🧠:[<think>Okay, so I need to solve this inequality x times f(x - 1) is less than or equal to 1. The function f is defined piecewise: it's -1 when x is less than 0, and 1 when x is greater than or equal to 0. Hmm, right. So first, I need to figure out what f(x - 1) does. Let me break it down step by step.The function f(x) is -1 for x < 0 and 1 for x >= 0. So when we have f(x - 1), that's just shifting the input by 1 unit to the right. So instead of checking if x is less than 0, we check if x - 1 is less than 0. Which means f(x - 1) will be -1 when x - 1 < 0, so x < 1, and 1 when x - 1 >= 0, so x >= 1. So, f(x - 1) is a piecewise function too:f(x - 1) = -1 if x < 1, 1 if x >= 1.So now, the inequality x * f(x - 1) <= 1 depends on the value of x. Since f(x - 1) changes its value at x = 1, we should consider two cases: x < 1 and x >= 1.Case 1: x < 1. In this case, f(x - 1) = -1. So the inequality becomes x*(-1) <= 1. That simplifies to -x <= 1, which is the same as x >= -1. But since we're in the case where x < 1, the solution here is x >= -1 and x < 1. So interval [-1, 1).Case 2: x >= 1. Here, f(x - 1) = 1. So the inequality is x*1 <= 1, which simplifies to x <= 1. But wait, in this case, x is already >= 1. So combining x >= 1 and x <= 1 gives x = 1. So the solution here is just {1}.Combining both cases, the solution set is [-1, 1) union {1} which is just [-1, 1]. Wait, that seems like the interval from -1 to 1, inclusive. Let me check again.In case 1: x < 1 and x >= -1, so from -1 to 1, not including 1. In case 2: x = 1. So combining both gives all x from -1 to 1, including both endpoints. Because at x = 1, we have x*f(x - 1) = 1*1 = 1, which is equal to 1, so that's included. So the total solution is [-1, 1]. So that's answer choice A. But let me check the answer options again.The options are:A: [-1, 1]B: [-1, 2]C: (-∞, 1]D: [-1, +∞)Hmm, so according to my calculation, the solution is [-1, 1], which is option A. But let me make sure I didn't make a mistake. Let's test some values.First, x = -1: Then x - 1 = -2, so f(-2) = -1. So x*f(x - 1) = (-1)*(-1) = 1, which is equal to 1, so it's included. Good.x = 0: x - 1 = -1, so f(-1) = -1. So x*f(x - 1) = 0*(-1) = 0 <=1. Included.x = 0.5: x -1 = -0.5, f(-0.5) = -1. So 0.5*(-1) = -0.5 <=1. True.x = 1: x -1 = 0, f(0) = 1. So 1*1 =1 <=1. Included.x = 2: Then x -1 =1, so f(1)=1. So 2*1=2 <=1? No, 2 is greater than 1. So x=2 is not in the solution. So B: [-1,2] is wrong.x = -2: x -1 = -3, f(-3) = -1. So x*f(x -1) = (-2)*(-1) =2 <=1? 2 is not <=1. So x=-2 is not in the solution. So the solution can't go beyond -1 on the left. So D is wrong. C is (-infty,1], but x=-2 is in C but not in solution. So C is wrong.Hence, the correct answer must be A: [-1,1]. Let me check the problem statement again. The function f(x) is defined as -1 for x <0 and 1 for x >=0. So f(x -1) is -1 when x -1 <0, which is x <1, and 1 when x >=1. Then x*f(x -1) <=1. So for x <1, -x <=1 => x >= -1. For x >=1, x <=1. So only x=1. So union is [-1,1]. So yes, A is correct. Wait, but let me check x=1/2. For example, x=0.5: 0.5*f(0.5 -1)=0.5*f(-0.5)=0.5*(-1)=-0.5 <=1, which is true. So included. Similarly, x=-1: (-1)*f(-2)=(-1)*(-1)=1 <=1. Good. x=1: 1*f(0)=1*1=1 <=1. So all points in [-1,1] work, and outside that, like x=2 or x=-2, they don't. Therefore, answer is A. Hmm, but the options given in the problem are A,B,C,D as above.Wait, but the original problem is Chinese, and sometimes the options might be different, but here the user has given the options as A to D. So based on this, the answer is A.But wait, I'm a bit confused because when I first thought about the cases, the first case was x <1 leading to x >= -1, so [-1,1), and x=1. So combined [-1,1]. But just to check if maybe there's a mistake here.Alternatively, maybe I should consider the inequality x*f(x -1) <=1. Let's re-express it.Since f(x -1) is -1 when x <1, so for x <1, the inequality is -x <=1, which is x >= -1. So x in [-1,1). For x >=1, f(x -1)=1, so inequality is x <=1. But since x >=1 here, the only solution is x=1. So combining both intervals, x ∈ [-1,1]. Therefore, answer A.Yes, that seems right. So I think the answer is A. But let me check the original problem again to ensure I didn't misread the function. The function f(x) is -1 when x <0, 1 when x>=0. Then f(x -1) is -1 when x -1 <0, which is x <1, and 1 when x >=1. So correct.Therefore, the solution set is [-1,1], option A. So the answer is boxed{A}Final Answerboxed{A}</think>Given the function ( f(x) = begin{cases} -1, & x < 0 1, & x geq 0 end{cases} ), we need to solve the inequality ( x f(x - 1) leq 1 ).First, we analyze ( f(x - 1) ). The function ( f(x - 1) ) is a piecewise function shifted by 1 unit to the right:- For ( x - 1 < 0 ) (i.e., ( x < 1 )), ( f(x - 1) = -1 ).- For ( x - 1 geq 0 ) (i.e., ( x geq 1 )), ( f(x - 1) = 1 ).This gives us two cases to consider for the inequality ( x f(x - 1) leq 1 ):Case 1: ( x < 1 )- Here, ( f(x - 1) = -1 ).- The inequality becomes ( x(-1) leq 1 ), which simplifies to ( -x leq 1 ) or ( x geq -1 ).- Combining the conditions ( x < 1 ) and ( x geq -1 ), we get ( -1 leq x < 1 ).Case 2: ( x geq 1 )- Here, ( f(x - 1) = 1 ).- The inequality becomes ( x(1) leq 1 ), which simplifies to ( x leq 1 ).- Combining the conditions ( x geq 1 ) and ( x leq 1 ), we get ( x = 1 ).Combining both cases, the solution set is ( [-1, 1) cup {1} = [-1, 1] ).Testing values within and outside this interval confirms the solution:- Values like ( x = -1 ), ( x = 0 ), ( x = 0.5 ), and ( x = 1 ) satisfy the inequality.- Values like ( x = 2 ) and ( x = -2 ) do not satisfy the inequality.Thus, the solution set of the inequality ( x f(x - 1) leq 1 ) is ( [-1, 1] ), which corresponds to option A.[boxed{A}],<think>Okay, let's see. I need to solve the inequality ( x f(x - 1) leq 1 ) where ( f(x) ) is defined piecewise. The function ( f(x) ) is -1 when ( x < 0 ) and 1 when ( x geq 0 ). The answer choices are A through D. Hmm, let me break this down step by step.First, I need to understand what ( f(x - 1) ) does. Since ( f ) is a piecewise function, substituting ( x - 1 ) into it will shift the input by 1. So, ( f(x - 1) ) will be -1 when ( x - 1 < 0 ), which is when ( x < 1 ), and 1 when ( x - 1 geq 0 ), which is when ( x geq 1 ). Therefore, the function ( f(x - 1) ) is:[f(x - 1) = begin{cases}-1, & x < 1 1, & x geq 1end{cases}]Now, the inequality we need to solve is ( x cdot f(x - 1) leq 1 ). Let's analyze this in two cases based on the piecewise definition of ( f(x - 1) ).Case 1: When ( x < 1 )In this case, ( f(x - 1) = -1 ), so the inequality becomes:[x cdot (-1) leq 1]Simplifying:[- x leq 1]Multiply both sides by -1, which reverses the inequality:[x geq -1]But we have to remember that this case is only for ( x < 1 ). So combining these two conditions, we get ( -1 leq x < 1 ).Case 2: When ( x geq 1 )Here, ( f(x - 1) = 1 ), so the inequality becomes:[x cdot 1 leq 1]Which simplifies to:[x leq 1]But this case is for ( x geq 1 ), so combining these gives ( x = 1 ). Because x has to be both greater than or equal to 1 and less than or equal to 1. So the only solution here is x = 1.Combining Both CasesPutting together the solutions from both cases:- From Case 1: ( -1 leq x < 1 )- From Case 2: ( x = 1 )So the entire solution set is ( -1 leq x leq 1 ), which is the interval ( [-1, 1] ). Let me check the answer choices. Option A is ( [-1, 1] ), so that seems to be the answer. But wait, let me verify again to be sure.Wait, but when x is between -1 and 1, in Case 1, x is less than 1, so f(x-1) is -1. Multiplying by x, which could be negative or positive. Let me test some values in this interval to make sure.Take x = 0: ( 0 cdot f(-1) = 0 cdot (-1) = 0 leq 1 ) ✅Take x = -1: ( (-1) cdot f(-2) = (-1) cdot (-1) = 1 leq 1 ) ✅Take x = 0.5: ( 0.5 cdot f(-0.5) = 0.5 cdot (-1) = -0.5 leq 1 ✅Take x approaching 1 from the left, say x = 0.9: 0.9 * (-1) = -0.9 ≤ 1 ✅At x = 1: 1 * f(0) = 1 * 1 = 1 ≤ 1 ✅What about x > 1? For example, x = 2: 2 * f(1) = 2 * 1 = 2 ≤ 1? No, 2 is not ≤ 1. So x = 2 is not a solution. Therefore, the solution set is indeed [-1,1], which is option A.But wait, hold on, the answer choices include option B: [-1,2]. Why would someone think it's [-1,2]? Maybe they made a mistake in combining the cases.Wait, perhaps I made an error in interpreting the inequality. Let me re-examine the cases.Case 1: x < 1. Then f(x -1) = -1. So the inequality is -x ≤ 1 → x ≥ -1. So x must be between -1 and 1. That's correct.Case 2: x ≥ 1. Then f(x -1) = 1. So the inequality is x ≤ 1. But x must also be ≥1. So x = 1. So that's correct.Therefore, combining [-1,1) and {1} gives [-1,1]. Therefore, A is correct.But let me check if there's a different interpretation. Suppose I considered the definition of f(x -1) again. Wait, f(x -1) is -1 when x -1 < 0, which is x <1, and 1 when x >=1. Correct.So the inequality x*f(x -1) <=1. Then:For x <1: x*(-1) <=1 → -x <=1 → x >= -1. So x in [-1,1).For x >=1: x*1 <=1 → x <=1. But x >=1, so x=1. So x=1.Therefore, combining them gives x in [-1,1]. So answer A.But the answer B is [-1,2], maybe someone thought that for x >=1, the inequality is x <=1, which is not true, because for x >=1, the inequality x <=1 would only include x=1, not up to 2.Alternatively, maybe a miscalculation in the case for x >=1. Let's check for x=2:xf(x-1) = 2*f(1) = 2*1=2, which is greater than 1. So 2 is not included. Similarly, x=1.5: 1.5*1=1.5>1. So indeed, only x=1 satisfies in the x >=1 case. So answer A is correct.But wait, let me check x < -1. For example, x = -2:xf(x -1) = (-2)*f(-3) = (-2)*(-1) = 2 <=1? No, 2 is not ≤1. So x=-2 is not included. So only x >= -1.Therefore, the solution is from -1 to 1, inclusive. So A is correct.Wait, but the options are A: [-1,1], B: [-1,2], C: (-infty,1], D: [-1, +infty). So the answer is A. But let me check once more.Suppose x is in [-1,1). Then for x in [-1,1), x <1, so f(x -1)= -1. Then x*(-1) <=1 → -x <=1 →x >= -1. So x >=-1 and x <1. So [-1,1).For x >=1, f(x-1)=1, so x*1 <=1 →x <=1. But x >=1, so x=1. Therefore, overall [-1,1]. Hence, answer A.Therefore, I think the correct answer is A.Final Answerboxed{A}</think>Given the function ( f(x) = begin{cases} -1, & x < 0 1, & x geq 0 end{cases} ), we need to solve the inequality ( x f(x - 1) leq 1 ).First, we determine ( f(x - 1) ):- When ( x - 1 < 0 ) (i.e., ( x < 1 )), ( f(x - 1) = -1 ).- When ( x - 1 geq 0 ) (i.e., ( x geq 1 )), ( f(x - 1) = 1 ).This gives us:[f(x - 1) = begin{cases} -1, & x < 1 1, & x geq 1 end{cases}]We analyze the inequality ( x f(x - 1) leq 1 ) in two cases:Case 1: ( x < 1 )- Here, ( f(x - 1) = -1 ), so the inequality becomes ( -x leq 1 ).- Solving this gives ( x geq -1 ).- Combining with ( x < 1 ), we get ( -1 leq x < 1 ).Case 2: ( x geq 1 )- Here, ( f(x - 1) = 1 ), so the inequality becomes ( x leq 1 ).- Combining with ( x geq 1 ), we get ( x = 1 ).Combining both cases, the solution set is ( [-1, 1] ).Thus, the correct answer is (boxed{A}).]

🔑:Since f(x)= begin{cases} -1, & x<0 1, & xgeq 0 end{cases}, f(x-1)= begin{cases} -1, & x<1 1, & xgeq 1 end{cases}, When x geq 1, xf(x-1) leq 1 Rightarrow x leq 1 Leftrightarrow x=1. When x < 1, -x leq 1 Leftrightarrow x geq -1, hence -1 leq x < 1. Combining the above, we know x in [-1, 1]. Therefore, the correct choice is boxed{A}. First, find the expression for xf(x-1), then solve the inequality by dividing it into segments. This problem tests the ability to solve piecewise inequalities, which is quite flexible. Finding the analytical expression of the function and using case analysis are key to solving this problem.

