🧠:[<think>Okay, so there are 5 students standing in a line, and I need to find the number of arrangements where a certain student isn't at the head of the line. Hmm, let me think. First, I remember that the total number of ways to arrange n distinct items is n factorial. Since there are 5 students, the total number of arrangements without any restrictions should be 5! which is 5 × 4 × 3 × 2 × 1 = 120. So, normally, there are 120 different ways they can line up.But the problem says we need to exclude the arrangements where a certain student is at the head of the line. Let me call that student A for simplicity. So, student A shouldn't be first in line. How do I calculate that?One approach is to subtract the number of arrangements where student A is at the head from the total number of arrangements. That way, I'm left with all the arrangements where A isn't first. So, first, how many arrangements have student A at the head? If A is fixed in the first position, then the remaining 4 students can be arranged in any order in the remaining 4 positions. The number of ways to arrange those 4 students is 4! = 24. Therefore, the number of valid arrangements where A isn't first should be total arrangements minus the ones with A first. So that's 120 - 24 = 96. Let me check if that makes sense.Alternatively, another way to think about it: For the first position in the line, we can't have student A. So, there are 4 choices (since there are 5 students total, minus A). Then, for the remaining positions, we can have any of the remaining students, including A. So, first position: 4 choices. Then, the second position: since one student is already placed in the first position (and it's not A), there are 4 students left (including A). Then the third position would have 3 students left, then 2, then 1. So, the total number of arrangements would be 4 × 4 × 3 × 2 × 1? Wait, hold on, that doesn't seem right. Wait, no, if the first position is fixed with one of the 4 students, then the remaining positions are permutations of the remaining 4 students, including A. So, the total would actually be 4 (for the first position) multiplied by 4! (for the rest). Wait, but 4 × 4! is 4 × 24 = 96. Which matches the previous result. So that checks out. So, either way, subtracting the restricted cases or calculating directly, I get 96. Let me think if there's another way to approach this, just to be thorough.Another approach: Each position in the line can be considered separately. The number of arrangements where A is not first. So, A can be in positions 2, 3, 4, or 5. For each of those positions, how many arrangements are there?If A is in position 2: Then the first position has 4 choices (since A can't be there), and the remaining 3 positions can be filled by the remaining 3 students. So, 4 × 1 (for A in position 2) × 3 × 2 × 1 = 4 × 3! = 24.Similarly, if A is in position 3: First position has 4 choices, then position 2 has 3 choices (since A is in position 3), then the remaining positions are 3 students. Wait, no. Wait, if A is in position 3, then positions 1, 2, 4, and 5 are filled by the other 4 students. So, positions 1 can be 4 students, position 2 can be 3 students, position 3 is A, position 4 is 2 students, and position 5 is 1 student. So, that's 4 × 3 × 1 × 2 × 1 = 24 again.Wait, same for A in position 4: 4 × 3 × 2 × 1 × 1 = 24. And A in position 5: 4 × 3 × 2 × 1 × 1 = 24. So, for each of the 4 positions that A can be in (positions 2-5), we have 24 arrangements each? But 4 positions times 24 would be 96. Wait, but that seems high. Wait, no, maybe I made a mistake here. Because if A is in position 2, the total number of arrangements is 4 (for first position) × 3! (for the remaining positions after placing A). But if A is in position 3, it's 4 × 3 × 2 × 1? Wait, no, if A is in position 3, then the first two positions can be any of the remaining 4 students, which is 4 × 3, then the last two positions are 2 × 1. So, that's 4 × 3 × 2 × 1 = 24. Similarly for A in position 4, it's 4 × 3 × 2 × 1 = 24. Wait, but that can't be right, because each of the different positions for A would lead to different counts. Wait, actually, maybe not. Because regardless of where A is placed (as long as it's not the first position), the number of arrangements is the same. Let's see.Suppose we fix A in a specific position. Let's say position k (where k ≠ 1). Then the number of arrangements would be the number of ways to arrange the other 4 students in the remaining 4 positions. However, the first position is among these remaining positions. Wait, no. If A is in position k, then the first position must be filled by one of the other students. So, actually, the number of arrangements with A in position k is equal to 4 × 4 × 3 × 2 × 1 divided by... Wait, maybe this is more complicated.Wait, maybe the better way is to recognize that when we fix A in a particular position (other than the first), the rest of the students can be arranged freely. For example, if A is in position 2, then the first position has 4 choices, and the remaining 3 positions have 3! = 6 arrangements. So, for A in position 2, it's 4 × 6 = 24. Similarly, if A is in position 3, the first position has 4 choices, the second position has 4 - 1 = 3 choices (since one student is already in the first position), and then the remaining positions are 2 × 1. Wait, no, that's not right. If A is in position 3, the first position can be any of 4 students, the second position can be any of the remaining 3 (excluding the one in the first position), the third position is A, the fourth position is 2 remaining students, and the fifth position is 1. So that's 4 × 3 × 1 × 2 × 1 = 24. Similarly for positions 4 and 5. So each time A is in a non-first position, there are 4 × 3 × 2 × 1 = 24 arrangements? Wait, that's 4! But 4! is 24. So, if A is fixed in any one of the 4 positions (positions 2-5), the remaining 4 students can be arranged in 4! ways. But wait, no, if A is fixed in position 2, then the first position is filled by 4 students, and the remaining 3 positions are filled by the remaining 3 students. Wait, that's 4 × 3! = 24. Similarly, if A is fixed in position 3, then the first two positions are filled by 4 and 3 students respectively, and the last two by 2 and 1. Wait, 4 × 3 × 2 × 1 = 24. So, same as 4! = 24. Hmm. So, regardless of where A is placed (positions 2-5), the number of arrangements is 4! each. Since there are 4 possible positions for A, does that mean 4 × 4! = 96? Wait, but that seems like overcounting. Because if A is in position 2, the total arrangements for that case are 4! (since the other 4 positions are arranged freely). Wait, no. Wait, if A is fixed in position 2, then the first position can be any of the remaining 4 students, and the remaining 3 positions can be any of the remaining 3 students. So, that's 4 × 3! = 24. Similarly, if A is fixed in position 3, then the first two positions are 4 × 3, and the remaining two are 2 × 1, which is 4 × 3 × 2 × 1 = 24. So, each time, fixing A in a different non-first position gives 24 arrangements. Since there are 4 positions where A can be (positions 2,3,4,5), that would be 4 × 24 = 96. But wait, that's not correct. Because when you fix A in different positions, you're not overlapping in arrangements. Each arrangement has A in exactly one position, so summing over all possible positions for A (excluding first) gives the total number of arrangements where A is not first, which is 4 × 24 = 96. But wait, this seems like 4 positions times 24 each, but 4 × 24 is 96. However, in reality, each arrangement where A is not first is counted exactly once in this sum, so 96 is correct. So, this method also gives 96. Therefore, all the methods give me the same answer: 96. So, I think that's the correct number of arrangements where student A is not at the head of the line.Final AnswerThe number of arrangements is boxed{96}.</think>To find the number of arrangements where a certain student does not stand at the head of the line, we start by calculating the total number of permutations of 5 students, which is (5! = 120).Next, we calculate the number of arrangements where the certain student (student A) is at the head of the line. If student A is fixed at the first position, the remaining 4 students can be arranged in (4! = 24) ways.Subtracting the number of restricted arrangements from the total number of arrangements gives us the number of valid arrangements where student A is not at the head:[5! - 4! = 120 - 24 = 96]Thus, the number of arrangements where a certain student does not stand at the head of the line is (boxed{96}).,<think>Okay, so I need to figure out how many ways there are to arrange 5 students in a line such that a certain student isn't at the front. Hmm. Let's see... First, without any restrictions, the number of arrangements for 5 students would just be 5 factorial, right? Because each position can be filled by any of the remaining students. So 5! = 5 × 4 × 3 × 2 × 1 = 120. That makes sense.But the problem says a certain student shouldn't be at the head of the line. Let me call that student A to make it easier. So we need all permutations where student A is not first. How do I calculate that? Well, maybe it's easier to calculate the total number of arrangements and then subtract the number of arrangements where student A is at the head. Yeah, that sounds like the complementary counting method.So total arrangements are 120. Now, how many arrangements have student A at the head? If A is fixed at the first position, then the remaining 4 students can be arranged in any order in the remaining 4 positions. So that would be 4! = 24.Therefore, the number of arrangements where A is not at the head would be total arrangements minus the ones with A at the front. So 120 - 24 = 96. Is that right?Wait, let me verify. Another way to think about it: for the first position, we can't have A, so there are 4 choices (since there are 5 students total, minus 1 which is A). Then, once the first position is filled by someone else, the remaining 4 positions can be filled by the remaining 4 students, including A. So that would be 4 × 4 × 3 × 2 × 1. Wait, no, that's not quite right. Because after choosing the first person (which is 4 choices), the next positions are permutations of the remaining 4 students. So actually, the number would be 4 (choices for first position) multiplied by 4! (arrangements of the rest). Which is 4 × 24 = 96. Same answer. So that checks out.Alternatively, if I think about the total number of permutations where A is not first, it's equivalent to 4 × 4! because the first position has 4 options and the rest can be permuted. Either way, 4 × 24 = 96. So that seems consistent.Therefore, the answer should be 96. But just to be thorough, let me think of a small example with fewer students to make sure this logic works.Suppose there are 2 students: A and B. How many arrangements where A is not first? Total permutations: 2! = 2. Arrangements with A first: 1 (A followed by B). So the answer should be 2 - 1 = 1. Which is correct, because the only valid arrangement is B followed by A. That works.Another example: 3 students, A, B, C. Total permutations: 6. If we exclude arrangements where A is first, then the allowed permutations are B-A-C, B-C-A, C-A-B, C-B-A. That's 4, which is 3! - 2! = 6 - 2 = 4. Wait, 3! - 2! = 6 - 2 = 4. Correct. So same logic: total minus the ones where A is first (which is 2! arrangements). So in the case of n students, it's n! - (n-1)!.So applying that formula here, with n=5: 5! - 4! = 120 - 24 = 96. So that formula works. Therefore, the answer is 96. I think that's solid. The key idea is recognizing that fixing a student in the first position reduces the problem to permuting the remaining students, and subtracting that from the total gives the desired count. Also, verifying with smaller numbers helps ensure the logic is sound.Final AnswerThe number of arrangements is boxed{96}.</think>To find the number of arrangements where a certain student does not stand at the head of the line, we start by calculating the total number of permutations of 5 students, which is (5! = 120).Next, we calculate the number of permutations where the certain student (let's call them A) is at the head of the line. If A is fixed at the first position, the remaining 4 students can be arranged in (4!) ways. This gives us (4! = 24) permutations.Subtracting the number of permutations where A is at the head from the total number of permutations gives us the desired count:[5! - 4! = 120 - 24 = 96]Thus, the number of arrangements where the certain student does not stand at the head of the line is (boxed{96}).]

🔑:To solve this problem, we first consider the total number of ways to arrange 5 students in a line, which is 5! = 120. Since there are 5 students, and we want to find the number of arrangements where a specific student does not stand at the head of the line, we subtract the arrangements where this student stands at the head. There are 4! = 24 such arrangements (as the remaining 4 students can be arranged in any order behind the specific student). Therefore, the number of arrangements where a certain student does not stand at the head of the line is 120 - 24 = 96.Thus, the number of arrangements is boxed{96}.

