💡Alright, so I have this problem where I need to find a complex number ( z ) that satisfies two conditions. The first condition is that ( frac{z - 3i}{z + i} ) is a negative real number, and the second condition is that ( frac{z - 3}{z + 1} ) is a purely imaginary number. Hmm, okay, let me try to unpack this step by step.First, I know that any complex number ( z ) can be written as ( a + bi ), where ( a ) and ( b ) are real numbers. So, I'll let ( z = a + bi ). That should make it easier to handle the real and imaginary parts separately.Starting with the first condition: ( frac{z - 3i}{z + i} ) is a negative real number. If this fraction is a negative real number, that means the imaginary part of the numerator and denominator must cancel each other out, right? So, let me write out the numerator and the denominator with ( z = a + bi ).The numerator is ( z - 3i = a + bi - 3i = a + (b - 3)i ).The denominator is ( z + i = a + bi + i = a + (b + 1)i ).So, the fraction becomes ( frac{a + (b - 3)i}{a + (b + 1)i} ). To simplify this, I can multiply the numerator and the denominator by the complex conjugate of the denominator. The complex conjugate of ( a + (b + 1)i ) is ( a - (b + 1)i ).Multiplying numerator and denominator by this conjugate, we get:Numerator: ( (a + (b - 3)i)(a - (b + 1)i) )Denominator: ( (a + (b + 1)i)(a - (b + 1)i) )Let me compute the denominator first because it's simpler. It's just the square of the modulus of the denominator, which is ( a^2 + (b + 1)^2 ).Now, the numerator: I need to expand this product. Let's do it step by step.First, multiply ( a ) by each term in the second complex number:( a times a = a^2 )( a times (- (b + 1)i) = -a(b + 1)i )Next, multiply ( (b - 3)i ) by each term:( (b - 3)i times a = a(b - 3)i )( (b - 3)i times (- (b + 1)i) = - (b - 3)(b + 1)i^2 )Since ( i^2 = -1 ), this becomes ( (b - 3)(b + 1) ).Putting it all together, the numerator is:( a^2 - a(b + 1)i + a(b - 3)i + (b - 3)(b + 1) )Combine like terms:The real parts: ( a^2 + (b - 3)(b + 1) )The imaginary parts: ( -a(b + 1)i + a(b - 3)i = a[(b - 3) - (b + 1)]i = a(-4)i = -4ai )So, the numerator simplifies to:( a^2 + (b - 3)(b + 1) - 4ai )Since the entire fraction ( frac{z - 3i}{z + i} ) is a negative real number, the imaginary part must be zero. Therefore, the coefficient of ( i ) in the numerator must be zero. So, we have:( -4a = 0 )Solving this gives ( a = 0 ). So, the real part of ( z ) is zero. That simplifies things a bit.Now, moving on to the second condition: ( frac{z - 3}{z + 1} ) is a purely imaginary number. That means the real part of this fraction must be zero. Let's write out the numerator and denominator with ( a = 0 ) and ( z = 0 + bi = bi ).The numerator is ( z - 3 = bi - 3 = -3 + bi ).The denominator is ( z + 1 = bi + 1 = 1 + bi ).So, the fraction becomes ( frac{-3 + bi}{1 + bi} ). Again, to simplify, I'll multiply numerator and denominator by the complex conjugate of the denominator, which is ( 1 - bi ).Numerator: ( (-3 + bi)(1 - bi) )Denominator: ( (1 + bi)(1 - bi) = 1 + b^2 )Let's compute the numerator:Multiply ( -3 ) by each term:( -3 times 1 = -3 )( -3 times (-bi) = 3bi )Multiply ( bi ) by each term:( bi times 1 = bi )( bi times (-bi) = -b^2 i^2 = -b^2 (-1) = b^2 )So, the numerator becomes:( -3 + 3bi + bi + b^2 = (-3 + b^2) + (3b + b)i = (-3 + b^2) + 4bi )Since the fraction ( frac{z - 3}{z + 1} ) is purely imaginary, the real part must be zero. Therefore, the real part of the numerator divided by the denominator must be zero. So, we have:( frac{-3 + b^2}{1 + b^2} = 0 )This implies that:( -3 + b^2 = 0 )Solving for ( b ):( b^2 = 3 )( b = sqrt{3} ) or ( b = -sqrt{3} )So, ( z = 0 + sqrt{3}i ) or ( z = 0 - sqrt{3}i ).But wait, we have to check if these solutions satisfy the first condition as well. Let me verify.First, let's take ( z = sqrt{3}i ):Compute ( frac{z - 3i}{z + i} = frac{sqrt{3}i - 3i}{sqrt{3}i + i} = frac{(sqrt{3} - 3)i}{(sqrt{3} + 1)i} = frac{sqrt{3} - 3}{sqrt{3} + 1} )This is a real number. Let's compute it:Multiply numerator and denominator by ( sqrt{3} - 1 ):( frac{(sqrt{3} - 3)(sqrt{3} - 1)}{(sqrt{3} + 1)(sqrt{3} - 1)} = frac{(3 - sqrt{3} - 3sqrt{3} + 3)}{3 - 1} = frac{6 - 4sqrt{3}}{2} = 3 - 2sqrt{3} )Which is a negative real number because ( 3 - 2sqrt{3} ) is approximately ( 3 - 3.464 = -0.464 ), which is negative.Now, let's check ( z = -sqrt{3}i ):Compute ( frac{z - 3i}{z + i} = frac{-sqrt{3}i - 3i}{-sqrt{3}i + i} = frac{(-sqrt{3} - 3)i}{(-sqrt{3} + 1)i} = frac{-sqrt{3} - 3}{-sqrt{3} + 1} )Simplify:Multiply numerator and denominator by ( -1 ):( frac{sqrt{3} + 3}{sqrt{3} - 1} )Multiply numerator and denominator by ( sqrt{3} + 1 ):( frac{(sqrt{3} + 3)(sqrt{3} + 1)}{(sqrt{3} - 1)(sqrt{3} + 1)} = frac{(3 + sqrt{3} + 3sqrt{3} + 3)}{3 - 1} = frac{6 + 4sqrt{3}}{2} = 3 + 2sqrt{3} )Which is a positive real number, but the first condition requires it to be negative. Therefore, ( z = -sqrt{3}i ) does not satisfy the first condition.Hence, the only solution is ( z = sqrt{3}i ).**Final Answer**The value of ( z ) is boxed{sqrt{3} mathrm{i}}.