🧠:[<think>Okay, let me try to figure out this problem step by step. So, we have the expansion of (2x + √3)^4, which is equal to a0 + a1x + a2x² + a3x³ + a4x⁴. We need to find the value of (a0 + a2 + a4)² - (a1 + a3)². Hmm, that seems like combining the even coefficients and the odd coefficients somehow. Maybe there's a trick using substitution of specific values for x to compute these sums?I remember that for polynomial expansions, substituting x = 1 and x = -1 can help find sums of coefficients. Let me recall: if I plug in x = 1, the value of the polynomial is the sum of all coefficients, right? So, f(1) = a0 + a1 + a2 + a3 + a4. Similarly, f(-1) would be a0 - a1 + a2 - a3 + a4. If I add these two, f(1) + f(-1) gives 2(a0 + a2 + a4), which is twice the sum of even coefficients. Similarly, subtracting them gives 2(a1 + a3), which is twice the sum of odd coefficients. So, that might be useful here.So, if I can compute f(1) and f(-1), then I can find a0 + a2 + a4 and a1 + a3. Then, the expression we need is ( (a0 + a2 + a4) )² - ( (a1 + a3) )². Which, by the way, is a difference of squares, so it factors into (a0 + a2 + a4 - a1 - a3)(a0 + a2 + a4 + a1 + a3). But maybe that's complicating things. Let me just compute the two sums first.First, compute f(1) = (2*1 + √3)^4 = (2 + √3)^4. Then f(-1) = (2*(-1) + √3)^4 = (-2 + √3)^4. Then, a0 + a2 + a4 = (f(1) + f(-1))/2, and a1 + a3 = (f(1) - f(-1))/2. So, once I compute those, I can square them and subtract.Alternatively, maybe there's a smarter substitution. Let me think. The expression (a0 + a2 + a4)^2 - (a1 + a3)^2. That looks like (sum_even)^2 - (sum_odd)^2, which factors into (sum_even - sum_odd)(sum_even + sum_odd). But sum_even + sum_odd is the total sum of coefficients, which is f(1). And sum_even - sum_odd would be f(-1). Because if you plug x = -1, then f(-1) = a0 - a1 + a2 - a3 + a4 = (a0 + a2 + a4) - (a1 + a3). So, indeed, sum_even - sum_odd = f(-1). Therefore, the expression is f(-1)*f(1). So, (sum_even)^2 - (sum_odd)^2 = f(-1)*f(1). So, instead of computing each part separately, maybe I can just compute f(1)*f(-1), which would be the answer.Wait, that's a great insight! So, (a0 + a2 + a4)^2 - (a1 + a3)^2 = f(1)*f(-1). Therefore, instead of calculating both sums and squaring them, etc., just compute f(1) and f(-1) and multiply them. That might be easier. Let me verify that.Yes, because:(sum_even + sum_odd) = f(1)(sum_even - sum_odd) = f(-1)Therefore, (sum_even)^2 - (sum_odd)^2 = (sum_even - sum_odd)(sum_even + sum_odd) = f(-1)*f(1)So, that's a much smarter approach. Therefore, the answer is f(1)*f(-1). Let's compute that.First, compute f(1) = (2 + √3)^4. Then f(-1) = (-2 + √3)^4. Let's compute these two values.First, (2 + √3)^4. Let's compute (2 + √3)^2 first, then square the result.(2 + √3)^2 = 2² + 2*2*√3 + (√3)^2 = 4 + 4√3 + 3 = 7 + 4√3.Then, (7 + 4√3)^2. Compute that:7² + 2*7*4√3 + (4√3)^2 = 49 + 56√3 + 16*3 = 49 + 56√3 + 48 = 97 + 56√3.So, f(1) = 97 + 56√3.Similarly, compute (-2 + √3)^4. Let's compute (-2 + √3)^2 first.(-2 + √3)^2 = (-2)^2 + 2*(-2)*√3 + (√3)^2 = 4 - 4√3 + 3 = 7 - 4√3.Then, (7 - 4√3)^2 = 7² - 2*7*4√3 + (4√3)^2 = 49 - 56√3 + 48 = 97 - 56√3.Therefore, f(-1) = 97 - 56√3.Now, multiply f(1) and f(-1):(97 + 56√3)(97 - 56√3) = 97² - (56√3)^2.Compute 97²: 97*97. Let's compute 100*100 = 10000, subtract 3*100 + 3*100 - 9 (wait, maybe straightforward). 97^2 = (100 - 3)^2 = 10000 - 600 + 9 = 9409.Then, (56√3)^2 = 56² * 3. 56^2 is 3136, so 3136*3 = 9408.Therefore, the product is 9409 - 9408 = 1.So, the answer is 1. Therefore, (a0 + a2 + a4)^2 - (a1 + a3)^2 = 1.Wait, so is it 1? Let me double-check the computations because that's a key point.First, f(1) = (2 + √3)^4. We computed (2 + √3)^2 = 7 + 4√3, then squared that: (7 + 4√3)^2 = 49 + 56√3 + 48 = 97 + 56√3. Correct.Similarly, f(-1) = (-2 + √3)^4. First, (-2 + √3)^2 = 7 - 4√3. Then squared: 49 - 56√3 + 48 = 97 - 56√3. Correct.Then, multiplying (97 + 56√3)(97 - 56√3) = 97² - (56√3)^2. 97² = 9409, (56√3)^2 = 56² * 3 = 3136*3=9408. 9409 - 9408 = 1. Yes, that's correct. So, the result is 1. Therefore, the answer is 1.Alternatively, if I didn't see the trick with f(1)*f(-1), I could compute a0 + a2 + a4 and a1 + a3 separately, then compute their squares and subtract. Let me check that as well to confirm.From before, we have:a0 + a2 + a4 = (f(1) + f(-1))/2 = ( (97 + 56√3) + (97 - 56√3) ) /2 = (194)/2 = 97.Similarly, a1 + a3 = (f(1) - f(-1))/2 = ( (97 + 56√3) - (97 - 56√3) ) /2 = (112√3)/2 = 56√3.Therefore, (a0 + a2 + a4)^2 - (a1 + a3)^2 = 97² - (56√3)^2. Which is the same as before, 9409 - 9408 = 1. So same answer. So regardless of the approach, we get 1. Therefore, the answer is 1. So, boxed, it's boxed{1}.Final Answerboxed{1}</think>To solve the problem, we need to find the value of ((a_0 + a_2 + a_4)^2 - (a_1 + a_3)^2) for the polynomial expansion ((2x + sqrt{3})^4).First, we use the substitution method by evaluating the polynomial at (x = 1) and (x = -1):1. Evaluate (f(1)): [ f(1) = (2 cdot 1 + sqrt{3})^4 = (2 + sqrt{3})^4 ] Compute ((2 + sqrt{3})^2): [ (2 + sqrt{3})^2 = 4 + 4sqrt{3} + 3 = 7 + 4sqrt{3} ] Then square the result: [ (7 + 4sqrt{3})^2 = 49 + 56sqrt{3} + 48 = 97 + 56sqrt{3} ] Thus, (f(1) = 97 + 56sqrt{3}).2. Evaluate (f(-1)): [ f(-1) = (2 cdot (-1) + sqrt{3})^4 = (-2 + sqrt{3})^4 ] Compute ((-2 + sqrt{3})^2): [ (-2 + sqrt{3})^2 = 4 - 4sqrt{3} + 3 = 7 - 4sqrt{3} ] Then square the result: [ (7 - 4sqrt{3})^2 = 49 - 56sqrt{3} + 48 = 97 - 56sqrt{3} ] Thus, (f(-1) = 97 - 56sqrt{3}).Next, we use these results to find the sums of the coefficients:- The sum of the even coefficients (a_0 + a_2 + a_4) is (frac{f(1) + f(-1)}{2}): [ frac{(97 + 56sqrt{3}) + (97 - 56sqrt{3})}{2} = frac{194}{2} = 97 ]- The sum of the odd coefficients (a_1 + a_3) is (frac{f(1) - f(-1)}{2}): [ frac{(97 + 56sqrt{3}) - (97 - 56sqrt{3})}{2} = frac{112sqrt{3}}{2} = 56sqrt{3} ]Finally, we compute the required expression:[(a_0 + a_2 + a_4)^2 - (a_1 + a_3)^2 = 97^2 - (56sqrt{3})^2]Calculate each term:- (97^2 = 9409)- ((56sqrt{3})^2 = 56^2 cdot 3 = 3136 cdot 3 = 9408)Thus, the result is:[9409 - 9408 = 1]Therefore, the final answer is (boxed{1}).,<think>Okay, so I have this problem where I need to expand (2x + √3)^4 and then compute (a0 + a2 + a4)^2 - (a1 + a3)^2. Hmm, let me start by recalling how to expand a binomial expression raised to a power. I think the binomial theorem is the way to go here. The binomial theorem states that (a + b)^n = Σ (from k=0 to n) C(n, k) * a^(n-k) * b^k, where C(n, k) is the combination of n things taken k at a time.So in this case, the expression is (2x + √3)^4. Let me identify a, b, and n here. a would be 2x, b is √3, and n is 4. So expanding that, each term will be C(4, k) * (2x)^(4 - k) * (√3)^k for k from 0 to 4.Let me compute each term step by step. Let's list out the coefficients and terms for each a0 to a4.First, when k = 0:C(4, 0) = 1(2x)^(4 - 0) = (2x)^4 = 16x^4(√3)^0 = 1So the term is 1 * 16x^4 * 1 = 16x^4. Therefore, a4 = 16.Wait, but hold on, actually, when k=0, the exponent on (2x) is 4 - 0 = 4, so the term is 16x^4, so the coefficient a4 is 16. Similarly, when k=1, the exponent on (2x) is 3, so that term would be C(4,1)*(2x)^3*(√3)^1. Let me compute each coefficient:For k=0:C(4,0) = 1(2x)^4 = 16x^4(√3)^0 = 1Term: 16x^4, so a4 = 16.For k=1:C(4,1) = 4(2x)^3 = 8x^3(√3)^1 = √3Term: 4 * 8x^3 * √3 = 32√3 x^3, so a3 = 32√3.For k=2:C(4,2) = 6(2x)^2 = 4x^2(√3)^2 = (√3)^2 = 3Term: 6 * 4x^2 * 3 = 6 * 12x^2 = 72x^2, so a2 = 72.For k=3:C(4,3) = 4(2x)^1 = 2x(√3)^3 = (√3)^3 = 3√3Term: 4 * 2x * 3√3 = 24√3 x, so a1 = 24√3.For k=4:C(4,4) = 1(2x)^0 = 1(√3)^4 = (√3)^4 = (3)^2 = 9Term: 1 * 1 * 9 = 9, so a0 = 9.So putting it all together, the expansion is:9 + 24√3 x + 72x^2 + 32√3 x^3 + 16x^4.Therefore, the coefficients are:a0 = 9,a1 = 24√3,a2 = 72,a3 = 32√3,a4 = 16.Now, the problem asks for (a0 + a2 + a4)^2 - (a1 + a3)^2. Let me compute that step by step.First, compute a0 + a2 + a4:9 + 72 + 16. Let's add those numbers. 9 + 72 is 81, and 81 + 16 is 97. So that's 97.Next, compute a1 + a3:24√3 + 32√3. Since both terms have √3, we can factor that out: (24 + 32)√3 = 56√3.So now, we need to compute 97^2 - (56√3)^2.Let's compute each part.First, 97 squared. 97^2. Hmm, 100^2 is 10000, so 97 is 3 less than 100. So (100 - 3)^2 = 10000 - 600 + 9 = 9409. Let me verify that: (a - b)^2 = a^2 - 2ab + b^2. So 100^2 - 2*100*3 + 3^2 = 10000 - 600 + 9 = 9409. Yep, correct.Next, (56√3)^2. Let's compute that. First, square 56 and square √3. 56^2 = 3136. (√3)^2 = 3. So the product is 3136 * 3 = 9408.Therefore, the expression becomes 9409 - 9408 = 1.Wait, that seems straightforward. But let me double-check all steps to make sure I didn't make a mistake.First, the expansion of (2x + √3)^4. Let's confirm each coefficient:For k=0: C(4,0)=1, (2x)^4=16x^4, (√3)^0=1. Term: 16x^4. So a4=16. Correct.k=1: C(4,1)=4, (2x)^3=8x^3, (√3)^1=√3. Term: 4*8*√3 x^3=32√3 x^3. So a3=32√3. Correct.k=2: C(4,2)=6, (2x)^2=4x^2, (√3)^2=3. Term: 6*4*3x^2=72x^2. So a2=72. Correct.k=3: C(4,3)=4, (2x)^1=2x, (√3)^3=3√3. Term: 4*2*3√3 x=24√3 x. So a1=24√3. Correct.k=4: C(4,4)=1, (2x)^0=1, (√3)^4=9. Term: 9. So a0=9. Correct.Therefore, the coefficients are correct.Then, summing a0 + a2 + a4: 9 + 72 + 16. 9+72=81, 81+16=97. Correct.Summing a1 + a3: 24√3 + 32√3 = 56√3. Correct.Calculating 97^2: 9409. Correct.Calculating (56√3)^2: 56^2=3136, times 3=9408. Correct.Subtracting: 9409 - 9408 = 1. So the answer is 1.Wait, that seems surprisingly simple. Let me see if there's another way to approach this problem that might confirm the answer. Maybe using substitution or properties of polynomials.Alternatively, the expression (a0 + a2 + a4)^2 - (a1 + a3)^2 can be factored as (a0 + a2 + a4 - a1 - a3)(a0 + a2 + a4 + a1 + a3). Maybe this factorization can be useful.Alternatively, if we consider evaluating the polynomial at x = 1 and x = -1.Because when x=1, the polynomial becomes (2*1 + √3)^4 = a0 + a1 + a2 + a3 + a4.Similarly, when x=-1, the polynomial becomes (2*(-1) + √3)^4 = a0 - a1 + a2 - a3 + a4.Then, if we compute (a0 + a2 + a4) + (a1 + a3) = sum when x=1.And (a0 + a2 + a4) - (a1 + a3) = sum when x=-1.Therefore, the product [(a0 + a2 + a4) + (a1 + a3)] * [(a0 + a2 + a4) - (a1 + a3)] = (sum at x=1) * (sum at x=-1) = (2 + √3)^4 * (-2 + √3)^4.Wait, that's an interesting approach. Let me verify that.If we let S1 = (2 + √3)^4 and S2 = (-2 + √3)^4, then the product S1 * S2 = [(2 + √3)(-2 + √3)]^4.Compute (2 + √3)(-2 + √3) first. Multiply them: (√3 + 2)(√3 - 2) = (√3)^2 - (2)^2 = 3 - 4 = -1.Therefore, [(2 + √3)(-2 + √3)]^4 = (-1)^4 = 1.Therefore, S1 * S2 = 1. Therefore, (a0 + a2 + a4)^2 - (a1 + a3)^2 = 1. So that's another way to see it. Therefore, the answer is 1. So this confirms the previous result.Therefore, even without expanding, we could substitute x = 1 and x = -1, compute the product, and get the answer as 1.That's a clever trick! So instead of expanding the entire expression, recognizing that the given expression is a difference of squares which can be factored, and then relating each factor to evaluating the polynomial at x=1 and x=-1. Then, multiplying those evaluations gives [(2 + √3)(-2 + √3)]^4 = (-1)^4 = 1.Therefore, regardless of the approach, the answer is 1. So that's pretty neat. It's good that there are two methods leading to the same answer, which gives me more confidence that it's correct.So to recap:Method 1: Expand the binomial, compute the coefficients, sum the even and odd indexed coefficients, square them, subtract. Result: 1.Method 2: Recognize the expression as a difference of squares, factor it into the product of (sum at x=1) and (sum at x=-1), compute those sums as (2 + √3)^4 and (-2 + √3)^4, multiply them which equals [(-1)^4] = 1. Result: 1.Both methods lead to the same answer. Therefore, the answer is 1.Final Answerboxed{1}</think>To solve the problem, we start by expanding ((2x + sqrt{3})^4) using the binomial theorem. The binomial theorem states that ((a + b)^n = sum_{k=0}^{n} binom{n}{k} a^{n-k} b^k). Here, (a = 2x), (b = sqrt{3}), and (n = 4).We compute each term of the expansion:- For (k = 0): (binom{4}{0} (2x)^4 (sqrt{3})^0 = 1 cdot 16x^4 cdot 1 = 16x^4), so (a_4 = 16).- For (k = 1): (binom{4}{1} (2x)^3 (sqrt{3})^1 = 4 cdot 8x^3 cdot sqrt{3} = 32sqrt{3}x^3), so (a_3 = 32sqrt{3}).- For (k = 2): (binom{4}{2} (2x)^2 (sqrt{3})^2 = 6 cdot 4x^2 cdot 3 = 72x^2), so (a_2 = 72).- For (k = 3): (binom{4}{3} (2x)^1 (sqrt{3})^3 = 4 cdot 2x cdot 3sqrt{3} = 24sqrt{3}x), so (a_1 = 24sqrt{3}).- For (k = 4): (binom{4}{4} (2x)^0 (sqrt{3})^4 = 1 cdot 1 cdot 9 = 9), so (a_0 = 9).The coefficients are (a_0 = 9), (a_1 = 24sqrt{3}), (a_2 = 72), (a_3 = 32sqrt{3}), and (a_4 = 16).Next, we compute ((a_0 + a_2 + a_4)^2 - (a_1 + a_3)^2):- Sum of even-indexed coefficients: (a_0 + a_2 + a_4 = 9 + 72 + 16 = 97).- Sum of odd-indexed coefficients: (a_1 + a_3 = 24sqrt{3} + 32sqrt{3} = 56sqrt{3}).Calculating the squares:- ((97)^2 = 9409)- ((56sqrt{3})^2 = 56^2 cdot 3 = 3136 cdot 3 = 9408)Subtracting these results:[9409 - 9408 = 1]We can also verify this using substitution. Evaluating the polynomial at (x = 1) and (x = -1):- At (x = 1): ((2 cdot 1 + sqrt{3})^4 = (2 + sqrt{3})^4)- At (x = -1): ((2 cdot (-1) + sqrt{3})^4 = (-2 + sqrt{3})^4)The product of these evaluations is:[(2 + sqrt{3})^4 cdot (-2 + sqrt{3})^4 = [(2 + sqrt{3})(-2 + sqrt{3})]^4 = (-1)^4 = 1]Thus, the final answer is (boxed{1}).]

🔑:Step 1: Set x=1. This gives us (2+sqrt{3})^{4} = a_{0} + a_{1} + a_{2} + a_{3} + a_{4}.Step 2: Set x=-1. This gives us (-2+sqrt{3})^{4} = a_{0} - a_{1} + a_{2} - a_{3} + a_{4}.Step 3: Add the two equations obtained in steps 1 and 2. This gives us 2(a_{0} + a_{2} + a_{4}) = (2+sqrt{3})^{4} + (-2+sqrt{3})^{4} = 194. Therefore, (a_{0} + a_{2} + a_{4}) = 97.Step 4: Subtract the equation obtained in step 2 from the equation obtained in step 1. This gives us 2(a_{1} + a_{3}) = (2+sqrt{3})^{4} - (-2+sqrt{3})^{4} = 112sqrt{3}. Therefore, (a_{1} + a_{3}) = 56sqrt{3}.Step 5: Calculate (a_{0} + a_{2} + a_{4})^{2} - (a_{1} + a_{3})^{2} = 97^{2} - (56sqrt{3})^{2} = 9409 - 9408 = 1.Step 6: Therefore, the final answer is boxed{1}.

🧠:[<think>Okay, so I need to find the negation of the proposition "There exists an x less than 0 such that x squared plus 2x minus m is greater than 0." The options are A, B, C, or D. Let me recall how negations work in logic. The original statement is an existential quantifier: ∃x < 0, P(x), where P(x) is "x² + 2x - m > 0". The negation of an existential quantifier is a universal quantifier with the negation of the predicate. So, ¬(∃x < 0, P(x)) becomes ∀x < 0, ¬P(x). So, first, let me write that out. The negation should be "For all x less than 0, x² + 2x - m is not greater than 0", which translates to "For all x < 0, x² + 2x - m ≤ 0". Looking at the options, that's option C. But let me double-check to make sure I didn't make a mistake. The original statement says there exists a negative x where the quadratic expression is greater than zero. The negation would mean that for all negative x, the expression is not greater than zero, so less than or equal to zero. That still sounds right. Let me also check the other options to see if they could be confused with the correct answer. Option A says "For all x ≥ 0, x² + 2x - m ≤ 0". That's changing the domain from x < 0 to x ≥ 0, which doesn't seem related to the negation. The negation shouldn't change the domain; it should stay within x < 0. So A is probably incorrect.Option B is "There exists x ≥ 0 such that x² + 2x - m > 0". Again, changing the domain to non-negative x and keeping the existence quantifier. That's not the negation; the negation would flip the quantifier and the inequality, but keep the domain the same. So B is out.Option D is "There exists x < 0 such that x² + 2x - m ≤ 0". This is just the negation of the predicate without flipping the quantifier. But the negation of "there exists" is "for all", so D is not correct either. So, yes, option C is the correct negation. Let me verify once more with an example. Suppose m is some number. If the original statement is that there's a negative x making the expression positive, then the negation would be that no negative x does that, which is equivalent to all negative x make the expression ≤0. That's exactly option C. I think that's solid. So the answer should be C.Final Answerboxed{C}</think>The original proposition is "exists x lt 0, x^{2}+2x-m gt 0". To find its negation, we need to negate both the existential quantifier and the inequality. The negation of an existential quantifier (exists) is a universal quantifier (forall), and the negation of the inequality > is leqslant. Thus, the negation of "exists x lt 0, x^{2}+2x-m gt 0" is "forall x lt 0, x^{2}+2x-m leqslant 0".Checking the options:- Option A changes the domain to x geqslant 0, which is incorrect.- Option B uses an existential quantifier and changes the domain, which is incorrect.- Option C correctly uses a universal quantifier and maintains the domain x < 0 with the negated inequality, which is correct.- Option D incorrectly uses an existential quantifier instead of a universal one.Therefore, the correct answer is (boxed{C}).,<think>Okay, let's see. The problem is about finding the negation of the proposition "There exists an x less than 0 such that x squared plus 2x minus m is greater than 0." And we have four options to choose from. Hmm, so I need to recall how negations work with quantifiers and inequalities.First, the original statement is an existential quantifier: ∃x < 0, P(x), where P(x) is "x² + 2x - m > 0." The negation of an existential quantifier is a universal quantifier with the negation of the predicate. So, ¬(∃x < 0, P(x)) becomes ∀x < 0, ¬P(x). Therefore, the negation should be "For all x less than 0, x² + 2x - m is not greater than 0," which translates to "For all x < 0, x² + 2x - m ≤ 0." Let me check the options. Option C says exactly that: "∀x < 0, x² + 2x - m ≤ 0." So that seems right. But wait, let me make sure I didn't skip any steps. The original statement is ∃x (x < 0 ∧ (x² + 2x - m > 0)), right? So the negation would be ∀x ¬(x < 0 ∧ (x² + 2x - m > 0)). Applying De Morgan's law, this becomes ∀x (x ≥ 0 ∨ (x² + 2x - m ≤ 0)). But that's equivalent to saying for all x, if x < 0, then x² + 2x - m ≤ 0. Which is exactly option C. So, the key here is that the negation of an existential quantifier is a universal quantifier with the negated statement. And since the original was about existence of some x < 0 with a certain property, the negation would be that for all x < 0, that property does not hold. Looking at the other options: A talks about x ≥ 0, which seems unrelated because the original was about x < 0. B is another existential statement but with x ≥ 0, which isn't the negation. D is an existential statement with x < 0 but the inequality flipped, which is not the negation since the negation requires a universal quantifier. So yeah, C is definitely the right answer.Final Answerboxed{C}</think>The original proposition is "exists x lt 0, x^{2}+2x-m gt 0". To find its negation, we need to negate both the existential quantifier and the inequality. 1. The negation of an existential quantifier (exists) is a universal quantifier (forall).2. The negation of the predicate x^{2} + 2x - m > 0 is x^{2} + 2x - m leq 0.Applying these rules, the negation of the proposition becomes "forall x lt 0, x^{2} + 2x - m leq 0".Checking the options:- Option C states "forall x lt 0, x^{2} + 2x - m leq 0", which matches our derived negation.- Other options involve different quantifiers or different domains, which do not correctly negate the original statement.Thus, the correct answer is (boxed{C}).]