🧠:[<think>Okay, so I need to solve this problem about chickens and rabbits in a cage. Let me read it again carefully. The total number of heads is 51. The total number of rabbit feet is 4 more than 3 times the total number of chicken feet. We need to find how many rabbits there are. Hmm, let's break it down.First, since both chickens and rabbits have one head each, the total number of animals is 51. So, if I let the number of chickens be C and the number of rabbits be R, then C + R = 51. That's straightforward.Now, the second part is about their feet. Chickens have 2 feet each, and rabbits have 4 feet each. The problem states that the total number of rabbit feet is 4 more than 3 times the total number of chicken feet. So, translating that into an equation: total rabbit feet = 4 + 3*(total chicken feet). Let me write that out.Total rabbit feet is 4R (since each rabbit has 4 feet), and total chicken feet is 2C. So, according to the problem, 4R = 4 + 3*(2C). Let me check that again: yes, "rabbit feet is 4 more than 3 times the chicken feet." So, 4R = 3*(2C) + 4. That simplifies to 4R = 6C + 4.So now we have two equations:1. C + R = 512. 4R = 6C + 4I can solve this system of equations. Let's see. From the first equation, maybe express C in terms of R or vice versa. Let's solve for C first. So, C = 51 - R. Then substitute that into the second equation.Substituting into the second equation: 4R = 6*(51 - R) + 4. Let's compute that step by step.First, expand the right side: 6*(51 - R) is 6*51 - 6*R. 6*51 is 306, so 306 - 6R. Then adding 4 gives 306 - 6R + 4 = 310 - 6R.So now the equation is 4R = 310 - 6R. Let's bring all the R terms to the left and constants to the right. So, 4R + 6R = 310. That becomes 10R = 310. Then dividing both sides by 10 gives R = 31.Wait, so R is 31? Let me check if that makes sense. Then C would be 51 - 31 = 20. Let's check the feet.Total rabbit feet: 31*4 = 124. Total chicken feet: 20*2 = 40. Now, 3 times the chicken feet is 3*40 = 120. 120 + 4 = 124. Which equals the rabbit feet. Perfect, that checks out.So the answer should be 31 rabbits. But let me just go through the steps again to make sure I didn't make a mistake.Starting with the two equations:1. C + R = 512. 4R = 6C + 4Express C as 51 - R, substitute into equation 2:4R = 6*(51 - R) + 44R = 306 - 6R + 44R = 310 - 6R4R + 6R = 31010R = 310R = 31Yes, that's correct. Then C = 51 - 31 = 20. Check the feet again: rabbits 31*4=124, chickens 20*2=40. Then 3*40 +4 = 124. Perfect. So the answer is 31 rabbits.I think that's solid. Maybe I can try another approach to verify. Suppose we didn't use substitution. Let's see.From equation 1: C = 51 - R.From equation 2: 4R - 6C = 4. Substitute C into this:4R -6*(51 - R) = 44R -306 +6R =4(4R +6R) -306 =410R -306 =410R =310R=31. Same result. Yep, that's consistent. So definitely 31 rabbits.Alternatively, maybe using ratios or something else, but substitution seems straightforward here. Since we have two equations and two variables, substitution is the way to go.Alternatively, think about the feet. Total number of feet: chicken feet + rabbit feet = 2C +4R. But the problem doesn't mention total feet, it relates rabbit feet to chicken feet. So 4R = 3*(2C) +4. Which again gives us the same equation. So yeah, same result.Alternatively, maybe using matrix algebra, but that's overcomplicating. Substitution is fine. So, conclusion: there are 31 rabbits.Final AnswerThe number of rabbits in the cage is boxed{31}.</think>Given a cage with chickens and rabbits, we know the following:1. The total number of heads is 51.2. The total number of rabbit feet is 4 more than 3 times the total number of chicken feet.We define the number of chickens as ( C ) and the number of rabbits as ( R ).From the first piece of information, we have the equation:[ C + R = 51 ]From the second piece of information, we know that the total number of rabbit feet (4R) is 4 more than 3 times the total number of chicken feet (2C). This gives us the equation:[ 4R = 3 times 2C + 4 ][ 4R = 6C + 4 ]We solve the first equation for ( C ):[ C = 51 - R ]Substituting this expression for ( C ) into the second equation:[ 4R = 6(51 - R) + 4 ][ 4R = 306 - 6R + 4 ][ 4R = 310 - 6R ][ 4R + 6R = 310 ][ 10R = 310 ][ R = 31 ]Thus, the number of rabbits in the cage is (boxed{31}).,<think>Okay, let's see. There are chickens and rabbits in a cage. Total heads are 51. The total number of rabbit feet is 4 more than 3 times the chicken feet. Need to find the number of rabbits. Hmm. Let me start by setting up variables.Let me denote the number of chickens as C and the number of rabbits as R. Since each chicken and each rabbit has one head, the total number of heads is C + R = 51. That's straightforward.Now, for the feet. Chickens have 2 feet each, rabbits have 4 feet each. The problem says the total number of rabbit feet is 4 more than 3 times the chicken feet. So, rabbit feet is 4R, chicken feet is 2C. According to the problem, 4R = 3*(2C) + 4. Let me write that down: 4R = 6C + 4.So now I have two equations:1) C + R = 512) 4R = 6C + 4I need to solve this system of equations. Let's solve equation 1 for one variable and substitute into equation 2. Let's solve for C. From equation 1, C = 51 - R. Then substitute that into equation 2.So, 4R = 6*(51 - R) + 4Let me compute the right side: 6*51 is 306, then 6*(-R) is -6R, so 306 - 6R + 4 = 310 - 6R.So equation becomes 4R = 310 - 6R.Bring all R terms to the left: 4R + 6R = 31010R = 310Divide both sides by 10: R = 31.Wait, so there are 31 rabbits. Let me check if that makes sense.If R = 31, then C = 51 - 31 = 20 chickens.Check the feet: Rabbit feet = 4*31 = 124.Chicken feet = 2*20 = 40.3 times chicken feet is 3*40 = 120. Then 120 + 4 = 124. Which equals rabbit feet. So that works.Therefore, the answer should be 31 rabbits.Final AnswerThe number of rabbits in the cage is boxed{31}.</think>Given a cage with chickens and rabbits, we know the total number of heads is 51. Let ( C ) be the number of chickens and ( R ) be the number of rabbits. We have the equation:[C + R = 51]Additionally, the total number of rabbit feet is 4 more than 3 times the total number of chicken feet. Chickens have 2 feet each and rabbits have 4 feet each, leading to the equation:[4R = 3(2C) + 4]Simplifying this, we get:[4R = 6C + 4]We solve the first equation for ( C ):[C = 51 - R]Substituting this into the second equation:[4R = 6(51 - R) + 4]Expanding and simplifying:[4R = 306 - 6R + 4][4R = 310 - 6R][10R = 310][R = 31]Verifying, if there are 31 rabbits, then the number of chickens is ( 51 - 31 = 20 ). Checking the feet condition:- Rabbit feet: ( 4 times 31 = 124 )- Chicken feet: ( 2 times 20 = 40 )- Three times chicken feet plus 4: ( 3 times 40 + 4 = 124 )The conditions are satisfied, so the number of rabbits is (boxed{31}).]

🔑:【Analysis】According to the problem, if one rabbit is less, then the total number of rabbit feet is three times the total number of chicken feet. Since the number of feet of one rabbit is 4 div 2=2 times the number of feet of one chicken, the total number of rabbit feet is three times the total number of chicken feet when the number of rabbits is 3 div 2=1.5 times the number of chickens. Therefore, the number of chickens is (51-1) div(1.5+1)=20. From this, we can find the number of rabbits.【Solution】Solution: 4 div 2=2begin{aligned}& (51-1) div(3 div 2+1) = & 50 div 2.5 = & 20 text { (heads) }end{aligned}51-20=31 (heads)Answer: There are 31 rabbits in the cage.Therefore, the answer is: 31.

🧠:[<think>Okay, so I have this problem here about sets A and B, and I need to find their intersection. Let me start by understanding what each set represents.First, set A is defined by the inequality (x - 3)(x + 1) ≤ 0. Hmm, quadratic inequalities can sometimes be tricky, but I remember that to solve them, you find the critical points where the expression equals zero and then test intervals around those points. Let me do that step by step.The inequality is (x - 3)(x + 1) ≤ 0. The critical points are x = 3 and x = -1 because those are the values that make each factor zero. So these points divide the number line into three intervals:1. x < -12. -1 < x < 33. x > 3Now, I need to test each interval to see where the product is less than or equal to zero.Starting with the first interval, x < -1. Let's pick a test point, say x = -2. Plugging into the expression: (-2 - 3)(-2 + 1) = (-5)(-1) = 5, which is positive. So this interval doesn't satisfy the inequality.Next interval: -1 < x < 3. Let's pick x = 0. (0 - 3)(0 + 1) = (-3)(1) = -3, which is negative. Since the inequality is ≤ 0, this interval is part of the solution.Third interval: x > 3. Let's choose x = 4. (4 - 3)(4 + 1) = (1)(5) = 5, positive again. So this interval doesn't work.Now, we also need to check the critical points themselves. At x = -1: (-1 - 3)(-1 + 1) = (-4)(0) = 0, which satisfies the inequality. At x = 3: (3 - 3)(3 + 1) = (0)(4) = 0, which also satisfies the inequality. So the solution set A is the closed interval [-1, 3].Alright, moving on to set B, which is defined by 2x > 2. Let me solve this inequality. Dividing both sides by 2 gives x > 1. So set B is all real numbers greater than 1, which in interval notation is (1, ∞).Now I need to find A ∩ B, the intersection of sets A and B. Set A is [-1, 3] and set B is (1, ∞). The intersection of two intervals is the overlap between them. Let me visualize both intervals on the number line.Set A starts at -1 and includes all numbers up to 3, including both endpoints. Set B starts just above 1 and goes to infinity. The overlap between these two would start just above 1 (since B doesn't include 1) and goes up to 3 (including 3 because A includes 3). So the intersection should be (1, 3].But let me verify this step by step. The intersection consists of all elements that are in both A and B. For a number to be in A, it must be between -1 and 3, inclusive. To be in B, it must be greater than 1. So the numbers that satisfy both are those greater than 1 but less than or equal to 3. That's exactly (1, 3].Looking at the answer choices provided:A: {x | -1 < x < 3}B: {x | 1 < x < 3}C: {x | x ≥ 3}D: {x | x > 2}E: {x | x > -1}Wait, hold on. Let me check the options again. The user mentioned options A through E, but the problem statement shows "A: {x|-12}" which seems truncated or mistyped. Wait, maybe there was a formatting error. Let me check again.Original problem statement:"Given the sets A={x|(x-3)(x+1) ≤ 0} and B={x|2x > 2}, then A ∩ B is (     ) A: {x|-12}"Hmm, the options seem to be cut off or misformatted. The first option is A: {x|-12}, which doesn't make sense. Wait, perhaps it's a typo? Maybe the original options were:A: {x | -1 < x < 3}B: {x | 1 < x < 3}C: {x | x ≥ 3}D: {x | x > 2}E: {x | x > -1}But the user might have had a formatting issue where the LaTeX wasn't rendered properly. Let me check the original problem again.Wait, in the original problem, the user wrote:"You will be given a problem. Please reason step by step, and put your final answer within boxed{}:Given the sets A={x|(x-3)(x+1) leq 0} and B={x|2x > 2}, then A cap B is (     )A: {x|-12}"So the options A to E are:A: {x | -1 < x < 2}B: {x | 1 < x < 3}C: {x | x ≥ 3}D: {x | x > 2}E: {x | x > -1}Wait, in the original problem, option A is written as {x|-12}, which might be a typo. Maybe it was supposed to be {x | -1 < x < 2}? The LaTeX code has "-12" which is odd. Alternatively, maybe the user intended to write something else, but there was a formatting error. Let me check.Looking at the LaTeX code for option A: A: {x|-12}. That would translate to A: {x | -12}, which is just the set containing -12, which doesn't make sense. Similarly, other options have similar issues? Let me check:Wait, no, in the original problem as written by the user, the options are:A: {x|-12}But looking at the LaTeX, this is written as {x|-12}, which would be {x | -12}, implying the set of x such that x is equal to -12? That can't be right.Alternatively, maybe the user made a typo and meant to write intervals. For example, maybe A was supposed to be {x | -1 < x < 2}, but due to a formatting mistake, it became {x | -12}.Similarly, maybe the other options have similar issues. However, given that the user provided the problem in Chinese and then translated it, perhaps there were some errors in the LaTeX conversion.Alternatively, maybe the original problem had numbers that were mistyped here. Since the user input shows:"A: {x|-12}" — perhaps "-12" was supposed to be "-1<x<2", but due to missing angle brackets or something. For example, if written as {x | -1 < x < 2}, but without the spaces, maybe it got combined into -12? That might be possible.Alternatively, maybe the original problem had the intervals written with hyphens instead of inequalities. For example, A: {x | -1 ≤ x ≤ 2} written as {x | -1-2}, but that seems unlikely.Wait, this is a bit confusing. Let me try to see if I can figure out the options based on the answer I found, which is (1, 3].Given that the answer should be {x | 1 < x ≤ 3}, but let's see the options. If the options are:A: {x | -1 < x < 2}B: {x | 1 < x < 3}C: {x | x ≥ 3}D: {x | x > 2}E: {x | x > -1}Then option B is {x | 1 < x < 3}, which is almost correct, but the original A ∩ B is (1, 3], which includes 3. If none of the options include 3, then the closest would be B. But if the options are miswritten, maybe C is {x | x ≥ 3}, which is not correct. Alternatively, maybe the options have a typo and option B was meant to be {x | 1 < x ≤ 3}, but written as {x |1 < x <3}.Given that, if the answer options are as above, with B being open interval (1,3), and the correct answer is (1,3], which is not exactly among the options. However, in the original problem, set A is [-1,3], and set B is (1, ∞). So the intersection is (1,3], but if the options do not have a closed interval at 3, then perhaps the answer is B: (1,3). But is 3 included in the intersection?Wait, set A includes 3 because the inequality is ≤ 0, so x=3 makes the expression zero, which is allowed. Set B is x >1, so 3 is in set B as well since 3 >1. Therefore, 3 is in both A and B, so it should be included in the intersection. Therefore, the correct interval is (1,3]. But if none of the options include 3, maybe there's a mistake. Wait, let me