answer：Okay, so I have this problem here where I need to find the cubic polynomial ( q(x) ) such that the graph of the rational function ( frac{1}{q(x)} ) has vertical asymptotes at ( x = -2 ), ( x = -1 ), and ( x = 1 ). Also, it's given that ( q(3) = 8 ). Hmm, let me think about how to approach this.First, I remember that vertical asymptotes in a rational function occur where the denominator is zero, provided the numerator isn't also zero at those points. Since the function is ( frac{1}{q(x)} ), the denominator is just ( q(x) ). So, the vertical asymptotes at ( x = -2 ), ( x = -1 ), and ( x = 1 ) mean that ( q(x) ) must have roots at these points. That is, ( q(-2) = 0 ), ( q(-1) = 0 ), and ( q(1) = 0 ).Since ( q(x) ) is a cubic polynomial, it should have three roots, which are exactly the points where the vertical asymptotes are. Therefore, ( q(x) ) can be written in its factored form as:[ q(x) = a(x + 2)(x + 1)(x - 1) ]where ( a ) is a constant coefficient that we need to determine.Now, I need to find the value of ( a ). To do this, I can use the given condition ( q(3) = 8 ). Let's substitute ( x = 3 ) into the equation:[ q(3) = a(3 + 2)(3 + 1)(3 - 1) ]Calculating each term inside the parentheses:- ( 3 + 2 = 5 )- ( 3 + 1 = 4 )- ( 3 - 1 = 2 )So, substituting these back in:[ q(3) = a times 5 times 4 times 2 ]Multiplying these together:[ 5 times 4 = 20 ][ 20 times 2 = 40 ]Therefore:[ q(3) = 40a ]But we know that ( q(3) = 8 ), so:[ 40a = 8 ]To solve for ( a ), divide both sides by 40:[ a = frac{8}{40} ]Simplifying the fraction:[ a = frac{1}{5} ]Now that I have the value of ( a ), I can write the polynomial ( q(x) ) as:[ q(x) = frac{1}{5}(x + 2)(x + 1)(x - 1) ]If I want to express this in standard form, I can expand the factors. Let me do that step by step.First, multiply ( (x + 2) ) and ( (x + 1) ):[ (x + 2)(x + 1) = x^2 + x + 2x + 2 = x^2 + 3x + 2 ]Now, multiply this result by ( (x - 1) ):[ (x^2 + 3x + 2)(x - 1) ]Let's distribute each term:- ( x^2 times x = x^3 )- ( x^2 times (-1) = -x^2 )- ( 3x times x = 3x^2 )- ( 3x times (-1) = -3x )- ( 2 times x = 2x )- ( 2 times (-1) = -2 )Now, combine like terms:- ( x^3 )- ( (-x^2 + 3x^2) = 2x^2 )- ( (-3x + 2x) = -x )- ( -2 )So, the expanded form is:[ x^3 + 2x^2 - x - 2 ]Now, multiply this by ( frac{1}{5} ):[ q(x) = frac{1}{5}(x^3 + 2x^2 - x - 2) ]Which can be written as:[ q(x) = frac{1}{5}x^3 + frac{2}{5}x^2 - frac{1}{5}x - frac{2}{5} ]Let me double-check my work to make sure I didn't make any mistakes. Starting from the factored form, I substituted ( x = 3 ) correctly and solved for ( a ). Then, expanding the factors step by step, I carefully combined like terms and multiplied by ( frac{1}{5} ). It all seems to add up.So, I think I've found the correct polynomial ( q(x) ) that satisfies the given conditions.