💡First, I recognize that a square has four sides of equal length.Given the side length of the square is 19 cm, I can use the formula for the perimeter of a square, which is four times the side length.So, I multiply 4 by 19 cm to find the perimeter.This calculation gives me a perimeter of 76 cm.

💡Okay, so I have this problem where there are several vectors drawn from a point O on a plane, and the sum of their lengths is 4. I need to prove that it's possible to choose some of these vectors (maybe just one) such that the length of their sum is greater than 1. Hmm, interesting.Let me start by understanding the problem better. We have vectors emanating from the same point O, and their total length adds up to 4. I need to show that among these vectors, there's a subset whose vector sum has a length greater than 1. So, it's not about the sum of their lengths, but the length of their vector sum.I remember that when dealing with vectors, the triangle inequality is often useful. The triangle inequality states that for any two vectors, the magnitude of their sum is less than or equal to the sum of their magnitudes. But here, I need something a bit different. Maybe I can use some sort of averaging or pigeonhole principle?Wait, the problem says "several vectors," so it's not specified how many. It could be two, three, or more. Maybe I can approach this by considering projections of these vectors onto a particular axis.Let me think. If I pick a direction, say the x-axis, and project all the vectors onto this axis, the sum of these projections will give me the x-component of the total vector sum. Similarly, if I project onto the y-axis, I get the y-component. The magnitude of the total vector sum would then be the square root of the sum of the squares of these components.But how does that help me? Well, if I can show that either the x-component or the y-component is greater than 1, then the magnitude of the total vector sum would be greater than 1. Because even if one component is greater than 1 and the other is zero, the magnitude is still greater than 1.So, maybe I can use some kind of averaging argument here. Since the total length of all vectors is 4, the sum of their projections onto any axis can't be more than 4, right? But I need to show that at least one of these projections must be greater than 1.Wait, actually, the sum of the projections onto a single axis can't exceed the total length of the vectors, which is 4. But if I consider both the x and y projections, maybe I can distribute the total length between them.Let me formalize this. Let’s denote the vectors as ( mathbf{a}_1, mathbf{a}_2, ldots, mathbf{a}_n ). The sum of their lengths is ( sum_{i=1}^n |mathbf{a}_i| = 4 ). Now, if I project each vector onto the x-axis, the sum of these projections is ( sum_{i=1}^n (mathbf{a}_i cdot hat{mathbf{i}}) ), where ( hat{mathbf{i}} ) is the unit vector along the x-axis. Similarly, the sum of the projections onto the y-axis is ( sum_{i=1}^n (mathbf{a}_i cdot hat{mathbf{j}}) ).The key idea here is that the sum of the absolute values of these projections can't be less than the total length. Wait, is that true? Let me think. For each vector, the projection onto the x-axis is ( |mathbf{a}_i| cos theta_i ), and similarly for the y-axis, it's ( |mathbf{a}_i| sin theta_i ). The sum of the absolute values of these projections would be ( sum_{i=1}^n (|mathbf{a}_i| |cos theta_i| + |mathbf{a}_i| |sin theta_i|) ). But since ( |cos theta_i| + |sin theta_i| geq 1 ) for any angle ( theta_i ), because the maximum value of ( |cos theta| + |sin theta| ) is ( sqrt{2} ), which is greater than 1. Therefore, the sum of the absolute values of the projections is at least equal to the total length of the vectors, which is 4.So, ( sum_{i=1}^n (|mathbf{a}_i| |cos theta_i| + |mathbf{a}_i| |sin theta_i|) geq 4 ). But since ( |cos