check again.Looking back at the problem statement, the user wrote the options as:A: {x|-12}But this is likely a formatting error. Given that the user might have intended to write something else, perhaps A through E are:A: {x | -1 < x < 2}B: {x | 1 < x < 3}C: {x | x ≥ 3}D: {x | x > 2}E: {x | x > -1}Assuming that, then the correct answer would be B: {x |1 < x <3}, even though technically 3 is included in the intersection. But in the options, if B is open interval, and the problem's answer choices don't have an option that includes 3, but B is the closest, then maybe the answer is B.Alternatively, maybe the original problem had a different definition for set B. Wait, set B is defined as 2x > 2, which simplifies to x >1, so B is (1, ∞). That's correct. Then A is [-1, 3]. The intersection is (1, 3].But the options given by the user are:A: {x | -12} [probably a typo]B: {x |1 <x <3}C: {x |x ≥3}D: {x |x >2}E: {x |x > -1}Assuming option B is (1,3), which is the open interval, but the actual intersection is (1,3]. However, if 3 is not included in any of the options, but option C is x ≥3, which is only the point 3 and beyond. But 3 is included in A, but in B, since B is x >1, 3 is in B. So 3 is in both A and B, so it should be included. Therefore, the intersection is (1,3], but none of the options have that. If I have to choose from the given options, B is the closest, but it's not entirely correct. However, maybe in the original problem, set A was defined with a strict inequality? Wait, let me check again.Original set A: (x -3)(x +1) ≤0. So it's a non-strict inequality, so endpoints are included. Therefore, 3 and -1 are in A. Set B: 2x > 2 ⇒ x >1, so 1 is not included. So the intersection is from greater than 1 to 3, including 3. So (1,3]. If none of the options include 3, but in the options given, B is {x |1 <x <3}, which is open at 3, and C is {x |x ≥3}, which starts at 3. So if we take the intersection, technically (1,3] is the correct answer, which is not exactly listed. But in some cases, depending on the test, they might consider [1,3], but here 1 is not included.Wait, perhaps the problem is in the interpretation of the original sets. Let me verify again:Set A: (x -3)(x +1) ≤0. The critical points at x = -1 and x =3. Testing intervals, as I did before, gives solution [-1,3].Set B: 2x >2 ⇒ x >1 ⇒ (1, ∞). So intersection is (1,3]. If the options are:A: (-1,2)B: (1,3)C: [3, ∞)D: (2, ∞)E: (-1, ∞)Then the correct answer would be B, but without including 3. However, since 3 is in both A and B, it should be included. If the options don't have that, then it's an error in the problem. Alternatively, maybe the problem was supposed to have a different inequality for set A or B.Alternatively, maybe I made a mistake in solving set A. Let me check again.For set A: (x -3)(x +1) ≤0.The roots are at x = -1 and x =3. The parabola opens upwards because the coefficient of x² is positive (1*1). Therefore, the parabola is positive outside the roots and negative between them. Hence, the inequality (x -3)(x +1) ≤0 is satisfied between the roots, including the roots. So solution is [-1,3]. That's correct.Set B: 2x >2 ⇒ x >1. So B is (1, ∞). Intersection is (1,3], which is not exactly listed. However, if the options are:A: {x | -1 <x <2}B: {x |1 <x <3}C: {x |x ≥3}D: {x |x >2}E: {x |x > -1}Then the correct answer is B: {x |1 <x <3}, even though it's missing the endpoint at 3. Maybe in the problem's context, they consider it as open interval. Alternatively, it's possible that there was a mistake in the problem options. However, given the choices, B is the closest.Alternatively, if we consider that in the original problem, maybe the inequality for set A was strict, (x -3)(x +1) <0, then the solution would be (-1,3), and the intersection with B would be (1,3), which matches option B. But the problem states ≤0, so it should include -1 and 3.Alternatively, perhaps a typo in the problem statement where set B is defined as 2x ≥2, which would make B [1, ∞), and then the intersection would be [1,3], but again, the options don't include that.Given the problem as stated, and assuming the options are as listed with possible typos, the most appropriate answer is B: {x |1 <x <3}. However, strictly speaking, the correct answer should include 3. Since the options don't have that, but B is the only one overlapping the interval, I'll go with B.But wait, looking back at the original problem's options again, the user wrote:"A: {x|-12}" which in LaTeX is {x | -12}, which is a set containing only the element -12, which is definitely not correct. Then option B is likely {x |1 <x <3}, but let me check:If the user's LaTeX code for the options is:A: {x|-12}B: {x|1<x<3}C: {x|x geq 3}D: {x|x >2}E: {x|x > -1}Then option B is correct. So assuming that in the original problem, option B is written correctly as {x |1 <x <3}, then that's the answer. The initial A option was probably a typo where "-1 <x <2" became "-12" due to missing angle brackets. So given that, the answer is B.Final Answerboxed{B}</think>Given the sets ( A = {x mid (x-3)(x+1) leq 0} ) and ( B = {x mid 2x > 2} ), we need to find ( A cap B ).1. Solving for set ( A ): - The inequality ((x-3)(x+1) leq 0) has critical points at ( x = -1 ) and ( x = 3 ). - Testing intervals around these points, we find the solution is the closed interval ([-1, 3]).2. Solving for set ( B ): - The inequality (2x > 2) simplifies to (x > 1), which is the interval ((1, infty)).3. Finding the intersection ( A cap B ): - Set ( A ) is ([-1, 3]) and set ( B ) is ((1, infty)). - The intersection of these intervals is the overlap between ([-1, 3]) and ((1, infty)), which is ((1, 3]).4. Considering the answer choices: - The correct answer should be ((1, 3]). However, the closest option provided is ( B: {x mid 1 < x < 3} ), which is the open interval. Despite the endpoint 3 being included in ( A ), the provided options do not include it. Therefore, the best match is option B.[boxed{B}],<think>Okay, let me try to figure out this problem step by step. So, we have two sets, A and B, defined by inequalities, and we need to find their intersection, right? Let me first understand each set separately before thinking about their intersection. Starting with set A: A is defined as all real numbers x such that (x - 3)(x + 1) ≤ 0. Hmm, this is a quadratic inequality. To solve this, I remember that the product of two terms is less than or equal to zero, which means the expression changes sign around its roots. So, first, I should find the roots of the quadratic equation (x - 3)(x + 1) = 0. The roots are x = 3 and x = -1. These are the critical points where the expression will change from positive to negative or vice versa. Now, to determine where the product is less than or equal to zero, I can use a sign chart or test intervals around these roots. The intervals to test are: 1. x < -12. -1 < x < 33. x > 3Let me test each interval.1. When x < -1, let's pick x = -2. Plugging into (x - 3)(x + 1): (-2 - 3)(-2 + 1) = (-5)(-1) = 5, which is positive. So, the product is positive here.2. When -1 < x < 3, let's choose x = 0. (0 - 3)(0 + 1) = (-3)(1) = -3, which is negative. So, the product is negative in this interval.3. When x > 3, let's take x = 4. (4 - 3)(4 + 1) = (1)(5) = 5, positive again.Since the inequality is ≤ 0, we include the points where the expression is equal to zero. That happens at x = -1 and x = 3. Therefore, the solution set A is the interval where the product is negative or zero, which is [-1, 3]. Okay, so set A is [-1, 3]. Got that.Now, moving on to set B: B is defined as all real numbers x such that 2x > 2. This seems simpler. Let me solve for x. Divide both sides by 2: x > 1. So, set B is (1, ∞). All numbers greater than 1. Now, the question is asking for A ∩ B, which is the intersection of the two sets. The intersection consists of all elements that are in both A and B. Since A is [-1, 3] and B is (1, ∞), we need to find the overlap between these two intervals. Let me visualize this. The interval A starts at -1 and goes up to 3, including both endpoints. The interval B starts just above 1 and goes to infinity. So, the overlapping part would be from just above 1 up to 3. But since intervals are continuous, the intersection should be (1, 3]. Wait, but let me check:Set A includes all numbers from -1 to 3, including -1 and 3. Set B includes all numbers greater than 1. So, the intersection should include numbers that are greater than 1 and less than or equal to 3. Because in set A, up to 3 is allowed, and in set B, anything greater than 1. So yes, the intersection is (1, 3]. But wait, let me verify. The point at 3 is included in A, and since B doesn't specify anything at 3, but since 3 is greater than 1, is 3 included in B? Wait, set B is x > 1, which is (1, ∞). So 3 is included in B because 3 > 1 is true, but in the inequality, it's just greater than, not greater than or equal. Wait, hold on. Wait, the original set B is 2x > 2. Solving that gives x > 1. So, x must be greater than 1, not greater than or equal. So, 1 itself is not included. Similarly, 3 is greater than 1, so 3 is included in B. Wait, no, in set B, since it's x > 1, any x greater than 1 is in B. So 3 is greater than 1, so 3 is in B, right? Because x > 1 includes all numbers larger than 1, no upper limit. So, even though 3 is the upper bound of A, since 3 is in A and also in B (because 3 > 1), then 3 is included in the intersection. Wait, hold on. Let me clarify. Set A includes 3 because the inequality was ≤ 0, and at x=3, (3 - 3)(3 + 1) = 0, which satisfies the inequality. So 3 is in A. Set B is x > 1, which is (1, ∞). So 3 is in B because 3 > 1. Therefore, 3 is in both A and B, so it's included in the intersection. However, the number 1 is not included in B, so even though in A, numbers from -1 to 3 are included, the intersection with B would start just above 1. Therefore, the intersection is (1, 3]. Let me double-check. If I take a number between 1 and 3, say 2. That's in A (since -1 ≤ 2 ≤ 3) and in B (since 2 > 1). So 2 is in the intersection. If I take 1.5, same thing. If I take 3, it's in both. If I take 1, it's in A (since 1 is between -1 and 3) but not in B (since 1 is not greater than 1), so 1 is excluded. If I take 0, it's in A but not in B. If I take 4, it's in B but not in A. So yes, the intersection is all numbers greater than 1 and up to and including 3. So (1, 3].Looking back at the answer choices:A: {x | -1 < x ≤ 3}B: {x | 1 < x ≤ 3}C: {x | x > 1}D: {x | x > 2}Wait, the options are labeled A, B, C, D. But in the original problem statement, the options are:A: {x | -1 < x ≤ 3}B: {x | 1 < x ≤ 3}C: {x | x > 1}D: {x | x > 2}Wait, but in the problem statement, the user wrote:"A: {x|-12}" — Wait, maybe there was a typo in the problem statement? Let me check again.Wait, the user wrote:"A: {x|-12}"Wait, that seems like a typo. Let me check the original problem again.Original problem:"Given the sets A={x|(x-3)(x+1) ≤ 0} and B={x|2x > 2}, then A ∩ B is (     ) A: {x|-12}"Wait, the options are presented as A, B, C, D, but in the original problem statement, the user wrote them all in one line. Let me parse it again."A: {x|-12}"Wait, perhaps there was a mistake in formatting. The original options might have been:A: {x | -1 < x ≤ 3}B: {x | 1 < x ≤ 3}C: {x | x > 1}D: {x | x > 2}But due to formatting issues, it got compressed. For example, maybe "-12" was actually "-1 < x ≤ 2" or similar. Wait, but "-12" is a number. Alternatively, maybe the user intended:A: {x | -1 < x ≤ 2}B: {x | 1 < x ≤ 3}C: {x | x > 1}D: {x | x > 2}But in the original problem statement, the user wrote:"A: {x|-12}"Hmm, perhaps that's a typo or formatting error. Wait, looking again, the user wrote:"A: {x|-12}"In LaTeX, that would be interpreted as {x | -12}, which doesn't make sense. Maybe there was a missing inequality. Alternatively, maybe the user intended to write:A: {x | -1 < x ≤ 2}B: {x | 1 < x ≤ 3}C: {x | x > 1}D: {x | x > 2}But in the problem statement, perhaps due to a typo, the LaTeX code got messed up. For example, if the user wrote:"A: {x mid -1 < x leq 3}" which would display as A: {x | -1 < x ≤ 3}But maybe in the original code, it was written as {x|-12} due to missing spaces or symbols. Alternatively, maybe the user mistyped the LaTeX.Alternatively, perhaps the options are:A: (-1, 3]B: (1, 3]C: (1, ∞)D: (2, ∞)But given the original problem statement's mention of A and B as sets defined by those inequalities, and the options given as A, B, C, D with those descriptions, perhaps the answer is B: {x | 1 < x ≤ 3}.But given that in the problem statement, the user wrote "A: {x|-12}", maybe the options were misformatted. For example, in LaTeX, if you write {x|-12}, it would show as {x|-12}, but maybe the user intended to write intervals. For example:A: (-1, 2]B: (1, 3]C: (1, ∞)D: (2, ∞)But given the original problem, the intersection we found is (1, 3], which would correspond to option B if that's one of the choices. However, given the original problem statement's options as presented, there's some ambiguity due to possible formatting errors.But assuming that the options are:A: {x | -1 < x ≤ 3}B: {x | 1 < x ≤ 3}C: {x | x > 1}D: {x | x > 2}Then the correct answer is B: {x | 1 < x ≤ 3}.Alternatively, if the options are:A: (-1, 3]B: (1, 3]C: (1, ∞)D: (2, ∞)Then again, the answer is B.Given the original problem statement's mention of "A: {x|-12}", which might be a typo for {x | -1 < x ≤ 2} or similar, but with the information we have, the intersection is (1, 3], so if option B is (1, 3], that's the answer.Therefore, the answer should be B.Final Answerboxed{B}</think>Given the sets ( A = {x mid (x-3)(x+1) leq 0} ) and ( B = {x mid 2x > 2} ), we need to find ( A cap B ).1. Solving for set ( A ): - The inequality ((x-3)(x+1) leq 0) has roots at ( x = 3 ) and ( x = -1 ). - Testing intervals around these roots, we find the solution is ([-1, 3]).2. Solving for set ( B ): - The inequality (2x > 2) simplifies to (x > 1), which gives the interval ((1, infty)).3. Finding the intersection ( A cap B ): - The interval for ( A ) is ([-1, 3]). - The interval for ( B ) is ((1, infty)). - The intersection of these intervals is ((1, 3]), which includes all numbers greater than 1 and up to 3.Thus, the correct answer is (boxed{B}).]