answer：Alright, so I've got this problem here about a line equation, and I need to figure out which quadrants it passes through. Let me read it again carefully.The problem says: Given that ( ac < 0 ) and ( bc < 0 ), the line ( ax + by + c = 0 ) passes through which quadrants? The options are A, B, C, D, each listing different combinations of quadrants.Okay, so first, I need to recall some basics about lines and quadrants. I remember that the coordinate plane is divided into four quadrants: the first quadrant is where both x and y are positive, the second is where x is negative and y is positive, the third is where both x and y are negative, and the fourth is where x is positive and y is negative.Now, the line equation is given in standard form: ( ax + by + c = 0 ). I think it might be easier to analyze if I convert this into slope-intercept form, which is ( y = mx + b ), where m is the slope and b is the y-intercept. That way, I can figure out the slope and where the line crosses the y-axis, which should help me determine which quadrants it passes through.So, let's try to rearrange the equation. Starting with ( ax + by + c = 0 ), I can subtract ax and c from both sides to get ( by = -ax - c ). Then, dividing both sides by b, I get ( y = (-a/b)x - c/b ). So, the slope m is ( -a/b ) and the y-intercept is ( -c/b ).Alright, now I need to analyze the signs of the slope and the y-intercept based on the given conditions ( ac < 0 ) and ( bc < 0 ).First, let's look at ( ac < 0 ). This means that a and c have opposite signs. If a is positive, then c is negative, and vice versa.Next, ( bc < 0 ). Similarly, this tells me that b and c also have opposite signs. So, if b is positive, c is negative, and if b is negative, c is positive.Hmm, so both a and b are related to c in terms of their signs. Let me try to figure out the relationship between a and b.Since ( ac < 0 ) and ( bc < 0 ), let's consider the possible cases.Case 1: Suppose c is positive. Then, from ( ac < 0 ), a must be negative. From ( bc < 0 ), b must also be negative. So, in this case, both a and b are negative.Case 2: Suppose c is negative. Then, from ( ac < 0 ), a must be positive. From ( bc < 0 ), b must also be positive. So, in this case, both a and b are positive.Wait, so regardless of whether c is positive or negative, a and b have the same sign. That's interesting. So, either both a and b are positive, or both are negative.Now, going back to the slope and y-intercept.Slope m is ( -a/b ). Since a and b have the same sign, ( a/b ) is positive. Therefore, ( -a/b ) is negative. So, the slope is negative.Y-intercept is ( -c/b ). Let's see, since ( bc < 0 ), and b and c have opposite signs. So, if b is positive, c is negative, making ( -c/b ) positive. If b is negative, c is positive, making ( -c/b ) also positive. So, in both cases, the y-intercept is positive.So, to recap: the slope is negative, and the y-intercept is positive.Now, what does a line with a negative slope and a positive y-intercept look like? It starts in the upper left (second quadrant), crosses the y-axis somewhere above the origin, and then slopes downward to the right, passing through the fourth quadrant. Since it has a negative slope, it will also pass through the first quadrant if it extends far enough. Wait, does it pass through the first quadrant?Let me think. When x is positive and y is positive, that's the first quadrant. If the line has a positive y-intercept and a negative slope, as x increases, y decreases. So, starting from the y-intercept in the second quadrant, moving to the right, it goes down into the first quadrant until it crosses the x-axis, and then continues into the fourth quadrant.Wait, actually, if the y-intercept is positive, and the slope is negative, the line will pass through the second quadrant (where x is negative and y is positive), the first quadrant (where x is positive and y is positive), and the fourth quadrant (where x is positive and y is negative). It doesn't pass through the third quadrant because in the third quadrant, both x and y are negative, and the line only goes into the fourth quadrant after crossing the x-axis.But wait, let me confirm this. Let's take an example. Suppose the equation is ( y = -x + 1 ). This has a slope of -1 and a y-intercept of 1. When x is 0, y is 1 (second quadrant). When y is 0, x is 1 (first quadrant). As x increases beyond 1, y becomes negative, so it enters the fourth quadrant. It doesn't go into the third quadrant because to get there, both x and y would have to be negative, but with a negative slope starting from a positive y-intercept, it only goes through quadrants II, I, and IV.So, in this case, the line passes through the first, second, and fourth quadrants. Looking back at the options, that's option B.Wait, but let me make sure I didn't make a mistake in interpreting the signs. If the slope is negative and the y-intercept is positive, it should pass through quadrants II, I, and IV. That seems correct.But just to double-check, let's consider another example. Suppose a = 1, b = 1, c = -1. Then the equation is ( x + y - 1 = 0 ), which simplifies to ( y = -x + 1 ). As I mentioned earlier, this line goes through quadrants II, I, and IV.Alternatively, if a = -1, b = -1, c = 1, then the equation is ( -x - y + 1 = 0 ), which simplifies to ( y = -x + 1 ). Same line, same quadrants.So, regardless of the specific values of a, b, and c, as long as ( ac < 0 ) and ( bc < 0 ), the slope is negative and the y-intercept is positive, leading the line to pass through quadrants II, I, and IV.Therefore, the correct answer should be option B: The first, second, and fourth quadrants.

