🧠:<think>Alright, so I'm trying to figure out how to get from 0 to 100 in exactly 4 seconds, where each second I can add or subtract any factorial number. Factorials are like 1!, 2!, 3!, and so on, which are 1, 2, 6, 24, 120, etc. Okay, so I need to use these numbers to reach 100 in four steps.First, I need to understand what exactly is allowed. Each second, I can choose any factorial number and either add it or subtract it from my current number. So, for example, in the first second, I could add 1! (which is 1) or subtract 1!, or add 2! (which is 2), subtract 2!, and so on. The same goes for the next three seconds.Since I need to reach 100 in four steps, I should probably think about the largest factorial that can get me close to 100. Let's see, 5! is 120, which is just above 100. 4! is 24, which is much smaller. So maybe using 5! is a good idea because it's the closest to 100.If I use 5! (120), then I need to adjust from 120 down to 100. That means I need to subtract 20. Now, how can I get 20 using the remaining three steps? Let's see, 24 (which is 4!) is larger than 20, so maybe I can subtract 4! and then add something else. If I subtract 24 from 120, I get 96. Then, I need to get from 96 to 100, which is adding 4. Hmm, 4 isn't a factorial, but I can get 4 by adding 2! twice (since 2! is 2). So, 96 + 2 + 2 = 100.So, one possible sequence is:1. Add 5! (120)2. Subtract 4! (24)3. Add 2! (2)4. Add 2! (2)But wait, I need to make sure that I can arrange these operations in any order, as long as it takes exactly four steps. So, the sequence could be different, but the total sum should still be 100.Alternatively, maybe I can use 5! and then adjust differently. For example, instead of subtracting 4!, maybe I can subtract smaller factorials. Let's see, 120 - 6 (which is 3!) = 114, then 114 - 12 (which is 4! + 2!) = 102, then 102 - 2 = 100. But that would require more steps, and I only have four steps.Wait, no, that's still four steps:1. Add 5! (120)2. Subtract 3! (6)3. Subtract 4! (24) and 2! (2) in the same step? No, each step is one addition or subtraction of a single factorial. So I can't subtract both 4! and 2! in one step. So that approach might not work.Let me think differently. Maybe I can use multiple smaller factorials to reach 100. For example, 4! is 24, so if I add 24 four times, that would be 96, which is close to 100. Then, I need to add 4 more, which I can do by adding 2! twice. But that would require six steps, which is too many.Alternatively, if I use 5! (120), then subtract 20. To get 20, I can use 4! (24) and subtract 4. But 4 isn't a factorial. Wait, 4 can be represented as 3! - 2! (6 - 2 = 4). So, maybe:1. Add 5! (120)2. Subtract 4! (24)3. Add 3! (6)4. Subtract 2! (2)That would give me 120 - 24 + 6 - 2 = 100. So that's another sequence.But I need to consider all possible sequences that can lead to 100 in four steps. Each step can be adding or subtracting any factorial, so I need to consider all combinations.Let me list out the factorials up to 5! because 6! is 720, which is way too big:1! = 12! = 23! = 64! = 245! = 120Now, I need to find all possible sequences of four operations (each being + or - of one of these factorials) that sum up to 100.Starting from 0, after four operations, I need to reach 100. So, the total sum of these four operations should be 100.Let me denote each operation as ±n!, where n is from 1 to 5.So, I need to find all combinations of four terms, each being ±1!, ±2!, ±3!, ±4!, or ±5!, such that their sum is 100.This seems like a combinatorial problem where I need to count the number of solutions to the equation:±a ±b ±c ±d = 100, where a, b, c, d are factorials from 1! to 5!.But since the order matters (each step is a separate operation), I need to consider permutations as well.However, this might be too broad. Maybe I can narrow it down by considering the largest factorial, 5! = 120. Since 120 is larger than 100, using 5! might be necessary because otherwise, the sum of four smaller factorials (maximum 4! = 24) would be 4*24=96, which is less than 100. So, to