🔑:B Key point: Intersection and its operations. Topic: Set. Analysis: To find the intersection of sets A and B, we need to solve the inequalities in A and B to determine the sets A and B, and then find their intersection. Solution: Solving the inequality in A gives -1 leq x leq 3, so A={x|-1 leq x leq 3}. Transforming the inequality in B gives 2x > 2 = 2 times 1, which leads to x > 1, so B={x|x > 1}. Therefore, A cap B = {x|1 < x leq 3}. Hence, the correct choice is: boxed{B}. Review: This problem tests the operation of intersection. Mastering the definition of intersection is key to solving this problem.

🧠:[<think>Okay, so I have this problem where I need to find all four-digit numbers formed using the digits 1, 9, 8, 5 without repeating any digits. Then, for parts a) to e), I need to determine how many of these numbers are divisible by 2, 3, 4, 5, and 6 respectively, and also list those numbers. Let me start by understanding the problem step by step.First, the digits allowed are 1, 9, 8, and 5. Each digit must be used exactly once in each four-digit number. So, since there are four digits and each is used once, the total number of possible four-digit numbers is 4! = 24. That makes sense because for a four-digit number, the first digit has 4 choices, the second has 3, the third 2, and the last 1, so 4*3*2*1 = 24.Now, the tasks are to find how many of these 24 numbers are divisible by 2, 3, 4, 5, 6. Let's tackle each part one by one.Part a) Divisible by 2A number is divisible by 2 if its last digit is even. The digits we have are 1, 9, 8, 5. Among these, the even digits are 8. The others (1, 9, 5) are odd. So, for a number to be divisible by 2, it must end with 8.Therefore, the last digit is fixed as 8. The remaining three digits (1, 9, 5) can be arranged in the first three positions in 3! = 6 ways. Therefore, there should be 6 numbers divisible by 2.But let me confirm by listing them. The numbers will be permutations of 1, 9, 5 in the first three digits followed by 8. Let's list all permutations:First three digits permutations:1, 9, 5:- 1 9 5 8- 1 5 9 8- 9 1 5 8- 9 5 1 8- 5 1 9 8- 5 9 1 8So those are 6 numbers. Let me check if they all end with 8 and have unique digits. Yes, each of them ends with 8, and the first three digits are all permutations of 1, 9, 5. So part a) answer is 6 numbers. Let me note them down:a) Numbers divisible by 2: 1958, 1598, 9158, 9518, 5198, 5918. Total 6.Part b) Divisible by 3A number is divisible by 3 if the sum of its digits is divisible by 3. Let's compute the sum of the digits 1 + 9 + 8 + 5. That is 1+9=10, 10+8=18, 18+5=23. Wait, 23? Hmm. But 23 divided by 3 is 7 with a remainder of 2. So the sum is 23, which is not divisible by 3. Therefore, none of the numbers formed by these digits will be divisible by 3. So the answer is 0 numbers.But wait, let me check again. Maybe I added wrong. 1 + 9 is 10, plus 8 is 18, plus 5 is 23. Yes, 23. 23 mod 3 is 2, since 3*7=21, 23-21=2. Therefore, sum is 23, which is not divisible by 3. Therefore, no number formed by these digits can be divisible by 3. Therefore, part b) answer is 0. So no numbers listed here.Part c) Divisible by 4A number is divisible by 4 if the number formed by its last two digits is divisible by 4. So, for each four-digit number, we need to check if the last two digits form a number that's divisible by 4.Given the digits 1, 9, 8, 5, each used exactly once. So, possible two-digit endings from these digits, without repetition. Let me list all possible two-digit endings and check which are divisible by 4.Possible two-digit endings (since the last two digits can be any permutation of two different digits from 1, 9, 8, 5):Possible endings:18, 19, 15, 81, 89, 85, 91, 98, 95, 51, 58, 59.Wait, but since the digits can't repeat, the possible endings are all permutations of two different digits. So from the four digits, the possible two-digit endings are 4P2 = 12, which are listed above.Now, let's check which of these two-digit numbers are divisible by 4.Check each:18: 18 ÷ 4 = 4.5 → Not divisible.19: 19 ÷ 4 = 4.75 → No.15: 15 ÷ 4 = 3.75 → No.81: 81 ÷ 4 = 20.25 → No.89: 89 ÷ 4 = 22.25 → No.85: 85 ÷ 4 = 21.25 → No.91: 91 ÷ 4 = 22.75 → No.98: 98 ÷ 4 = 24.5 → Wait, 24*4=96, 98-96=2. So remainder 2. Not divisible.95: 95 ÷ 4 = 23.75 → No.51: 51 ÷ 4 = 12.75 → No.58: 58 ÷ 4 = 14.5 → 14*4=56, 58-56=2. Remainder 2. Not divisible.59: 59 ÷ 4 = 14.75 → No.Wait a second, none of these two-digit endings are divisible by 4? That can't be right. Let me check again.Wait, 12 possible two-digit endings. Let me list them:From digits 1,9,8,5:Possible endings:18, 19, 15, 81, 89, 85, 91, 98, 95, 51, 58, 59.Divisible by 4?Divisible by 4 numbers must be even and the last two digits as a number divisible by 4. However, in our case, even numbers would end with 8 or a digit that is even. But our digits have only one even digit: 8. So the last digit must be 8 to be even, but in that case, the second last digit can be 1,9,5. So possible endings with 8 as the last digit:18, 98, 58.Wait, but if the last digit is 8, then the previous digit can be 1,9, or 5. So possible endings are 18, 98, 58.So, check these:18 ÷ 4 = 4.5 → Not divisible.98 ÷ 4 = 24.5 → Not divisible.58 ÷ 4 = 14.5 → Not divisible.So, even though 8 is the only even digit, the two-digit numbers ending with 8 (18, 98, 58) are not divisible by 4. Therefore, there are no four-digit numbers formed with these digits that are divisible by 4. Hence, part c) answer is 0.But wait, wait. Let me check again. Maybe I missed a possible ending. But the possible endings with 8 are only those three: 18, 98, 58, none of which are divisible by 4. The other possible endings with other digits would have an odd last digit, so the number formed by the last two digits would be odd, which can't be divisible by 4. Therefore, yes, no numbers here are divisible by 4. Therefore, part c) is 0.Part d) Divisible by 5A number is divisible by 5 if its last digit is 0 or 5. Since our digits are 1,9,8,5, and 0 is not among them. Therefore, the last digit must be 5. So, similar to part a), fix the last digit as 5, and the first three digits are permutations of 1,9,8.Number of permutations: 3! = 6. Therefore, 6 numbers.Let me list them:Permutations of 1,9,8 in the first three digits, ending with 5:1985, 1895, 9185, 9815, 8195, 8915.Check each:1985: ends with 5 – yes.1895: same.9185, 9815, 8195, 8915. All end with 5. Each of these numbers uses digits 1,9,8,5 exactly once. So part d) answer is 6 numbers.Part e) Divisible by 6A number is divisible by 6 if it is divisible by both 2 and 3. From earlier, we saw that numbers divisible by 2 must end with 8, and numbers divisible by 3 require the sum of digits to be divisible by 3. However, the sum of the digits is 23, which isn't divisible by 3. Therefore, there are no numbers that are divisible by both 2 and 3. Hence, part e) answer is 0.But let me verify this. Since the sum of digits is 23, which isn't divisible by 3, so even if a number is divisible by 2, it can't be divisible by 3. Therefore, no numbers are divisible by 6. So answer is 0.Wait, but 6 is 2 and 3, but since none are divisible by 3, regardless of being even, they can't be divisible by 6. So yes, part e) is 0.Summary:a) Divisible by 2: 6 numbers (ending with 8)b) Divisible by 3: 0 numbers (sum is 23)c) Divisible by 4: 0 numbers (no valid two-digit endings)d) Divisible by 5: 6 numbers (ending with 5)e) Divisible by 6: 0 numbers (not divisible by 3)Listing the numbers:For a) Divisible by 2:1958, 1598, 9158, 9518, 5198, 5918For d) Divisible by 5:1985, 1895, 9185, 9815, 8195, 8915Others (b, c, e) have no numbers.Let me double-check the numbers listed for a) and d) to ensure they are unique and use each digit once.For a):1958: digits 1,9,5,8 – all unique.1598: 1,5,9,8 – unique.9158: 9,1,5,8 – unique.9518: 9,5,1,8 – unique.5198: 5,1,9,8 – unique.5918: 5,9,1,8 – unique. All good.For d):1985: 1,9,8,5 – unique.1895: 1,8,9,5 – unique.9185: 9,1,8,5 – unique.9815: 9,8,1,5 – unique.8195: 8,1,9,5 – unique.8915: 8,9,1,5 – unique. All correct.Therefore, the answers are as above.Final Answera) boxed{6} b) boxed{0} c) boxed{0} d) boxed{6} e) boxed{0}The numbers are: a) 1958, 1598, 9158, 9518, 5198, 5918 d) 1985, 1895, 9185, 9815, 8195, 8915</think>To determine the number of four-digit numbers formed using the digits 1, 9, 8, 5 (each exactly once) that are divisible by 2, 3, 4, 5, and 6, we analyze each case:a) Divisible by 2 A number is divisible by 2 if its last digit is even. The only even digit is 8. Fixing 8 as the last digit, the remaining digits 1, 9, 5 can be arranged in 3! = 6 ways. Numbers: 1958, 1598, 9158, 9518, 5198, 5918 Answer: boxed{6}b) Divisible by 3 A number is divisible by 3 if the sum of its digits is divisible by 3. The sum of the digits 1 + 9 + 8 + 5 = 23, which is not divisible by 3. Answer: boxed{0}c) Divisible by 4 A number is divisible by 4 if its last two digits form a number divisible by 4. Checking all possible endings, none of the two-digit combinations (18, 19, 15, 81, 89, 85, 91, 98, 95, 51, 58, 59) are divisible by 4. Answer: boxed{0}d) Divisible by 5 A number is divisible by 5 if its last digit is 5. Fixing 5 as the last digit, the remaining digits 1, 9, 8 can be arranged in 3! = 6 ways. Numbers: 1985, 1895, 9185, 9815, 8195, 8915 Answer: boxed{6}e) Divisible by 6 A number is divisible by 6 if it is divisible by both 2 and 3. Since no numbers are divisible by 3, none can be divisible by 6. Answer: boxed{0} Final Answera) boxed{6} b) boxed{0} c) boxed{0} d) boxed{6} e) boxed{0}The numbers are: a) 1958, 1598, 9158, 9518, 5198, 5918 d) 1985, 1895, 9185, 9815, 8195, 8915,<think>Okay, let's tackle this problem step by step. We need to form all possible four-digit numbers using the digits 1, 9, 8, 5 without repeating any digits, and then determine how many of those numbers are divisible by 2, 3, 4, 5, and 6. Also, we need to list these numbers. Let me start by understanding the problem and breaking it down.First, since we're using four distinct digits (1, 9, 8, 5), the total number of four-digit numbers we can form is 4! = 24. So, there are 24 unique four-digit numbers. Each of these numbers uses each digit exactly once. Now, for each part (a to e), we need to find how many of these 24 numbers are divisible by the given divisor and list them.Let me handle each part one by one, but maybe there's a way to approach them systematically.a) Divisible by 2:A number is divisible by 2 if its last digit is even. The digits we have are 1, 9, 8, 5. Among these, the even digits are only 8. So, the last digit must be 8. The remaining three digits (1, 9, 5) can be arranged in the first three positions in any order. So, the number of such numbers is 3! = 6. Let me list them:The last digit is fixed as 8. The first three digits can be permutations of 1, 9, 5.Possible permutations for the first three digits:1. 1, 9, 5 → 19582. 1, 5, 9 → 15983. 9, 1, 5 → 91584. 9, 5, 1 → 95185. 5, 1, 9 → 51986. 