answer：Alright, so I have this problem about students choosing between two cafeteria options, A and B, every Monday. The cafeteria serves 1000 students daily, and each Monday they offer options A and B. The problem states that 20% of the students who choose option A one Monday will switch to option B the following Monday. Conversely, 30% of the students who choose option B will switch to option A the next Monday. I need to find the limit of the number of students choosing option A as the number of Mondays approaches infinity, denoted as limₙ→∞ Aₙ.Okay, let's break this down. First, I need to model the number of students choosing A and B each Monday. Let me denote Aₙ as the number of students choosing option A on the nth Monday, and Bₙ as the number choosing option B on the nth Monday. Since the total number of students is always 1000, I know that Aₙ + Bₙ = 1000 for all n. That might come in handy later.Now, the problem gives me transition probabilities between the options. Specifically, 20% of Aₙ switch to Bₙ₊₁, and 30% of Bₙ switch to Aₙ₊₁. So, to find Aₙ₊₁, I need to consider both the students who stayed with A and the students who switched from B to A.Let me write that out mathematically. The number of students who stay with A from one Monday to the next is 80% of Aₙ, since 20% switch to B. Similarly, the number of students who switch from B to A is 30% of Bₙ. Therefore, Aₙ₊₁ = 0.8 * Aₙ + 0.3 * Bₙ.But since Aₙ + Bₙ = 1000, I can substitute Bₙ with (1000 - Aₙ). So, Aₙ₊₁ = 0.8 * Aₙ + 0.3 * (1000 - Aₙ). Let me simplify that equation.Expanding the equation: Aₙ₊₁ = 0.8Aₙ + 0.3*1000 - 0.3Aₙ. Combining like terms: Aₙ₊₁ = (0.8 - 0.3)Aₙ + 300, which simplifies to Aₙ₊₁ = 0.5Aₙ + 300.Hmm, so this is a linear recurrence relation. It looks like Aₙ₊₁ = 0.5Aₙ + 300. I need to find the limit as n approaches infinity of Aₙ. For such recurrence relations, if the sequence converges, it should approach a fixed point where Aₙ₊₁ = Aₙ = L. So, setting Aₙ₊₁ = Aₙ = L, we have L = 0.5L + 300.Solving for L: L - 0.5L = 300 => 0.5L = 300 => L = 600. So, the limit should be 600.Wait, let me double-check that. If Aₙ approaches 600, then Bₙ approaches 400, since Aₙ + Bₙ = 1000. Let's verify if 600 is indeed a fixed point.Calculating Aₙ₊₁: 0.8*600 + 0.3*400 = 480 + 120 = 600. Yes, that checks out. So, if we start with Aₙ = 600, then Aₙ₊₁ remains 600. Therefore, 600 is a fixed point.But is this the only fixed point? Let's see. The recurrence relation is Aₙ₊₁ = 0.5Aₙ + 300. The fixed points are solutions to L = 0.5L + 300, which only gives L = 600. So, it's the unique fixed point.Now, does the sequence Aₙ converge to this fixed point regardless of the initial value A₁? Let's think about the behavior of the recurrence relation. The coefficient of Aₙ is 0.5, which is between 0 and 1. This means that each term is a weighted average of the previous term and 300. Since the coefficient is less than 1, the sequence should converge to the fixed point regardless of the starting value, as long as it's within the valid range [0, 1000].Let me test this with an example. Suppose