theta_i| + |sin theta_i| geq 1 ), this inequality holds.Now, if I consider the sum of the projections on the x-axis and the sum on the y-axis separately, their total is at least 4. Therefore, at least one of these sums must be at least 2, right? Because if both were less than 2, their total would be less than 4, which contradicts the previous statement.Wait, no. Actually, the sum of the absolute values of the projections on x and y is at least 4, but the actual sums (without absolute values) could be less. Hmm, maybe I need to think differently.Alternatively, perhaps I can use the pigeonhole principle. Since the total length is 4, if I consider the projections onto two perpendicular axes, the sum of the projections on at least one of these axes must be at least 2. Because if both were less than 2, their total would be less than 4, which contradicts the total length.But wait, the sum of the projections on each axis can be positive or negative, so their actual sum might not add up directly. Hmm, maybe I need to take absolute values into account.Let me try this approach. For each vector, the projection on the x-axis is ( mathbf{a}_i cdot hat{mathbf{i}} ), which can be positive or negative. Similarly for the y-axis. If I take the absolute value of each projection, the sum of these absolute values is at least 4, as I mentioned earlier.Therefore, either the sum of the absolute projections on the x-axis is at least 2, or the sum on the y-axis is at least 2. Because if both were less than 2, their total would be less than 4, which is not possible.So, without loss of generality, assume that the sum of the absolute projections on the x-axis is at least 2. Now, among these projections, some are positive and some are negative. If I can find a subset of vectors whose projections on the x-axis are all positive (or all negative), then their sum would be at least 2, and thus the magnitude of their vector sum would be at least 2, which is greater than 1.But wait, how do I ensure that such a subset exists? Maybe I can sort the vectors based on their angles and pick those that lie in a particular half-plane.Alternatively, perhaps I can use the idea that if the sum of the absolute values is at least 2, then there must be a subset whose sum is at least 1. But I'm not sure about that.Wait, another approach: consider the vectors as complex numbers. The sum of their magnitudes is 4. I need to find a subset whose vector sum has magnitude greater than 1. Maybe I can use some probabilistic method or expectation.Alternatively, think about the average. The average magnitude of the vectors is 4/n, but that might not directly help.Wait, going back to the projection idea. If the sum of the absolute projections on the x-axis is at least 2, then there must be a subset of vectors whose projections on the x-axis sum to at least 1. Because if all subsets had projections less than 1, then the total sum would be less than 2, which contradicts the earlier statement.But how do I formalize this? Maybe using the pigeonhole principle again. If I have a total sum of absolute projections on the x-axis of at least 2, then by dividing these projections into two groups, one group must have a sum of at least 1.Wait, actually, that's a good point. If the total absolute projection on the x-axis is S, then S ≥ 2. If I consider the positive and negative projections separately, the sum of the positive projections minus the sum of the negative projections is the total projection. But the sum of the absolute values is S = sum of positive projections + sum of negative projections.If S ≥ 2, then either the sum of positive projections is ≥1 or the sum of negative projections is ≥1 (in absolute value). Therefore, either the sum of positive projections is ≥1, or the sum of negative projections is ≥1. So, if I take all the vectors with positive projections, their sum on the x-axis is at least 1, and similarly for the negative ones.Therefore, the magnitude of the vector sum of these selected vectors would be at least 1, because the x-component is at least 1, and the y-component could be anything, but the magnitude would be at least 1.Wait, but the problem says "greater than 1," not "at least 1." So, is it possible that the magnitude is exactly 1? Hmm, but in reality, the sum of the projections is at least 1, but the actual vector sum could have a magnitude slightly more than 1 due to the y-components.Alternatively, maybe I can argue that the