + v_2) times frac{3}{2} = text{Distance}]Equating the two distances:[(v_1 + v_2) times frac{5}{3} = (v_1 + v_2 + 10) times frac{3}{2}]Solving this, we get:[10(v_1 + v_2) = 9(v_1 + v_2) + 90 implies v_1 + v_2 = 90]Xiao Chen mentioned that if he started half an hour earlier, they would have met 20 minutes earlier (9:20 AM). Xiao Chen's travel time is (frac{11}{6}) hours and Xiao Cheng's travel time is (frac{4}{3}) hours. The distance equation is:[v_2 times frac{11}{6} + v_1 times frac{4}{3} = text{Distance}]Substituting the distance:[90 times frac{5}{3} = frac{11}{6}v_2 + frac{4}{3}v_1]Solving this, we get:[150 = frac{11}{6}v_2 + frac{4}{3}v_1 implies 900 = 11v_2 + 8v_1]Using (v_1 + v_2 = 90):[900 = 11v_2 + 8(90 - v_2) implies 900 = 3v_2 + 720 implies 180 = 3v_2 implies v_2 = 60 implies v_1 = 30]The distance between A and B is:[(v_1 + v_2) times frac{5}{3} = 90 times frac{5}{3} = 150]The distance between A and B is (boxed{150}) kilometers.,<think>Okay, let's try to figure out this problem. So, we have two people, Xiao Cheng and Xiao Chen, starting from points A and B respectively, heading towards each other. They leave at 8:00 AM and meet at 9:40 AM. That means they traveled for 1 hour and 40 minutes before meeting. Converting that to hours, that's 1 + 40/60 = 1 + 2/3 = 5/3 hours. So, their meeting time is 5/3 hours after 8:00 AM.Now, both of them make statements about hypothetical scenarios where they either speed up or leave earlier, resulting in meeting earlier. Both statements are correct, and we need to find the distance between A and B.Let's denote:- Let v1 be Xiao Cheng's original speed (from A).- Let v2 be Xiao Chen's original speed (from B).- Let D be the distance between A and B.Since they meet after 5/3 hours, the total distance D must be covered by their combined speeds. So, D = v1*(5/3) + v2*(5/3). That simplifies to D = (v1 + v2)*(5/3). That's our first equation.Now, let's parse Xiao Cheng's statement: If he had traveled 10 km/h faster, they would have met 10 minutes earlier. So, in this scenario, his speed is v1 + 10, and the meeting time is 5/3 hours - 10 minutes. 10 minutes is 1/6 of an hour, so the new meeting time would be 5/3 - 1/6 = (10/6 - 1/6) = 9/6 = 3/2 hours. So, in this case, the time taken to meet would have been 3/2 hours.But wait, in this hypothetical scenario, Xiao Cheng is going faster, but Xiao Chen's speed is the same? Or does Xiao Chen also adjust? Wait, the problem says "If I had traveled 10 kilometers more per hour," implying only Xiao Cheng's speed changes. Xiao Chen's speed remains v2. So, their combined speed in this scenario is (v1 + 10) + v2. The distance D remains the same, so:D = (v1 + 10 + v2)*(3/2)But we already have D = (v1 + v2)*(5/3). So, setting them equal:(v1 + v2)*(5/3) = (v1 + v2 + 10)*(3/2)Hmm, that's interesting. Let me write that down:Equation 1: D = (v1 + v2)*(5/3)Equation 2: D = (v1 + v2 + 10)*(3/2)So, equate them:(v1 + v2)*(5/3) = (v1 + v2 + 10)*(3/2)Let me solve for (v1 + v2). Let's let S = v1 + v2 for simplicity.Then:S*(5/3) = (S + 10)*(3/2)Multiply both sides by 6 to eliminate denominators:6*(5/3)*S = 6*(3/2)*(S + 10)Simplify:10S = 9(S + 10)10S = 9S + 9010S - 9S = 90S = 90So, v1 + v2 = 90 km/hTherefore, the original combined speed is 90 km/h, so the distance D = 90*(5/3) = 150 km.Wait, but hold on. That seems straightforward, but we also have Xiao Chen's statement to consider. The problem says "If both of them are correct," so we need to check that this answer also satisfies Xiao Chen's statement.Xiao Chen says: “If I had set off half an hour earlier, we would have met 20 minutes earlier.”Let's analyze this scenario. If Xiao Chen had set off half an hour earlier, that means he starts at 7:30 AM instead of 8:00 AM. The original meeting time is 9:40 AM. If they meet 20 minutes earlier, that would be at 9:20 AM.So, in this scenario, Xiao Chen starts at 7:30 AM, and Xiao Cheng starts at 8:00 AM as usual. Let's denote the meeting time as t hours after 7:30 AM.But Xiao Cheng only starts at 8:00 AM, so he travels for (t - 0.5) hours until the meeting time. Wait, let me clarify:If Xiao Chen starts at 7:30 AM, and they meet at 9:20 AM, then Xiao Chen has been traveling for 1 hour and 50 minutes, which is 1 + 50/60 = 1.833... hours, or 11/6 hours. Xiao Cheng started at 8:00 AM, so he traveled for 1 hour and 20 minutes, which is 1 + 20/60 = 1.333... hours, or 4/3 hours.So, in this scenario:Distance covered by Xiao Chen: v2*(11/6)Distance covered by Xiao Cheng: v1*(4/3)Total distance D = v2*(11/6) + v1*(4/3)But D is also equal to 150 km from the previous calculation. So:v1*(4/3) + v2*(11/6) = 150But we already know that v1 + v2 = 90, so maybe we can substitute.From Equation 1, D = 150 = (v1 + v2)*(5/3). Since v1 + v2 = 90, that checks out.Now, let's use Xiao Chen's scenario equation:(4/3)v1 + (11/6)v2 = 150Multiply both sides by 6 to eliminate denominators:8v1 + 11v2 = 900We also know that v1 + v2 = 90. Let's solve the system of equations:Equation A: v1 + v2 = 90Equation B: 8v1 + 11v2 = 900From Equation A: v1 = 90 - v2. Substitute into Equation B:8*(90 - v2) + 11v2 = 900720 - 8v2 + 11v2 = 900720 + 3v2 = 9003v2 = 180v2 = 60Then v1 = 90 - 60 = 30So, Xiao Cheng's original speed is 30 km/h, Xiao Chen's original speed is 60 km/h.But let's check if these values satisfy both scenarios.First, Xiao Cheng's scenario: If he travels at v1 + 10 = 40 km/h, then combined speed is 40 + 60 = 100 km/h. Time to meet is D / (v1 + v2 + 10) = 150 / 100 = 1.5 hours = 1 hour 30 minutes. Original meeting time was 1 hour 40 minutes, so 10 minutes earlier. That matches.Xiao Chen's scenario: He starts half an hour earlier. So he travels from 7:30 AM, while Xiao Cheng starts at 8:00 AM. Let's denote the meeting time as t hours after 7:30 AM. Then Xiao Chen has traveled t hours, and Xiao Cheng has traveled (t - 0.5) hours when they meet.The distance covered by Xiao Chen: 60tThe distance covered by Xiao Cheng: 30(t - 0.5)Total distance: 60t + 30(t - 0.5) = 60t + 30t - 15 = 90t - 15 = 150So, 90t = 165 => t = 165/90 = 11/6 hours ≈ 1.833... hours = 1 hour 50 minutes. So meeting time is 7:30 AM + 1 hour 50 minutes = 9:20 AM, which is 20 minutes earlier than the original 9:40 AM. That also matches.So both statements are satisfied. Therefore, the distance between A and B is 150 km.Final AnswerThe distance between A and B is boxed{150} kilometers.