🔑:To negate the given proposition, we follow the logical rules of negation for quantifiers and inequalities. The original proposition is "exists x lt 0, x^{2}+2x-m gt 0". Negating this involves two steps:1. The existential quantifier exists (there exists) becomes the universal quantifier forall (for all).2. The inequality gt (greater than) becomes leqslant (less than or equal to).Applying these rules, the negation of the original proposition is: - Change exists x lt 0 to forall x lt 0.- Change x^{2}+2x-m gt 0 to x^{2}+2x-mleqslant 0.Therefore, the negated proposition is "forall x lt 0, x^{2}+2x-mleqslant 0". Matching this with the given options, we find that the correct answer is boxed{text{C}}.

🧠:[<think>Okay, so I need to find the value of f(5) for this piecewise function. Let me start by understanding the function definition. The function f(x) is defined in two parts: when x is less than 4, it's 2 raised to the power of x. But when x is greater than or equal to 4, it's defined recursively as f(x-1). Hmm, recursion can sometimes be tricky, but maybe I can work through it step by step.First, let's note the cases clearly. For any x < 4, f(x) = 2^x. That part seems straightforward. For example, if x is 3, then f(3) = 2^3 = 8. But when x is 4 or more, instead of directly computing something, we have to refer back to the function value at x-1. So, f(4) = f(3), f(5) = f(4), and so on. That suggests that for x >= 4, each value depends on the previous one, kind of like a sequence where each term is based on the term before it.Since I need to find f(5), let's break it down. Since 5 is greater than or equal to 4, according to the definition, f(5) = f(5 - 1) = f(4). But then f(4) is also greater than or equal to 4, right? Because 4 is equal to 4. So f(4) = f(4 - 1) = f(3). Now, 3 is less than 4, so here we can use the first part of the function definition. Therefore, f(3) = 2^3 = 8. Wait, so f(4) = f(3) = 8, and then f(5) = f(4) = 8. Therefore, f(5) should be 8? That seems straightforward, but let me verify to make sure I didn't miss anything.Let me write out each step methodically:1. Start with f(5). Since 5 >= 4, use the second case: f(5) = f(5 - 1) = f(4).2. Now compute f(4). Since 4 >= 4, again use the second case: f(4) = f(4 - 1) = f(3).3. Compute f(3). Here, 3 < 4, so use the first case: f(3) = 2^3 = 8.4. Substitute back up: f(4) = f(3) = 8.5. Substitute back again: f(5) = f(4) = 8.Yes, that seems right. Each recursive call reduces x by 1 until we reach x = 3, which is in the base case (x < 4). Then, we just compute 2^3 and pass that value back up the chain.Let me check if there's another way this could be interpreted. Sometimes, recursive functions can have multiple interpretations, especially if the base case isn't clear. But in this problem, the base case is clearly x < 4, and for x >= 4, it's defined in terms of x-1. So, as long as we keep subtracting 1 from x each time, we should eventually land in the base case.For example, if we had f(6), following the same logic: f(6) = f(5) = f(4) = f(3) = 8. Similarly, f(7) would also be 8. In fact, for any x >= 4, it seems like f(x) will always evaluate to f(3) = 8, since each subsequent call just decrements x by 1 until it reaches 3. So, once x is 4 or higher, the function's value is always 8. Is that correct?Wait, let's verify with x=4. f(4) = f(3) = 8. x=5: f(5) = f(4) = 8. x=6: f(6) = f(5) = 8. Yep, that seems consistent. So, regardless of how large x is, as long as it's 4 or above, it'll just keep referring back until it hits f(3), which is 8. Therefore, f(x) for x >= 3? Wait, no. Wait, for x >= 4, it refers back to x-1. So, when x=4, it's f(3); when x=5, f(4), which is f(3), etc. So, all x >=4 will resolve to f(3)=8.But let me make sure there isn't a different base case. The function is defined as 2^x when x <4, and f(x-1) when x >=4. So, the base case is all x <4. So, for any x starting at 4 or higher, we subtract 1 repeatedly until x is 3, then apply 2^x. Therefore, yes, for x >=4, f(x) is equal to f(3) = 8.Therefore, f(5) is 8. Hmm. That seems straightforward, but let me think if there's a possible misunderstanding here. For example, sometimes recursive functions can have different behavior if they refer to different indices or if there's a step I'm missing. But in this case, since each time x >=4, we just go back one step, and the recursion stops once we hit x=3. So, there's no infinite recursion here, because each recursive call reduces x by 1, and eventually x will be less than 4.Another way to look at it is that the function f(x) for x >=4 is equivalent to 2^(x - (x -3))? Wait, maybe not. Let's see. If we unroll the recursion:f(5) = f(4)f(4) = f(3)f(3) = 2^3 =8So, f(5) = f(4) = f(3) =8.Alternatively, we can think of f(x) as 2^3 for all x >=3. Wait, but the function is defined as 2^x for x <4. So, for x=3, it's 2^3, but for x=4, it's f(3)=8. Then x=5 is f(4)=8, and so on. Therefore, the function f(x) is 2^x when x <4, and 8 when x >=4. So, essentially, once x reaches 4 or higher, the function value remains constant at 8.Is that the case? Let's check x=4. f(4)=f(3)=8. x=5=f(4)=8. x=6=f(5)=8, etc. So yes, for all x >=4, the function is 8. So, the function f(x) is 2^x for x <4, and 8 otherwise. Therefore, f(5) is 8.Alternatively, if we consider the general formula, for x >=4, f(x) = 2^(3), because we subtract 1 (x - (x - 3)) times? Wait, that might not be the right way to think about it. Let me see.Suppose x is 4: subtract 1 once to get to 3. x=5: subtract 1 twice to get to 3. x=6: subtract 1 three times to get to 3. So, in general, for x >=4, f(x) = f(x -1) = f(x -2) = ... = f(3) =8. Therefore, for x >=4, f(x) =8. Hence, regardless of how large x is, as long as it's 4 or above, f(x)=8.So, putting this all together, f(5)=8.I think that's solid. Let me just check with an example. Suppose we calculate f(4). Since 4 >=4, f(4)=f(3)=2^3=8. Correct. Then f(5)=f(4)=8. f(6)=f(5)=8. So, all these give 8. So, yes, that seems consistent.Another thing to check: what if x is a non-integer? Wait, the problem doesn't specify whether x is an integer or not. Hmm. The function is defined as 2^x for x <4, and f(x-1) otherwise. If x is, say, 4.5, then f(4.5)=f(4.5 -1)=f(3.5). Then, since 3.5 <4, f(3.5)=2^3.5. But the question here is about f(5), which is an integer. So, maybe the problem assumes x is an integer? The question is in Chinese, but the original problem statement here is given in English, and it just says "x <4" and "x >=4". So, maybe x is any real number. But in that case, for non-integer x >=4, we would subtract 1 each time until we get to a value less than 4. But for x=5, which is an integer, subtracting 1 enough times gets us to 3. But for a non-integer, like 4.5, subtracting 1 once gives 3.5, which is still less than 4, so f(4.5)=2^3.5.But in this problem, the question is about f(5), which is an integer. So, for x=5, an integer, the recursion steps down through integers until it hits 3. Therefore, f(5)=8. If x were a non-integer greater than 4, like 5.2, then f(5.2)=f(4.2)=f(3.2)=2^3.2. But since the problem asks for f(5), which is an integer, the answer is 8.So, in conclusion, the answer is 8. I think that's solid. Let me just recap once more to make sure.Starting with f(5):- Since 5 >=4, f(5) = f(5-1) = f(4)- f(4) = f(4-1) = f(3)- f(3) = 2^3 = 8- Therefore, f(4) = 8- Therefore, f(5) = 8Yep, that's consistent. All steps check out, no infinite loops, and the logic holds. The key was recognizing that once x is 4 or more, we decrement x until we hit 3, which is the base case. Then, the value is 8. So, f(5) is 8.Final AnswerThe value of ( f(5) ) is boxed{8}.</think>To find the value of ( f(5) ) for the given piecewise function ( f(x) ), we start by analyzing the function definition. The function is defined as follows:- ( f(x) = 2^x ) for ( x < 4 )- ( f(x) = f(x-1) ) for ( x geq 4 )We need to compute ( f(5) ):1. Since ( 5 geq 4 ), we use the recursive case: ( f(5) = f(5-1) = f(4) ).2. Next, we compute ( f(4) ). Again, since ( 4 geq 4 ), we use the recursive case: ( f(4) = f(4-1) = f(3) ).3. Now, we compute ( f(3) ). Since ( 3 < 4 ), we use the base case: ( f(3) = 2^3 = 8 ).Substituting back, we find:- ( f(4) = f(3) = 8 )- ( f(5) = f(4) = 8 )Thus, the value of ( f(5) ) is (boxed{8}).,<think>Okay, so I need to find f(5) for this piecewise function. Let me read the problem again to make sure I understand it. The function f(x) is defined as 2^x when x is less than 4, and for x greater than or equal to 4, it's defined as f(x-1). So, for x >= 4, the function is recursive, meaning it calls itself with the argument x-1. Alright, let's break this down.First, let's recall what a piecewise function is. It's a function that has different definitions depending on the value of the input. In this case, if x is less than 4, we just calculate 2 raised to the power of x. But if x is 4 or more, we have to compute f(x-1) instead. That recursion part is key here. So, to get f(5), since 5 is greater than or equal to 4, we have to find f(5-1) which is f(4). But then f(4) is also greater than or equal to 4, so we have to go down another level to f(3). Now, 3 is less than 4, so we can finally compute 2^3. Let me check if that's correct.Wait, let's go step by step. Starting with f(5):f(5) = f(5 - 1) because 5 >= 4. So that's f(4).Now compute f(4). Again, 4 is equal to 4, so we use the second case: f(4) = f(4 - 1) = f(3).Now, f(3): here, 3 is less than 4, so we use the first case. So f(3) = 2^3 = 8.Therefore, going back up, f(4) = f(3) = 8, and f(5) = f(4) = 8. So f(5) should be 8.Wait, but let me make sure that recursion isn't going deeper. For example, if the function had a different recursive step, maybe we would have to go further. But in this case, when x is 4, we subtract 1 to get 3, which is less than 4, so we can compute it directly. So, yes, this seems correct.Alternatively, maybe I can think of it as f(x) = f(x - 1) for x >= 4. So, for any x >= 4, the function's value is the same as the value at x - 1. This is like a recursive sequence where each term is equal to the previous term. So, starting from x = 4, f(4) = f(3). Then f(3) is 2^3, which is 8. Therefore, f(4) is 8, and f(5) is the same as f(4), which is 8. Similarly, f(6) would be f(5) which is 8, and so on. Wait, so does this mean that for any x >= 4, the function f(x) is 8? Because each time you subtract 1 until you get to 3, which is 2^3, and then all subsequent values just repeat 8?Let me test that. Let's compute f(4):f(4) = f(3) = 8.f(5) = f(4) = 8.f(6) = f(5) = 8.So, indeed, once you hit x = 4, the function stays at 8 for all higher x values. Therefore, f(5) is 8.But let me double-check. Maybe I made a mistake in the recursion. Let's write out each step explicitly:1. Compute f(5): Since 5 >= 4, f(5) = f(5 - 1) = f(4).2. Compute f(4): Since 4 >= 4, f(4) = f(4 - 1) = f(3).3. Compute f(3): Since 3 < 4, f(3) = 2^3 = 8.4. Therefore, f(4) = 8.5. Therefore, f(5) = 8.Yep, that seems right. Let's see if there's another way to look at it. Maybe expanding the recursion:f(5) = f(4) = f(3) = 2^3 = 8.So, the recursion stops when the argument becomes less than 4, which happens at x=3. Therefore, every time x is 4 or higher, you subtract 1 until you get down to 3, then compute 2^3.Therefore, the answer should be 8.But just to be thorough, let's check with another value. Let's compute f(6):f(6) = f(5) = f(4) = f(3) = 8. So f(6) is also 8. Similarly, f(4) is 8, f(5) is 8, etc. So the function becomes constant at 8 for x >= 3? Wait, no, for x >=4. Wait, when x is 4, you go to 3, compute 8, then for x=4,5,6,... it's all 8. So the function is 2^x for x <4 and 8 otherwise. So, f(5) is 8.Alternatively, maybe there's a pattern here. Let's write out the function for several values:x | f(x)---|---0 | 2^0=11 | 2^1=22 | 2^2=43 | 2^3=84 | f(3)=85 | f(4)=86 | f(5)=8... | ...Yes, that's consistent. So starting from x=4 onwards, all f(x) values are 8. Hence, f(5) is 8.Is there any other way this could be interpreted? Maybe if the recursion was f(x-1) + something, but in this case, it's just f(x-1). So, it's a direct recursion without any modification. Therefore, the value propagates down until it hits the base case (x <4). So, yes, 8 is correct.Wait, but what if someone thinks that for x >=4, the function is f(x-1) + some other term? But no, the problem states it's just f(x-1). So, it's a direct recursive call. Therefore, the answer is 8.I think that's solid. No steps seem missing here. So, I feel confident that the answer is 8.Final Answerboxed{8}</think>To find ( f(5) ) for the given piecewise function ( f(x) ):[f(x) = begin{cases} 2^x & text{if } x < 4 f(x-1) & text{if } x geq 4 end{cases}]1. Evaluate ( f(5) ): - Since ( 5 geq 4 ), we use the recursive case: ( f(5) = f(5-1) = f(4) ).2. Evaluate ( f(4) ): - Since ( 4 geq 4 ), we use the recursive case: ( f(4) = f(4-1) = f(3) ).3. Evaluate ( f(3) ): - Since ( 3 < 4 ), we use the base case: ( f(3) = 2^3 = 8 ).4. Backtrack the values: - ( f(4) = f(3) = 8 ) - ( f(5) = f(4) = 8 )Thus, the value of ( f(5) ) is (boxed{8}).]

🔑:AnalysisThis problem involves evaluating a piecewise function. When xgeqslant 4, the given expression represents a recursive relationship with a step size of 1. Therefore, this relationship can be used to progressively transform and calculate the value of f(5). The key point of this problem is to evaluate a piecewise function, and a recursive relationship with a step size of 1 is given in the analytical expression. When solving, it is necessary to evaluate based on the analytical expressions in different intervals of the function. Careful deduction is required during this transformation due to the generally large absolute values of the related numbers, to avoid careless mistakes.SolutionSince when xgeqslant 4, f(x)=f(x-1) Therefore, f(5)=f(4)=f(3) And when x < 4, f(x)=2^{x} Therefore, f(5)=f(3)=2^{3}=8 Hence, the answer is boxed{8}.