reach 100, I probably need to use 5! at least once.So, let's assume that one of the operations is +5! (120). Then, the remaining three operations need to sum up to 100 - 120 = -20.So, now the problem reduces to finding the number of ways to sum three operations (each being ±1!, ±2!, ±3!, ±4!) to get -20.Let me denote the remaining three operations as x, y, z, where each is ±1!, ±2!, ±3!, or ±4!.So, x + y + z = -20.Now, I need to find all possible combinations of x, y, z such that their sum is -20.Again, considering the factorials:1! = 12! = 23! = 64! = 24So, the possible values for x, y, z are ±1, ±2, ±6, ±24.I need to find all possible combinations of three numbers from this set that add up to -20.Let me think about the possible combinations.First, note that 24 is quite large. If I include -24, then the remaining two numbers need to sum up to -20 + 24 = 4.So, let's consider cases where one of the operations is -24.Case 1: One of the operations is -24.Then, the remaining two operations need to sum to 4.Possible pairs that sum to 4:- 6 + (-2) = 4- 2 + 2 = 4- 1 + 3! (6) - but 3! is 6, so 1 + 6 = 7, which is too big.Wait, actually, the factorials are 1, 2, 6, 24. So, the possible numbers are ±1, ±2, ±6, ±24.So, to get 4, possible combinations:- 6 + (-2) = 4- 2 + 2 = 4- 1 + 3! (but 3! is 6, so 1 + 6 = 7, which is too big)- 24 - 20 = 4, but 20 isn't a factorial.So, only two possibilities: 6 + (-2) and 2 + 2.Therefore, in this case, the three operations would be -24, 6, -2 or -24, 2, 2.Now, considering the order, since each operation is a separate step, the number of permutations for each combination.For the combination -24, 6, -2:There are 3 distinct numbers, so the number of permutations is 3! = 6.For the combination -24, 2, 2:There are two 2's and one -24, so the number of permutations is 3 (since the two 2's are identical).So, total for Case 1: 6 + 3 = 9 ways.Case 2: None of the operations is -24.Then, the maximum sum of three operations is 6 + 6 + 6 = 18, and the minimum is -6 -6 -6 = -18. But we need the sum to be -20, which is outside this range. Therefore, it's impossible to get -20 without using -24.So, Case 2 yields 0 ways.Therefore, the total number of ways to get -20 in three operations is 9.But wait, I need to consider that each operation can be any factorial, not just up to 4!. But since we already used 5! in the first step, the remaining operations can still use 5! if needed. However, using 5! again would complicate things because 5! is 120, and we're already trying to reach -20. So, using 5! again would make it harder to reach -20.But technically, the problem allows using any factorial in each step, so I need to consider if using 5! again is possible.Wait, but 5! is 120, and if I use it again, say, subtract 120, then the remaining two operations would need to sum to -20 + 120 = 100, which is too big because the maximum sum of two operations is 24 + 24 = 48. So, it's impossible to reach 100 with two operations.Therefore, using 5! again in the remaining three operations is not feasible because it would require the other two operations to sum to 100, which is impossible. So, we can safely ignore using 5! in the remaining three operations.Therefore, the only way to get -20 in three operations is by using -24 and then getting 4 with the remaining two operations, which we've already calculated as 9 ways.So, for the case where we use +5! in the first step, we have 9 possible sequences for the remaining three steps.But wait, the first step doesn't have to be +5!. It could be any of the four steps. So, the +5! could be in any of the four positions.Therefore, for each of the 9 sequences, we need to consider where the +5! is placed.Since the +5! can be in any of the four positions, and the remaining three operations are fixed in their sequence, the total number of sequences would be 9 * 4 = 36.But wait, is that correct? Let me think.Actually, no. Because the remaining three operations are already considering the order. So, if I fix the +5! in one position, the remaining three operations have 