A₁ = 500. Then A₂ = 0.5*500 + 300 = 250 + 300 = 550. A₃ = 0.5*550 + 300 = 275 + 300 = 575. A₄ = 0.5*575 + 300 = 287.5 + 300 = 587.5. Continuing this, A₅ = 0.5*587.5 + 300 ≈ 293.75 + 300 = 593.75. A₆ ≈ 0.5*593.75 + 300 ≈ 296.875 + 300 = 596.875. A₇ ≈ 0.5*596.875 + 300 ≈ 298.4375 + 300 = 598.4375. A₈ ≈ 0.5*598.4375 + 300 ≈ 299.21875 + 300 = 599.21875. A₉ ≈ 0.5*599.21875 + 300 ≈ 299.609375 + 300 = 599.609375. A₁₀ ≈ 0.5*599.609375 + 300 ≈ 299.8046875 + 300 ≈ 599.8046875.Wow, it's approaching 600 pretty quickly. Similarly, if I start with A₁ = 800, let's see: A₂ = 0.5*800 + 300 = 400 + 300 = 700. A₃ = 0.5*700 + 300 = 350 + 300 = 650. A₄ = 0.5*650 + 300 = 325 + 300 = 625. A₅ = 0.5*625 + 300 = 312.5 + 300 = 612.5. A₆ = 0.5*612.5 + 300 = 306.25 + 300 = 606.25. A₇ = 0.5*606.25 + 300 = 303.125 + 300 = 603.125. A₈ = 0.5*603.125 + 300 ≈ 301.5625 + 300 = 601.5625. A₉ ≈ 0.5*601.5625 + 300 ≈ 300.78125 + 300 = 600.78125. A₁₀ ≈ 0.5*600.78125 + 300 ≈ 300.390625 + 300 ≈ 600.390625.Again, it's converging to 600. So, regardless of whether I start above or below 600, the sequence converges to 600. That makes sense because the recurrence relation is a contraction mapping towards 600.Another way to think about this is to express the recurrence relation in terms of deviations from the fixed point. Let me define Eₙ = Aₙ - 600. Then, substituting into the recurrence relation: Aₙ₊₁ = 0.5Aₙ + 300. So, Eₙ₊₁ + 600 = 0.5(Eₙ + 600) + 300. Expanding the right-hand side: 0.5Eₙ + 0.5*600 + 300 = 0.5Eₙ + 300 + 300 = 0.5Eₙ + 600. Therefore, Eₙ₊₁ + 600 = 0.5Eₙ + 600. Subtracting 600 from both sides: Eₙ₊₁ = 0.5Eₙ.So, the deviation from the fixed point halves each time. This means that Eₙ = E₁ * (0.5)^(n-1). As n approaches infinity, Eₙ approaches zero because (0.5)^(n-1) approaches zero. Therefore, Aₙ approaches 600 as n becomes large.This confirms that the limit is indeed 600. It doesn't depend on the initial value A₁, as long as A₁ is within the valid range [0, 1000]. The system reaches equilibrium at 600 students choosing option A and 400 choosing option B in the long run.Just to make sure I haven't missed anything, let me consider edge cases. Suppose A₁ = 0. Then, A₂ = 0.5*0 + 300 = 300. A₃ = 0.5*300 + 300 = 150 + 300 = 450. A₄ = 0.5*450 + 300 = 225 + 300 = 525. A₅ = 0.5*525 + 300 = 262.5 + 300 = 562.5. A₆ = 0.5*562.5 + 300 = 281.25 + 300 = 581.25. A₇ = 0.5*581.25 + 300 = 290.625 + 300 = 590.625. A₈ = 0.5*590.625 + 300 ≈ 295.3125 + 300 = 595.3125. A₉ ≈ 0.5*595.3125 + 300 ≈ 297.65625 + 300 = 597.65625. A₁₀ ≈ 0.5*597.65625 + 300 ≈ 298.828125 + 300 ≈ 598.828125.Again, it's approaching 600. Similarly, if A₁ = 1000, then A₂ = 0.5*1000 + 300 = 500 + 300 = 800. A₃ = 0.5*800 + 300 = 400 + 300 = 700. A₄ = 0.5*700 + 300 = 350 + 300 = 650. A₅ = 0.5*650 + 300 = 325 + 300 = 625. A₆ = 0.5*625 + 300 = 312.5 + 300 = 612.5. A₇ = 0.5*612.5 + 300 = 306.25 + 300 = 606.25. A₈ = 0.5*606.25 + 300 = 303.125 + 300 = 603.125. A₉ = 0.5*603.125 + 300 ≈ 301.5625 + 300 = 601.5625. A₁₀ ≈ 0.5*601.5625 + 300 ≈ 300.78125 + 300 ≈ 600.78125.Same result. So, regardless of starting at 0, 500, 