5, 9, 1 → 5918So, the numbers are: 1958, 1598, 9158, 9518, 5198, 5918. That's six numbers. So part a) answer is 6.b) Divisible by 3:A number is divisible by 3 if the sum of its digits is divisible by 3. Let's compute the sum of the digits: 1 + 9 + 8 + 5 = 23. Wait, 23 divided by 3 is 7 with remainder 2. So the sum is 23, which is not divisible by 3. Therefore, none of the numbers formed by these digits will be divisible by 3. Therefore, the count is 0. But let me confirm the sum again. 1 + 9 is 10, 10 + 8 is 18, 18 +5 is 23. Yes, 23. Since the sum is 23, which isn't divisible by 3, there can't be any numbers here divisible by 3. So part b) answer is 0.c) Divisible by 4:A number is divisible by 4 if its last two digits form a number that's divisible by 4. Since we have digits 1,9,8,5, all used exactly once, the possible two-digit endings are the permutations of two different digits from the set. Let me list all possible two-digit endings and check which are divisible by 4.Possible two-digit endings (last two digits):Since all digits are used once, the last two digits can be any two distinct digits from 1,9,8,5. Let's list all possible permutations:From the digits 1,9,8,5, the possible two-digit numbers (permutations where order matters):14 possible pairs (since 4 digits, taken 2 at a time, with order):19, 18, 15, 91, 98, 95, 81, 89, 85, 51, 59, 58, (Wait, but actually, since we're considering the last two digits of a four-digit number, the last two digits can be any permutation of two different digits from the four. So total of 4*3=12 possibilities.Wait, let's list them:1 in the tens place:19, 18, 15,9 in the tens place:91, 98, 95,8 in the tens place:81, 89, 85,5 in the tens place:51, 59, 58.So total 12 possible two-digit endings. Now check which of these are divisible by 4.Let's check each:19: 19 ÷ 4 = 4.75 → no.18: 18 ÷ 4 = 4.5 → no.15: 15 ÷4=3.75→no.91: 91 ÷4=22.75→no.98: 98 ÷4=24.5→no (24*4=96, 25*4=100, so 98-96=2, remainder 2→not divisible.95: 95 ÷4=23.75→no.81: 81 ÷4=20.25→no.89:89 ÷4=22.25→no.85:85 ÷4=21.25→no.51:51 ÷4=12.75→no.59:59 ÷4=14.75→no.58:58 ÷4=14.5→ no (Wait, 58 divided by 4 is 14.5, but since we need an integer result, 14*4=56, 15*4=60, so remainder 2. So 58 is not divisible by 4.Wait, but wait, 12 possibilities and none are divisible by 4? That seems odd. Let me double-check.Wait, 8 is one of the digits. Let's check the endings with 8:Possible endings where the last two digits include 8:18, 98, 58, 81, 89, 85, 48, 84? Wait, but our digits are 1,9,8,5. So possible endings with 8 are:18, 98, 58, 81, 89, 85. Wait, no, in the two-digit endings, if the last digit is 8, then the tens place could be 1,9,5. So numbers ending with 8 would have last two digits as:18, 98, 58. Let's check these:18 ÷4=4.5→no.98 ÷4=24.5→no.58 ÷4=14.5→no.So none of the endings with 8 in the units place (since 8 is even, but the two-digit number needs to be divisible by 4). Hmm. Then, any other two-digit combinations?Looking at all 12, none are divisible by 4. That suggests that there are zero numbers in this set divisible by 4. But that can't be right, can it? Wait, let me check again.Wait, maybe I made a mistake in listing. Let's check again each two-digit number:19: 19 mod4= 3.18: 18 mod4=2.15:15 mod4=3.91:91 mod4= 91-88=3, so 3.98:98 mod4=2.95:95 mod4=3.81:81 mod4=1.89:89 mod4=1.85:85 mod4=1.51:51 mod4=3.59:59 mod4=3.58:58 mod4=2.So, none of them have remainder 0. Therefore, none of the two-digit endings are divisible by 4. Therefore, none of the four-digit numbers formed with these digits can be divisible by 4. Therefore, part c) answer is 0.But wait, that seems surprising. Let me confirm with an example. Suppose a number ends with 12, which is divisible by 4, but our digits don't include 2. Wait, the digits are 1,9,8,5. So possible endings. Let me think if 8 is in the tens place. For example, 84 is divisible by 4, but we don't have 4. We have 8 in the digits, but the other digits are 1,9,5. So the possible endings with 8 in the tens place would be 81, 89, 85, none of which are divisible by 4. Similarly, 8 in the units place would be endings like 18, 98, 58, which we've already checked. So indeed, none. Therefore, part c) answer is 0.d) Divisible by 5:A number is divisible by 5 if its last digit is 0 or 5. Since we don't have 0 in the digits, the last digit must be 5. Then, the remaining digits (1,9,8) can be arranged in the first three positions. The number of such numbers is 3! = 6. Let's list them:The last digit is 5. The first three digits are permutations of 1,9,8.Possible permutations:1. 1,9,8 → 19852. 1,8,9 → 18953. 9,1,8 → 91854. 9,8,1 → 98155. 8,1,9 → 81956. 8,9,1 → 8915So the numbers are: 1985, 1895, 9185, 9815, 8195, 8915. That's six numbers. So part d) answer is 6.e) Divisible by 6:A number is divisible by 6 if it is divisible by both 2 and 3. From part a), numbers divisible by 2 are those ending with 8. From part b), the sum of digits is 23, which isn't divisible by 3, so there are no numbers divisible by 3. Therefore, there can't be any numbers divisible by both 2 and 3. Hence, the answer for part e) is 0.Wait, but let me confirm again. Divisible by 6 requires divisible by 2 and 3. Since none are divisible by 3, even if they are divisible by 2, they won't be divisible by 6. Hence, yes, 0.Summary:a) 6 numbers: 1958, 1598, 9158, 9518, 5198, 5918.b) 0 numbers.c) 0 numbers.d) 6 numbers: 1985, 1895, 9185, 9815, 8195, 8915.e) 0 numbers.I need to make sure that I listed all numbers correctly for parts a) and d). Let me cross-verify.For part a), numbers ending with 8:Possible first three digits permutations of 1,9,5:1-9-5: 19581-5-9: 15989-1-5: 91589-5-1: 95185-1-9: 51985-9-1: 5918Yes, that's six numbers. Correct.For part d), numbers ending with 5:First three digits permutations of 1,9,8:1-9-8: 19851-8-9: 18959-1-8: 91859-8-1: 98158-1-9: 81958-9-1: 8915Yes, that's six numbers. Correct.For parts b), c), and e), as reasoned, there are none. So the answers are 0.I think this covers all the parts. Let me just check if there's any mistake in logic.For part c), divisibility by 4 depends on the last two digits. Since none of the possible two-digit combinations from the digits 1,9,8,5 (each used once) are divisible by 4, the answer is indeed 0. Similarly, for part e), since no numbers are divisible by 3, even those divisible by 2 won't make them divisible by 6. Hence, correct.</think>a) Divisible by 2:Numbers must end with 8. The remaining digits (1, 9, 5) form the first three places:- 1958, 1598, 9158, 9518, 5198, 5918 Total: boxed{6}b) Divisible by 3:Sum of digits (1 + 9 + 8 + 5 = 23) is not divisible by 3. Total: boxed{0}c) Divisible by 4:None of the possible two-digit endings are divisible by 4. Total: boxed{0}d) Divisible by 5:Numbers must end with 5. The remaining digits (1, 9, 8) form the first three places:- 1985, 1895, 9185, 9815, 8195, 8915 Total: boxed{6}e) Divisible by 6:No numbers are divisible by both 2 and 3. Total: boxed{0}]

🔑:a) Among the four-digit numbers mentioned, exactly those ending in an even digit, i.e., 8, are divisible by 2. With the remaining 3 digits, 6 different numbers can be formed, so there are exactly 6 such numbers divisible by 2, namely 1598, 1958, 5198, 5918, 9158, and 9518.b, e) Since the cross sum of 1985 is not divisible by 3, and this also applies to all numbers formed by rearrangement, none of the numbers are divisible by 3 and thus none are divisible by 7.c) A number is divisible by 4 if and only if the number formed by the last two digits is divisible by 4. However, from the digits 1, 5, 8, and 9, no such two-digit number can be formed; it would have to be even, i.e., end in 8, and the numbers 18, 58, and 98 are not divisible by 4. Therefore, there are no such numbers divisible by 4.d) Among these numbers, exactly those ending in 5 are divisible by 5. With the remaining 3 digits, 6 different numbers can be formed. Thus, exactly 6 such numbers are divisible by 5, namely 1895, 1985, 8195, 8915, 9185, and 9815.

🧠:[<think>Okay, let's try to figure out how to solve these problems about the number of divisors. So, the question is asking for how many divisors a number has when it's composed of different prime numbers raised to various powers. Let's take each part step by step.Starting with part a) which is just pq, where p and q are different primes. Hmm, I remember that the number of divisors of a number can be found by looking at its prime factorization and then using a formula. But wait, pq is already the prime factorization since p and q are primes. So maybe there's a rule here.Let me recall, if a number is factored into primes like p₁^a * p₂^b * ... * pₙ^k, then the number of divisors is (a+1)(b+1)...(k+1). Is that right? Let me check with a simple example. Take 6, which is 2*3. The divisors are 1, 2, 3, 6. That's 4 divisors. Applying the formula, the exponents of 2 and 3 are both 1, so (1+1)(1+1) = 2*2 = 4. That works. So maybe the formula is correct.So for part a) pq, the exponents of p and q are both 1. Therefore, the number of divisors should be (1+1)(1+1) = 4. Let me list them for pq: 1, p, q, pq. Yep, that's four. So part a) should be 4. Got that.Moving on to part b) p²q. Again, primes p and q, different. So the prime factorization is p squared times q. Using the same formula, the exponent of p is 2 and exponent of q is 1. So the number of divisors would be (2+1)(1+1) = 3*2 = 6. Let me verify with an example. Take p=2, q=3. So the number is 4*3=12. Divisors of 12 are 1, 2, 3, 4, 6, 12. That's six divisors. Perfect, the formula holds. So part b) is 6.Part c) is p²q². So both primes squared. The exponents here are 2 for p and 2 for q. Applying the formula, (2+1)(2+1) = 3*3 = 9. Let's check with p=2, q=3. The number is 4*9=36. Divisors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36. That's nine divisors. Yep, matches. So part c) is 9.Finally, part d) is the general case: p^m q^n. So exponents m and n. Then, following the formula, the number of divisors should be (m+1)(n+1). Let me verify with an example. Let m=3, n=2, p=2, q=3. The number is 8*9=72. The divisors of 72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72. That's 12 divisors. Applying the formula: (3+1)(2+1)=4*3=12. Correct. So the general formula seems to hold.Therefore, the answers should be:a) 4b) 6c) 9d) (m+1)(n+1)Wait, but let me make sure I didn't miss anything. The problem states that p and q are different primes, so in all cases, the prime factorization is unique and has two distinct primes. So the formula applies here because each prime is separate. If there were multiple primes, the formula would extend accordingly, but since it's only two primes, the number of divisors is indeed (exponent of p +1) multiplied by (exponent of q +1). Another check for part d). Suppose m=0 and n=0, but wait, primes can't have exponent 0 because the number would be 1, but 1 has only 1 divisor. However, in the problem statement, p and q are primes, but if m or n are 0, then the formula would still work. For example, if m=0, then p^0 is 1, so the number is q^n, which has (0+1)(n+1)=1*(n+1)=n+1 divisors. But in the problem, are m and n allowed to be zero? The question says "p^m q^n", but since p and q are primes, m and n are positive integers? Wait, actually, in the previous parts, they used exponents 1, 2, etc., so maybe m and n are positive integers. But the formula (m+1)(n+1) works even if m or n are zero, but in the context of the problem, since we are given p and q as primes, maybe m and n are at least 1. But the problem doesn't specify, so maybe we should assume m and n are non-negative integers, which would include zero. However, if m=0 and n=0, the number is 1, which has 1 divisor, and the formula (0+1)(0+1)=1 gives that. So the formula works regardless. Therefore, the answer for part d) is indeed (m+1)(n+1). I think that's solid. All the examples check out, and the formula seems consistent.Final Answera) boxed{4}b) boxed{6}c) boxed{9}d) boxed{(m+1)(n+1)}</think>To determine the number of divisors of a number given its prime factorization, we use the formula: if a number is factored into primes as ( p_1^{a} cdot p_2^{b} cdot ldots cdot p_n^{k} ), then the number of divisors is ( (a+1)(b+1)ldots(k+1) ).a) For ( pq ), where ( p ) and ( q ) are different primes:- The exponents of ( p ) and ( q ) are both 1.