9 possible sequences. But since the +5! can be in any of the four positions, we need to multiply by 4.But wait, in the previous calculation, we considered the three operations after the +5!. But actually, the +5! could be in any of the four positions, so the remaining three operations could be before or after.Therefore, the total number of sequences would be 9 * 4 = 36.But let me verify this.Suppose the +5! is in the first step: then the remaining three steps have 9 possibilities.If the +5! is in the second step: then the first step is one of the remaining operations, and the remaining two steps are the other two operations. But wait, no, because the remaining three operations are fixed in their sequence. So, actually, it's not just placing +5! in any position, but considering that the remaining three operations are a sequence that sums to -20.Therefore, for each of the 9 sequences of three operations that sum to -20, we can insert the +5! in any of the four positions, leading to 9 * 4 = 36 sequences.But wait, is that overcounting? Because if the +5! is in different positions, but the remaining three operations are the same, it's a different sequence.Yes, because the order of operations matters. So, each different position of +5! with the same three operations is a different sequence.Therefore, the total number of sequences is 36.But let me think again. Is there any other way to reach 100 without using +5!?Earlier, I thought that without using 5!, the maximum sum is 4 * 24 = 96, which is less than 100. So, it's impossible to reach 100 without using 5!.Therefore, all sequences must include +5! exactly once, and the remaining three operations must sum to -20.Therefore, the total number of ways is 36.Wait, but earlier I thought that using -24 and then getting 4 with the remaining two operations gives 9 ways, and then multiplying by 4 for the position of +5! gives 36.Yes, that seems correct.But let me double-check if there are other ways to get -20 without using -24.As I thought earlier, without using -24, the maximum sum of three operations is 6 + 6 + 6 = 18, and the minimum is -6 -6 -6 = -18. So, -20 is outside this range. Therefore, it's impossible to get -20 without using -24.Therefore, the only way is to use -24 and then get 4 with the remaining two operations, which gives 9 sequences, and then multiply by 4 for the position of +5!, giving 36.So, the final answer is 36.</think>

🧠:<think>Alright, so I've got this vector problem here. Let me try to wrap my head around it step by step. First, the problem says that the angle between vectors m and n is π/6, which is 30 degrees. The magnitudes of these vectors are given: |m| is √3 and |n| is 2. Okay, so that's some basic vector stuff.Now, we're dealing with triangle ABC. The vectors AB and AC are given in terms of m and n. Specifically, vector AB is 2m + 2n, and vector AC is 2m - 6n. Point D is the midpoint of side BC, and we need to find the length of vector AD.Hmm, okay. So, since D is the midpoint of BC, I remember that in vectors, the midpoint can be found by averaging the position vectors of B and C. But wait, do I have the position vectors of B and C? Not directly, but I have vectors AB and AC, which are from point A.Let me recall that vector AB is the vector from A to B, and vector AC is from A to C. So, if I can find vectors AB and AC, maybe I can find vectors B and C relative to A, and then find the midpoint D.Wait, actually, in vector terms, if I consider point A as the origin, then vector AB would be the position vector of B, and vector AC would be the position vector of C. Is that right? Because if A is the origin, then the position vector of B is just AB, and similarly for C. So, yeah, I think that's a valid approach.So, if A is the origin, then:- Position vector of B = AB = 2m + 2n- Position vector of C = AC = 2m - 6nThen, the midpoint D of BC would be the average of the position vectors of B and C. So, vector AD would be (vector AB + vector AC)/2. Let me write that down:AD = (AB + AC)/2Substituting the given vectors:AD = (2m + 2n + 2m - 6n)/2Let me compute that:First, combine like terms:2m + 2m = 4m2n - 6n = -4nSo, AD = (4m - 4n)/2 = 2m - 2nOkay, so vector AD is 2m - 2n. Now, we need to find the magnitude of this vector, |AD|.To find the magnitude of a vector expressed in terms of two other vectors, we can use the formula:|AD| = sqrt( (2m - 2n) • (2m - 2n) )Where "•" denotes the dot product. So, let's compute this dot product.First, let's expand the dot product:(2m - 2n) • (2m - 2n) = (2m) • (2m) + (2m) • (-2n) + (-2n) • (2m) + (-2n) • (-2n)Simplify each term:= 4(m • m) - 4(m • n) - 4(n • m) + 4(n • n)Now, since the dot product is commutative, m • n = n • m, so we can combine the middle terms:= 4(m • m) - 8(m • n) + 4(n • n)Now, let's compute each of these dot products.First, m • m is |m|², which is (√3)² = 3.Similarly, n • n is |n|², which is 2² = 4.Next, m • n is |m||n|cosθ, where θ is the angle between them, which is π/6.So, m • n = (√3)(2)cos(π/6)We know that cos(π/6) is √3/2, so:m • n = (√3)(2)(√3/2) = (√3 * √3) * (2/2) = 3 * 1 = 3Wait, that seems too high. Let me double-check:(√3)(2) = 2√3Then, 2√3 * (√3/2) = (2√3 * √3)/2 = (2*3)/2 = 6/2 = 3Yes, that's correct. So, m • n = 3.Now, plugging these back into our expression:= 4(3) - 8(3) + 4(4)Compute each term:4*3 = 128*3 = 244*4 = 16So, substituting:= 12 - 24 + 16Compute that:12 - 24 = -12-12 + 16 = 4So, the dot product (2m - 2n) • (2m - 2n) = 4Therefore, |AD| = sqrt(4) = 2Wait, that can't be right because 2 is one of the options, but I feel like I might have made a mistake somewhere.Let me go back and check my steps.First, when I found vector AD, I did (AB + AC)/2:AB = 2m + 2nAC = 2m - 6nAdding them: 2m + 2n + 2m - 6n = 4m - 4nDivide by 2: 2m - 2nThat seems correct.Then, computing |AD|:|AD| = sqrt( (2m - 2n) • (2m - 2n) )Expanding the dot product:= 4(m • m) - 8(m • n) + 4(n • n)Computed:m • m = 3n • n = 4m • n = 3So:4*3 = 12-8*3 = -244*4 = 16Adding: 12 -24 +16 = 4sqrt(4) = 2Hmm, so according to this, |AD| is 2, which is option A.But wait, I feel like the answer should be larger because vectors AB and AC are combinations of m and n, and D is the midpoint. Maybe I missed a step somewhere.Wait, another approach: Maybe instead of treating A as the origin, I should consider vectors AB and AC as vectors from A, not as position vectors. So, vector AB is from A to B, and vector AC is from A to C. Then, to find vector AD, which is from A to D, the midpoint of BC.In that case, vector AD can be expressed as (vector AB + vector AC)/2, because D is the midpoint.Wait, that's exactly what I did earlier. So, that part seems correct.Alternatively, maybe I should compute vectors AB and AC, find vectors BC, then find vector AD.Let me try that.Given:AB = 2m + 2nAC = 2m - 6nSo, vector BC = AC - AB = (2m - 6n) - (2m + 2n) = -8nSo, vector BC = -8nThen, since D is the midpoint of BC, vector BD = vector DC = (-8n)/2 = -4nSo, vector AD = vector AB + vector BDVector AB is 2m + 2nVector BD is -4nSo, vector AD = 2m + 2n - 4n = 2m - 2nSame result as before. So, vector AD is indeed 2m - 2nThen, |AD| = sqrt( (2m - 2n) • (2m - 2n) ) = sqrt(4) = 2So, according to this, the answer is 2, which is option A.But wait, the options are A)2, B)4, C)6, D)8But I feel like 2 is too small. Let me check my calculations again.Wait, when I computed m • n, I got 3. Let me verify that.m • n = |m||n|cosθ = (√3)(2)cos(π/6)cos(π/6) is √3/2, so:= (√3)(2)(√3/2) = (√3 * √3) * (2/2) = 3 * 1 = 3Yes, that's correct.So, m • n = 3Then, the dot product:(2m - 2n) • (2m - 2n) = 4(m • m) - 8(m • n) + 4(n • n) = 4*3 -8*3 +4*4 = 12 -24 +16 = 4So, |AD| = sqrt(4) = 2Hmm, seems consistent. Maybe the answer is indeed 2.But wait, let me think differently. Maybe I should compute the coordinates of points A, B, C, and then find D.Assume point A is at the origin (0,0). Then, vector AB is 2m + 2n, so point B is at (2m + 2n). Similarly, point C is at (2m - 6n).But to find coordinates, I need to express m and n in terms of unit vectors. Since m and n are vectors with given magnitudes and angle between them, I can represent them in a coordinate