</think>Given that Xiao Cheng and Xiao Chen set off from locations A and B respectively at 8:00 AM and met at 9:40 AM, we know they traveled for ( frac{5}{3} ) hours (1 hour and 40 minutes). Let ( v_1 ) be Xiao Cheng's speed, ( v_2 ) be Xiao Chen's speed, and ( D ) be the distance between A and B. The distance ( D ) can be expressed as:[D = (v_1 + v_2) cdot frac{5}{3}]Xiao Cheng's statement implies that if he traveled 10 km/h faster, they would have met 10 minutes earlier (i.e., in ( frac{3}{2} ) hours):[D = (v_1 + 10 + v_2) cdot frac{3}{2}]Equating the two expressions for ( D ):[(v_1 + v_2) cdot frac{5}{3} = (v_1 + v_2 + 10) cdot frac{3}{2}]Let ( S = v_1 + v_2 ):[S cdot frac{5}{3} = (S + 10) cdot frac{3}{2}]Solving for ( S ):[10S = 9(S + 10) implies 10S = 9S + 90 implies S = 90]Thus, ( v_1 + v_2 = 90 ) km/h, and substituting back, we find ( D ):[D = 90 cdot frac{5}{3} = 150 text{ km}]For Xiao Chen's statement, if he started half an hour earlier, they would meet 20 minutes earlier. Xiao Chen travels for ( frac{11}{6} ) hours (1 hour 50 minutes) and Xiao Cheng travels for ( frac{4}{3} ) hours (1 hour 20 minutes):[frac{4}{3}v_1 + frac{11}{6}v_2 = 150]Using ( v_1 + v_2 = 90 ):[8v_1 + 11v_2 = 900]Solving the system:[v_1 = 30 text{ km/h}, quad v_2 = 60 text{ km/h}]Both scenarios are verified to satisfy the conditions. Therefore, the distance between A and B is:[boxed{150}]]

🔑:【Solution】Solution: 9: 40-8: 00=1 hour 40 minutes =frac{5}{3} hours 1 hour 40 minutes -10 minutes =1 hour 30 minutes =frac{3}{2} hours10 times frac{3}{2}=15 text { (km) }Let the sum of the two people's speeds be x km per hourbegin{array}{c}frac{5}{3} x-frac{3}{2} x=15 frac{1}{6} x=15 x=15 times 6 x=90 90 times frac{5}{3}=150 text { (km) }end{array}Answer: The distance between location A and location B is 150 km.The answer is: 150 .

🧠:[<think>Okay, so I need to find the unit digit of 2023 raised to the power of 2022 to the 2021st power. That is, the unit digit of 2023^(2022^2021). Hmm, that's a massive exponent, but since we're only interested in the last digit, maybe there's a pattern or a shortcut we can use instead of trying to compute the whole thing. Let me think.First, I remember that the unit digit of a number raised to a power cycles in a pattern. For example, 3^1 is 3, 3^2 is 9, 3^3 is 27 (unit digit 7), 3^4 is 81 (unit digit 1), and then it repeats every 4 powers: 3, 9, 7, 1, 3, 9, etc. So if we can figure out where the exponent falls in this cycle, we can determine the unit digit. In this case, the base is 2023, whose unit digit is 3. So, the unit digit of 2023^(anything) will follow the same cycle as 3^(anything). Therefore, the problem reduces to finding 3 raised to the power of (2022^2021) modulo 4, since the cycle length is 4. Wait, why modulo 4? Because the cycle repeats every 4, so the exponent modulo 4 will tell us the position in the cycle. If the exponent is congruent to 1 mod 4, it's the first term (3), 2 mod 4 is the second (9), and so on. But if the exponent is 0 mod 4, that would correspond to the fourth term (1). So, to find 3^(2022^2021) mod 10 (since we want the unit digit, which is equivalent to mod 10), we can use Euler's theorem or just the cyclicity. Since 3 and 10 are coprime, Euler's theorem tells us that 3^φ(10) ≡ 1 mod 10, where φ(10) is Euler's totient function. φ(10) = 4, so 3^4 ≡ 1 mod 10. That confirms the cycle length of 4. Therefore, 3^k mod 10 = 3^(k mod 4) mod 10. However, if k mod 4 is 0, then it's 3^4 ≡ 1 mod 10.Therefore, we need to compute the exponent 2022^2021 mod 4. Let's calculate that. First, 2022 divided by 4: 2022 ÷ 4 is 505.5, which is 505 * 4 = 2020, so 2022 = 4*505 + 2. Therefore, 2022 ≡ 2 mod 4. So, 2022^2021 ≡ 2^2021 mod 4. Now, 2^1 = 2 mod 4, 2^2 = 0 mod 4, and any higher power of 2 is 0 mod 4. Therefore, 2^2021 mod 4. Wait, 2^1 mod 4 = 2, 2^2 mod 4 = 0, 2^3 mod 4 = 0, etc. So, for exponents greater than or equal to 2, 2^exponent is 0 mod 4. Therefore, since 2021 is greater than 1, 2^2021 ≡ 0 mod 4. So, 2022^2021 ≡ 0 mod 4.But hold on, when we reduce the exponent modulo 4, if the result is 0, we need to use 4 instead (since 3^4 ≡ 1 mod 10, so if the exponent is 0 mod 4, it's equivalent to 4). Therefore, the exponent 2022^2021 is congruent to 0 mod 4, which would imply that 3^(2022^2021) ≡ 3^4 ≡ 1 mod 10. But wait, let me check that again. If the exponent is 0 mod 4, then 3^(4k) ≡ (3^4)^k ≡ 1^k ≡ 1 mod 10. So, yes, the unit digit would be 1. But the options given are (A) 3, (B) 9, (C) 7, (D) 1, (E) 5. So, is the answer D) 1?Wait, but let me recheck my steps to be sure. 1. The base is 2023, unit digit 3. So, we need to find 3^(2022^2021) mod 10.2. The cycle of 3 mod 10 is 4: 3, 9, 7, 1, 3, 9, etc.3. Therefore, 3^k mod 10 = 3^(k mod 4) mod 10, with the exception when k mod 4 = 0, then it's 1.4. So, compute exponent k = 2022^2021 mod 4.5. 