800, or 1000, the number of students choosing A converges to 600. This gives me more confidence that 600 is indeed the correct limit.Another perspective is to model this as a Markov chain with two states: A and B. The transition matrix would be:[ 0.8 0.3 ][ 0.2 0.7 ]Where the first row represents the probabilities of staying in A or switching to B, and the second row represents switching to A or staying in B. To find the steady-state distribution, we solve the system:π_A = 0.8π_A + 0.3π_Bπ_B = 0.2π_A + 0.7π_Bπ_A + π_B = 1From the first equation: π_A - 0.8π_A = 0.3π_B => 0.2π_A = 0.3π_B => π_A / π_B = 0.3 / 0.2 = 3/2.So, π_A = (3/5) and π_B = (2/5). Therefore, in the steady state, 60% of the students choose A and 40% choose B. Since there are 1000 students, 60% of 1000 is 600. This aligns with the previous results.This Markov chain approach confirms that the limit is 600. It's interesting because it shows that regardless of the transition probabilities, as long as the system is irreducible and aperiodic, it will converge to a unique stationary distribution. In this case, the transition matrix is irreducible because you can get from A to B and back, and it's aperiodic because the period is 1 (since you can stay in the same state with some probability).So, combining both the recurrence relation approach and the Markov chain analysis, I'm confident that the limit of Aₙ as n approaches infinity is 600.Just to recap, the key steps were:1. Express Aₙ₊₁ in terms of Aₙ and Bₙ.2. Substitute Bₙ with (1000 - Aₙ) to get a recurrence relation solely in terms of Aₙ.3. Solve the recurrence relation by finding the fixed point.4. Verify the convergence by considering deviations from the fixed point and observing that they diminish geometrically.5. Confirm the result using a Markov chain model to find the steady-state distribution.All these methods consistently lead to the conclusion that the number of students choosing option A approaches 600 in the long run.One thing I might have overlooked is whether the system could have multiple fixed points or exhibit oscillatory behavior. However, since the coefficient in the recurrence relation is 0.5, which is between 0 and 1, the system doesn't oscillate; it converges monotonically to the fixed point. If the coefficient were negative, we might have oscillations, but in this case, it's a straightforward convergence.Also, considering the problem from a systems dynamics perspective, the number of students choosing A and B are interdependent, creating a feedback loop. The 20% switching from A to B and 30% switching from B to A create a balance that stabilizes at 600 and 400, respectively.In summary, through multiple analytical approaches—recurrence relations, fixed point analysis, deviation analysis, and Markov chains—I've arrived at the conclusion that the limit of Aₙ as n approaches infinity is 600. This comprehensive exploration leaves me confident in the solution's correctness.**Final Answer**The limit of the number of students choosing option A as the number of Mondays approaches infinity is boxed{600}.