🧠:[<think>Okay, so I have this problem here: Find all positive integers x, y, z such that xy mod z ≡ yz mod x ≡ zx mod y = 2. Hmm, that seems a bit complicated at first glance, but let me break it down step by step.First, let's parse the problem statement. It says that when we compute xy modulo z, yz modulo x, and zx modulo y, all of these congruences are equal to 2. So, we need to find all triples (x, y, z) of positive integers where each of these three expressions is congruent to 2 modulo the third variable. Let me write that out more formally. The conditions are:1. xy ≡ 2 mod z2. yz ≡ 2 mod x3. zx ≡ 2 mod yAnd all of x, y, z must be positive integers.Alright, so each of these congruences gives us some constraints on x, y, z. Let me think about how to approach this. Maybe starting with small values? Since we are dealing with moduli and remainders, perhaps the variables can't be too large. Also, since the remainder is 2, each modulus (z, x, y respectively) must be greater than 2, right? Because if you take mod something, the remainder has to be less than the modulus. So if xy mod z = 2, then z must be greater than 2. Similarly, x and y must also be greater than 2. So x, y, z ≥ 3. That's a starting point.So all three variables must be at least 3. Now, perhaps we can look for symmetric solutions? Or maybe some permutations of numbers where these conditions hold. Let me consider first if there's a possibility that x, y, z are all equal. Suppose x = y = z = k. Then the conditions become:1. k*k ≡ 2 mod k → k² mod k = 0 ≡ 2 mod k. But 0 ≡ 2 mod k implies that k divides 2. But since k ≥ 3, this is impossible. So there are no solutions where all three variables are equal.Alright, so the variables must be distinct or maybe two equal and one different? Let's test that. Suppose two variables are equal. Let's say x = y. Then the conditions become:1. x*x ≡ 2 mod z2. x*z ≡ 2 mod x3. z*x ≡ 2 mod xWait, conditions 2 and 3 become the same if x = y. Let's see. If x = y, then:First condition: x² ≡ 2 mod zSecond condition: x*z ≡ 2 mod x. But x*z mod x is 0, since x divides x*z. So 0 ≡ 2 mod x, which implies that x divides 2. But x ≥ 3, so this is impossible. Therefore, if two variables are equal, we get a contradiction. Therefore, all three variables must be distinct.So, x, y, z are distinct integers greater than or equal to 3. Now, let's think about each congruence.Starting with the first one: xy ≡ 2 mod z. This means that when you divide xy by z, the remainder is 2. So, xy = k*z + 2 for some integer k ≥ 0. Similarly, yz = m*x + 2 and zx = n*y + 2 for some integers m, n ≥ 0.Since all variables are positive integers, k, m, n must be non-negative integers. But since x, y, z ≥ 3, the products xy, yz, zx are at least 9, so z must be such that z ≤ (xy - 2)/k, but this might not be straightforward. Maybe instead, we can consider the inequalities.From xy ≡ 2 mod z, since the remainder is 2, we know that z must divide xy - 2. So z | (xy - 2). Similarly, x | (yz - 2) and y | (zx - 2). So z is a divisor of xy - 2, x is a divisor of yz - 2, and y is a divisor of zx - 2.Therefore, we can write:1. xy - 2 = a*z2. yz - 2 = b*x3. zx - 2 = c*yFor some positive integers a, b, c. Since z, x, y ≥ 3, then the right-hand sides are at least 3*a, 3*b, 3*c. So the left-hand sides must be at least 3. So:xy - 2 ≥ 3*ayz - 2 ≥ 3*bzx - 2 ≥ 3*cBut since a, b, c are positive integers (since z, x, y are positive integers and xy - 2, yz - 2, zx - 2 must be positive as well because x, y, z ≥ 3, so xy ≥ 9, so xy - 2 ≥ 7, etc.), so a, b, c are positive integers.But this seems a bit abstract. Maybe we can look for small values. Let's try small x, y, z starting from 3.Let me start by assuming the smallest possible values for x, y, z. Let's try x = 3. Then, since x = 3, we need yz ≡ 2 mod 3. So yz ≡ 2 mod 3. Similarly, z*3 ≡ 2 mod y. Let's see.Suppose x = 3. Then:From the second condition: yz ≡ 2 mod 3. So yz mod 3 = 2.From the third condition: 3z ≡ 2 mod y. So 3z - 2 is divisible by y.Also, from the first condition: 3y ≡ 2 mod z. So 3y - 2 is divisible by z.So, with x = 3, we have two conditions on y and z.Let me try to fix x = 3 and look for possible y and z.First, from yz ≡ 2 mod 3. Since 3 is prime, we can say that y and z are numbers such that their product is congruent to 2 modulo 3. The possible residues modulo 3 are 0, 1, 2. So y and z can't be 0 mod 3, because then their product would be 0 mod 3, which isn't 2. So y and z must be either 1 or 2 mod 3. But 1*2 = 2 mod 3, so one of y or z must be 1 mod 3 and the other 2 mod 3.So either y ≡ 1 mod 3 and z ≡ 2 mod 3, or y ≡ 2 mod 3 and z ≡ 1 mod 3.Let me consider possible y and z starting from the smallest values.Case 1: y = 4 (since y must be at least 3, and 4 is the next after 3; but y can't be 3 because if y = 3, then from the first condition 3*3 ≡ 0 mod z ≡ 2 mod z, which would require z divides 9 - 2 = 7. So z divides 7. Since z ≥3, z=7. Let's check if that works. If x=3, y=3, z=7. Wait, but we assumed x=3 and trying y=3. But earlier we saw that if two variables are equal, there's a problem. Let me check:Wait, if x=3, y=3, then from the second condition: yz mod x = 3z mod 3 = 0 ≡ 2 mod 3. Which would require 0 ≡2 mod3, which is false. So y cannot be 3 if x=3. Therefore, y must be at least 4.So y starts from 4. Let's try y=4, x=3.From the first condition: 3*4 = 12 ≡ 2 mod z. So 12 mod z =2. Therefore, z divides 12 -2 =10. So z must be a divisor of 10. The divisors of 10 are 1,2,5,10. But z must be at least 3. So possible z=5 or 10.Let me check z=5.Check second condition: yz mod x = 4*5 mod3 =20 mod3=2. Which matches the required 2. Good.Third condition: zx mod y =5*3 mod4=15 mod4=3. But we need this to be 2. So 3≡2 mod4? No, that's not true. So z=5 does not satisfy the third condition.Next, z=10.Check second condition: yz modx=4*10 mod3=40 mod3=1. But we need 2. So that's bad. So z=10 is invalid.So for y=4, x=3, possible z=5 or10, but neither works.Next, y=5, x=3.First condition: 3*5=15≡2 mod z. So 15 mod z=2. Therefore, z divides 15-2=13. 13 is prime, so z=13.Check second condition: yz modx=5*13 mod3=65 mod3=65-63=2. Good.Third condition: zx mody=13*3 mod5=39 mod5=4. Need 2, so no good.Thus, y=5, x=3, z=13 fails the third condition.Next, y=6, x=3.First condition: 3*6=18≡2 mod z. So z divides 18-2=16. Divisors of16 greater than2: 4,8,16. But z must be at least3, so z=4,8,16.Check z=4:Second condition: yz modx=6*4 mod3=24 mod3=0≡2 mod3? No.z=8:Second condition:6*8=48 mod3=0≡2? No.z=16:6*16=96 mod3=0≡2? No. So none work.Next, y=7, x=3.First condition:3*7=21≡2 modz. Thus z divides 21-2=19. z=19.Check second condition:7*19=133 mod3=133-132=1≡2? No.Third condition:19*3=57 mod7=57-56=1≡2? No. So invalid.y=8, x=3:3*8=24≡2 mod z. So z divides24-2=22. Divisors:11,22. z=11 or22.Check z=11:Second condition:8*11=88 mod3=88-87=1≡2? No.z=22:8*22=176 mod3=176-174=2≡2. Good.Third condition:22*3=66 mod8=66-64=2≡2. Perfect! So for x=3, y=8, z=22, all conditions are satisfied?Wait, let's check all three:1. xy mod z =3*8 mod22=24 mod22=2. Correct.2. yz modx=8*22 mod3=176 mod3=2. Correct.3. zx mody=22*3 mod8=66 mod8=2. Correct.Yes! So (3,8,22) is a solution.Wait, that's a valid solution. Let me note that down. So (3,8,22). Also, since the problem is symmetric in x, y, z? Wait, no, actually the problem isn't symmetric. Wait, let's check.Wait, the problem states that xy mod z ≡ yz mod x ≡ zx mod y ≡2. So all three congruences must equal 2. So in the solution (3,8,22), all three congruences give 2. So this is a valid solution.Now, since variables are ordered, permutations might not work here. For example, if we permute (3,8,22), say (8,22,3), let's check if that's a solution.xy mod z =8*22 mod3=176 mod3=2. Good.yz modx=22*3 mod8=66 mod8=2. Good.zx mody=3*8 mod22=24 mod22=2. Good. So (8,22,3) is also a solution. Similarly, other permutations would be (22,3,8), etc. So all permutations of (3,8,22) are solutions.But wait, are these permutations distinct? Since x, y, z are ordered triples, yes. So each permutation is a distinct solution. So we have six permutations, but since 3,8,22 are distinct, all six permutations would be different. However, let's check if other permutations can be formed with different numbers.But first, let's see if there are more solutions beyond this. Let's continue with x=3.Next, y=9, x=3:First condition:3*9=27≡2 mod z. So z divides27-2=25. z=5,25.Check z=5:Second condition:9*5=45 mod3=0≡2? No.z=25:9*25=225 mod3=0≡2? No. Doesn't work.y=10, x=3:3*10=30≡2 mod z. z divides30-2=28. Divisors of28 greater than2:4,7,14,28.Check z=4:yz modx=10*4=40 mod3=1≡2? No.z=7:10*7=70 mod3=1≡2? No.z=14:10*14=140 mod3=140-138=2≡2. Good.Third condition:zx mody=14*3 mod10=42 mod10=2. Perfect. So (3,10,14) is a solution.Check all three conditions:1. 3*10 mod14=30 mod14=2. Correct.2.10*14 mod3=140 mod3=2. Correct.3.14*3 mod10=42 mod10=2. Correct.So that's another solution: (3,10,14). Similarly, permutations of this would also work. Let's check (10,14,3):xy modz=10*14 mod3=140 mod3=2. Correct.yz modx=14*3 mod10=42 mod10=2. Correct.zx mody=3*10 mod14=30 mod14=2. Correct. So permutations are also solutions.Continuing with x=3, y=11:3*11=33≡2 mod z. z divides33-2=31. z=31.Check second condition:11*31=341 mod3=341-339=2≡2. Good.Third condition:31*3=93 mod11=93-88=5≡2? No. 5≡2 mod11? No. So fails.y=12, x=3:3*12=36≡2 mod z. z divides36-2=34. z=17,34.Check z=17:Second condition:12*17=204 mod3=0≡2? No.z=34:12*34=408 mod3=0≡2? No.So no solutions here.y=13, x=3:3*13=39≡2 mod z. z divides39-2=37. z=37.Check second condition:13*37=481 mod3=481-480=1≡2? No.Third condition:37*3=111 mod13=111-104=7≡2? No.Not working.y=14, x=3:3*14=42≡2 mod z. z divides42-2=40. z=5,8,10,20,40.Check z=5:yz modx=14*5=70 mod3=1≡2? No.z=8:14*8=112 mod3=112-111=1≡2? No.z=10:14*10=140 mod3=2. Good.Third condition:10*3 mod14=30 mod14=2. Correct. Wait, but this is the same as the previous solution (3,10,14). So when y=14, x=3, z=10. So that's just another permutation. So perhaps we can consider that for x=3, y=10 gives z=14, and y=14 gives z=10.But since variables are ordered, these are different solutions.Continuing y=15, x=3:3*15=45≡2 mod z. z divides45-2=43. z=43.Check second condition:15*43=645 mod3=0≡2? No.So, seems like with x=3, we have two solutions: (3,8,22) and (3,10,14), along with their permutations. Let's see if there are other solutions with x=4.Trying x=4.So, starting with x=4. Then:From the second condition: yz ≡2 mod4. So yz ≡2 mod4.From the third condition: z*4 ≡2 mod y.From the first condition:4y ≡2 modz.So let's analyze yz ≡2 mod4. The product yz mod4 is 2. Let's recall that modulo4, the possible residues are 0,1,2,3. So possible pairs (y mod4, z mod4) such that their product ≡2 mod4.Possible pairs:- (1,2): 1*2=2- (2,1): 2*1=2- (3,2): 3*2=6≡2 mod4- (2,3): 2*3=6≡2 mod4So possible cases:1. y≡1 mod4 and z≡2 mod42. y≡2 mod4 and z≡1 mod43. y≡3 mod4 and z≡2 mod44. y≡2 mod4 and z≡3 mod4So y and z must be in one of these cases.Also, from the third condition:4z ≡2 mod y →4z -2 is divisible by y. So y divides (4z -2).From the first condition:4y ≡2 modz → z divides (4y -2).So with x=4, similar to x=3, we have these constraints. Let's try small values.Starting with y=5 (since y must be at least3, but x=4, so y must be different from x? Not necessarily, but variables can be equal? Wait, earlier we saw that if two variables are equal, there's a problem. Let me check if y=4:If y=4, then from the second condition: yz modx=4z mod4=0≡2 mod4. But 0≡2 mod4 is false. So y can't be 4. Therefore, y must be at least5.Wait, variables can be equal only if all three are equal, which we saw is impossible. Wait, no, two variables could be equal but we saw that leads to a problem. Wait, if two variables are equal, say x=y=4, then as above, yz modx=4z mod4=0≡2 mod4, which is impossible. So even two variables equal is impossible. Therefore, with x=4, y must be at least5, and z must be at least3, and different from x and y.Let's start with y=5, x=4.First condition:4*5=20≡2 mod z. So 20 modz=2→z divides20-2=18. So z=3,6,9,18.But z must be at least3. Let's check each:z=3:Check second condition: yz modx=5*3=15 mod4=15-12=3≡2 mod4? No.z=6:yz modx=5*6=30 mod4=30-28=2≡2. Good.Third condition:zx mody=6*4=24 mod5=24-20=4≡2 mod5? No. 4≡2 mod5? No.z=9:yz modx=5*9=45 mod4=1≡2? No.z=18:yz modx=5*18=90 mod4=90-88=2≡2. Good.Third condition:18*4=72 mod5=72-70=2≡2. Perfect. So (4,5,18) is a solution.Check all conditions:1.4*5=20 mod18=20-18=2. Correct.2.5*18=90 mod4=2. Correct.3.18*4=72 mod5=2. Correct.So (4,5,18) works. Similarly, permutations of this will also work. Let's check (5,18,4):xy modz=5*18=90 mod4=2. Correct.yz modx=18*4=72 mod5=2. Correct.zx mody=4*5=20 mod18=2. Correct. So permutations are valid.Next, y=6, x=4:First condition:4*6=24≡2 mod z. So z divides24-2=22. z=11,22.Check z=11:Second condition:6*11=66 mod4=66-64=2≡2. Good.Third condition:11*4=44 mod6=44-42=2≡2. Perfect. So (4,6,11) is a solution.Check all:1.4*6=24 mod11=24-22=2. Correct.2.6*11=66 mod4=2. Correct.3.11*4=44 mod6=2. Correct.Similarly, permutations here would work.Next, y=7, x=4:First condition:4*7=28≡2 modz. So z divides28-2=26. z=13,26.Check z=13:Second condition:7*13=91 mod4=91-88=3≡2? No.z=26:7*26=182 mod4=182-180=2≡2. Good.Third condition:26*4=104 mod7=104-98=6≡2? No. 6≡2 mod7? No. So fails.y=8, x=4:First condition:4*8=32≡2 mod z. So z divides32-2=30. z=5,6,10,15,30.Check z=5:Second condition:8*5=40 mod4=0≡2? No.z=6:8*6=48 mod4=0≡2? No.z=10:8*10=80 mod4=0≡2? No.z=15:8*15=120 mod4=0≡2? No.z=30:8*30=240 mod4=0≡2? No. All no.So no solutions here.y=9, x=4:4*9=36≡2 modz. z divides36-2=34. z=17,34.Check z=17:Second condition:9*17=153 mod4=153-152=1≡2? No.z=34:9*34=306 mod4=306-304=2≡2. Good.Third condition:34*4=136 mod9=136-126=10≡2 mod9? 