2022 mod 4 = 2. So, k = 2^2021 mod 4.6. Since 2^2 = 4 ≡ 0 mod 4, any exponent higher than or equal to 2 for 2 will be 0 mod 4. So, 2^2021 mod 4 = 0, since 2021 ≥ 2.7. Therefore, k ≡ 0 mod 4.8. Therefore, 3^k mod 10 = 3^(4) mod 10 = 81 mod 10 = 1.Therefore, the unit digit is 1, which is option D. But wait, let me confirm once more because sometimes when reducing exponents in Euler's theorem, you have to be careful with the modulus. Wait, φ(10) is 4, so 3^4 ≡ 1 mod 10. Therefore, 3^(4m) ≡ 1 mod 10. Since k = 2022^2021 is 0 mod 4, i.e., 4m, so 3^(4m) ≡ 1 mod 10. Therefore, yes, 1 is the correct answer. So, option D.But let me check with a different approach to be absolutely sure. Maybe using the concept of cyclicity directly.The unit digit cycles of 3 are every 4. So, exponent cycles:- Exponent 1: 3- Exponent 2: 9- Exponent 3: 7- Exponent 4: 1- Exponent 5: 3, and so on.So, the exponent in question is 2022^2021. Let's find what 2022^2021 is mod 4. As before, 2022 mod 4 is 2. So, 2^2021 mod 4. Since 2^1=2 mod4, 2^2=0 mod4, 2^3=0 mod4, etc. So, for any exponent greater than or equal to 2, 2^exponent is 0 mod4. Therefore, 2^2021 mod4=0. So, the exponent is 0 mod4, which corresponds to the 4th position in the cycle, which is 1. Therefore, unit digit is 1. So, answer D.Alternatively, maybe we can write out the exponent 2022^2021 as 2^2021 + 4*505^2021 or something, but that seems unnecessary. Alternatively, note that 2022 is even, so 2022^2021 is divisible by 2^2021, which is certainly divisible by 4, since 2021 >=2. So, 2022^2021 is a multiple of 4. Therefore, the exponent is a multiple of 4. So, 3^(multiple of 4) ends with 1. Therefore, answer D.Alternatively, let's take smaller exponents to test the pattern. For example, 3^1=3, 3^2=9, 3^3=27, 3^4=81, 3^5=243 (unit digit 3), so cycle length 4. So, exponent mod4:If exponent ≡1 mod4, unit digit 3.If exponent≡2 mod4, unit digit 9.If exponent≡3 mod4, unit digit 7.If exponent≡0 mod4, unit digit 1.Therefore, if we can confirm that 2022^2021 ≡0 mod4, then unit digit is 1.2022 mod4=2, so 2^2021. Now, 2 mod4=2.2^1=2 mod4.2^2=4≡0 mod4.2^3=8≡0 mod4.Therefore, for exponents n≥2, 2^n≡0 mod4. Since 2021≥2, 2^2021≡0 mod4.Therefore, exponent is 0 mod4, so 3^0 mod4? Wait, no. Wait, exponent is 0 mod4, so 3^(4k)= (3^4)^k=81^k≡1^k=1 mod10. Therefore, unit digit 1. So, answer D.Wait, but let me take an example with smaller numbers to verify.Suppose we have 3^(2^2). That is 3^4=81, unit digit 1. Which is 3^(4) as exponent, which is 0 mod4, so unit digit 1.Another example: 3^(2^1) =3^2=9, unit digit 9. Here, exponent 2≡2 mod4, so unit digit 9. Which matches.Another example: 3^(2^3). 2^3=8, so 3^8. 3^4=81, 3^8=(3^4)^2=81^2=6561, unit digit 1. Here, exponent 8≡0 mod4, unit digit 1. Correct.Another example: 3^(2^4). 2^4=16, so 3^16. Since 3^4=81, 3^16=(3^4)^4=81^4. Regardless, unit digit is 1. So, same as before. So, if exponent is multiple of 4, unit digit is 1.Therefore, in our problem, since exponent 2022^2021≡0 mod4, the unit digit is 1. Therefore, answer D.Wait, but wait a second. Let me check another angle. Let's suppose the exponent was 2 instead of 2021. Then, 2022^2. 2022 mod4=2, so 2^2=4≡0 mod4. So, 3^0 mod10=1. Wait, but 3^4=81≡1 mod10. So, exponent 4 gives 1. So, even if the exponent is 0 mod4, you have to compute 3^4 mod10=1, not 3^0. Because exponent is 4k, which is 3^(4k)= (3^4)^k=1^k=1. So, the key is that the exponent on 3 is a multiple of 4. Therefore, even if the exponent is 0 mod4, which is 4k, then it's 1. So, the conclusion holds.Therefore, I think the answer is D) 1. But let me check once more with different numbers. For example, let's take 3^(4). That's 81, unit digit 1. So, exponent 4. Then, if we have 3^(2^2), which is 3^4, same result. If we have 3^(2^3)=3^8, which is 6561, unit digit 1. So, same. Therefore, any exponent that's a multiple of 4 on 3 gives unit digit 1. Therefore, since 2022^2021 is a multiple of 4, the unit digit is 1. So, answer D.But wait, the options given are A)3, B)9, C)7, D)1, E)5. So, D is 1. Therefore, the correct answer should be D.But just to make sure, maybe the problem is with my initial assumption. Let me check the steps again:1. Unit digit of 2023^(2022^2021) is same as unit digit of 3^(2022^2021).2. The cycle for 3's unit digit is 4: 3,9,7,1,...3. Therefore, need to compute 2022^2021 mod4.4. 2022 mod4=2.5. So, 2^2021 mod4. Since 2^2=0 mod4, and 2021 is >=2, 2^2021 mod4=0.6. Therefore, 2022^2021 mod4=0.7. Therefore, exponent is 0 mod4, so 3^(4k) mod10=1.Hence, unit digit 1. So, answer D. Yes, that seems correct. Therefore, I think the answer is D)1.Final Answerboxed{D}</think>To find the unit digit of (2023^{left(2022^{2021}right)}), we need to consider the unit digit of (3^{left(2022^{2021}right)}) because the unit digit of 2023 is 3. The unit digit of powers of 3 follows a cycle: 3, 9, 7, 1, repeating every 4 terms. Therefore, we need to determine the exponent (2022^{2021} mod 4).1. Calculate (2022 mod 4): [ 2022 div 4 = 505 text{ remainder } 2 implies 2022 equiv 2 mod 4 ]2. Now, we need to find (2^{2021} mod 4): - (2^1 equiv 2 mod 4) - (2^2 equiv 0 mod 4) - For any exponent (n geq 2), (2^n equiv 0 mod 4) Since (2021 geq 2), we have: [ 2^{2021} equiv 0 mod 4 ]3. Therefore, (2022^{2021} equiv 0 mod 4).4. This means we need to find (3^0 mod 10). However, since the cycle length is 4, we use (3^4 equiv 1 mod 10). Hence, any exponent that is a multiple of 4 will result in a unit digit of 1.Thus, the unit digit of (2023^{left(2022^{2021}right)}) is (boxed{D}).,<think>Alright, so I need to find the unit digit of 2023 raised to the power of (2022^2021). Hmm, that seems like a big exponent. But I remember that unit digits often repeat in cycles, so maybe I can figure out the pattern here. Let me think step by step.First, the unit digit of a number raised to a power depends only on the unit digit of the base. So even though the base here is 2023, which has a unit digit of 3, I can just focus on 3 raised to the power of (2022^2021). That simplifies the problem a bit.Now, let me recall the pattern of the unit digits when raising 3 to different powers. Let's list them out:- 3^1 = 3 --> unit digit 3- 3^2 = 9 --> unit digit 9- 3^3 = 27 --> unit digit 7- 3^4 = 81 --> unit digit 1- 3^5 = 243 --> unit digit 3- 3^6 = 729 --> unit digit 9Oh, so the unit digits cycle every 4 exponents: 3, 9, 7, 1, and then repeats. So the cycle length here is 4. That means to find the unit digit of 3^k, I need to find k modulo 4. If k is a multiple of 4, the unit digit is 1. If k ≡ 1 mod 4, it's 3; if k ≡ 2 mod 4, it's 9; and if k ≡ 3 mod 4, it's 7.Therefore, in our problem, we need to compute the exponent (2022^2021) modulo 4. Once we have that result, say it's r, then the unit digit of 3^r will be the answer.So let's compute (2022^2021) mod 4. Let's break that down.First, 2022 divided by 4. 