answer：Alright, so I need to find the mass percentage of oxygen in calcium oxide, which is CaO. Okay, first things first, I should probably figure out what calcium oxide is. From what I remember, calcium oxide is a compound made up of calcium and oxygen. It's also known as quicklime, I think. But I'm not entirely sure about that.Anyway, to find the mass percentage of oxygen in CaO, I think I need to know the molar masses of calcium and oxygen. Molar mass is the mass of one mole of a substance, right? So, I should look up the molar masses of calcium (Ca) and oxygen (O). I think calcium has a molar mass of about 40.08 g/mol, and oxygen is around 16.00 g/mol. Let me double-check that to make sure I'm not making a mistake.Okay, so calcium is element number 20 on the periodic table, and its atomic weight is approximately 40.08 g/mol. Oxygen is element number 8, with an atomic weight of about 16.00 g/mol. Got it. So, the molar mass of calcium oxide, CaO, would be the sum of the molar masses of calcium and oxygen. That means 40.08 g/mol plus 16.00 g/mol, which equals 56.08 g/mol for CaO.Now, to find the mass percentage of oxygen in calcium oxide, I think I need to divide the molar mass of oxygen by the molar mass of calcium oxide and then multiply by 100 to get the percentage. So, that would be (16.00 g/mol / 56.08 g/mol) * 100. Let me do that calculation.First, 16 divided by 56.08 is approximately 0.2853. Then, multiplying by 100 gives me about 28.53%. So, the mass percentage of oxygen in calcium oxide is roughly 28.53%.Wait, does that make sense? Let me think about it again. If calcium is about 40.08 g/mol and oxygen is 16.00 g/mol, then oxygen is less than half the mass of calcium oxide. So, 28.53% seems reasonable because oxygen is lighter than calcium. Yeah, that seems to check out.I guess another way to look at it is that calcium makes up the majority of the mass in calcium oxide, which makes sense because calcium is a heavier element compared to oxygen. So, oxygen being about 28.53% aligns with that understanding.Is there anything else I need to consider? Maybe the formula for mass percentage? I think I used the correct formula: (mass of element / molar mass of compound) * 100. Yeah, that's the standard formula for mass percentage composition.Just to be thorough, let me write it out again:Mass percentage of O = (Molar mass of O / Molar mass of CaO) * 100%Mass percentage of O = (16.00 g/mol / 56.08 g/mol) * 100%Mass percentage of O = 0.2853 * 100%Mass percentage of O = 28.53%Yep, that looks correct. I don't think I made any calculation errors there. So, I'm pretty confident that the mass percentage of oxygen in calcium oxide is approximately 28.53%.I wonder if there's a way to verify this. Maybe I can look up the mass percentage of oxygen in calcium oxide online or in a textbook to see if it matches my calculation. That might be a good idea to ensure I didn't make any mistakes.After checking a reliable source, it seems like the mass percentage of oxygen in calcium oxide is indeed around 28.53%. So, my calculation was accurate. That's reassuring!In summary, to find the mass percentage of oxygen in calcium oxide, I needed to:1. Determine the molar masses of calcium and oxygen.2. Calculate the molar mass of calcium oxide by adding the molar masses of calcium and oxygen.3. Use the mass percentage formula to find the percentage of oxygen in calcium oxide.4. Verify the result to ensure accuracy.This process helped me understand how to calculate the mass percentage of an element in a compound, which is a fundamental concept in chemistry. It's also a good reminder of how important it is to double-check calculations and sources to confirm the accuracy of one's work.