10≡1 mod9. No.So fails.y=10, x=4:4*10=40≡2 modz. z divides40-2=38. z=19,38.Check z=19:Second condition:10*19=190 mod4=190-188=2≡2. Good.Third condition:19*4=76 mod10=6≡2? No. 6≡2 mod10? No.z=38:10*38=380 mod4=0≡2? No.So no.y=11, x=4:4*11=44≡2 mod z. z divides44-2=42. z=6,7,14,21,42.Check z=6:Second condition:11*6=66 mod4=2. Good.Third condition:6*4=24 mod11=24-22=2. Perfect. Wait, so (4,11,6) is a solution.But that's same as (4,6,11) which we already found earlier. So permutations.Similarly, z=7:Second condition:11*7=77 mod4=1≡2? No.z=14:11*14=154 mod4=154-152=2≡2. Good.Third condition:14*4=56 mod11=56-55=1≡2? No.z=21:11*21=231 mod4=231-228=3≡2? No.z=42:11*42=462 mod4=462-460=2≡2. Good.Third condition:42*4=168 mod11=168-154=14≡3≡2? No. 14 mod11=3.So only z=6 and z=42. z=6 works but is a permutation. z=42 doesn't work.Continuing y=12, x=4:4*12=48≡2 mod z. z divides48-2=46. z=23,46.Check z=23:Second condition:12*23=276 mod4=0≡2? No.z=46:12*46=552 mod4=0≡2? No.So no solutions.y=13, x=4:4*13=52≡2 modz. z divides52-2=50. z=5,10,25,50.Check z=5:Second condition:13*5=65 mod4=1≡2? No.z=10:13*10=130 mod4=130-128=2≡2. Good.Third condition:10*4=40 mod13=40-39=1≡2? No.z=25:13*25=325 mod4=1≡2? No.z=50:13*50=650 mod4=650-648=2≡2. Good.Third condition:50*4=200 mod13=200-182=18≡5≡2? No.So no.So with x=4, we found two solutions: (4,5,18) and (4,6,11) and their permutations. Let's check if there are more solutions.Trying x=5.So x=5.From the second condition: yz ≡2 mod5.From the third condition:5z ≡2 mody.From the first condition:5y ≡2 modz.So similar approach. Let's try small y.y=3:First condition:5*3=15≡2 modz. So z divides15-2=13. z=13.Second condition:yz modx=3*13=39 mod5=4≡2 mod5? No.y=4:First condition:5*4=20≡2 modz. z divides20-2=18. z=3,6,9,18.Check z=3:Second condition:4*3=12 mod5=2. Good.Third condition:5*3=15 mod4=15-12=3≡2 mod4? No.z=6:Second condition:4*6=24 mod5=4≡2? No.z=9:4*9=36 mod5=1≡2? No.z=18:4*18=72 mod5=2≡2. Good.Third condition:5*18=90 mod4=90-88=2≡2. Perfect. So (5,4,18) is a solution. Which is a permutation of (4,5,18).Similarly, check permutations.y=6, x=5:First condition:5*6=30≡2 modz. z divides30-2=28. z=4,7,14,28.Check z=4:Second condition:6*4=24 mod5=4≡2? No.z=7:6*7=42 mod5=2≡2. Good.Third condition:5*7=35 mod6=35-30=5≡2? No.z=14:6*14=84 mod5=4≡2? No.z=28:6*28=168 mod5=3≡2? No.No solutions here.y=7, x=5:First condition:5*7=35≡2 modz. z divides35-2=33. z=3,11,33.Check z=3:Second condition:7*3=21 mod5=1≡2? No.z=11:7*11=77 mod5=2≡2. Good.Third condition:5*11=55 mod7=55-49=6≡2? No.z=33:7*33=231 mod5=1≡2? No.y=8, x=5:First condition:5*8=40≡2 modz. z divides40-2=38. z=19,38.Check z=19:Second condition:8*19=152 mod5=152-150=2≡2. Good.Third condition:5*19=95 mod8=95-88=7≡2? No.z=38:8*38=304 mod5=4≡2? No.y=9, x=5:First condition:5*9=45≡2 modz. z divides45-2=43. z=43.Second condition:9*43=387 mod5=2≡2. Good.Third condition:5*43=215 mod9=215-207=8≡2? No.y=10, x=5:First condition:5*10=50≡2 modz. z divides50-2=48. z=3,4,6,8,12,16,24,48.Check z=3:Second condition:10*3=30 mod5=0≡2? No.z=4:10*4=40 mod5=0≡2? No.z=6:10*6=60 mod5=0≡2? No.z=8:10*8=80 mod5=0≡2? No.z=12:10*12=120 mod5=0≡2? No.z=16:10*16=160 mod5=0≡2? No.z=24:10*24=240 mod5=0≡2? No.z=48:10*48=480 mod5=0≡2? No.All no.y=11, x=5:First condition:5*11=55≡2 modz. z divides55-2=53. z=53.Second condition:11*53=583 mod5=3≡2? No.y=12, x=5:5*12=60≡2 modz. z divides60-2=58. z=29,58.Check z=29:Second condition:12*29=348 mod5=348-345=3≡2? No.z=58:12*58=696 mod5=1≡2? No.So x=5 seems to only give permutations of previous solutions.Proceeding similarly for x=6.But this process might take a long time. Let me think if there's a pattern here.Looking at the solutions found so far: (3,8,22), (3,10,14), (4,5,18), (4,6,11). Their permutations also work. Let's check if these solutions have some relation.Looking at (3,8,22):From first condition:3*8=24≡2 mod22. 24-22=2. So yes.Second condition:8*22=176≡176 mod3. 176 divided by3 is 58*3 +2. So 176≡2 mod3.Third condition:22*3=66≡66 mod8=66-8*8=66-64=2. So 66≡2 mod8.Similarly, (3,10,14):3*10=30≡2 mod14. 30-2*14=30-28=2.10*14=140≡140 mod3=140-3*46=140-138=2.14*3=42≡42 mod10=42-4*10=42-40=2.Same for (4,5,18):4*5=20≡2 mod18. 20-18=2.5*18=90≡90 mod4=90-22*4=90-88=2.18*4=72≡72 mod5=72-14*5=72-70=2.And (4,6,11):4*6=24≡2 mod11. 24-2*11=24-22=2.6*11=66≡66 mod4=66-16*4=66-64=2.11*4=44≡44 mod6=44-7*6=44-42=2.So in each case, the product is exactly 2 more than the modulus. So, for example, in (3,8,22), 3*8=22 +2=24. Similarly, 8*22=176=3*58 +2. Wait, but 8*22=176, which is 3*58=174, so 176=174+2. Similarly, 22*3=66=8*8 +2=64+2=66. So in each case, the product is exactly two more than a multiple of the modulus.Therefore, in general, if we have xy = kz +2, yz = lx +2, zx = my +2 for some integers k, l, m. Then, substituting each equation into the next, perhaps we can find a general solution or find relations between variables.Alternatively, maybe these solutions are part of a family. Let me check if there are other solutions with larger numbers.Trying x=5, y=7, z=?From the first condition:5*7=35≡2 modz→ z divides35-2=33→z=3,11,33.Check z=3:Second condition:7*3=21 mod5=1≡2? No.z=11:7*11=77 mod5=77-75=2≡2. Good.Third condition:5*11=55 mod7=55-49=6≡2? No.So z=11 fails third condition.z=33:7*33=231 mod5=231-230=1≡2? No.No solution.Alternatively, trying x=5, y=9:First condition:5*9=45≡2 modz→z divides43, which is prime. So z=43.Check second condition:9*43=387 mod5=2. Good.Third condition:5*43=215 mod9=215-207=8≡2? No.So no.Alternatively, x=6, y=4:First condition:6*4=24≡2 modz→z divides22. So z=11,22.Check z=11:Second condition:4*11=44 mod6=44-36=8≡2? No.z=22:4*22=88 mod6=88-84=4≡2? No.So no.Alternatively, x=6, y=5:First condition:6*5=30≡2 modz→z divides28. z=4,7,14,28.Check z=4:Second condition:5*4=20 mod6=20-18=2≡2. Good.Third condition:6*4=24 mod5=24-20=4≡2? No.z=7:5*7=35 mod6=35-30=5≡2? No.z=14:5*14=70 mod6=70-66=4≡2? No.z=28:5*28=140 mod6=140-138=2≡2. Good.Third condition:6*28=168 mod5=168-165=3≡2? No.So z=4 and z=28. Neither works.Hmm, so seems like after x=4, we don't get any new solutions, only permutations of the existing ones. Let's check x=7.x=7.First condition:7y ≡2 modz.Second condition:yz ≡2 mod7.Third condition:7z ≡2 mody.Starting with y=3:First condition:7*3=21≡2 modz→z divides19. z=19.Check second condition:3*19=57 mod7=57-56=1≡2? No.y=4:7*4=28≡2 modz→z divides26. z=13,26.Check z=13:Second condition:4*13=52 mod7=52-49=3≡2? No.z=26:4*26=104 mod7=104-98=6≡2? No.y=5:7*5=35≡2 modz→z divides33. z=3,11,33.Check z=3:Second condition:5*3=15 mod7=15-14=1≡2? No.z=11:5*11=55 mod7=55-49=6≡2? No.z=33:5*33=165 mod7=165-161=4≡2? No.y=6:7*6=42≡2 modz→z divides40. z=5,8,10,20,40.Check z=5:Second condition:6*5=30 mod7=30-28=2≡2. Good.Third condition:7*5=35 mod6=35-30=5≡2? No.z=8:6*8=48 mod7=48-42=6≡2? No.z=10:6*10=60 mod7=60-56=4≡2? No.z=20:6*20=120 mod7=120-112=8≡1≡2? No.z=40:6*40=240 mod7=240-238=2≡2. Good.Third condition:7*40=280 mod6=280-276=4≡2? No.y=8:7*8=56≡2 modz→z divides54. z=6,9,18,27,54.Check z=6:Second condition:8*6=48 mod7=48-42=6≡2? No.z=9:8*9=72 mod7=72-63=9≡2? No.z=18:8*18=144 mod7=144-140=4≡2? No.z=27:8*27=216 mod7=216-210=6≡2? No.z=54:8*54=432 mod7=432-427=5≡2? No.y=9:7*9=63≡2 modz→z divides61. z=61.Check second condition:9*61=549 mod7=549-546=3≡2? No.y=10:7*10=70≡2 modz→z divides68. z=4,17,34,68.Check z=4:Second condition:10*4=40 mod7=40-35=5≡2? No.z=17:10*17=170 mod7=170-161=9≡2? No.z=34:10*34=340 mod7=340-336=4≡2? No.z=68:10*68=680 mod7=680-679=1≡2? No.So x=7 doesn't yield any new solutions.Given that as x increases, the possible z becomes larger, and the third condition becomes harder to satisfy, maybe there are only a finite number of solutions. So far, we have:1. (3,8,22) and permutations2. (3,10,14) and permutations3. (4,5,18) and permutations4. (4,6,11) and permutationsLet me check if there's another solution with x=5, y= something.Wait, x=5, y=7, z=?From first condition:5*7=35≡2 modz→z=33 (since 35-2=33). Checked earlier, didn't work.Alternatively, x=5, y=12:First condition:5*12=60≡2 modz→z divides58. z=29,58.Check z=29:Second condition:12*29=348 mod5=348-345=3≡2? No.z=58:12*58=696 mod5=696-695=1≡2? No.x=5, y=13:5*13=65≡2 modz→z=63. z=63.Second condition:13*63=819 mod5=819-815=4≡2? No.Hmm. Perhaps these are the only solutions. Let's check x=6, y=7.x=6, y=7:First condition:6*7=42≡2 modz→z divides40. z=5,8,10,20,40.Check z=5:Second condition:7*5=35 mod6=35-30=5≡2? No.z=8:7*8=56 mod6=56-54=2≡2. Good.Third condition:6*8=48 mod7=48-42=6≡2? No.z=10:7*10=70 mod6=70-66=4≡2? No.z=20:7*20=140 mod6=140-138=2≡2. Good.Third condition:6*20=120 mod7=120-119=1≡2? No.z=40:7*40=280 mod6=280-276=4≡2? No.No good.Alternatively, x=6, y=11:First condition:6*11=66≡2 modz→z divides64. z=16,32,64.Check z=16:Second condition:11*16=176 mod6=176-174=2≡2. Good.Third condition:6*16=96 mod11=96-88=8≡2? No.z=32:11*32=352 mod6=352-348=4≡2? No.z=64:11*64=704 mod6=704-702=2≡2. Good.Third condition:6*64=384 mod11=384-363=21≡10≡2? No.So no solutions here.Given that we have found solutions for x=3,4, and permutations, but not for higher x, perhaps these are the only solutions.To confirm, let's check another possible solution. Suppose we take (5,18,4), which is a permutation of (4,5,18):Check all conditions:1.5*18=90 mod4=90-88=2. Correct.2.18*4=72 mod5=72-70=2. Correct.3.4*5=20 mod18=2. Correct.Similarly, (6,11,4):6*11=66 mod4=2. Correct.11*4=44 mod6=2. Correct.4*6=24 mod11=2. Correct.Yes, all permutations work.Now, considering the solutions found, they seem to follow a pattern where each variable is part of two multiplicative pairs that are 2 more than a multiple of the third variable. For example, in (3,8,22):3*8 = 22 + 28*22 = 3*58 + 222*3 = 8*8 + 2But these multiples aren't consistent. Alternatively, perhaps there's a chain of dependencies.Alternatively, we can look at the equations:From the three conditions:xy = kz + 2yz = lx + 2zx = my + 2We can substitute these equations into each other. Let's try substituting.From the first equation, z = (xy - 2)/kFrom the second equation, x = (yz - 2)/l = (y*(xy - 2)/k - 2)/lThis seems complicated. Let's attempt to express variables in terms.Alternatively, multiply all three equations:(xy -2)(yz -2)(zx -2) = k l m z x yBut this seems too vague.Alternatively, note that from each equation, the variable is expressed in terms of the other two. Maybe we can find ratios or relations.Alternatively, suppose that all variables are pairwise coprime? Not necessarily. For example, in (3,8,22), gcd(3,8)=1, gcd(8,22)=2, so not coprime.Alternatively, see if the variables follow a particular sequence or are part of a number pattern.But given the time I've spent and the solutions found, perhaps these are the only solutions. To ensure completeness, let's check x=7, y=8, z=?First condition:7*8=56≡2 modz→z divides54. z=6,9,18,27,54.Check z=6:Second condition:8*6=48 mod7=48-42=6≡2? No.z=9:8*9=72 mod7=72-63=9≡2? No.z=18:8*18=144 mod7=144-140=4≡2? No.z=27:8*27=216 mod7=216-210=6≡2? No.z=54:8*54=432 mod7=432-427=5≡2? No.No solutions.Similarly, trying x=8, y=3:First condition:8*3=24≡2 modz→z divides22. z=11,22.Check z=11:Second condition:3*11=33 mod8=1≡2? No.z=22:3*22=66 mod8=66-64=2≡2. Good.Third condition:8*22=176 mod3=176-174=2≡2. Good.Wait, but this is the solution (8,3,22), which is a permutation of (3,8,22). So that works.Similarly, for x=8, y=22, z=3:First condition:8*22=176≡176 mod3=176-174=2. Correct.Second condition:22*3=66 mod8=66-64=2. Correct.Third condition:3*8=24 mod22=2. Correct.So permutations are all valid. Given that, and given that I can't find any more solutions beyond permutations of the ones already found, perhaps these are the only solutions.Therefore, the positive integer solutions are all permutations of (3,8,22), (3,10,14), (4,5,18), and (4,6,11).To confirm, let's check if there's a solution like (5, something, something).Wait, previously tried x=5, y=12:First condition:5*12=60≡2 modz→z=58. Second condition:12*58=696 mod5=1≡2? No. Doesn't work.x=5, y=7, z=11:From previous checks, third condition fails.x=5, y=13, z=... nothing works.So, yes, it seems these are all the solutions.Final AnswerThe positive integer solutions are all permutations of boxed{(3, 8, 22)}, boxed{(3, 10, 14)}, boxed{(4, 5, 18)}, and boxed{(4, 6, 11)}.</think>To solve the problem of finding all positive integers (x, y, z) such that (xy pmod{z} equiv yz pmod{x} equiv zx pmod{y} = 2), we need to satisfy the following conditions:1. (xy equiv 2 pmod{z})2. (yz equiv 2 pmod{x})3. (zx equiv 2 pmod{y})We start by noting that (x, y, z geq 3) since the remainder when divided by any number must be less than that number. We then systematically check small values and permutations to find solutions. Key Steps:1. Checking Small Values: - For (x = 3), we find solutions ((3, 8, 22)) and ((3, 10, 14)). - For (x = 4), we find solutions ((4, 5, 18)) and ((4, 6, 11)).2. Verifying Permutations: - Each solution found is checked for permutations. For example, ((3, 8, 22)) permutes to ((8, 22, 3)), ((22, 3, 8)), etc., all of which satisfy the conditions.3. Generalizing the Pattern: - Each product (xy, yz, zx) is exactly 2 more than a multiple of the third variable. This pattern is consistent across all found solutions. Solutions Found:- The solutions are all permutations of the following triples: - ((3, 8, 22)) - ((3, 10, 14)) - ((4, 5, 18)) - ((4, 6, 11)) Final AnswerThe positive integer solutions are all permutations of boxed{(3, 8, 22)}, boxed{(3, 10, 14)}, boxed{(4, 5, 18)}, and boxed{(4, 6, 11)}.,<think>Okay, so I need to find all positive integers x, y, z such that each of the three expressions xy mod z, yz mod x, and zx mod y is equal to 2. Hmm. Let me start by recalling what the modulo operation means. When we say a mod b = c, it means that when a is divided by b, the remainder is c. So, in this case, each of these products (xy, yz, zx) when divided by the third variable (z, x, y respectively) leaves a remainder of 2. First, let me note that since the remainders are 2, the divisors (z, x, y) must be greater than 2. Because if, for example, z was 2, then xy mod 2 would have to be 2, but the possible remainders when dividing by 2 are 0 and 1. So, 2 can't be a remainder mod 2. Therefore, z, x, y must all be at least 3. That's a useful starting point.So, x, y, z ≥ 3.Now, let's write down the three congruences:1. xy ≡ 2 mod z2. yz ≡ 2 mod x3. zx ≡ 2 mod yEach of these congruences can be rewritten in terms of division. For example, the first one says that xy = kz + 2 for some integer k ≥ 0. Similarly for the others. So, we have:1. xy = kz + 22. yz = lx + 23. zx = my + 2Where k, l, m are non-negative integers. Since x, y, z are positive integers ≥3, and the remainders are 2, then k, l, m must also be positive integers? Wait, not necessarily. Let me check.Take the first equation: xy = kz + 2. Since x, y, z are at least 3, then xy is at least 9, so kz must be at least 7 (since 9 - 2 = 7). Therefore, k and z must satisfy that kz ≥7. Since z ≥3, k could be 0? But then kz would be 0, which can't be since xy = 0 + 2 = 2, but x and y are at least 3, so their product is at least 9. Therefore, k must be at least 1, and similarly for l and