2022 ÷ 4 = 505.5, so 4×505 = 2020, so 2022 = 4×505 + 2. Therefore, 2022 ≡ 2 mod 4. Therefore, 2022^2021 ≡ 2^2021 mod 4.Now, 2^1 mod 4 = 22^2 mod 4 = 02^3 mod 4 = 0...So once the exponent is 2 or higher, 2^k ≡ 0 mod 4. Since 2021 is definitely greater than 2, 2^2021 mod 4 = 0. Therefore, the exponent (2022^2021) ≡ 0 mod 4.But wait, in the cycle of 4, we can't have a remainder of 0. Wait, in modulus terms, 0 mod 4 is equivalent to 4 mod 4. So in the cycle, if the exponent is 0 mod 4, that would correspond to the 4th term in the cycle, which is 1. So if the exponent is congruent to 0 mod 4, the unit digit is 1. Let me confirm that.Yes, 3^4 is 81, unit digit 1. 3^8 is 6561, unit digit 1. So when the exponent is a multiple of 4, the unit digit is 1.So, putting it all together:Unit digit of 2023^(2022^2021) is the same as unit digit of 3^(2022^2021). Since 2022^2021 ≡ 0 mod 4, 3^(0 mod 4) is 3^4, which has unit digit 1. Therefore, the answer should be 1, which is option D.Wait, but let me double-check. Maybe I made a mistake in computing the exponent.Wait, exponent is 2022^2021. 2022 is congruent to 2 mod 4, so 2^2021 mod 4. As mentioned before, 2^1 mod 4 = 2, 2^2 mod 4 = 0, and for any exponent higher than or equal to 2, 2^k ≡ 0 mod 4. So 2^2021 ≡ 0 mod 4, since 2021 >= 2. Therefore, 2022^2021 ≡ 0 mod 4. So then, 3^(0 mod 4) = 3^4, which is 81, unit digit 1.But wait, let me check with a smaller exponent. Let's say the exponent is 2, so 2022^2. Then 2022 mod 4 is 2, so 2^2=4, which is 0 mod 4. So exponent is 0 mod 4, and 3^0 mod 4? Wait, no. Wait, in modular arithmetic, exponents are not directly modular unless we use Euler's theorem or something. Wait, but in this case, the exponent itself is computed modulo φ(n), where φ is Euler's totient function. Wait, but here, to compute 3^k mod 10 (since we care about the unit digit), we can use Euler's theorem. But since 3 and 10 are coprime, φ(10)=4. So 3^4 ≡ 1 mod 10. Therefore, 3^k ≡ 3^(k mod 4) mod 10. So yes, the exponent k = 2022^2021, so k mod 4.We found that k = 2022^2021 ≡ 0 mod 4. Therefore, 3^k ≡ 3^0 ≡ 1 mod 10. Wait, but 3^0 is 1. Wait, but if k ≡ 0 mod 4, then 3^(4m) ≡ (3^4)^m ≡ 1^m ≡ 1 mod 10. So yes, that works. So the unit digit is 1.Therefore, the answer is D) 1.But wait, the options are (A)3, (B)9, (C)7, (D)1, (E)5. So yeah, D is 1. But wait, let me check with another approach.Alternatively, maybe I can compute the exponent modulo φ(10)=4, but since 10 is composite, φ(10)=4. Then 3^4 ≡ 1 mod 10, so 3^k mod 10 cycles every 4. So, exponent k is 2022^2021. So k mod 4.2022 mod 4 is 2, so 2022 ≡ 2 mod 4. Therefore, 2022^2021 ≡ 2^2021 mod 4. As 2^2=4≡0 mod 4, so any higher power of 2 will be 0 mod 4. Therefore, 2^2021 ≡ 0 mod 4. So k ≡ 0 mod 4. Then 3^k ≡ 3^4 ≡ 1 mod 10. So the unit digit is 1.Hmm, same answer. So it's 1. Therefore, D.But let me check with an example. Let's take a smaller exponent. Suppose instead of 2022^2021, let's take 2^1. Then the exponent is 2, so 3^2=9, unit digit 9. If exponent is 2^2=4, then 3^4=81, unit digit 1. If exponent is 2^3=8, then 3^8=6561, unit digit 1. Wait, so 2^3=8, which is 0 mod 4, so 3^8 mod 10=1. Then, even exponent 8 (which is 0 mod 4), the result is 1.Similarly, if exponent is 4, which is 0 mod 4, 3^4=81, unit digit 1. So seems consistent. So if the exponent is 0 mod 4, the unit digit is 1. Therefore, the answer is 1, option D.But wait, let me check another angle. Maybe I made a mistake in the modulus when dealing with exponents.Wait, the key steps are:1. The unit digit of 2023^n is the same as the unit digit of 3^n.2. The unit digit of 3^n cycles every 4: 3,9,7,1,3,9,7,1...3. Therefore, we need to compute n mod 4, where n = 2022^2021.4. 2022 mod 4 is 2, so 2022^2021 mod 4 is equivalent to 2^2021 mod 4.5. Since 2^2=4 ≡ 0 mod 4, and 2^1=2 mod 4, for exponents >=2, 2^k ≡ 0 mod 4. Therefore, 2^2021 ≡ 0 mod 4.6. Therefore, n ≡ 0 mod 4. Hence, 3^n ≡ 3^0 ≡ 1 mod 10 (since the cycle is 4, so 0 mod 4 corresponds to the 4th term, which is 1).7. Therefore, the unit digit is 1.All steps seem solid. So yes, the answer is D) 1.Final Answerboxed{D}</think>To find the unit digit of (2023^{left(2022^{2021}right)}), we focus on the unit digit of the base, which is 3. The unit digits of powers of 3 follow a cycle: 3, 9, 7, 1, repeating every 4 exponents. Therefore, we need to determine the exponent (2022^{2021} mod 4).1. Calculate (2022 mod 4): [ 2022 div 4 = 505 text{ remainder } 2 implies 2022 equiv 2 mod 4 ]2. Simplify (2022^{2021} mod 4): [ 2022^{2021} equiv 2^{2021} mod 4 ] Since (2^2 equiv 0 mod 4), any higher power of 2 (exponent ≥ 2) will be 0 modulo 4. Thus: [ 2^{2021} equiv 0 mod 4 ]3. Determine the unit digit of (3^{left(2022^{2021}right)}): Since the exponent (2022^{2021} equiv 0 mod 4), we use the cycle of unit digits for powers of 3: [ 3^4 equiv 1 mod 10 ] Therefore, the unit digit of (3^0) (which is equivalent to (3^4) in the cycle) is 1.Thus, the unit digit of (2023^{left(2022^{2021}right)}) is (boxed{D}).]

🔑:Problem 1. The correct answer is (D).Since 3^{4}=81, if n is any multiple of 4, the units digit of 3^{n} is 1. The same will be true for any t^{n}, if t is a natural number where the units digit is 3. Since the exponent of 2023 in the given expression, namely 2022^{2021}, is a multiple of 4, it follows that the required digit is 1.