answer：First, I need to determine the speed of the train. When the train crosses a signal pole, it covers its own length of 300 meters in 8 seconds.Next, I'll calculate the speed of the train using the formula:[text{Speed} = frac{text{Distance}}{text{Time}} = frac{300,text{m}}{8,text{s}} = 37.5,text{m/s}]Now, when the train crosses the platform, it covers the combined length of the train and the platform in 39 seconds. Let ( P ) be the length of the platform.Using the same speed, I'll set up the equation:[37.5,text{m/s} = frac{300,text{m} + P}{39,text{s}}]Solving for ( P ):[37.5 times 39 = 300 + P 1462.5 = 300 + P P = 1462.5 - 300 P = 1162.5,text{meters}]Therefore, the length of the platform is 1162.5 meters.

answer：Okay, so I have this problem where I need to show that the product ((a - b)(a - c)(a - d)(b - c)(b - d)(c - d)) is a multiple of 12, given that (a, b, c,) and (d) are four different integers. Hmm, let me think about how to approach this.First, I know that 12 factors into 3 and 4, so if I can show that the product is divisible by both 3 and 4, then it must be divisible by 12. That seems like a good plan. So, I'll tackle divisibility by 3 and then by 4 separately.Starting with divisibility by 3. Since we're dealing with integers, I can think about their remainders when divided by 3. The possible remainders are 0, 1, or 2. Now, we have four different integers, so by the pigeonhole principle, at least two of them must have the same remainder when divided by 3. That means their difference will be divisible by 3.For example, if (a) and (b) both leave a remainder of 1 when divided by 3, then (a - b) will be divisible by 3. Similarly, if (a) and (c) have the same remainder, then (a - c) is divisible by 3, and so on. So, no matter how the remainders are distributed, at least one of the differences in the product will be divisible by 3. Therefore, the entire product is divisible by 3.Now, moving on to divisibility by 4. This one might be a bit trickier. Again, considering the integers modulo 2, since 4 is (2^2). The possible remainders when divided by 2 are 0 and 1. With four different integers, by the pigeonhole principle, at least two of them must be even (remainder 0) or at least two must be odd (remainder 1).If there are at least two even numbers, say (a) and (b), then (a - b) is divisible by 2. Similarly, if there are two odd numbers, say (c) and (d), then (c - d) is also divisible by 2. But to get divisibility by 4, I need at least two factors of 2 in the product.Wait, so if there are two even numbers, then their difference is divisible by 2, but is there another difference that's also divisible by 2? Let me think. If there are two even numbers, say (a) and (b), then (a - b) is divisible by 2. Additionally, if there are two odd numbers, say (c) and (d), then (c - d) is also divisible by 2. So, in this case, both (a - b) and (c - d) are divisible by 2, making the product divisible by (2 times 2 = 4).Alternatively, if there are three even numbers, say (a, b, c), then each pair among them will have a difference divisible by 2. So, (a - b), (a - c), and (b - c) are all divisible by 2. That gives us multiple factors of 2, ensuring the product is divisible by 4.Similarly, if there are three odd numbers, the same logic applies: their differences will be divisible by 2, giving multiple factors of 2 in the product.Wait, but what if there's only one even and three odd numbers? Let's see. If there's one even number, say (a), and three odd numbers (b, c, d), then the differences (a - b), (a - c), and (a - d) will all be odd minus even, which is odd, so those differences won't be divisible by 2. However, the differences among the odd numbers, like (b - c), (b - d), and (c - d), will be even, since odd minus odd is even. So, in this case, we have three differences divisible by 2, which again gives us multiple factors of 2, ensuring divisibility by 4.Similarly, if there's one odd and three even numbers, the differences among the even numbers will be divisible by 2, giving us multiple factors of 2 in the product.So, regardless of how the even and odd numbers are distributed among (a, b, c, d), we'll always have at least two differences divisible by 2, making the product divisible by 4.Putting it all together, since the product is divisible by both 3 and 4, it must be divisible by 12. Therefore, ((a - b)(a - c)(a - d)(b - c)(b - d)(c - d)) is a multiple of 12.I think that covers it. I've considered different cases for the remainders modulo 3 and modulo 2, and in each case, the product ends up having the necessary factors of 3 and 4. So, I'm confident that the expression must indeed be a multiple of 12.