system.Let me set m along the x-axis for simplicity. So, m = (√3, 0)Then, n makes an angle of π/6 with m, so n can be represented as:n = |n|*(cosθ, sinθ) = 2*(cos(π/6), sin(π/6)) = 2*(√3/2, 1/2) = (√3, 1)So, n = (√3, 1)Now, vector AB = 2m + 2n = 2*(√3, 0) + 2*(√3, 1) = (2√3 + 2√3, 0 + 2*1) = (4√3, 2)Similarly, vector AC = 2m - 6n = 2*(√3, 0) - 6*(√3, 1) = (2√3 - 6√3, 0 - 6*1) = (-4√3, -6)So, coordinates:Point A: (0,0)Point B: (4√3, 2)Point C: (-4√3, -6)Now, midpoint D of BC:D_x = (4√3 + (-4√3))/2 = 0/2 = 0D_y = (2 + (-6))/2 = (-4)/2 = -2So, point D is at (0, -2)Therefore, vector AD is from A(0,0) to D(0, -2), which is (0, -2)So, |AD| = sqrt(0² + (-2)²) = sqrt(4) = 2So, again, the length of AD is 2, which is option A.Wait, but earlier I thought 2 was too small, but now with coordinates, it's confirmed. So, maybe 2 is correct.Alternatively, maybe I made a mistake in representing m and n. Let me double-check.I set m along the x-axis as (√3, 0). That's correct because |m| = √3.Then, n makes an angle of π/6 with m, so its components are |n|*(cosθ, sinθ) = 2*(cos(π/6), sin(π/6)) = 2*(√3/2, 1/2) = (√3, 1). That seems correct.Then, vector AB = 2m + 2n = 2*(√3, 0) + 2*(√3, 1) = (2√3 + 2√3, 0 + 2) = (4√3, 2). Correct.Vector AC = 2m - 6n = 2*(√3, 0) - 6*(√3, 1) = (2√3 - 6√3, 0 - 6) = (-4√3, -6). Correct.Midpoint D:x-coordinate: (4√3 + (-4√3))/2 = 0y-coordinate: (2 + (-6))/2 = -2So, D is (0, -2). Vector AD is from (0,0) to (0, -2), which is (0, -2). Length is 2.So, yes, 2 is correct. Therefore, the answer is A) 2.But wait, in the initial vector approach, I got |AD| = 2, and in the coordinate approach, I also got |AD| = 2. So, seems consistent.But the options include 4, which is double of 2. Maybe I missed a factor somewhere?Wait, in the vector approach, I had vector AD = 2m - 2n, and then |AD| = sqrt(4) = 2.But in the coordinate approach, vector AD is (0, -2), which has length 2.So, both methods agree. Therefore, I think the answer is indeed 2, which is option A.But wait, the problem says "In triangle ABC", so maybe I should visualize it. Points A, B, C with vectors AB and AC given. D is the midpoint of BC. So, AD is a median.Given that, and the calculations, I think 2 is correct.Alternatively, maybe the problem expects the answer in terms of the vectors without breaking them into coordinates, but my vector approach already confirmed it's 2.So, I think I'm confident that the answer is 2, option A.</think>

🧠:<think>First, I need to determine the relationship between speed, distance, and time. I know that distance equals speed multiplied by time.Given that the worker's usual time to reach the office is 24 minutes, I can express the distance ( D ) as ( D = S times 24 ), where ( S ) is her normal speed.When she walks at 3/4 of her normal speed, her new speed becomes ( frac{3}{4}S ). To find the time taken at this slower speed, I use the formula ( T = frac{D}{frac{3}{4}S} ).Substituting the expression for ( D ) from the first equation into the second equation gives ( T = frac{S times 24}{frac{3}{4}S} ).Simplifying this equation by canceling out ( S ), I get ( T = frac{24}{frac{3}{4}} ).Calculating further, ( T = 24 times frac{4}{3} = 32 ) minutes.Finally, to find out how many minutes later she is, I subtract the usual time (24 minutes) from the slower time (32 minutes), resulting in ( 32 - 24 = 8 ) minutes.</think>

🧠:<think>Given that ( x ) and ( y ) are distinct nonzero real numbers such that ( x + frac{3}{x} = y + frac{3}{y} ), we need to find ( xy ).First, we denote the common value by ( B ). Thus, we have:[B = x + frac{3}{x} = y + frac{3}{y}]Next, we consider the equation for any ( u ):[u + frac{3}{u} = B]Multiplying through by ( u ) to clear the fraction, we get:[u^2 - Bu + 3 = 0]Since ( x ) and ( y ) are distinct solutions to this quadratic equation, by Vieta's formulas, we know:- The sum of the roots ( x + y ) is equal to ( B ).- The product of the roots ( xy ) is equal to the constant term, which is ( 3 ).Therefore, the value of ( xy ) is ( 3 ).The final answer is (boxed{textbf{(C)} 3}).</think>