🧠:[<think>Okay, let me try to figure out this problem step by step. So, the universal set is all real numbers, U = ℝ. We have two sets, A and B. The question is asking for A intersect the complement of B. The options are given as intervals, so I need to find the interval representing A ∩ Bᶜ.First, let's parse what set A is. The problem says A is the set of all x such that x(x - 2) < 0. Hmm, that's a quadratic inequality. To solve x(x - 2) < 0, I can find the critical points where the expression equals zero, which are x = 0 and x = 2. Then, I can test intervals around these critical points to see where the product is negative.So, the critical points divide the real number line into three intervals: (-∞, 0), (0, 2), and (2, ∞). Let's pick test points in each interval.For (-∞, 0), let's take x = -1. Plugging into x(x - 2): (-1)(-1 - 2) = (-1)(-3) = 3, which is positive. So this interval is not part of the solution.For (0, 2), let's take x = 1. Then, 1(1 - 2) = 1(-1) = -1, which is negative. So this interval is part of the solution.For (2, ∞), let's take x = 3. Then, 3(3 - 2) = 3(1) = 3, positive again. So not part of the solution.Since the inequality is strict (< 0), the endpoints 0 and 2 are not included. Therefore, set A is the interval (0, 2).Wait, let me confirm. The quadratic x(x - 2) is a parabola opening upwards with roots at 0 and 2. The graph is below zero between the roots, so yes, the solution is (0, 2). So A = (0, 2).Alright, moving on to set B. The problem states B is the set of all x such that y = ln(1 - x). Hmm, I need to figure out the domain of the function y = ln(1 - x). Since the natural logarithm is only defined for positive arguments, the expression inside the logarithm, 1 - x, must be greater than 0.Therefore, 1 - x > 0 → -x > -1 → x < 1. So the domain of the function y = ln(1 - x) is all real numbers x where x < 1. Therefore, set B is (-∞, 1).So, B is (-∞, 1). Then, the complement of B in U (which is ℝ) is all real numbers not in (-∞, 1), which is [1, ∞). Because the complement of everything less than 1 is everything from 1 onwards, including 1 itself.So, complement of B, denoted as Bᶜ, is [1, ∞).Now, we need to find A ∩ Bᶜ. Since A is (0, 2) and Bᶜ is [1, ∞), the intersection of these two intervals is the overlap between them.Let me visualize this. A is from 0 to 2 (not including 0 and 2), and Bᶜ is from 1 to infinity, including 1. So the overlap would start at 1 (included because Bᶜ includes it) and go up to 2 (but not including 2, since A stops there). Therefore, the intersection is [1, 2).Wait, but A is (0, 2), which doesn't include 1, right? Wait, no. A is (0, 2), which does include 1. Wait, hold on. A is all numbers between 0 and 2, not including 0 and 2. So 1 is included in A. However, Bᶜ is [1, ∞), so 1 is included in Bᶜ. Therefore, the intersection should be [1, 2). Because A includes numbers up to 2, but not including 2, and Bᶜ includes 1 and onwards. So starting at 1 (included) and ending at 2 (excluded). Therefore, [1, 2). So that's option B.Wait, but let me double-check. If A is (0, 2) and Bᶜ is [1, ∞), then the intersection is the set of x such that x is in both (0,2) and [1, ∞). So x must be >=1 and <2. Therefore, [1, 2). Yes, that makes sense. Because in A, 1 is included (since it's greater than 0 and less than 2), and in Bᶜ, 1 is included. So their intersection includes 1 and goes up to 2, but doesn't include 2. So the answer should be [1, 2), which is option B.But wait, let me check the options again. The options are:A: (-2,1)B: [1,2)C: (-2,1]D: (1,2)Hmm, so the answer is B: [1,2). Wait, but I need to make sure that I didn't make a mistake in interpreting set A or B.Let me recheck set A. The inequality was x(x - 2) < 0. We solved it by testing intervals. The critical points at 0 and 2. The product is negative when x is between 0 and 2, which is correct. So A is indeed (0,2).Set B: the domain of ln(1 - x). Since ln(z) is defined when z > 0, so 1 - x > 0 → x < 1. So B is (-∞, 1). Therefore, complement is [1, ∞). So A intersect Bᶜ is [1, 2). So the answer is B. That seems right.But let me check why option D is (1,2). Maybe someone might think that since A is open at 0 and 2, and Bᶜ is closed at 1, the intersection is open at 1? But no, because Bᶜ includes 1, and A includes 1 (since 1 is between 0 and 2). Therefore, the intersection includes 1. So it's [1,2). So option B is correct.Wait, but the options given in the problem: A is (-2,1), B is [1,2), C is (-2,1], D is (1,2). So according to our reasoning, the answer is B. But let me check again if there's a mistake in the problem's initial statements.Wait, the universal set is ℝ. So complement of B is indeed [1, ∞). Then, intersecting with A, which is (0,2), gives [1,2). So yes, the answer is B.But let me check once more. Maybe there's a typo in the problem or I misread something.Wait, set A is defined as {x | x(x - 2) < 0}, which is (0,2). Set B is {x | y = ln(1 - x)}. The domain here is x such that 1 - x > 0, so x < 1. So B is (-∞,1). The complement is [1, ∞). So A ∩ Bᶜ is [1,2). That's exactly option B. So I think that's the right answer.But just to make sure, let's check the options again. The options are:A: (-2,1) – that's from -2 to 1, open both ends. But our answer is from 1 to 2.B: [1,2) – that's what we have.C: (-2,1] – again, different interval.D: (1,2) – open at 1 and 2.So unless there's a mistake in my reasoning, option B is correct.Wait, but why does the original problem mention the universal set U = ℝ? Well, the complement of B is with respect to U, so Bᶜ is U B, which is ℝ (-∞,1) = [1, ∞). That's correct.So, A is (0,2). So intersection with [1, ∞) is [1,2). So yes, B is the answer. Therefore, I think the answer is B.Final Answerboxed{B}</think>To determine the set ( A cap (complement_{U} B) ), we start by analyzing the sets ( A ) and ( B ).1. Set ( A ): - Defined by the inequality ( x(x - 2) < 0 ). - The critical points are ( x = 0 ) and ( x = 2 ). - Testing intervals around these points, we find the solution is ( (0, 2) ).2. Set ( B ): - Defined by the function ( y = ln(1 - x) ). - The argument of the logarithm must be positive: ( 1 - x > 0 ). - This gives ( x < 1 ), so ( B = (-infty, 1) ).3. Complement of ( B ): - The complement of ( B ) in ( mathbb{R} ) is ( complement_{U} B = [1, infty) ).4. Intersection ( A cap (complement_{U} B) ): - ( A = (0, 2) ) - ( complement_{U} B = [1, infty) ) - The intersection is the overlap of these intervals, which is ( [1, 2) ).Thus, the answer is (boxed{B}).,<think>Okay, so I need to solve this problem where the universal set is all real numbers, U = ℝ. The sets A and B are defined as follows: A is the set of all x such that x(x - 2) < 0, and B is the set of all x where y = ln(1 - x). Then, I have to find A ∩ (complement of B) and choose the correct option from A to D. Let me break this down step by step.First, let me figure out what set A is. The inequality given is x(x - 2) < 0. To solve this, I can analyze the critical points where the expression equals zero, which