m. So k, l, m are positive integers.Therefore, we can say:1. xy = kz + 2, k ≥12. yz = lx + 2, l ≥13. zx = my + 2, m ≥1Now, perhaps I can manipulate these equations. Let's see. If I can express z from the first equation, x from the second, and y from the third, and substitute them into each other.From the first equation: z = (xy - 2)/kFrom the second equation: x = (yz - 2)/lFrom the third equation: y = (zx - 2)/mBut substituting these into each other might get complicated. Let me see. For instance, substitute z from the first equation into the second equation:x = (y * [(xy - 2)/k] - 2)/lThat's quite messy. Maybe another approach is needed.Alternatively, since all three variables are symmetric in the problem, maybe there are solutions where x = y = z? Let's check that possibility.Suppose x = y = z = n. Then the three conditions become:n*n mod n = 0, but we need it to be 2. But n*n mod n is 0, so that can't work. Therefore, x, y, z cannot all be equal.So, maybe two of them are equal? Let's say x = y, but z different. Then let's see:First, if x = y, then the conditions:xy mod z = x^2 mod z = 2yz mod x = xz mod x = 0 (since xz is divisible by x). But the problem states that yz mod x = 2. Therefore, 0 ≡ 2 mod x, which implies that x divides 2. But x ≥3, which would mean 3 divides 2, which is impossible. Therefore, x cannot equal y. Similarly, if any two variables are equal, then the modulus involving the third variable would result in 0 mod the first, which can't be 2. Therefore, x, y, z must all be distinct.Wait, hold on. Let me verify this. Suppose x = y. Then:xy mod z = x^2 mod z = 2yz mod x = xz mod x = 0, which should be 2. Therefore, 0 ≡ 2 mod x ⇒ x divides 2, but x ≥3, which is impossible. Similarly, if x = z, then:xy mod z = xy mod x = (x*y) mod x = (0) mod x = 0, but it should be 2. So 0 ≡ 2 mod x ⇒ x divides 2, which is again impossible. Same with y = z. Therefore, all variables must be distinct.Therefore, x, y, z are distinct integers ≥3.Now, since they are distinct, perhaps I can assume an order, say, x < y < z. Then try to find possible triples. But the problem is symmetric, so the solution could be in any order. However, modulo conditions might not be symmetric. Let's see.Alternatively, maybe we can look for small values. Since x, y, z ≥3, maybe start with x=3, then see possible y and z.Let me try x=3. Then, from the second equation: yz ≡2 mod 3. So yz ≡2 mod3. Also, from the third equation: z*3 ≡2 mod y ⇒ 3z ≡2 mod y. And from the first equation: 3y ≡2 mod z.So let's write down:1. 3y ≡2 mod z ⇒ 3y = kz + 2, k≥12. yz ≡2 mod3 ⇒ yz ≡2 mod33. 3z ≡2 mod y ⇒ 3z = my + 2, m≥1Let me see if I can find integers y, z ≥3 satisfying these.First, since x=3, let's consider equation 2: yz ≡2 mod3. Since 3 is prime, this implies that y and z cannot both be multiples of 3. So, possible cases:Either y ≡1 mod3 and z ≡2 mod3, or y≡2 mod3 and z≡1 mod3, or y≡0 mod3 and z≡2/y≡2 mod3 and z≡0 mod3. Wait, but if y≡0 mod3, then z must satisfy 0*z ≡2 mod3 ⇒ 0≡2 mod3, which is impossible. Similarly if z≡0 mod3, then y*0 ≡2 mod3 ⇒0≡2 mod3, impossible. Therefore, neither y nor z can be multiples of3. Therefore, y and z are each either 1 or 2 mod3. Then, possible combinations:Case 1: y ≡1 mod3, z≡2 mod3Case 2: y≡2 mod3, z≡1 mod3So let's take case 1: y ≡1 mod3, z≡2 mod3.From equation1: 3y ≡2 mod z ⇒3y -2 is divisible by z. Since z≥3, and z≡2 mod3.From equation3: 3z ≡2 mod y ⇒3z -2 is divisible by y. Since y≥3, and y≡1 mod3.So, let me try small y and z. Let's suppose y is 4, since y must be at least3, but y can't be 3 (since z cannot be a multiple of3). So next is y=4 (which is 1 mod3: 4≡1 mod3). Let's see if y=4 works.If y=4, then from equation2: yz≡2 mod3 ⇒4*z≡2 mod3 ⇒ (since 4≡1 mod3) ⇒ z≡2 mod3. So z≡2 mod3. So possible z=5,8,11,... but z≥3. Let's try z=5.Check equation1: 3y =3*4=12. 12≡2 mod z=5. 12 mod5=2. Yes, that works. So equation1 is satisfied.Check equation3:3z=3*5=15. 15≡2 mod y=4. 15 mod4=3. Not 2. So that fails.So z=5 doesn't work. Next z=8.Check equation1: 3*4=12. 12≡2 mod8. 12 mod8=4≠2. So that fails.Wait, equation1:3y=12≡2 mod z. So 12-2=10 must be divisible by z. So z divides10. Since z≥3, z could be 5 or10. But z≡2 mod3. z=5 is 2 mod3 (5-2=3). z=10 is 1 mod3 (10-9=1). Therefore, only z=5 is possible here. But we saw z=5 gives equation3 failing. So with y=4, z=5 is the only candidate, but it doesn't work.Next y=7 (which is 1 mod3). Then z≡2 mod3.From equation1:3y=21. 21≡2 mod z ⇒21-2=19 must be divisible by z. So z divides19. 19 is prime. So z=19. But z≡2 mod3. 19 mod3=1. So no, z=19≡1 mod3. Not suitable. Thus no solution here.Next y=10 (1 mod3). Then z≡2 mod3.Equation1:3*10=30≡2 mod z ⇒30-2=28 divisible by z. So z divides28, z≥3, z≡2 mod3. Divisors of28:4,7,14,28. Check which are ≡2 mod3:4≡1,7≡1,14≡2,28≡1. So z=14. Let's check.z=14. Check equation2: y*z=10*14=140. 140 mod3. 140/3=46*3 + 2. So 140≡2 mod3. Correct. Equation3:3z=42. 42≡2 mod y=10. 42 mod10=2. Yes! 42=4*10 +2. So equation3 is satisfied. So (x,y,z)=(3,10,14). Let's check all three conditions:1. xy=3*10=30. 30 mod14=2. Correct.2. yz=10*14=140. 140 mod3=140-46*3=140-138=2. Correct.3. zx=14*3=42. 42 mod10=2. Correct.Wow, that works! So (3,10,14) is a solution. Are there others with x=3?Let's try next z=14 and y=10, but that's already found. Let's check another y. Let's go back.y=4 didn't work with z=5. y=7 didn't work. y=10 worked with z=14. Maybe y=13 (1 mod3). Then z≡2 mod3.Equation1:3*13=39. 39≡2 mod z ⇒ z divides37. 37 is prime. z=37. Check z≡2 mod3:37 mod3=1. Not 2. Doesn't work.Similarly, y=16 (1 mod3). Then z≡2 mod3. Equation1:3*16=48≡2 mod z ⇒ z divides46. Divisors of46≥3:23,46. z≡2 mod3. 23 mod3=2, 46 mod3=1. So z=23.Check equation2: y*z=16*23=368. 368 mod3: 3+6+8=17→1+7=8≡2 mod3. Correct.Check equation3:3z=69≡2 mod y=16. 69 divided by16 is4*16=64, remainder5. 5≠2. So no good. Therefore, z=23 doesn't work.Hmm. So maybe (3,10,14) is the only solution with x=3. Let's check y=10, z=14.Alternatively, maybe there are other solutions with x=3 if we consider different y's. Let's check another approach.Alternatively, let's take equation3:3z ≡2 mod y. So 3z = my +2. So z=(my +2)/3. Since z must be integer, (my +2) must be divisible by3. So my ≡1 mod3. Since m is a positive integer. So m ≡ y^{-1} mod3. But y is ≡1 or2 mod3. If y≡1 mod3, then m≡1 mod3. If y≡2 mod3, then m≡2 mod3 (since 2*2=4≡1 mod3). So depending on y's residue, m has to be a certain residue.But maybe this is getting too abstract. Let's try x=3 and another y. Let's try y=7 (1 mod3). Then z=(m*7 +2)/3. To make z integer, m*7 +2 must be divisible by3. 7≡1 mod3, so m*1 +2 ≡0 mod3 ⇒ m≡1 mod3. So m=1,4,7,...Take m=1: z=(7 +2)/3=9/3=3. But z must be at least3. So z=3. But then z=3, which would mean yz=7*3=21. 21 mod x=3 is 0, which should be2. So invalid.m=4: z=(4*7 +2)/3=(28 +2)/3=30/3=10. z=10. Check equation2: yz=7*10=70. 70 mod3=1. But should be2. Not valid.m=7: z=(7*7 +2)/3=51/3=17. Check equation2:7*17=119. 119 mod3: 1+1+9=11→1+1=2→2 mod3. Correct. Equation1:3y=21. 21 mod z=17=21-17=4≠2. So invalid.Hmm. So no luck here. So perhaps only y=10 gives a valid solution for x=3.Alternatively, let's try x=4. Let's see if that can give a solution.If x=4, then the equations become:1.4y ≡2 mod z2.yz ≡2 mod43.4z ≡2 mod yStarting with equation2:yz ≡2 mod4. Since y and z are integers ≥3. So yz mod4=2.Possible cases for y and z mod4. Let's see:yz ≡2 mod4.Possible pairs (y mod4, z mod4):(1,2), (2,1), (3,2), (2,3), (2,2) would give 2 mod4.But let's check:1*2=2 mod42*1=2 mod43*2=6≡2 mod42*3=6≡2 mod42*2=4≡0 mod4So possible residues:Either y≡1 mod4 and z≡2 mod4,y≡2 mod4 and z≡1 mod4,y≡3 mod4 and z≡2 mod4,y≡2 mod4 and z≡3 mod4.So four possible cases.Let me try y=3 (3 mod4). Then z≡2 mod4.From equation1:4y=12≡2 mod z. So 12-2=10 must be divisible by z. So z divides10. z≥3, z≡2 mod4. Possible z=10, but 10≡2 mod4. z=10. Check equation3:4z=40≡2 mod y=3. 40 mod3=1. Not 2. Doesn't work.Next, z=2 mod4. z=6. But 6 divides10? 10 divided by6 is1 with remainder4, so no. So z must divide10 and be ≡2 mod4. 10 is possible, but as above, doesn't work.Alternatively, take y=5 (1 mod4). Then z≡2 mod4.From equation1:4*5=20≡2 mod z. So 20-2=18 must be divisible by z. So z divides18, z≥3, z≡2 mod4. Possible z=6, 18. 6≡2 mod4? 6 mod4=2. Yes. z=6 and z=18.Check z=6. Equation3:4*6=24≡2 mod y=5. 24 mod5=4≠2. Doesn't work.z=18. Equation3:4*18=72≡2 mod5. 72 mod5=2. Yes! So 72=14*5 +2. Correct. Now check equation2:y*z=5*18=90. 90 mod4=2. Correct. So (x,y,z)=(4,5,18). Check all conditions:1.4*5=20. 20 mod18=2. Correct.2.5*18=90. 90 mod4=2. Correct.3.4*18=72. 72 mod5=2. Correct.So (4,5,18) is another solution. Nice.Let me check another case. Suppose y=2 mod4 and z=1 mod4.Take y=6 (2 mod4), z=5 (1 mod4). Check equation1:4*6=24≡2 mod5. 24 mod5=4≠2. Nope.Take y=6, z=9 (1 mod4). Equation1:24≡2 mod9. 24 mod9=6≠2.Take y=10 (2 mod4), z=1 mod4. Let's see z=13. Equation1:4*10=40≡2 mod13. 40 mod13=1≡2? No.z=5: Equation1:40≡2 mod5. 40 mod5=0≡2? No.Alternatively, z=9: 40 mod9=4≡2? No.Not working. Maybe another pair.Take y=2 mod4, z=3 mod4. Wait, no, from equation2, if y≡2 mod4, z≡3 mod4, then y*z≡2*3=6≡2 mod4. So that's valid.So another case is y≡2 mod4, z≡3 mod4. For example, y=6 (2 mod4), z=7 (3 mod4).Check equation1:4*6=24≡2 mod7. 24 mod7=3≠2. Nope.y=10 (2 mod4), z=7 (3 mod4). Equation1:4*10=40≡2 mod7. 40 mod7=5≠2.z=11 (3 mod4). 40≡2 mod11. 40 mod11=7≠2.Hmm.Alternatively, y=2 mod4, z=3 mod4. Let me pick y=14 (2 mod4), z=3 mod4. Let's take z=7.Equation1:4*14=56≡2 mod7. 56 mod7=0≡2? No.z=11: 56≡2 mod11. 56 mod11=1≡2? No.z=15:56 mod15=11≡2? No.Not working. Maybe this case doesn't yield solutions.Alternatively, take y=3 mod4 and z=2 mod4. Let's try y=7 (3 mod4), z=6 (2 mod4).Equation1:4*7=28≡2 mod6. 28 mod6=4≡2? No.y=7, z=10 (2 mod4). 28≡2 mod10. 28 mod10=8≡2? No.y=11 (3 mod4), z=6. 4*11=44≡2 mod6. 44 mod6=2. Yes! So equation1 is satisfied.Now equation3:4z=24≡2 mod y=11. 24 mod11=2. Correct. So (4,11,6). Wait, but z=6 is less than y=11. But in our initial assumption, x=4, but we didn't assume any order. Let's check all conditions:1. xy=4*11=44. 44 mod6=44-7*6=44-42=2. Correct.2. yz=11*6=66. 66 mod4=2. Correct.3. zx=6*4=24. 24 mod11=2. Correct.So (4,11,6) is also a solution. But since the problem didn't specify an order, but x,y,z are positive integers, this is valid. However, z=6, which is less than y=11 and x=4. So permutations might also be solutions.Wait, but in this case, if we consider (x,y,z)=(4,11,6), but z=6 is less than x=4? Wait, no, z=6 is greater than x=4. x=4, y=11, z=6. Wait, but z=6 is less than y=11. Anyway, the variables don't need to be ordered. So this is a valid solution. However, let's check if (4,6,11) is also a solution. Let's permute.If (x,y,z)=(4,6,11):1.xy=24 mod11=24-2*11=2. Correct.2.yz=66 mod4=2. Correct.3.zx=44 mod6=44-7*6=2. Correct. So (4,6,11) is also a solution. So depending on the ordering, these are different triples. But since the problem asks for all positive integers x,y,z, so all permutations would be considered distinct if the variables are different. Wait, but maybe some of them are the same solution under permutation. However, since the problem states "all positive integers x,y,z", and not "unordered triples", so we need to list all permutations as separate solutions. But let's confirm.Wait, for instance, (3,10,14) and (3,14,10): would both be solutions? Let's check.Take (3,14,10):1.xy=3*14=42 mod10=2. Correct.2.yz=14*10=140 mod3=2. Correct.3.zx=10*3=30 mod14=2. Correct. So yes, (3,14,10) is also a solution.Similarly, (10,3,14):1.xy=10*3=30 mod14=2. Correct.2.yz=3*14=42 mod10=2. Correct.3.zx=14*10=140 mod3=2. Correct.So all permutations are solutions. Therefore, the solutions are all permutations of (3,10,14), (4,5,18), and (4,6,11). Wait, but (4,11,6) is same as (4,6,11). Similarly, (4,5,18) and permutations. Let's check whether there are other solutions with x=4.For example, in equation2, when x=4, y=5, z=18. Are there other y and z for x=4?Earlier, we found (4,5,18) and (4,11,6). Let's see if there are more.Take y=9 (1 mod4). Then z≡2 mod4.Equation1:4*9=36≡2 mod z. So z divides34 (36-2=34). z≥3, z≡2 mod4. Divisors of34:2,17. z=34 (but 34≡2 mod4). Wait, 34 is a divisor? 34 divides34. z=34. Check equation3:4z=136≡2 mod y=9. 136 mod9:1+3+6=10→1+0=1≡1≠2. Doesn't work.z=17: 17≡1 mod4. Not required here. So z=34, but equation3 fails. So no.Another y=13 (1 mod4). z≡2 mod4. Equation1:4*13=52≡2 mod z ⇒ z divides50 (52-2=50). z≥3, z≡2 mod4. Divisors of50:2,5,10,25,50. z=10,25,50. z=10≡2 mod4. Let's check z=10.Equation3:4*10=40≡2 mod13. 40 mod13=1≡2? No. z=25: 25≡1 mod4. Not required. z=50≡2 mod4. Equation3:4*50=200≡2 mod13. 200/13=15*13=195, remainder5. 5≠2. Not good.No solution here.Alternatively, y=2 mod4. Let's take y=6 (2 mod4). Then z≡3 mod4 or z≡1 mod4. Wait, from equation2: yz≡2 mod4. If y≡2 mod4, then z≡1 or3 mod4. Let's take z=7 (3 mod4). Equation1:4*6=24≡2 mod7. 24 mod7=3≠2. Nope.z=5 (1 mod4). Equation1:24≡2 mod5. 24 mod5=4≠2. Nope.Similarly, y=10 (2 mod4). z≡3 mod4: z=7. Equation1:4*10=40≡2 mod7. 40 mod7=5≡2? No.z=11 (3 mod4). Equation1:40≡2 mod11. 40 mod11=7≡2? No.Not working. So maybe only (4,5,18) and (4,11,6) are solutions for x=4.Now, let's try x=5.Equations become:1.5y ≡2 mod z2.yz ≡2 mod53.5z ≡2 mod yStarting with equation2: yz ≡2 mod5. So possible cases:Either y≡1 mod5 and z≡2 mod5,y≡2 mod5 and z≡1 mod5,y≡3 mod5 and z≡4 mod5,y≡4 mod5 and z≡3 mod5,or y≡0 mod5 and z≡2/y≡2 mod5 and z≡0 mod5. But similar to earlier reasoning, if y≡0 mod5, then z must satisfy 0*z≡2 mod5⇒0≡2 mod5, impossible. Same with z≡0 mod5. So y and z cannot be multiples of5.Therefore, possible residues for y and z mod5:(1,2),(2,1),(3,4),(4,3).Let's pick y=6 (1 mod5). Then z≡2 mod5.Equation1:5*6=30≡2 mod z. So 30-2=28 must be divisible by z. So z divides28, z≥3, z≡2 mod5. Divisors of28:4,7,14,28. Which are ≡2 mod5:7≡2 mod5 (7-5=2). 14≡4 mod5, 28≡3 mod5. So z=7.Check equation3:5z=35≡2 mod y=6. 35 mod6=5≡2? No. Doesn't work.Next, y=11 (1 mod5). z≡2 mod5. z=7 (already checked). z=12 (2 mod5? 12≡2 mod5). Equation1:5*11=55≡2 mod12. 55 mod12=7≡2? No.z=17 (2 mod5). 55≡2 mod17. 55-3*17=55-51=4≡2? No.Not working.Another case: y=2 mod5. Let's take y=7 (2 mod5). Then z≡1 mod5.Equation1:5*7=35≡2 mod z. So z divides33 (35-2=33). z≥3, z≡1 mod5. Divisors of33:3,11,33. Check which are≡1 mod5:11≡1 mod5 (11-10=1). So z=11.Check equation3:5z=55≡2 mod y=7. 55 mod7=6≡2? No.Next, z=33. Equation3:5*33=165≡2 mod7. 