- Using the formula, the number of divisors is ( (1+1)(1+1) = 2 times 2 = 4 ).- Example: ( pq = 6 ) has divisors 1, 2, 3, 6.b) For ( p^2 q ):- The exponent of ( p ) is 2 and the exponent of ( q ) is 1.- Using the formula, the number of divisors is ( (2+1)(1+1) = 3 times 2 = 6 ).- Example: ( p^2 q = 12 ) has divisors 1, 2, 3, 4, 6, 12.c) For ( p^2 q^2 ):- The exponents of ( p ) and ( q ) are both 2.- Using the formula, the number of divisors is ( (2+1)(2+1) = 3 times 3 = 9 ).- Example: ( p^2 q^2 = 36 ) has divisors 1, 2, 3, 4, 6, 9, 12, 18, 36.d) For ( p^m q^n ):- The exponents of ( p ) and ( q ) are ( m ) and ( n ) respectively.- Using the formula, the number of divisors is ( (m+1)(n+1) ).- Example: ( p^3 q^2 = 72 ) has divisors 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72 (12 divisors). Final Answera) boxed{4}b) boxed{6}c) boxed{9}d) boxed{(m+1)(n+1)},<think>Okay, let's see. I need to figure out how many divisors each of these numbers has, given that p and q are different prime numbers. Hmm. Alright, starting with part a) which is just pq. So, first, primes have only two divisors, 1 and themselves. But when you multiply two different primes together, how many divisors does that product have?Let me think. If I take p times q, then the divisors should be 1, p, q, and pq itself. Right? Because 1 times pq is pq, p times q is pq, and q times p is pq. So that's four divisors. Let me check with an example. Let's take p=2 and q=3. Then pq=6. The divisors of 6 are 1, 2, 3, 6. Yep, four divisors. So part a) should be 4. Okay, that makes sense.Moving on to part b) p squared times q. So p²q. Let's take the same example, p=2, q=3. Then p²q=4*3=12. The divisors of 12 are 1, 2, 3, 4, 6, 12. That's six divisors. Wait, how does that break down? Let's see. The prime factorization is 2²*3¹. The exponents are 2 and 1. To find the number of divisors, I remember there's a formula where you take the exponents of the primes, add one to each, and multiply them together. So for p²q, the exponents are 2 and 1. Add one to each: 3 and 2. Multiply them: 3*2=6. That's the number of divisors. Let me verify with another example. Suppose p=3 and q=5. Then p²q=9*5=45. Divisors of 45: 1, 3, 5, 9, 15, 45. Six divisors. Yep, that checks out. So the formula seems to work. So the answer for part b) is 6.Part c) is p squared times q squared, so p²q². Let's take p=2, q=3 again. Then p²q²=4*9=36. Divisors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36. That's nine divisors. Using the formula again, exponents are 2 and 2. Add one to each: 3 and 3. Multiply: 3*3=9. Perfect. Another example: p=3, q=5. p²q²=9*25=225. Divisors of 225: 1, 3, 5, 9, 15, 25, 45, 75, 225. Nine divisors. So part c) is 9.Now part d) is the general case: p^m q^n. So exponents m and n. Using the same logic as before, if we have a number N = p^m * q^n, then the number of divisors is (m+1)(n+1). Because for each prime, the exponent can range from 0 up to m for p, and 0 up to n for q. So each exponent choice gives a unique divisor. For example, if m=3 and n=2, then the number of divisors is (3+1)(2+1)=12. Let's test this. Take p=2, q=3, m=3, n=2. Then N=8*9=72. Divisors of 72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72. Twelve divisors. Yep, that works. So the formula is (m+1)(n+1).Therefore, for each part:a) pq: exponents 1 and 1. So (1+1)(1+1)=4.b) p²q: exponents 2 and 1. (2+1)(1+1)=6.c) p²q²: exponents 2 and 2. (2+1)(2+1)=9.d) p^m q^n: (m+1)(n+1).I think that covers all the parts. Let me just make sure there isn't a case where p and q being different primes affects something. But in the formula for the number of divisors, it's based on the exponents in the prime factorization. Since p and q are distinct primes, their exponents are independent, so the formula holds. If they were the same prime, like p*p, then it would be p², but since they are different, the factorization is unique. So even if p and q are different, the number of divisors only depends on the exponents. So yeah, the answers should be 4, 6, 9, and (m+1)(n+1) for each part respectively.Final Answera) boxed{4}b) boxed{6}c) boxed{9}d) boxed{(m+1)(n+1)}</think>To determine the number of divisors for each given number, we use the formula for the number of divisors of a number based on its prime factorization. If a number ( N ) has a prime factorization ( N = p_1^{k_1} p_2^{k_2} cdots p_r^{k_r} ), then the number of divisors of ( N ) is given by ( (k_1 + 1)(k_2 + 1) cdots (k_r + 1) ).a) For ( pq ), where ( p ) and ( q ) are different primes:- The prime factorization is ( p^1 q^1 ).- The exponents are 1 and 1.- Adding 1 to each exponent gives ( (1+1)(1+1) = 2 times 2 = 4 ).- Therefore, the number of divisors is (boxed{4}).b) For ( p^2 q ):- The prime factorization is ( p^2 q^1 ).- The exponents are 2 and 1.- Adding 1 to each exponent gives ( (2+1)(1+1) = 3 times 2 = 6 ).- Therefore, the number of divisors is (boxed{6}).c) For ( p^2 q^2 ):- The prime factorization is ( p^2 q^2 ).- The exponents are 2 and 2.- Adding 1 to each exponent gives ( (2+1)(2+1) = 3 times 3 = 9 ).- Therefore, the number of divisors is (boxed{9}).d) For ( p^m q^n ):- The prime factorization is ( p^m q^n ).- The exponents are ( m ) and ( n ).- Adding 1 to each exponent gives ( (m+1)(n+1) ).- Therefore, the number of divisors is (boxed{(m+1)(n+1)}).]

🔑:г) Each divisor also has the form p^{k} q^{l}, and the exponent k can be chosen in m+1 ways (from 0 to m), while the exponent l can be chosen in n+1 ways.## Answera) 4 divisors;b) 6;c) 9;d) (m+1)(n+1) divisors.Send a comment

🧠:[<think>Okay, so I have this problem here about triangle ABC with points K on AB and N on BC. The conditions are that BK is twice AK, and CN is twice BN. Then Q is the intersection point of AN and CK. I need to find the ratio of the area of triangle ABC to the area of triangle BCQ. Hmm, let me try to visualize this first.First, let me sketch triangle ABC in my mind. Let's label the vertices A, B, and C. Point K is on AB such that BK = 2AK. So, if I consider AB as a line segment, then K divides AB into two parts where the part from K to B is twice as long as the part from A to K. Similarly, point N is on BC such that CN = 2BN. So N divides BC into BN and NC, with NC being twice BN.Q is the intersection of AN and CK. So lines AN and CK meet at point Q inside the triangle. The question is asking for the ratio of the area of the entire triangle ABC to the area of triangle BCQ. That is, S_ABC / S_BCQ. So I need to compute how many times bigger the area of ABC is compared to BCQ.I remember that in geometry problems involving ratios of segments and areas, coordinate geometry or mass point geometry can be helpful. Alternatively, using vectors or area ratios based on similar triangles. Maybe I can try coordinate geometry here since I can assign coordinates to the points and compute the intersection.Let me assign coordinates to the triangle. Let's place point A at (0,0), B at (3,0) because BK = 2AK, so if AB is divided into three equal parts, AK is 1 part and KB is 2 parts. So if AB is from 0 to 3 on the x-axis, then K would be at (1,0). Similarly, for BC, we need to set coordinates such that CN = 2BN. Let's see. If I place point B at (3,0), then where should point C be? Maybe let me set coordinates for C as (3,3) for simplicity? Wait, but then BC would be from (3,0) to (3,3), which is vertical, so N would be a point dividing BC such that CN = 2BN. If BC is vertical from (3,0) to (3,3), then BN would be from B(3,0) to N(3,1) and NC from N(3,1) to C(3,3), so NC is twice BN. That works. So point N would be at (3,1). Then point K is at (1,0). Then I need to find the intersection Q of AN and CK.Wait, but if I fix coordinates this way, maybe the calculations are easier. Let's confirm:Points:A(0,0)B(3,0)K is on AB such that BK = 2AK. Since AB is from (0,0) to (3,0), the length is 3 units. If AK is x, BK is 2x, so x + 2x = 3 => x=1. So AK =1, so K is at (1,0).Similarly, BC is from (3,0) to (3,3). Let's check N. If CN = 2BN, then BN is y, CN is 2y. So total BC length is y + 2y = 3y. If BC is 3 units (from 0 to 3 on y-axis), then 3y = 3 => y=1. So BN =1, so N is at (3,1). So coordinates are:C(3,3)N(3,1)Now, need to find equations of lines AN and CK.First, line AN connects A(0,0) to N(3,1). The slope of AN is (1 - 0)/(3 - 0) = 1/3. So equation is y = (1/3)x.Second, line CK connects C(3,3) to K(1,0). The slope here is (0 - 3)/(1 - 3) = (-3)/(-2) = 3/2. So equation is y - 3 = (3/2)(x - 3). Let's compute that:y = (3/2)x - (9/2) + 3 = (3/2)x - (9/2) + (6/2) = (3/2)x - (3/2).So equations:AN: y = (1/3)xCK: y = (3/2)x - 3/2Find intersection point Q. Solve the system:(1/3)x = (3/2)x - 3/2Multiply both sides by 6 to eliminate denominators:2x = 9x - 92x - 9x = -9-7x = -9 => x = 9/7Then y = (1/3)(9/7) = 3/7So point Q is at (9/7, 3/7).Now, need to compute the area of triangle BCQ. Points B(3,0), C(3,3), Q(9/7, 3/7).One way to compute the area is using the shoelace formula.Coordinates:B(3,0)C(3,3)Q(9/7, 3/7)Compute area:Arrange the points:First point: B(3,0)Second point: C(3,3)Third point: Q(9/7, 3/7)Back to first point: B(3,0)Shoelace formula:Area = 1/2 |sum(x_i y_{i+1} - x_{i+1} y_i)|Calculations:x1 y2 - x2 y1: 3*3 - 3*0 = 9 - 0 = 9x2 y3 - x3 y2: 3*(3/7) - (9/7)*3 = 9/7 - 27/7 = -18/7x3 y1 - x1 y3: (9/7)*0 - 3*(3/7) = 0 - 9/7 = -9/7Sum: 9 - 18/7 - 9/7 = 9 - (27/7) = (63/7 - 27/7) = 36/7Take absolute value and multiply by 1/2: (1/2)*(36/7) = 18/7So area of BCQ is 18/7.Now, area of ABC. Since ABC is a right triangle? Wait, in my coordinate system, ABC is a right triangle if I set A(0,0), B(3,0), C(3,3). So yes, ABC is a right-angled