are x = 0 and x = 2. These points divide the real number line into intervals. I can test each interval to see where the product x(x - 2) is negative.The intervals are:1. (-∞, 0)2. (0, 2)3. (2, ∞)Let me pick test points in each interval:1. For x = -1 (in (-∞, 0)): (-1)(-1 - 2) = (-1)(-3) = 3 > 02. For x = 1 (in (0, 2)): (1)(1 - 2) = (1)(-1) = -1 < 03. For x = 3 (in (2, ∞)): (3)(3 - 2) = (3)(1) = 3 > 0So the inequality x(x - 2) < 0 holds in the interval (0, 2). Therefore, set A is (0, 2).Next, let's figure out set B. B is defined as the set of all x where y = ln(1 - x) is defined. The natural logarithm function ln(z) is defined when z > 0. Therefore, 1 - x must be greater than 0. Solving the inequality:1 - x > 0 => -x > -1 => x < 1So the domain of the function y = ln(1 - x) is all real numbers x where x < 1. Therefore, set B is (-∞, 1).Now, the question asks for A ∩ (complement of B). Let's recall that the complement of B in the universal set U = ℝ is all real numbers not in B. Since B is (-∞, 1), the complement of B is [1, ∞). So complement of B, which we can write as B', is [1, ∞).Now, A is (0, 2) and B' is [1, ∞). The intersection of these two sets is the set of all x that are in both A and B'. Let's visualize both intervals:- A is from 0 to 2, not including 0 and 2.- B' is from 1 to infinity, including 1.The overlap between these two intervals is from 1 to 2. Since A includes up to but not including 2, and B' includes 1 and beyond. Therefore, the intersection is [1, 2). Wait, but A is (0, 2), so at x = 1, since B' includes 1, but A does include numbers approaching 1 from the right. Wait, hold on. Let me check.Wait, no. A is (0, 2), which is open at both 0 and 2. B' is [1, ∞), which is closed at 1. So when taking the intersection, the lower bound is 1 (included because B' includes it), and upper bound is 2 (excluded because A excludes it). Therefore, the intersection is [1, 2). So that would be option B.But let me verify once again to make sure I didn't make a mistake.Set A: x(x - 2) < 0. Solved as (0, 2). Correct.Set B: ln(1 - x) is defined when 1 - x > 0 => x < 1. So B is (-∞, 1). Complement of B is [1, ∞). Then A ∩ [1, ∞) is [1, 2). Yes, that seems right.Looking at the options:A: (-2,1) – That's from -2 to 1, which seems unrelated.B: [1,2) – This matches our conclusion.C: (-2,1] – Also seems unrelated.D: (1,2) – This is open at 1, but since the complement of B includes 1, the intersection should include 1. So D is wrong because it's open at 1, while B is correct.Therefore, the answer should be B: [1,2).Wait, but let me check if the complement of B is correctly calculated. Since B is all x < 1, then the complement is x ≥ 1. So yes, [1, ∞). Therefore, intersecting with A (0,2) gives [1, 2). So option B is correct.Wait, but the options given are labeled A to D. Let me check again the problem statement:Options:A: (-2,1)B: [1,2)C: (-2,1]D: (1,2)Yes, so B is [1,2). So the answer is B. But wait, the original problem might have a typo? Let me just check again.Wait, the universal set is U = ℝ, so complement of B is ℝ B = ℝ (-∞,1) = [1, ∞). A is (0,2). So their intersection is [1,2). So yes, option B. Therefore, I think the correct answer is B.But hold on, when the problem says "A ∩ (complement U B)", that is, complement with respect to U, which is ℝ, so yes, complement of B is [1, ∞). So intersection with A is [1,2). So B is correct.Therefore, the answer is option B.Final Answerboxed{B}</think>To determine the set ( A cap (complement_{U} B) ), we start by analyzing the given sets ( A ) and ( B ).1. Set ( A ): - The inequality ( x(x - 2) < 0 ) is solved by finding the critical points ( x = 0 ) and ( x = 2 ). - Testing intervals around these points, we find the solution is ( (0, 2) ). Thus, ( A = (0, 2) ).2. Set ( B ): - The function ( y = ln(1 - x) ) is defined when the argument ( 1 - x > 0 ), which simplifies to ( x < 1 ). - Therefore, ( B = (-infty, 1) ).3. Complement of ( B ): - The complement of ( B ) with respect to ( mathbb{R} ) is ( complement_{U} B = [1, infty) ).4. Intersection ( A cap (complement_{U} B) ): - ( A = (0, 2) ) - ( complement_{U} B = [1, infty) ) - The intersection of these sets is the overlap between ( (0, 2) ) and ( [1, infty) ), which is ( [1, 2) ).Thus, the correct answer is (boxed{B}).]

🔑:From the given information:because x(x-2)<0therefore 0<x<2, thus A={x|x(x-2)<0}={x|0<x<2},because B={x|y=ln(1-x)}={x|x<1},therefore complement_{U}B={x|xgeq1}Since A={x|0<x<2}, therefore Acapcomplement_{U}B=[1,2)Hence, the correct answer is boxed{text{B}}.

🧠:[<think>Okay, so I need to find the probability that a point P with coordinates (m, n) lies outside the circle defined by the equation x² + y² = 16. The coordinates m and n are obtained by rolling two dice consecutively, which means each die roll is independent, right? Let me break this down step by step.First, let me recall that when you roll a standard die, there are 6 possible outcomes for each roll, numbered from 1 to 6. Since we're rolling two dice consecutively, the total number of possible coordinate pairs (m, n) should be 6 * 6 = 36. So, the sample space has 36 equally likely outcomes.Now, the problem is asking for the probability that the point P(m, n) lies outside the circle x² + y² = 16. To be outside the circle, the distance from the origin to the point P must be greater than the radius of the circle. The equation x² + y² = 16 is a circle centered at the origin with radius √16 = 4. Therefore, the point P(m, n) will lie outside the circle if m² + n² > 16.So, the key here is to count how many of the 36 possible (m, n) pairs satisfy the inequality m² + n² > 16. Then, the probability would be that number divided by 36. That makes sense.Let me list all possible values of m and n (each from 1 to 6) and check for each pair whether m² + n² > 16. Maybe I can make a table or systematically go through each possible m and n.Alternatively, since m and n are both integers from 1 to 6, perhaps I can compute m² + n² for all possible pairs and count how many results are greater than 16.But listing them all might take some time, but since there are only 36 pairs, it's manageable. Let's proceed.First, for m = 1 (the first die shows 1):For each n from 1 to 6, compute 1² + n² = 1 + n². So:n=1: 1 + 1 = 2 (which is ≤16)n=2: 1 + 4 = 5 (≤16)n=3: 1 + 9 = 10 (≤16)n=4: 1 + 16 = 17 (>16)n=5: 1 + 25 = 26 (>16)n=6: 1 + 36 = 37 (>16)So, for m=1, the points where n=4,5,6 will be outside the circle. That's 3 points.Next, m=2:Compute 2² + n² = 4 + n².n=1: 4 + 1 = 5 (≤16)n=2: 4 + 4 = 8 (≤16)n=3: 4 + 9 = 13 (≤16)n=4: 4 + 16 = 20 (>16)n=5: 4 + 25 = 29 (>16)n=6: 4 + 36 = 40 (>16)Again, n=4,5,6: 3 points outside.Wait, same as m=1. Hmm.m=3:3² + n² = 9 + n².n=1: 9 + 1 = 10 (≤16)n=2: 9 + 4 = 13 (≤16)n=3: 9 + 9 = 18 (>16)n=4: 9 + 16 = 25 (>16)n=5: 9 + 25 = 34 (>16)n=6: 9 + 36 = 45 (>16)So, for m=3, n=3,4,5,6: 4 points outside.m=4:4² + n² = 16 + n².n=1: 16 + 1 = 17 (>16)n=2: 16 + 4 = 20 (>16)n=3: 16 + 9 = 25 (>16)n=4: 16 + 16 = 32 (>16)n=5: 16 + 25 = 41 (>16)n=6: 16 + 36 = 52 (>16)So for m=4, all n from 1 to 6 result in points outside the circle. That's 6 points.m=5:5² + n² = 25 + n².n=1: 25 + 1 = 26 (>16)n=2: 25 + 4 = 29 (>16)n=3: 25 + 9 = 34 (>16)n=4: 25 + 16 = 41 (>16)n=5: 25 + 25 = 50 (>16)n=6: 25 + 36 = 61 (>16)All n from 1-6, so 6 points outside.m=6:6² + n² = 36 + n².Similarly, n=1: 36 +1=37 (>16)n=2:36 +4=40 (>16)...All n will result in 36 + n² which is definitely greater than 16. So, all 6 points here.Now, let's summarize the counts:For m=1: 3 points outsidem=2: 3 pointsm=3: 4 pointsm=4: 6 pointsm=5: 6 pointsm=6: 6 pointsTotal points outside: 3+3+4+6+6+6 = let's compute that:3 + 3 = 66 + 4 = 1010 + 6 = 1616 + 6 = 2222 + 6 = 28Wait, 3+3+4+6+6+6 = 3+3=6; 6+4=10; 10+6=16; 16+6=22; 22+6=28. So 28 points.But wait, total possible outcomes are 36. So the probability is 28/36, which simplifies to 7/9.But wait, let me double-check the counts to be sure. Sometimes when adding up in steps, there might be a mistake.Let me recheck each m:m=1: n=4,5,6 → 3 points. Correct.m=2: n=4,5,6 → 3. Correct.m=3: n=3,4,5,6 → 4. Correct.m=4: all 6 n. Correct.m=5: all 6. Correct.m=6: all 6. Correct.So 3+3+4+6+6+6. Let's add again:First, 3 (m=1) + 3 (m=2) = 6.6 + 4 (m=3) = 10.10 + 6 (m=4) = 16.16 +6 (m=5) =22.22 +6 (m=6)=28. Yes, that's 28.So 28 out of 36. Then 28/36 can be reduced by dividing numerator and denominator by 4: 28 ÷4=7, 36 ÷4=9. So 7/9. So the probability is 7/9.But wait, hold on. Let me confirm once more that all these points are correctly counted. Maybe for m=3, n=3,4,5,6:m=3, n=3: 3² +3²=9+9=18>16, correct.n=4:9+16=25>16, yes. Similarly, n=5 and 6. So 4 points. Correct.For m=4, even n=1: 16 +1=17>16, so all of them. Correct.Similarly, m=5 and 6, all n. So total 28 points. So 28/36=7/9.Hmm. Alternatively, maybe the answer is different. Wait, let me think again. Maybe I made a mistake in the counting?Wait, let's verify each m again:For m=1: n=4,5,6. That's 3.m=2: same. 3.m=3: n=3,4,5,6. 4.m=4: n=1,2,3,4,5,6. 6.m=5: same as m=4. 6.m=6: same. 6.Total: 3+3+4+6+6+6=28. So 28. So 28/36=7/9.But, wait, let's check for m=3, n=3: 3,3. 3²+3²=18, which is outside. So yes, 18>16. Correct.Similarly, m=3, n=2: 9 +4=13, which is inside. So only starting from n=3 for m=3.Similarly, m=2, n=4:4+16=20>16. Correct. So n=4,5,6 for m=2.Similarly, m=1, n=4:1+16=17>16. Correct.So, all counts seem accurate.Alternatively, maybe it's easier to compute the number of points inside the circle and subtract that from 36. Let's try that approach to verify.Number of points inside or on the circle x² + y² ≤16. Then subtract that from 36 to get the number outside.So, let's compute how many (m,n) pairs satisfy m² +n² ≤16.Again, for each m from 1 to6, count the n's such that n² ≤16 - m².For m=1: n² ≤16 -1=15. So n² ≤15. Since n is integer from1-6, n² can be 1,4,9,16,25,36. So n² ≤15 means n can be 1,2,3 because 1²=1, 2²=4, 3²=9, which are all ≤15. n=4 would be 16 which is >15. So n=1,2,3: 3 points.So, for m=1, 3 points inside.Similarly, m=2: n² ≤16 -4=12. So n² ≤12. n=1,2,3 (1,4,9 ≤12). n=4 would be 16>12. So again 3 points.m=3: n² ≤16 -9=7. So n² ≤7. n can be 1,2 since 1²=1, 2²=4, and 3²=9>7. So n=1,2: 2 points.Wait, but earlier when counting points outside, for m=3, we had 4 points outside. Since m=3 has n=1-6, so 6 total. If 2 points are inside, then 6-2=4 outside. Correct.m=4: n² ≤16 -16=0. So n² ≤0. But n is at least 1, so no solutions. So 0 points inside. Therefore, all 6 points outside.m=5: n² ≤16 -25= -9. Impossible. So 0 points inside. All 6 outside.m=6: Similarly, n² ≤16 -36= -20. Impossible. 0 points inside. All 6 outside.So total points inside: for m=1:3, m=2:3, m=3:2, m=4:0, m=5:0, m=6:0. Total inside=3+3+2=8. Therefore, points outside=36 -8=28. Which matches the previous count. So 28/36=7/9. So the probability is 7/9.Therefore, the answer should be 7/9. Wait, but I just want to make sure that in the problem statement, it's "outside the circle", so strictly greater than 16. Not greater or equal. But the equation of the circle is x² + y²=16. So the circle includes points where x² + y²=16. But the problem says "outside the circle", which would be points where x² + y² >16. So in that case, the points on the circle are not counted as outside. So in our counting, when we considered m² +n² >16, that is correct. So in the previous calculation, points on the circle (i.e., m² +n²=16) are part of the circle and thus not counted as outside. So if any of our pairs (m,n) result in m² +n²=16, they should be excluded from the "outside" count.Wait a second, let's check: for m=4, n=0: but n can't be 0. The possible points are m and n from 1 to6. So, can any of the pairs (m,n) satisfy m² +n²=16?Check m=4, n=0: nope. n=1: m=4, n=1: 16 +1=17>16. Similarly, m=0, n=4: m is at least 1. So there are actually no points (m,n) where m and n are from 1 to6 that lie exactly on the circle x² + y²=16. Because 16 can only be achieved by (4,0) or (0,4), but m and n are from 1-6. Therefore, in this problem, the set of points where m² +n²=16 is empty. Therefore, the count remains the same. So all the points where m² +n² >16 are strictly outside, and the points where m² +n² ≤16 are strictly inside. So in that case, our count of 28 points outside and 8 inside is correct.Therefore, the probability is 28/36=7/9.Hence, the answer is 7/9.Final AnswerThe probability that point ( P ) lies outside the circle ( x^{2} + y^{2} = 16 ) is boxed{dfrac{7}{9}}.</think>To find the probability that a point ( P(m, n) ) lies outside the circle ( x^2 + y^2 = 16 ), we need to count the number of coordinate pairs ((m, n)) obtained from rolling two dice that satisfy ( m^2 + n^2 > 16 ).1. Total possible outcomes: Rolling two dice results in ( 6 times 6 = 36 ) possible pairs ((m, n)).2. Counting pairs outside the circle: - For ( m = 1 ): ( n = 4, 5, 6 ) (3 pairs) - For ( m = 2 ): ( n = 4, 5, 6 ) (3 pairs) - For ( m = 3 ): ( n = 3, 4, 5, 6 ) (4 pairs) - For ( m = 4 ): All ( n ) (6 pairs) - For ( m = 5 ): All ( n ) (6 pairs) - For ( m = 6 ): All ( n ) (6 pairs)3. Summing the valid pairs: - ( 3 + 3 + 4 + 6 + 6 + 6 = 28 )4. Calculating the probability: - The probability is the number of valid pairs divided by the total outcomes: ( frac{28}{36} = frac{7}{9} ).Thus, the probability that point ( P ) lies outside the circle ( x^2 + y^2 = 16 ) is (boxed{dfrac{7}{9}}).,<think>Okay, so I need to find the probability that a point P with coordinates (m, n), where m and n are numbers obtained from rolling two dice consecutively, lies outside the circle defined by the equation x² + y² = 16. Hmm, let me think through this step by step.First, let me recall that when you roll two dice, each die has numbers from 1 to 6. Since they're rolled consecutively, m and n can each be any integer from 1 to 6, right? So the coordinates (m, n) are all the possible ordered pairs where both m and n are between 1 and 6, inclusive. That should give us a total of 6 × 6 = 36 possible outcomes, right? So the sample space has 36 points.Now, the problem is asking for the probability that P lies outside the