165 mod7=165-23*7=165-161=4≡2? No.No good.Next case: y=3 mod5, z≡4 mod5.Take y=8 (3 mod5). z≡4 mod5.Equation1:5*8=40≡2 mod z. So z divides38 (40-2=38). z≥3, z≡4 mod5. Divisors of38:2,19. z=19 (19≡4 mod5). Check equation3:5*19=95≡2 mod8. 95 mod8=15*8=120, 95-120=-25≡7 mod8≠2. Doesn't work.Next y=13 (3 mod5). z≡4 mod5. Equation1:5*13=65≡2 mod z. So z divides63 (65-2=63). z≥3, z≡4 mod5. Divisors of63:3,7,9,21,63. Which are ≡4 mod5:9≡4 mod5, 21≡1 mod5, 63≡3 mod5. So z=9.Check equation3:5*9=45≡2 mod13. 45 mod13=6≡2? No.Not working.Last case: y≡4 mod5, z≡3 mod5.Take y=9 (4 mod5). z≡3 mod5.Equation1:5*9=45≡2 mod z. So z divides43 (45-2=43). Prime number. z=43≡3 mod5 (43-40=3). Check equation3:5*43=215≡2 mod9. 215 mod9:2+1+5=8≡8≠2. Nope.Another y=14 (4 mod5). z≡3 mod5. Equation1:5*14=70≡2 mod z. So z divides68. Divisors of68≥3:4,17,34,68. z≡3 mod5. 17≡2 mod5, 34≡4 mod5, 68≡3 mod5. So z=68.Check equation3:5*68=340≡2 mod14. 340 mod14=340-24*14=340-336=4≡2? No.No luck here. So maybe x=5 doesn't give any solutions.Trying x=6.Equations:1.6y ≡2 mod z2.yz ≡2 mod63.6z ≡2 mod yStarting with equation2: yz ≡2 mod6. So possible combinations where yz ≡2 mod6. Let's factor 6 into 2 and3. Then yz ≡2 mod2 and yz≡2 mod3.yz mod2=0 if either y or z even, 1 if both odd. But 2 mod2=0. Therefore, yz≡0 mod2. But 2 mod2=0. So equation2 implies yz≡0 mod2 and yz≡2 mod3. Therefore, yz must be even (so at least one of y or z even) and yz≡2 mod3. So yz≡2 mod3.Therefore, since yz≡2 mod3, similar to before, neither y nor z can be multiples of3. So y,z ≡1 or2 mod3. And their product ≡2 mod3. Which requires either y≡1 and z≡2, or y≡2 and z≡1 mod3.Additionally, yz must be even.So possible cases:Case1: y even, z≡2 mod3.Case2: z even, y≡2 mod3.Case3: y even and z≡1 mod3.But yz≡2 mod3 requires:If y even (so y can be 0,2,4,... mod3). If y is even and ≡1 mod3, then z must≡2 mod3.If y even and ≡2 mod3, then z≡1 mod3.Similarly, if z even and y≡2 mod3.This is getting complex. Let's pick x=6 and try some y.Take y=4 (even, and 4≡1 mod3). Then z≡2 mod3.From equation1:6*4=24≡2 mod z. So z divides22 (24-2=22). z≥3, z≡2 mod3. Divisors of22:2,11. z=11 (11≡2 mod3). Check equation3:6*11=66≡2 mod4. 66 mod4=2. Yes. So (x,y,z)=(6,4,11). Check all conditions:1.6*4=24 mod11=24-2*11=2. Correct.2.4*11=44 mod6=44-7*6=44-42=2. Correct.3.6*11=66 mod4=2. Correct.So (6,4,11) is a solution. Since variables can be permuted, permutations like (6,11,4) also work. Let's check:(6,11,4):1.6*11=66 mod4=2. Correct.2.11*4=44 mod6=2. Correct.3.4*6=24 mod11=2. Correct. Yes.Another possibility with x=6. Take y=5 (which is 2 mod3, but y is odd). Then z needs to be≡1 mod3. From equation2: yz=5z≡2 mod6. Since5z≡2 mod6⇒5z≡2 mod6⇒Multiply both sides by5^{-1} mod6. Since5*5=25≡1 mod6, inverse is5. So z≡5*2=10≡4 mod6. So z≡4 mod6. Also, z≡1 mod3. Let's see:z≡4 mod6 implies z≡1 mod3 (since4 mod3=1). So z=6k+4. Take z=10 (k=1). Check equation1:6*5=30≡2 mod10. 30 mod10=0≡2? No.z=16 (6*2+4=16). Equation1:6*5=30≡2 mod16. 30-16=14≡14≡2 mod16? No. 30 mod16=14≠2.z=22:6*5=30≡2 mod22. 30 mod22=8≡2? No.Not working. So y=5 doesn't work.Take y=7 (1 mod3). If y=7, which is odd, then z must be even (from equation2:yz even) and z≡2 mod3. So z≡2 mod3 and even. Therefore, z≡2 mod6. Let's take z=8 (≡2 mod6). Check equation2:7*8=56≡2 mod6. 56 mod6=2. Correct.Check equation1:6*7=42≡2 mod8. 42 mod8=42-5*8=42-40=2. Correct.Check equation3:6*8=48≡2 mod7. 48 mod7=6≡2? No.So z=8: equation3 fails. Next z=14 (≡2 mod6). Equation1:6*7=42≡2 mod14. 42 mod14=0≡2? No.z=20:6*7=42≡2 mod20. 42 mod20=2. Correct. Check equation3:6*20=120≡2 mod7. 120 mod7=120-17*7=120-119=1≡2? No.z=26:6*7=42≡2 mod26. 42 mod26=16≡2? No.Hmm.Alternatively, take y=8 (even, 8≡2 mod3). Then z≡1 mod3. Equation2: yz=8z≡2 mod6. 8z≡2 mod6 ⇒8z≡2 mod6 ⇒2z≡2 mod6 (divided both sides by2) ⇒z≡1 mod3. Which aligns with our earlier requirement. So z≡1 mod3. Now, equation1:6*8=48≡2 mod z. So z divides46 (48-2=46). z≥3, z≡1 mod3. Divisors of46:2,23. z=23≡2 mod3. Not 1 mod3. So no solution.Another y=10 (even,10≡1 mod3). Then z≡2 mod3. Equation1:6*10=60≡2 mod z. So z divides58 (60-2=58). z≥3, z≡2 mod3. Divisors of58:2,29. z=29≡2 mod3. Check equation3:6*29=174≡2 mod10. 174 mod10=4≡2? No.No good. So (6,4,11) is the only solution for x=6.Continuing this process for x=7,8,... but this might take a long time. Let me see if there's a pattern or a way to generalize.Looking at the solutions found so far:(3,10,14), (3,14,10), (10,3,14), etc. (all permutations)(4,5,18), permutations(4,6,11), permutations(6,4,11), permutationsSo each solution is a triple where each pair multiplies to 2 mod the third. Let's see if these solutions follow a particular structure.Looking at (3,10,14):3*10=30≡2 mod14 (30-2*14=2)10*14=140≡2 mod3 (140-46*3=2)14*3=42≡2 mod10 (42-4*10=2)Similarly, (4,5,18):4*5=20≡2 mod18 (20-1*18=2)5*18=90≡2 mod4 (90-22*4=2)18*4=72≡2 mod5 (72-14*5=2)And (4,6,11):4*6=24≡2 mod11 (24-2*11=2)6*11=66≡2 mod4 (66-16*4=2)11*4=44≡2 mod6 (44-7*6=2)Notice that in each case, the product of two numbers is exactly 2 more than a multiple of the third. So, for example, in (3,10,14), 3*10 = 2 + 2*14 ⇒30=2+28. Similarly, 10*14=140=2 +46*3 ⇒140=2+138, and 14*3=42=2 +4*10 ⇒42=2+40.So each product is equal to 2 plus a multiple of the third variable. That seems to be a necessary condition. Let's check for another solution. Take (4,5,18):4*5=20=2 +1*185*18=90=2 +22*418*4=72=2 +14*5Similarly, (6,4,11):6*4=24=2 +2*114*11=44=2 +7*611*6=66=2 +16*4So each time, the multiplier k in the first equation is equal to the smaller number in the next equation. For example, in (3,10,14), 3*10=2 +2*14, where the multiplier is 2, and the next equation is 10*14=2 +46*3, where the multiplier is 46. Not sure if there's a relation here.Alternatively, maybe these solutions are part of a family where each variable is 2 more than a multiple of another variable. For example, in (3,10,14):14 = 4*3 + 210 = 3*3 + 1 (not sure)Alternatively, 10 = 14 - 4, but not sure.Alternatively, the numbers are part of a chain where each is related to the next by some formula.Alternatively, these solutions can be generated by a certain recursive method. For example, starting with 3,10,14:If we take 3, then 10 is such that 10 ≡ (2/3) mod14, but this is abstract.Alternatively, notice that in (3,10,14):From 3*10 ≡2 mod14 ⇒30 ≡2 mod14 ⇒30-2=28 which is 2*14. So 3*10=2 +2*14.Similarly, 10*14=140=2 +46*3, so 10*14=2 +46*3.And 14*3=42=2 +4*10 ⇒42=2 +40.So, the multipliers here are 2,46,4. Not sure.Alternatively, perhaps these triples are part of a Pythagorean-like triplet but for modulo conditions. It's unclear.Alternatively, maybe all solutions are the permutations of the triples we found: (3,10,14), (4,5,18), (4,6,11), and (6,4,11). Wait, but (6,4,11) is a permutation of (4,6,11).Alternatively, maybe there are larger solutions. Let's check x=5 again.Wait, when x=5, we tried several y and z without success, but maybe there's a solution with larger numbers.Alternatively, think of the equations as:From the three equations:xy = kz +2yz = lx +2zx = my +2We can multiply all three equations together:(xy)(yz)(zx) = (kz +2)(lx +2)(my +2)But this might not be helpful. Alternatively, adding the equations:But this seems messy.Alternatively, divide the first equation by z: (xy -2)/z =kSimilarly, (yz -2)/x =lAnd (zx -2)/y =mSo, k = (xy -2)/zl = (yz -2)/xm = (zx -2)/yTherefore, k,l,m must be positive integers.If we multiply all three: klm = [(xy -2)(yz -2)(zx -2)]/(xyz)But not sure.Alternatively, perhaps take the ratios. For example:From the first and second equation:xy -2 =kzyz -2 =lxDivide these two: (xy -2)/(yz -2)=kz/(lx)Not helpful.Alternatively, consider the case where k=l=m=2. Let's see:If k=2, then xy=2z +2Similarly, l=2: yz=2x +2m=2: zx=2y +2So we have the system:1.xy=2z +22.yz=2x +23.zx=2y +2Let's try solving this system.From equation1: z=(xy -2)/2From equation2: yz=2x +2 ⇒ y*(xy -2)/2=2x +2 ⇒ y(xy -2)=4x +4Expand: x y^2 -2y =4x +4 ⇒ x y^2 -4x =2y +4 ⇒x(y^2 -4)=2(y +2)Thus, x=2(y +2)/(y^2 -4)=2(y +2)/[(y -2)(y +2)]=2/(y -2)Since x must be a positive integer, 2 must be divisible by(y -2). Therefore, y -2 divides2.Possible y -2=1,2. Thus y=3,4.If y=3: x=2/(3 -2)=2. But x must be ≥3. Disregarded.If y=4: x=2/(4 -2)=1. But x≥3. Disregarded.Therefore, no solutions with k=l=m=2.Alternatively, suppose k=1. Let's take k=1 in the first equation: xy =z +2. Then z=xy -2.Substitute into the second equation: y z= l x +2 ⇒y(xy -2)=l x +2 ⇒x y^2 -2y =l x +2.Then, l=(x y^2 -2y -2)/x= y^2 - (2y +2)/x.Since l must be integer, (2y +2) must be divisible byx.Similarly, from the third equation: z x=my +2 ⇒(xy -2)x=my +2 ⇒x^2 y -2x=my +2 ⇒my= x^2 y -2x -2 ⇒m=(x^2 y -2x -2)/y= x^2 - (2x +2)/y.So (2x +2) must be divisible byy.Therefore, x divides(2y +2) and y divides(2x +2).Given that x,y ≥3.Let me take x=3.Then, from above:x=3:From z=3y -2.From second equation: y z= l*3 +2. z=3y -2. So y*(3y -2)=3l +2 ⇒3y^2 -2y -2=3l. Thus l=(3y^2 -2y -2)/3.Which needs to be integer. Therefore, 3 divides(3y^2 -2y -2).So 3y^2 -2y -2 ≡0 mod3 ⇒0 -2y -2 ≡0 mod3 ⇒-2y -2 ≡0 mod3 ⇒2y +2≡0 mod3 ⇒2y≡-2≡1 mod3 ⇒y≡2^{-1}*1≡2*1=2 mod3.So y≡2 mod3. Since y≥3, possible y=5,8,11,...From third equation: m=(3^2 y -2*3 -2)/y=(9y -6 -2)/y=(9y -8)/y=9 -8/y.Thus 8/y must be integer. Therefore, y divides8. But y≥3 and y≡2 mod3. Divisors of8≥3:4,8. Check if 4≡2 mod3? 4≡1 mod3. 8≡2 mod3. So y=8.Check y=8:Then z=3*8 -2=24-2=22.Check second equation: l=(3*8^2 -2*8 -2)/3=(192 -16 -2)/3=174/3=58. Integer.Check third equation: m=9 -8/8=9 -1=8. So m=8.Therefore, (x,y,z)=(3,8,22). Check all conditions:1.xy=24≡2 mod22. 24 mod22=2. Correct.2.yz=8*22=176≡2 mod3. 176 mod3=176-58*3=176-174=2. Correct.3.zx=22*3=66≡2 mod8. 66 mod8=2. Correct.So (3,8,22) is a solution. But wait, didn't we miss this earlier?Wait, when we tried x=3, y=4,5,7,10, etc., but didn't try y=8. Let me see:From earlier, when x=3, we tried y=4,7,10, etc., but didn't check y=8. But according to this approach, with k=1, we get a solution. So (3,8,22) is another solution.Similarly, permutations of this triple would also be solutions. Let's check:(3,22,8):1.3*22=66≡2 mod8. 66 mod8=2. Correct.2.22*8=176≡2 mod3. Correct.3.8*3=24≡2 mod22. Correct.Yes.So, this suggests there are more solutions than initially found. Therefore, maybe there's an infinite family of solutions.Wait, how was this solution found? By setting k=1 in the first equation, leading to z=xy -2, then enforcing divisibility conditions on l and m. Similarly, if we set k=1, l=1, or m=1, perhaps we can find more solutions.Alternatively, take k=1, and other variables.For example, let's take x=4.From z=xy -2.From second equation: yz= l x +2 ⇒y(xy -2)=4l +2 ⇒x y^2 -2y =4l +2 ⇒4l= x y^2 -2y -2 ⇒l=(x y^2 -2y -2)/4.From third equation: z x=my +2 ⇒(xy -2)x=my +2 ⇒x^2 y -2x=my +2 ⇒my= x^2 y -2x -2 ⇒m=(x^2 y -2x -2)/y= x^2 - (2x +2)/y.Therefore, (2x +2) must be divisible by y.So for x=4:l=(4 y^2 -2y -2)/4 must be integer. So 4 y^2 -2y -2 must be divisible by4.Which implies that -2y -2 ≡0 mod4 ⇒-2y ≡2 mod4 ⇒2y ≡-2≡2 mod4 ⇒y≡1 mod2. So y is odd.Also, from m=(16 y -8 -2)/y=16 -10/y. So y must divide10. Since y is odd, y divides10 and is odd: y=5.Check y=5:z=4*5 -2=20-2=18.Check l=(4*25 -10 -2)/4=(100 -12)/4=88/4=22. Integer.Check m=16 -10/5=16-2=14. Integer.Thus, (x,y,z)=(4,5,18). Which we found earlier. So this approach recovers that solution.Similarly, if we take x=5:k=1 ⇒z=5y -2.Then from second equation: y z= l*5 +2 ⇒y(5y -2)=5l +2 ⇒5y² -2y -2=5l ⇒l=(5y² -2y -2)/5.For l to be integer, 5 divides(5y² -2y -2). Therefore:5y² -2y -2 ≡0 mod5 ⇒0 -2y -2 ≡0 mod5 ⇒-2y ≡2 mod5 ⇒2y≡-2≡3 mod5 ⇒y≡3*2^{-1}≡3*3≡9≡4 mod5.So y≡4 mod5. Thus y=5k +4.From third equation: m=(25 y -10 -2)/y=(25y -12)/y=25 -12/y. So y divides12. Since y≡4 mod5 and y≥3, possible y=4,9,14,... But y divides12. So y=4. But y≡4 mod5. 4≡4 mod5. y=4.Check y=4:z=5*4 -2=18.Check l=(5*16 -8 -2)/5=(80 -10)/5=70/5=14. Integer.Check m=25 -12/4=25-3=22. Integer.Thus, (x,y,z)=(5,4,18). Check all conditions:1.5*4=20≡2 mod18. Correct.2.4*18=72≡2 mod5. 72 mod5=2. Correct.3.18*5=90≡2 mod4. 90 mod4=2. Correct.Yes, another solution. But since variables can be permuted, this is a permutation of (4,5,18).So, this suggests that by setting k=1 in the first equation, we can generate solutions where z=xy -2, and y divides2x +2, and x divides2y +2. This seems to generate solutions like (3,8,22), (4,5,18), (5,4,18), etc.Similarly, if we set l=1 in the second equation, we might generate other solutions, and similarly for m=1 in the third equation.Let's try l=1 in the second equation: yz =x +2. Then x=yz -2.Substitute into first equation: xy =kz +2 ⇒(yz -2)y =kz +2 ⇒y² z -2y =kz +2 ⇒z(y² -k) =2y +2 ⇒z=(2y +2)/(y² -k).Since z must be a positive integer, denominator y² -k must divide2y +2 and be positive. So y² -k divides2y +2 and y² -k >0 ⇒k <y².Let’s take k=1 (smallest possible). Then z=(2y +2)/(y² -1). So denominator y² -1.z=(2(y +1))/((y -1)(y +1))=2/(y -1).Thus z=2/(y -1). Since z must be integer ≥3, denominator y -1 must divide2 and y -1≤2/3 (since z≥3 ⇒2/(y -1)≥3 ⇒y -1≤2/3). But y ≥3, so y -1≥2, which would mean 2/(y -1)≤1, contradicting z≥3. Therefore, no solution with l=1, k=1.Take k=2. Then z=(2y +2)/(y² -2). For z to be integer, y² -2 divides2y +2. Let's check small y:y=3: denominator=9-2=7. 2*3+2=8. 8/7 not integer.y=4: denominator=16-2=14. 2*4+2=10. 10/14=5/7. No.y=5: denominator=25-2=23. 2*5+2=12. 12/23 no.Not working.Similarly, k=3: z=(2y +2)/(y² -3).y=3: denominator=9-3=6. numerator=8. 8/6=4/3. No.y=4: denominator=13. numerator=10. 10/13. No.No solution. So maybe l=1 doesn't give solutions.Alternatively, set m=1 in the third equation: zx = y +2. Then y=zx -2.Substitute into second equation: yz = l x +2 ⇒(zx -2)z = l x +2 ⇒z² x -2z =l x +2 ⇒x(z² -l) =2z +2 ⇒x=(2z +2)/(z² -l).Take l=1: x=(2z +2)/(z² -1)=2(z +1)/[(z -1)(z +1)]=2/(z -1). Thus x=2/(z -1). Since x≥3, denominator z -1≤2/3⇒z -1≤0. Not possible.Similarly, no solution.Thus, perhaps the only way to generate solutions is by setting k=1 in the first equation and following through.Similarly, we can set k=2 in the first equation:If k=2, then xy=2z +2 ⇒z=(xy -2)/2.Substitute into second equation: yz= l x +2 ⇒y*(xy -2)/2= l x +2 ⇒(xy² -2y)/2= l x +2 ⇒xy² -2y=2 l x +4 ⇒xy² -2 l x =2y +4 ⇒x(y² -2 l)=2y +4.This is getting complicated. Perhaps this trial and error approach is not the best.Alternatively, noticing that in all solutions found, one of the variables is 3,4,5,6, etc., and others are derived accordingly. It might be that the only solutions are permutations of (3,10,14), (4,5,18), (4,6,11), (3,8,22), and their respective permutations.But how to confirm there are no others? Well, for x=3, we found (3,10,14) and (3,8,22). For x=4, (4,5,18) and (4,6,11). For x=5, (5,4,18) which is a permutation. For x=6, (6,4,11) permutation. For x=8, permutations of existing solutions.Continuing, let's try x=7 with k=1.From z=7y -2.From second equation: y z= l x +2 ⇒y(7y -2)=7l +2 ⇒7 y² -2y -2=7 l ⇒l=(7 y² -2y -2)/7.Which must be integer. So 7 divides7 y² -2y -2 ⇒7 divides(-2y -2) ⇒-2y -2≡0 mod7 ⇒2y ≡-2≡5 mod7 ⇒y≡5*4≡20≡6 mod7 (since 2^{-1}≡4 mod7). So y≡6 mod7. y≥3, possible y=6,13,20,...From third equation: m=(49 y -14 -2)/y=(49 y -16)/y=49 -16/y. So y must divide16. Since y≡6 mod7, possible y=6 (divides16? 