triangle at B, with legs of length 3 each. So area is (3*3)/2 = 9/2.Wait, but in my coordinate system, ABC is from (0,0) to (3,0) to (3,3). So base is 3, height is 3, area is (3*3)/2 = 9/2.So ratio S_ABC / S_BCQ = (9/2) / (18/7) = (9/2)*(7/18) = (63/36) = 7/4. Wait, 63 divided by 36 simplifies to 7/4? Wait 63/36 is 7/4? No, 63 divided by 9 is 7, 36 divided by 9 is 4. Yes, 7/4. So the ratio is 7/4? So 1.75.Wait, but is this correct? Let me double-check.First, verify the coordinates:A(0,0), B(3,0), C(3,3). Then K is at (1,0), N at (3,1). Equations:AN from (0,0) to (3,1): slope 1/3, equation y = x/3.CK from (3,3) to (1,0): slope (0-3)/(1-3) = (-3)/(-2) = 3/2. Equation: y - 3 = (3/2)(x - 3), so y = (3/2)x - 9/2 + 3 = (3/2)x - 3/2. Correct.Intersection at (9/7, 3/7). Then coordinates of Q are (9/7, 3/7).Shoelace formula for BCQ:Points B(3,0), C(3,3), Q(9/7, 3/7):Compute the terms:First term: x1 y2 = 3*3 = 9x2 y3 = 3*(3/7) = 9/7x3 y1 = (9/7)*0 = 0Sum: 9 + 9/7 + 0 = 72/7Second terms: y1 x2 = 0*3 = 0y2 x3 = 3*(9/7) = 27/7y3 x1 = (3/7)*3 = 9/7Sum: 0 + 27/7 + 9/7 = 36/7Subtract: 72/7 - 36/7 = 36/7Take absolute value and divide by 2: 18/7. Correct.Area of ABC: 9/2. Ratio: (9/2)/(18/7) = (9/2)*(7/18) = (63/36) = 7/4. So 7/4. Therefore, the answer is 7/4.Wait, but in my coordinate system, ABC is a right-angled triangle. The problem didn't specify ABC is right-angled. So does this ratio hold for any triangle ABC? Or is this answer dependent on the coordinates I chose?Hmm, that's a good point. In the problem, ABC is a general triangle, not necessarily right-angled. But I assumed coordinates where ABC is right-angled. Wait, is this ratio preserved under affine transformations?Because areas ratios are preserved under affine transformations. Since the problem is purely about ratios of areas and ratios of segments, which are affine invariants, then the answer should be the same regardless of the specific triangle. Therefore, even if I choose a right-angled triangle for simplicity, the ratio would be valid for any triangle.Therefore, the answer is 7/4. So 7 over 4, which is 1.75, so the ratio is 7:4.Wait, but let me confirm with another approach to be sure.Alternative approach using mass point geometry.Let me recall mass point. Since BK = 2AK, so AK/KB = 1/2. Then masses at A and B would be 2 and 1 respectively. Similarly, CN = 2BN, so BN/NC = 1/2, masses at B and C would be 2 and 1.But maybe it's better to set up masses for points A, B, C to find the ratios.Alternatively, using coordinate geometry in a general triangle.Let me try barycentric coordinates.Alternatively, use vectors.Let me assign vectors to the points.Let me denote vectors with position vectors relative to point A as the origin.Let me set point A as the origin. Let vector AB = b, and vector AC = c. Then coordinates:A: 0B: bC: cPoint K is on AB such that BK = 2AK. Since AK + KB = AB, and BK = 2AK, so AK = (1/3)AB, hence K is located at 1/3 of AB from A. Therefore, the position vector of K is (1/3)b.Similarly, point N is on BC such that CN = 2BN. So BN + NC = BC, and CN = 2BN. Therefore, BN = (1/3)BC. Since BC = c - b (since from point B to point C is c - b). Therefore, position vector of N is B + BN = b + (1/3)(c - b) = (2/3)b + (1/3)c.Now, line AN is from A (0) to N ( (2/3)b + (1/3)c ). So parametric equation for AN: t*( (2/3)b + (1/3)c ), where t ∈ [0,1].Similarly, line CK is from C (c) to K ( (1/3)b ). Parametric equation for CK: c + s*( (1/3)b - c ) = c + s*( (1/3)b - c ) = (1 - s)c + (s/3)b, where s ∈ [0,1].Find the intersection point Q of AN and CK. So set the parametric equations equal:t*( (2/3)b + (1/3)c ) = (1 - s)c + (s/3)bThis gives:(2t/3)b + (t/3)c = (s/3)b + (1 - s)cEquate coefficients for b and c:For b:2t/3 = s/3 => 2t = sFor c:t/3 = 1 - s => t/3 = 1 - 2t => Multiply both sides by 3: t = 3 - 6t => 7t = 3 => t = 3/7Therefore, s = 2t = 6/7Therefore, point Q is located at t = 3/7 on AN:Q = (3/7)*( (2/3)b + (1/3)c ) = (3/7)*(2/3 b + 1/3 c ) = (2/7 b + 1/7 c )So position vector of Q is (2b + c)/7.Now, to compute the area of BCQ.Points B, C, Q.Vectors:B: bC: cQ: (2b + c)/7Expressed as vectors relative to A:Vector AB = b, AC = c.But to compute the area of triangle BCQ, perhaps express vectors BC and BQ, then take half the magnitude of their cross product.First, compute vectors BC and BQ.Vector BC = C - B = c - bVector BQ = Q - B = (2b + c)/7 - b = (2b + c - 7b)/7 = (-5b + c)/7Area of triangle BCQ is (1/2) | BC × BQ |.Compute cross product BC × BQ:(c - b) × ( (-5b + c)/7 ) = (1/7)(c - b) × (-5b + c)Compute cross product (c - b) × (-5b + c )= (c × (-5b)) + (c × c ) + (-b × (-5b)) + (-b × c )Cross product is anti-commutative, so c × c = 0, b × b = 0= -5(c × b) + 0 + 5(b × b) - (b × c )= -5(c × b) + 0 + 0 - (b × c )Note that b × c = - (c × b )So:= -5(c × b ) - ( - (c × b )) = -5(c × b ) + (c × b ) = -4(c × b )But cross product is anti-commutative, so c × b = - (b × c )Therefore:= -4*(- (b × c )) = 4(b × c )Therefore, the cross product (c - b ) × (-5b + c ) = 4(b × c )Therefore, the magnitude is 4 |b × c |, and the area is (1/2)*(1/7)*4 |b × c | = (2/7)|b × c |.But the area of triangle ABC is (1/2)|b × c |.Therefore, the ratio S_ABC / S_BCQ = ( (1/2)|b × c | ) / ( (2/7)|b × c | ) = (1/2)/(2/7) = (1/2)*(7/2) = 7/4.So the ratio is 7/4, which matches the coordinate geometry approach.Therefore, regardless of the coordinate system or the specific triangle, the ratio is 7/4. So the answer should be 7/4.Final Answerboxed{dfrac{7}{4}}</think>Given a triangle (ABC) with points (K) on (AB) and (N) on (BC) such that (BK = 2AK) and (CN = 2BN). We need to compute the ratio (dfrac{S_{triangle ABC}}{S_{triangle BCQ}}) where (Q) is the intersection of (AN) and (CK).1. Assigning Coordinates: - Place (A) at ((0,0)), (B) at ((3,0)), and (C) at ((3,3)). - (K) divides (AB) such that (BK = 2AK), so (K) is at ((1,0)). - (N) divides (BC) such that (CN = 2BN), so (N) is at ((3,1)).2. Finding Equations of Lines: - Line (AN) has a slope of (frac{1}{3}), so its equation is (y = frac{1}{3}x). - Line (CK) has a slope of (frac{3}{2}), so its equation is (y = frac{3}{2}x - frac{3}{2}).3. Finding Intersection (Q): - Solve the system (y = frac{1}{3}x) and (y = frac{3}{2}x - frac{3}{2}): - (frac{1}{3}x = frac{3}{2}x - frac{3}{2}) - Solving gives (x = frac{9}{7}) and (y = frac{3}{7}) - Intersection point (Q) is (left(frac{9}{7}, frac{3}{7}right)).4. Calculating Areas: - Area of (triangle BCQ) using the shoelace formula: - Coordinates: (B(3,0)), (C(3,3)), (Qleft(frac{9}{7}, frac{3}{7}right)) - Area = (frac{1}{2} left| 3 cdot 3 + 3 cdot frac{3}{7} + frac{9}{7} cdot 0 - 0 cdot 3 - 3 cdot frac{9}{7} - frac{3}{7} cdot 3 right| = frac{18}{7}) - Area of (triangle ABC) = (frac{1}{2} times 3 times 3 = frac{9}{2})5. Ratio Calculation: - (dfrac{S_{triangle ABC}}{S_{triangle BCQ}} = dfrac{frac{9}{2}}{frac{18}{7}} = dfrac{9}{2} times dfrac{7}{18} = dfrac{7}{4})Thus, the ratio is (boxed{dfrac{7}{4}}).,<think>Okay, so I have this geometry problem here. Let me try to visualize it first. There's a triangle ABC, and two points K on AB and N on BC. The conditions are that BK is twice AK, meaning K divides AB in the ratio 2:1, right? Similarly, CN is twice BN, so N divides BC in the ratio 2:1 as well. Then Q is the intersection point of lines AN and CK. We need to find the ratio of the area of triangle ABC to the area of triangle BCQ.Hmm, so maybe coordinate geometry could work here. I can assign coordinates to the points and use coordinate methods to find the intersection point Q, then calculate the areas. Alternatively, maybe using mass point geometry or similar triangles. Let me think.First, let me sketch the triangle. Let's denote point A at the top, B at the left, and C at the right. Then K is on AB such that BK = 2AK. So if AB is a line segment, starting at A and going to B, then K is closer to A because BK is longer than AK. Wait, actually, BK is twice AK, so if AK is x, then BK is 2x, so total AB is 3x. Therefore, K divides AB in the ratio AK:KB = 1:2. So starting from A, moving towards B, K is 1/3 of the way. Similarly, N is on BC such that CN = 2BN. So starting from B, moving towards C, BN is x, CN is 2x, so BC is 3x. Thus, N divides BC in the ratio BN:NC = 1:2. So N is 1/3 of the way from B to C.Now, lines AN and CK intersect at Q. We need to find the area ratio S(ABC)/S(BCQ). Coordinate geometry might be straightforward here. Let me assign coordinates to the triangle. Let's choose coordinates such that point B is at (0, 0), C is at (3, 0) because BC is divided by N into BN:NC = 1:2, so if BC is length 3, then N is at (1, 0). Wait, but coordinates can be arbitrary, right? Maybe to make it easier, set coordinates with B at (0,0), C at (3,0), so N is at (1*2 + 3*1)/3? Wait, no. If BN:NC = 1:2, then from B to C, the coordinates of N can be calculated as follows: B is (0,0), C is (3,0). The point dividing BC in the ratio BN:NC = 1:2. So using the section formula, coordinates of N would be [(2*0 + 1*3)/ (1+2), (2*0 + 1*0)/ (1+2)] = (3/3, 0) = (1, 0). Wait, that doesn't make sense. Wait, if BN:NC = 1:2, then N is closer to B. So the formula is ( (mx2 + nx1)/(m + n), (my2 + ny1)/(m + n) ). Here, m:n is BN:NC = 1:2. So starting from B to C, the coordinates of N would be ( (1*3 + 2*0)/3, (1*0 + 2*0)/3 ) = (3/3, 0) = (1, 0). So N is at (1, 0). Similarly, K is on AB. Let's assign coordinates to A. Let me set A at (0, a) so that AB is from (0, a) to (0,0). Wait, but then AB is vertical. If K divides AB such that BK = 2AK, then AK:KB = 1:2. So from A to B, K is 1/3 of the way. So if A is at (0, a), then K would be at (0, a - (a)/3 ) = (0, 2a/3). Wait, but maybe choosing coordinates where AB is horizontal? Let's see. Alternatively, perhaps assign coordinates with A at (0,0), B at (3,0), but then the points might get confusing. Maybe it's better to assign coordinates so that calculations are easier.Alternatively, use barycentric coordinates. But maybe I'm overcomplicating. Let's try to assign coordinates systematically.Let me place point B at (0, 0). Let me place point C at (3, 0) because if BC is divided in a 1:2 ratio, then N is at (1, 0). Let me place point A somewhere in the plane, say at (0, a), so AB is from (0, a) to (0, 0). Then point K is on AB such that BK = 2AK. Since AB is from (0, a) to (0, 0), the length of AB is a. Then AK = (1/3)AB = a/3, so K is at (0, a - a/3) = (0, 2a/3).Then line AN connects A(0, a) to N(1, 0). Let me find the equation of line AN. The slope of AN is (0 - a)/(1 - 0) = -a. So the equation is y = -a x + a.Line CK connects C(3, 0) to K(0, 2a/3). The slope here is (2a/3 - 0)/(0 - 3) = (2a/3)/(-3) = -2a/9. So the equation is y - 0 = (-2a/9)(x - 3), which simplifies to y = (-2a/9)x + (6a/9) = (-2a/9)x + (2a/3).Now, to find point Q, the intersection of AN and CK. So set the two equations equal:From AN: y = -a x + aFrom CK: y = (-2a/9)x + (2a/3)Set them equal:-a x + a = (-2a/9)x + (2a/3)Let's solve for x:Multiply both sides by 9 to eliminate denominators:-9a