circle x² + y² = 16. So, to restate that, we need to find how many of these 36 points (m, n) satisfy the inequality m² + n² > 16. Then, the probability would be the number of such points divided by 36.So, my task is to count all pairs (m, n) where m and n are integers from 1 to 6, and m² + n² > 16.Let me approach this systematically. Maybe list all possible values of m and n and check which ones satisfy the condition. But since there are 36 pairs, that might take a while, but perhaps manageable.Alternatively, maybe I can find for each value of m, how many n's satisfy the inequality. Let's try that.First, m can be from 1 to 6, same with n. Let's start with m = 1.For m = 1: The inequality becomes 1² + n² > 16 → 1 + n² > 16 → n² > 15. Since n is an integer from 1 to 6, n² must be greater than 15. Let's compute n² for n=1 to 6:n=1: 1, n=2:4, n=3:9, n=4:16, n=5:25, n=6:36. So n² >15 would be n=4,5,6. So for m=1, n can be 4,5,6. That's 3 values.Next, m=2: 2² + n² >16 → 4 + n² >16 → n² >12. So n² must be greater than 12. Let's check the squares again:n=1:1, n=2:4, n=3:9, n=4:16, etc. So n² >12 would be n=4,5,6. So again, 3 values.Wait, n=4:16 >12, yes. So same as before, 3 values.m=3: 3² + n² >16 → 9 + n² >16 → n² >7. So n² must be greater than 7. Let's see:n=1:1, n=2:4, n=3:9, which is 9>7. So n=3,4,5,6. So 4 values.Wait, n=3 gives 9, which is greater than 7. So n=3,4,5,6: 4 values.m=4: 4² + n² >16 →16 + n² >16 →n² >0. Since n is at least 1, n² is always at least 1, which is greater than 0. So for m=4, all n from 1 to 6 satisfy the inequality. So 6 values.Wait, hold on. Wait, 16 + n² >16 is equivalent to n² >0. But n is an integer from 1 to 6, so n² is always ≥1, which is indeed >0. Therefore, all 6 values of n work when m=4. Similarly for m=5 and m=6?Wait, let's check m=5: 5² + n² =25 + n² >16. Since 25 is already greater than 16, adding n² (which is at least 1) will make it even larger. So regardless of n, 25 + n² will always be greater than 16. Therefore, for m=5, all n from 1 to 6 satisfy the inequality. Similarly for m=6: 36 + n², which is definitely greater than 16. So all n=1-6.Wait, so for m=4,5,6, all n are valid. So that's 6 (for m=4) +6 (m=5) +6 (m=6) =18.Then, for m=1, 3 values; m=2, 3 values; m=3, 4 values. So total for m=1,2,3: 3+3+4=10.So total number of valid (m,n) pairs is 10 +18=28.Therefore, the probability would be 28/36, which simplifies to 7/9. Is that right? Wait, let me check again.Wait, m=1: n=4,5,6 →3m=2: same →3m=3: n=3,4,5,6 →4m=4: all 6m=5: all 6m=6: all 63+3+4+6+6+6=28. Yes, 28. So 28/36=7/9. So the probability is 7/9.But wait, let me verify this again. Maybe I made a mistake in the calculation for m=3.Wait, m=3: inequality is 9 +n² >16 →n²>7. Since n is integer, n²>7 implies n²≥9 (since 7 is between 4 and 9). So n≥3. Therefore, n=3,4,5,6. So that's 4 values. That seems correct.Similarly, m=1: n²>15 →n≥4. m=2: n²>12 →n≥4. So that's correct.For m=4,5,6, all n are allowed. So that's 3×6=18.So total is 3+3+4+6+6+6=28. 28/36=7/9. So probability is 7/9. Hmm.But wait, let me check specific points. For example, when m=3 and n=3: 3² +3²=9+9=18>16, so that's outside. Similarly, m=3 and n=2: 9+4=13<16, which is inside. So n=3 is the first n where it's outside. So that seems right.Similarly, m=1 and n=4: 1+16=17>16. Yes. So that's correct.Wait, but what about m=4 and n=1: 16 +1=17>16, so yes. So all points where m is 4,5,6, regardless of n, are outside. Because even if n=1, m=4: 16 +1=17>16. Similarly, m=5, n=1:25 +1=26>16. So yes, all of them are outside.Similarly, m=3 and n=3:9 +9=18>16. But m=3, n=2:9 +4=13<16. So n=2 is inside.Therefore, the count for m=3 is n=3,4,5,6. Correct.So adding up, 3+3+4+6+6+6=28. So 28 out of 36. Simplify 28/36: divide numerator and denominator by 4: 7/9. So probability is 7/9. So the answer should be boxed{dfrac{7}{9}}.Wait, but let me just cross-verify by another approach. Instead of counting the points outside, maybe count the points inside or on the circle and subtract from total.Total points:36. So if I compute the number of points where m² +n² ≤16, then subtract that from 36 to get the points outside.Let's see. For m=1:m=1: n² ≤16 -1=15. So n² ≤15. So n=1,2,3 (since 16 is 4², but 15 is between 3²=9 and 4²=16). So n=1,2,3. So 3 points.m=2: n² ≤16 -4=12. So n² ≤12. So n=1,2,3. Because 3²=9, 4²=16 which is over 12. So again, 3 points.m=3: n² ≤16 -9=7. So n² ≤7. So n=1,2 (since 2²=4, 3²=9>7). So n=1,2. Wait, that's 2 points. Wait, but earlier I thought for m=3, n=3,4,5,6 are outside. So if inside is n=1,2, then 2 points. Therefore, outside would be 6-2=4 points. Which matches the previous count.Wait, so m=3: inside is 2 points, so outside is 4. So total inside points for m=1,2,3: 3+3+2=8.Then for m=4: inside is when 16 +n² ≤16 →n² ≤0. But n is at least 1, so n² ≥1. Therefore, no points for m=4,5,6 are inside. So inside points are 8. Therefore, outside points are 36 -8=28. Which is same as before. So 28/36=7/9. Therefore, same answer.So this cross-verification confirms that the number of points inside or on the circle is 8, so outside is 28. So probability 28/36=7/9.Therefore, the answer is 7/9.Final Answerboxed{dfrac{7}{9}}</think>To find the probability that the point ( P(m, n) ) lies outside the circle ( x^2 + y^2 = 16 ) where ( m ) and ( n ) are numbers obtained from rolling two dice consecutively, we need to count the number of pairs ((m, n)) such that ( m^2 + n^2 > 16 ).1. Total possible outcomes: Since each die has 6 faces, there are ( 6 times 6 = 36 ) possible outcomes.2. Counting pairs outside the circle: - For ( m = 1 ): ( 1 + n^2 > 16 Rightarrow n^2 > 15 ). Valid ( n ) values: 4, 5, 6 (3 values). - For ( m = 2 ): ( 4 + n^2 > 16 Rightarrow n^2 > 12 ). Valid ( n ) values: 4, 5, 6 (3 values). - For ( m = 3 ): ( 9 + n^2 > 16 Rightarrow n^2 > 7 ). Valid ( n ) values: 3, 4, 5, 6 (4 values). - For ( m = 4, 5, 6 ): Since ( 16 + n^2 > 16 ), ( 25 + n^2 > 16 ), and ( 36 + n^2 > 16 ) are always true for any ( n ). Valid ( n ) values: 1, 2, 3, 4, 5, 6 (6 values each).3. Summing up valid pairs: - ( m = 1 ): 3 pairs - ( m = 2 ): 3 pairs - ( m = 3 ): 4 pairs - ( m = 4 ): 6 pairs - ( m = 5 ): 6 pairs - ( m = 6 ): 6 pairs Total valid pairs: ( 3 + 3 + 4 + 6 + 6 + 6 = 28 ).4. Calculating probability: The probability is the number of valid pairs divided by the total outcomes: [ frac{28}{36} = frac{7}{9} ]Thus, the probability that point ( P ) lies outside the circle ( x^2 + y^2 = 16 ) is (boxed{dfrac{7}{9}}).]

🔑:First, let's find all possible points (m, n) based on the problem statement. Since a die has 6 faces, there are 6 times 6 = 36 possible coordinate pairs.Next, we need to find the number of points that lie outside the circle x^{2}+y^{2}=16. These points are:(1, 4), (1, 5), (1, 6),(2, 4), (2, 5), (2, 6),(3, 3), (3, 4), (3, 5), (3, 6),(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6).There are 28 such points.The probability that point P lies outside the circle is the ratio of favorable outcomes to total outcomes, which is frac{28}{36} = boxed{frac{7}{9}}.