16/6≈2.666), no. y=13:16/13≈1.23. y=20:16/20=0.8. No. So no solution here.Thus, no solution for x=7 with k=1.Similarly, x=8 with k=1:z=8y -2.From second equation: y z= l*8 +2 ⇒y(8y -2)=8l +2 ⇒8 y² -2y -2=8 l ⇒l=(8 y² -2y -2)/8.Which requires 8 divides(8 y² -2y -2) ⇒8 divides(-2y -2) ⇒-2y -2≡0 mod8 ⇒2y ≡-2≡6 mod8 ⇒y≡3 mod4.From third equation: m=(64 y -16 -2)/y=(64 y -18)/y=64 -18/y. So y divides18. y≥3, y≡3 mod4. Possible y=3,7,11,… which divide18. y=3: divides18 yes. y=3.Check y=3:z=8*3 -2=22.Check l=(8*9 -6 -2)/8=(72-8)/8=64/8=8. Integer.Check m=64 -18/3=64 -6=58. Integer.Thus, (x,y,z)=(8,3,22). Check all conditions:1.8*3=24≡2 mod22. Correct.2.3*22=66≡2 mod8. Correct.3.22*8=176≡2 mod3. Correct.Yes, this works. Which is a permutation of (3,8,22).Similarly, y=7: divides18? 18/7 not integer. y=9:9≡1 mod4. Doesn't fit.Thus, the only solution here is y=3, giving a permutation.Continuing this pattern, it seems that all solutions are permutations of the triples we found:- (3,10,14) and permutations- (3,8,22) and permutations- (4,5,18) and permutations- (4,6,11) and permutationsTherefore, the complete set of solutions are all permutations of these four triples.To ensure there are no other solutions, we can try x=9 with k=1:z=9y -2.From second equation: y z= l*9 +2 ⇒y(9y -2)=9l +2 ⇒9 y² -2y -2=9 l ⇒l=(9 y² -2y -2)/9.Requires 9 divides(9 y² -2y -2) ⇒9 divides(-2y -2) ⇒-2y -2≡0 mod9 ⇒2y≡-2≡7 mod9 ⇒y≡7*5≡35≡8 mod9 (since 2^{-1}≡5 mod9). So y≡8 mod9.From third equation: m=(81 y -18 -2)/y=(81 y -20)/y=81 -20/y. So y divides20. y≥3, y≡8 mod9. Possible y=8,17,… which divide20. y=8: divides20? 20/8=2.5. No. y=17: no. Thus, no solution.Thus, no solutions for x=9.Similarly, higher x would likely follow the same pattern, requiring y to satisfy conflicting divisibility conditions, leading to no solutions.Therefore, the complete set of solutions are all permutations of the triples:(3,8,22), (3,10,14), (4,5,18), and (4,6,11).Final AnswerThe positive integer solutions are all permutations of boxed{(3, 10, 14)}, boxed{(3, 8, 22)}, boxed{(4, 5, 18)}, and boxed{(4, 6, 11)}.</think>To find all positive integers (x, y, z) such that (xy pmod{z} equiv yz pmod{x} equiv zx pmod{y} = 2), we start by noting that each of (x, y, z) must be at least 3. We then analyze the congruences:1. (xy equiv 2 pmod{z})2. (yz equiv 2 pmod{x})3. (zx equiv 2 pmod{y})We rewrite these congruences as equations involving integer quotients:1. (xy = kz + 2)2. (yz = lx + 2)3. (zx = my + 2)By substituting and solving these equations, we find several solutions through trial and error, considering permutations and different values for (x, y, z). Key steps include checking small values for (x) and deriving possible (y) and (z) values, ensuring they satisfy all three congruences.Through detailed analysis and verification, we find the following solutions and their permutations:1. For (x = 3), we find solutions ((3, 10, 14)) and ((3, 8, 22)).2. For (x = 4), we find solutions ((4, 5, 18)) and ((4, 6, 11)).These solutions are verified to satisfy all three congruences. The permutations of these triples are also solutions due to the symmetric nature of the problem. Final AnswerThe positive integer solutions are all permutations of boxed{(3, 10, 14)}, boxed{(3, 8, 22)}, boxed{(4, 5, 18)}, and boxed{(4, 6, 11)}.]

🔑:11. Solution: From the conditions, we know that x|y z-2, z| x y-2, y mid x z-2,then x y z mid(y z-2)(x y-2)(x z-2),and (x y-2)(x z-2)(y z-2)=x^{2} y^{2} z^{2}-2left(x^{2} y z+x y^{2} z+x y z^{2}right)+4(x y+x z+y z)-8, so x y z mid 4(x y+x z+y z)-8.Let 4(x y+x z+y z)-8=k x y z, then 4left(frac{1}{x}+frac{1}{y}+frac{1}{z}right)=k+frac{8}{x y z}.From the conditions, we know that x, y, z>2, if two of x, y, z are equal. Without loss of generality, let x=y, then y z equiv 0(bmod x), which is a contradiction! Therefore, x, y, z are distinct.Without loss of generality, let x leqslant y leqslant z, then 4left(frac{1}{x}+frac{1}{y}+frac{1}{z}right) leqslant 4left(frac{1}{3}+frac{1}{4}+frac{1}{5}right)frac{4}{x y z}, thus frac{4}{x}+frac{4}{y}+frac{4}{z}>1+frac{8}{x y z}, which is a contradiction!If x geqslant 12, then4left(frac{1}{x}+frac{1}{y}+frac{1}{z}right) leqslant 4left(frac{1}{12}+frac{1}{13}+frac{1}{14}right)<1, which is a contradiction!Therefore, 5 leqslant x leqslant 11.(¡) x=5, solving yields (x, y, z)=(5,21,412),(5,22,216),(5,24,118),(5,28,69),(5,27,76),(5,34,48), none of which satisfy the requirements upon inspection.(ii) x=6, substituting yields (x, y, z)=(6,13,152),(6,14,82),(6,16,47),(6,17,40),(6,19,32),(6,22,26), upon inspection (6,14,82),(6,22,26) satisfy the requirements.(iii) x=7, solving yields (x, y, z)=(7,10,136),(7,11,60),(7,12,41),(7,16,22), none of which satisfy the requirements upon inspection.(iv) x=8, solving yields (x, y, z)=(8,9,70),(8,10,39), none of which satisfy the requirements upon inspection.Repeating the above process, it is easy to see that when x=9,10,11 none satisfy the requirements.In summary, the (x, y, z) that satisfy the conditions are (3,8,22),(3,10,14),(4,5,18),(4,6,11),(6,14,82),(6,22,26) and their symmetric arrays.

🧠:[<think>Okay, so I need to figure out how to transform the graph of y = sin(2x) into y = sin(2x - π/3). The question is asking whether this transformation is a shift left or right by π/3 or π/6 units. Let me recall how phase shifts work with trigonometric functions.First, the general form of a sine function with phase shift is y = sin(Bx - C) + D, where the phase shift is C/B. The phase shift tells us how much the graph is shifted horizontally. If the phase shift is positive, the graph shifts to the right; if it's negative, it shifts to the left.In this case, the original function is y = sin(2x), and we want to transform it to y = sin(2x - π/3). Let's compare this to the general form. Here, B is 2, and C is π/3. So the phase shift should be C/B, which is (π/3)/2 = π/6. Since it's in the form sin(Bx - C), the phase shift is to the right by π/6 units.Wait, let me make sure. Let me break it down step by step. Suppose we have y = sin(2x - π/3). Let me factor out the 2 from inside the sine function to make it clearer. So, sin(2x - π/3) can be written as sin[2(x - π/6)]. Because if you distribute the 2 inside, it becomes 2x - π/3. So, that's equivalent to shifting the graph of sin(2x) to the right by π/6 units. Alternatively, another way to think about it is, when you have y = sin(B(x - C)), then the shift is C units to the right. So here, after factoring out the 2, we have x - π/6, which means the shift is π/6 to the right. So that's answer D.But wait, sometimes I get confused whether it's left or right when the coefficient is inside the argument. Let me verify with a simple example. Suppose we have y = sin(x - π/3). That should be a shift to the right by π/3, right? Because replacing x with x - π/3 shifts the graph right. Similarly, if we have y = sin(2x - π/3), which is the same as sin[2(x - π/6)], so the shift is π/6 to the right.Yes, that makes sense. So the answer should be D: to the right by π/6 units.Alternatively, let's consider the order of transformations. The original function is sin(2x). If we want to shift it horizontally, we need to counteract the horizontal compression. Since the period of sin(2x) is π, compared to the regular 2π period. So, a horizontal shift would be scaled by 1/B, where B is the coefficient of x. Here, B is 2, so any horizontal shift is divided by 2. Therefore, if we have a phase shift of π/3, then the actual shift is π/3 divided by 2, which is π/6. Since it's subtracted inside the argument, it's a shift to the right.Another way to check: Let's consider a point on the original graph. For example, when x = 0, sin(2*0) = 0. So the point (0,0) is on the original graph. For the transformed function, when does 2x - π/3 equal 0? Solving 2x - π/3 = 0 gives x = π/6. So the point that was originally at x=0 is now at x=π/6. That means the graph has shifted to the right by π/6 units. Similarly, the maximum of sin(2x) occurs at x = π/4 (since sin(π/2) = 1). For the transformed function, set 2x - π/3 = π/2. Solving for x: 2x = π/2 + π/3 = (3π + 2π)/6 = 5π/6, so x = 5π/12. The original maximum at x=π/4 (which is 3π/12) is now at 5π/12, which is a shift of 5π/12 - 3π/12 = 2π/12 = π/6 to the right. So that confirms it.Therefore, the correct answer is D: to the right by π/6 units.Final Answerboxed{D}</think>To determine the horizontal shift required to transform the graph of ( y = sin(2x) ) into ( y = sin(2x - frac{pi}{3}) ), we need to analyze the phase shift. The general form of a sine function with a phase shift is ( y = sin(Bx - C) ), where the phase shift is given by ( frac{C}{B} ). For the function ( y = sin(2x - frac{pi}{3}) ), we identify ( B = 2 ) and ( C = frac{pi}{3} ). The phase shift is calculated as:[text{Phase shift} = frac{C}{B} = frac{frac{pi}{3}}{2} = frac{pi}{6}]Since the term inside the sine function is ( 2x - frac{pi}{3} ), which can be rewritten as ( 2left(x - frac{pi}{6}right) ), the graph of ( y = sin(2x) ) is shifted to the right by ( frac{pi}{6} ) units.To verify, consider a key point on the original graph. For example, the point where ( sin(2x) = 0 ) occurs at ( x = 0 ). For the transformed function, solving ( 2x - frac{pi}{3} = 0 ) gives ( x = frac{pi}{6} ), indicating a right shift of ( frac{pi}{6} ).Thus, the correct answer is (boxed{D}).,<think>Okay, so I need to figure out how the graph of y = sin(2x - π/3) relates to the graph of y = sin(2x). The question is asking whether it's a shift left or right by π/3 or π/6 units. Let me recall how phase shifts work in trigonometric functions.First, the general form for a sine function with a phase shift is y = sin(Bx - C) + D. Here, the phase shift is calculated as C/B. The phase shift tells us how much the graph is shifted horizontally. If the phase shift is positive, it's a shift to the right, and if it's negative, it's a shift to the left.In this problem, we're comparing y = sin(2x - π/3) to y = sin(2x). So let's break down the given function:The original function is y = sin(2x). The transformed function is y = sin(2x - π/3). Comparing this to the general form, B is 2, and C is π/3. Therefore, the phase shift should be C/B, which is (π/3)/2 = π/6. Since the phase shift is positive, that means it's a shift to the right by π/6 units. Let me verify that because sometimes the direction can be confusing. If we have y = sin(B(x - D)), then the shift is D units to the right. Let's rewrite the given function in that form. Starting with y = sin(2x - π/3), factor out the 2 from inside the sine function:y = sin[2(x - π/6)]Yes, because if you distribute the 2 inside, you get 2x - π/3. So, this is equivalent to shifting x by π/6 units to the right. Therefore, the phase shift is π/6 to the right. So looking at the options:A: left by π/3 – Hmm, that seems too much and in the wrong direction.B: right by π/3 – The shift amount is wrong here.C: left by π/6 – Direction is incorrect.D: right by π/6 – That's exactly what we found.Wait, let me double-check. Sometimes when there's a coefficient inside the argument, the shift can be counterintuitive. For example, if you have y = sin(Bx - C), the phase shift is C/B. So in this case, C is π/3, so phase shift is π/3 divided by B which is 2. So yes, π/6. And since it's sin(Bx - C), that's equivalent to shifting right by C/B. If it were sin(Bx + C), that would be shifting left by C/B. So here, since it's minus π/3 inside the argument, it's a shift to the right. Alternatively, another way to think about it: if you replace x with x - h, then the graph shifts h units to the right. So in this case, starting from sin(2x), to get sin(2x - π/3), you can write it as sin[2(x - π/6)]. Therefore, h is π/6, so shifting right by π/6. That confirms the answer D.I think that's solid. Let me just visualize it. Suppose we have the original function sin(2x), which has a period of π. If we shift each point to the right by π/6, then the graph will start at π/6 instead of 0. For example, the original function has a zero crossing at x = 0. The shifted function should have a zero crossing when 2x - π/3 = 0 ⇒ x = π/6. So indeed, the first zero crossing is shifted from 0 to π/6, which is a shift to the right by π/6. So that seems correct.Therefore, the answer is D.Final Answerboxed{D}</think>To determine how the graph of ( y = sin(2x - frac{pi}{3}) ) relates to the graph of ( y = sin(2x) ), we need to analyze the phase shift. The general form of a sine function with a phase shift is ( y = sin(Bx - C) ). The phase shift is calculated as ( frac{C}{B} ). For the given function ( y = sin(2x - frac{pi}{3}) ), we have ( B = 2 ) and ( C = frac{pi}{3} ). The phase shift is:[frac{C}{B} = frac{frac{pi}{3}}{2} = frac{pi}{6}]Since the phase shift is positive, it indicates a shift to the right. Rewriting the function in the form ( sin(2(x - frac{pi}{6})) ) confirms that the shift is ( frac{pi}{6} ) units to the right.Thus, the correct answer is (boxed{D}).]

🔑:AnalysisThis question examines the knowledge of the translation transformation rule of trigonometric function graphs. Mastering the principle of "left add, right subtract" for graph translation is key to solving this problem.According to the principle of "left add, right subtract" for graph translation, combined with the analytical expressions of the functions before and after translation, the answer can be obtained.SolutionGiven the function y=sin left(2x- frac{pi}{3}right)=sin 2left(x- frac{pi}{6}right),Thus, by shifting the graph of the function y=sin 2x to the right by frac{pi}{6} units,the graph of the function y=sin left(2x- frac{pi}{3}right) can be obtained.Therefore, the correct choice is boxed{text{D}}.