x + 9a = -2a x + 6aBring all terms to left side:-9a x + 9a + 2a x - 6a = 0(-7a x) + 3a = 0-7a x = -3aDivide both sides by -7a:x = (-3a)/(-7a) = 3/7Then substitute x = 3/7 into AN's equation:y = -a*(3/7) + a = (-3a/7 + 7a/7) = 4a/7Therefore, point Q has coordinates (3/7, 4a/7)Now, we need to compute the areas of triangles ABC and BCQ.First, area of triangle ABC. Since ABC has coordinates A(0, a), B(0, 0), and C(3, 0). The base is BC, which is 3 units, and the height is the vertical distance from A to BC, which is a. So area S(ABC) = (1/2)*base*height = (1/2)*3*a = 3a/2.Next, area of triangle BCQ. Points B(0,0), C(3,0), and Q(3/7, 4a/7). Let's use the formula for the area of a triangle given coordinates:S = (1/2)| (x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)) |Plugging in B(0,0), C(3,0), Q(3/7, 4a/7):S(BCQ) = (1/2)| 0*(0 - 4a/7) + 3*(4a/7 - 0) + (3/7)*(0 - 0) | = (1/2)| 0 + 12a/7 + 0 | = (1/2)*(12a/7) = 6a/7.Therefore, the ratio S(ABC)/S(BCQ) = (3a/2)/(6a/7) = (3a/2)*(7/(6a)) = (21a)/(12a) = 21/12 = 7/4.Wait, 7/4? That seems a bit high. Let me check my calculations again.First, coordinates:- A(0, a)- B(0, 0)- C(3, 0)- N is (1, 0)- K is (0, 2a/3)Equations of AN and CK:AN: from (0, a) to (1, 0). Slope is -a. Equation: y = -a x + a.CK: from (3,0) to (0, 2a/3). Slope is (2a/3 - 0)/(0 - 3) = -2a/9. Equation: y = (-2a/9)(x - 3) = (-2a/9)x + 6a/9 = (-2a/9)x + 2a/3.Intersection when -a x + a = (-2a/9)x + 2a/3.Multiply both sides by 9: -9a x + 9a = -2a x + 6a-9a x + 9a = -2a x + 6aBring -2a x to left and 9a to right: -9a x + 2a x = 6a -9a-7a x = -3ax = (-3a)/(-7a) = 3/7. Then y = -a*(3/7) + a = 4a/7. So Q is (3/7, 4a/7). That seems correct.Area of ABC: (1/2)*3*a = 3a/2. Correct.Area of BCQ: Using coordinates B(0,0), C(3,0), Q(3/7,4a/7). Using the formula:(1/2)| x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) |.Plugging in:(1/2)| 0*(0 - 4a/7) + 3*(4a/7 - 0) + (3/7)*(0 - 0) | = (1/2)|0 + 12a/7 + 0| = 6a/7. That's correct.Therefore, the ratio is (3a/2)/(6a/7) = (3/2)/(6/7) = (3/2)*(7/6) = 21/12 = 7/4. So 7/4 is 1.75.Wait, but does this ratio make sense? Let's see. The area of BCQ is 6a/7, and the area of ABC is 3a/2. 6a/7 is roughly 0.857a, and 3a/2 is 1.5a. So the ratio is 1.5 / 0.857 ≈ 1.75, which is 7/4. That seems okay.Alternatively, maybe using mass point geometry. Let's try that approach to verify.In mass point geometry, we assign masses to the vertices such that the masses are inversely proportional to the lengths.First, for point K on AB with BK = 2AK. The masses at A and B should be in the ratio 2:1. So mass at A is 2, mass at B is 1, making the mass at K equal to 3.Similarly, for point N on BC with CN = 2BN. The masses at B and C should be in the ratio 2:1. So mass at B is 1, mass at C is 2, making mass at N equal to 3.Now, we need to find the mass at Q, the intersection of AN and CK. Let's analyze line AN. On line AN, masses at A is 2 and at N is 3 (since N's mass is 3 from BC). So the masses at A and N would affect the mass at Q. Similarly, on line CK, masses at C is 2 and K is 3. Hmm, mass point might need more careful consideration.Alternatively, maybe use the concept of lever arms. For line AN: masses at A and N. The masses should balance the lever. For line CK: masses at C and K. The intersection Q would then have masses combining both.Alternatively, think of it as systems of masses. Let me see.In mass point geometry, to find the mass at Q, we can assign masses such that:For line AN: masses at A and N. Let the mass at A be m and at N be n. Then the mass at Q along AN is m + n.Similarly, for line CK: masses at C and K. Let the mass at C be p and at K be q. Then the mass at Q along CK is p + q.But this might get complicated. Alternatively, use the area ratios directly.Another approach: use Ceva's theorem. But since we have two cevians intersecting, maybe not directly applicable here. Alternatively, use the formula for the area ratios based on the intersection point.Alternatively, think in terms of vectors. Let's try that.Let me assign vectors to the points. Let’s take B as the origin. Let vector BA = a and vector BC = c. Then coordinates:- Point B: 0 (origin)- Point A: a- Point C: cPoint K is on AB such that BK = 2AK. Since AB is from A to B, the vector from A to B is -a. Then K divides AB such that BK = 2AK. So the position vector of K is A + (1/3)AB = a + (1/3)(- a) = (2/3)a.Wait, no. If BK = 2AK, then from point A to K is 1 part, and from K to B is 2 parts. So the ratio AK:KB = 1:2. So starting from A, moving towards B, K is 1/3 of the way. Therefore, position vector of K is A + (1/3)(B - A) = a + (1/3)(- a) = (2/3)a.Similarly, point N is on BC such that CN = 2BN. So BN:NC = 1:2. So starting from B towards C, N is 1/3 of the way. Therefore, position vector of N is B + (1/3)(C - B) = 0 + (1/3)c = (1/3)c.Now, line AN connects A(a) to N( (1/3)c ). The parametric equation for AN is a + t( (1/3)c - a ), where t ∈ [0,1].Similarly, line CK connects C(c) to K( (2/3)a ). The parametric equation for CK is c + s( (2/3)a - c ), where s ∈ [0,1].To find the intersection Q, set the parametric equations equal:a + t( (1/3)c - a ) = c + s( (2/3)a - c )Expanding both sides:Left side: a - ta + (t/3)cRight side: c + (2s/3)a - scCollect like terms:For a: (1 - t)a on left, (2s/3)a on right.For c: (t/3)c on left, (1 - s)c on right.Therefore, equating coefficients:(1 - t) = 2s/3 ...(1)t/3 = 1 - s ...(2)From equation (2): s = 1 - t/3Substitute into equation (1):1 - t = 2*(1 - t/3)/3Multiply both sides by 3:3(1 - t) = 2(1 - t/3)3 - 3t = 2 - (2t)/3Bring all terms to left side:3 - 3t - 2 + (2t)/3 = 01 - 3t + (2t)/3 = 0Multiply all terms by 3 to eliminate denominator:3 - 9t + 2t = 03 - 7t = 07t = 3t = 3/7Then from equation (2): s = 1 - (3/7)/3 = 1 - 1/7 = 6/7Therefore, point Q is on AN at t = 3/7:Position vector of Q = a + (3/7)( (1/3)c - a ) = a - (3/7)a + (3/7)(1/3)c = (4/7)a + (1/7)cSo Q is (4/7)a + (1/7)cNow, we need to compute the area of triangle BCQ. Points B(0), C(c), Q( (4/7)a + (1/7)c )The area of triangle BCQ can be computed using the cross product in vector terms. The area is (1/2)| (C - B) × (Q - B) |.Since B is the origin, vectors C and Q are c and (4/7)a + (1/7)c respectively.So (C - B) × (Q - B) = c × ( (4/7)a + (1/7)c ) = (4/7)(c × a) + (1/7)(c × c). But c × c = 0, so this simplifies to (4/7)(c × a).Therefore, the area S(BCQ) = (1/2)*(4/7)|a × c| = (2/7)|a × c|.The area of triangle ABC is (1/2)|a × c|. Therefore, the ratio S(ABC)/S(BCQ) = (1/2)/(2/7) = (1/2)*(7/2) = 7/4.Same result as before. So this confirms the ratio is 7/4.Alternatively, using coordinate geometry with different coordinates. Let me try with A at (0, 0), B at (1, 0), C at (0, 1). Wait, but then the ratios might change. Let's see.Wait, maybe another coordinate system for verification.Let me set point B at (0,0), point A at (0,3), so that AB is from (0,3) to (0,0). Then K is on AB such that BK = 2AK. Since AB is length 3, AK = 1, BK = 2. So K is at (0, 3 - 1) = (0, 2).Point C can be at (3,0). Then BC is from (0,0) to (3,0). N is on BC such that CN = 2BN. So BN:NC =1:2. So N is at (1, 0).Then line AN is from A(0,3) to N(1,0). The slope is (0 - 3)/(1 - 0) = -3. Equation: y = -3x + 3.Line CK is from C(3,0) to K(0,2). The slope is (2 - 0)/(0 - 3) = -2/3. Equation: y = (-2/3)x + 2.Find intersection Q of y = -3x + 3 and y = (-2/3)x + 2.Set -3x + 3 = (-2/3)x + 2Multiply both sides by 3:-9x + 9 = -2x + 6-9x + 9 = -2x + 6-7x = -3x = 3/7Then y = -3*(3/7) + 3 = -9/7 + 21/7 = 12/7So Q is (3/7, 12/7).Compute areas.Area of ABC: (1/2)*base*height. Base BC is 3 units (from (0,0) to (3,0)), height is 3 (from A(0,3)). So area = (1/2)*3*3 = 9/2.Area of BCQ: Points B(0,0), C(3,0), Q(3/7, 12/7). Using the formula:(1/2)| x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) |Plug in:(1/2)| 0*(0 - 12/7) + 3*(12/7 - 0) + (3/7)*(0 - 0) | = (1/2)| 0 + 36/7 + 0 | = 18/7.So the ratio is (9/2)/(18/7) = (9/2)*(7/18) = (63/36) = 7/4. Same result. So regardless of the coordinate system, the ratio is 7/4.Thus, the answer should be 7/4.Final Answerboxed{dfrac{7}{4}}</think>Given a triangle (ABC) with points (K) on (AB) and (N) on (BC) such that (BK = 2AK) and (CN = 2BN), we need to find the ratio (dfrac{S_{triangle ABC}}{S_{triangle BCQ}}) where (Q) is the intersection of (AN) and (CK).1. Assigning Coordinates: - Place (B) at ((0, 0)), (C) at ((3, 0)), and (A) at ((0, a)). - (K) divides (AB) in the ratio (1:2), so (K) is at ((0, frac{2a}{3})). - (N) divides (BC) in the ratio (1:2), so (N) is at ((1, 0)).2. Equations of Lines: - Line (AN) through (A(0, a)) and (N(1, 0)) has the equation (y = -ax + a). - Line (CK) through (C(3, 0)) and (K(0, frac{2a}{3})) has the equation (y = -frac{2a}{9}x + frac{2a}{3}).3. Finding Intersection (Q): - Solve the equations of (AN) and (CK): [ -ax + a = -frac{2a}{9}x + frac{2a}{3} ] - Solving gives (x = frac{3}{7}) and (y = frac{4a}{7}). Thus, (Q) is (left(frac{3}{7}, frac{4a}{7}right)).4. Calculating Areas: - Area of (triangle ABC) is (frac{1}{2} times 3 times a = frac{3a}{2}). - Area of (triangle BCQ) using coordinates (B(0, 0)), (C(3, 0)), (Qleft(frac{3}{7}, frac{4a}{7}right)): [ S_{triangle BCQ} = frac{1}{2} left| 0 left(0 - frac{4a}{7}right) + 3 left(frac{4a}{7} - 0right) + frac{3}{7} left(0 - 0right) right| = frac{1}{2} left| frac{12a}{7} right| = frac{6a}{7} ]5. Ratio Calculation: - The ratio (dfrac{S_{triangle ABC}}{S_{triangle BCQ}} = dfrac{frac{3a}{2}}{frac{6a}{7}} = dfrac{3a/2}{6a/7} = dfrac{21}{12} = dfrac{7}{4}).Thus, the final answer is (boxed{dfrac{7}{4}}).]

🔑:1. Extend line (BQ) to intersect (AC) at point (M). - By Ceva's Theorem, for concurrent lines (AN), (CK), and (BM) in triangle (ABC), we have: [ frac{BK}{KA} cdot frac{AM}{MC} cdot frac{CN}{NB} = 1 ] - Given (BK = 2AK) and (CN = 2BN), we substitute these ratios: [ frac{2}{1} cdot frac{AM}{MC} cdot frac{2}{1} = 1 ] - Simplifying, we get: [ 4 cdot frac{AM}{MC} = 1 implies frac{AM}{MC} = frac{1}{4} ]2. Apply Menelaus' Theorem on triangle (ABM) with transversal (KQC). - Menelaus' Theorem states: [ frac{AM}{MC} cdot frac{CQ}{QK} cdot frac{BK}{BA} = 1 ] - Substituting the known ratios (frac{AM}{MC} = frac{1}{4}) and (frac{BK}{BA} = frac{2}{3}): [ frac{1}{4} cdot frac{CQ}{QK} cdot frac{2}{3} = 1 ] - Solving for (frac{CQ}{QK}): [ frac{1}{4} cdot frac{2}{3} cdot frac{CQ}{QK} = 1 implies frac{CQ}{QK} = 6 ]3. Determine the area ratios. - Since (frac{CQ}{QK} = 6), the ratio of the areas of triangles (BCQ) and (BCK) is: [ frac{[BCQ]}{[BCK]} = frac{CQ}{CQ + QK} = frac{6}{6+1} = frac{6}{7} ] - Given (frac{BK}{BA} = frac{2}{3}), the ratio of the areas of triangles (BCK) and (ABC) is: [ frac{[BCK]}{[ABC]} = frac{2}{3} ] - Combining these ratios, we find: [ frac{[BCQ]}{[ABC]} = frac{[BCQ]}{[BCK]} cdot frac{[BCK]}{[ABC]} = frac{6}{7} cdot frac{2}{3} = frac{4}{7} ] - Therefore, the ratio (frac{S_{triangle ABC}}{S_{triangle BCQ}}) is: [ frac{S_{triangle ABC}}{S_{triangle BCQ}} = frac{1}{frac{4}{7}} = frac{7}{4} ]The final answer is (boxed{frac{7}{4}})