magnitude is strictly greater than 1. Because if the x-component is at least 1, and there's some y-component, then the magnitude would be greater than 1.Wait, no. If all the vectors have their projections on the x-axis exactly 1, and their y-components cancel out, then the magnitude could be exactly 1. But in reality, since the total projection on x is at least 2, and we're selecting a subset whose projection is at least 1, the actual vector sum might have a magnitude greater than 1.Hmm, maybe I need to refine this argument. Let me think again.Suppose I have a set of vectors with total length 4. I project them onto the x-axis, and the sum of the absolute projections is at least 4. Therefore, the sum of the positive projections plus the sum of the negative projections is at least 4.But the total projection on the x-axis is the sum of positive projections minus the sum of negative projections. Let me denote the sum of positive projections as P and the sum of negative projections as N. Then, P + N ≥ 4, and the total projection is P - N.Now, if I consider the subset of vectors with positive projections, their total projection is P. Similarly, the subset with negative projections has a total projection of -N.If P ≥ 2, then the magnitude of the vector sum of these positive vectors is at least P, which is ≥2, so their magnitude is ≥2 >1.Similarly, if N ≥2, then the magnitude of the vector sum of the negative vectors is at least N, which is ≥2 >1.But what if P <2 and N <2? Then P + N <4, which contradicts P + N ≥4. Therefore, at least one of P or N must be ≥2.Wait, that seems to be a solid argument. So, either the sum of positive projections P is ≥2, or the sum of negative projections N is ≥2. Therefore, selecting all vectors with positive projections (if P ≥2) or all vectors with negative projections (if N ≥2) gives a subset whose vector sum has a magnitude of at least 2, which is greater than 1.Therefore, such a subset exists.Wait, but the problem says "it is possible to choose several vectors (or perhaps just one vector) such that the length of their sum is greater than 1." So, in this case, we've shown that there exists a subset whose vector sum has a magnitude of at least 2, which is definitely greater than 1.Therefore, the proof is complete.But let me double-check if there's any case where this might not hold. Suppose all vectors are aligned in such a way that their projections on both axes are less than 1. But if the total projection on one axis is at least 2, then selecting the subset with positive projections would give a sum of at least 2 on that axis, making the magnitude of the vector sum at least 2.Alternatively, if all vectors are aligned in a way that their projections on both axes are exactly 1, but that would require each vector to have a magnitude of √2, and the total length would be n√2 =4, so n=4/√2=2√2≈2.828, which is not an integer. So, it's not possible for all vectors to have projections exactly 1 on both axes.Therefore, the argument holds.Another way to think about it is using the concept of vector addition. If you have vectors adding up to a total length of 4, then their vector sum must have a magnitude less than or equal to 4, but we need to show that at least one subset has a magnitude greater than 1.But actually, the argument above shows that there's a subset with magnitude at least 2, which is stronger than needed.So, in conclusion, by considering the projections onto two perpendicular axes and using the pigeonhole principle, we can show that there must be a subset of vectors whose vector sum has a magnitude greater than 1.

💡Okay, so I have this problem here: I need to find how many pairs (a, b) of non-zero real numbers satisfy the equation 2/a + 2/b = 1/(a + b). The answer choices are none, 1, 2, one pair for each b ≠ 0, or two pairs for each b ≠ 0. Hmm, let me try to figure this out step by step.First, I'll write down the equation again to make sure I have it right:2/a + 2/b = 1/(a + b)Alright, so I need to find all pairs (a, b) where neither a nor b is zero, and they satisfy this equation. Let me think about how to approach this. Maybe I can combine the fractions on the left side to simplify the equation.So, the left side is 2/a + 2/b. To combine these, I need a common denominator, which would be ab. Let me rewrite that:2/a + 2/b = (2b + 2a)/abSo now the equation becomes:(2a + 2b)/ab = 1/(a + b)Hmm, okay. Let me write that again:(2(a + b))/ab = 1/(a + b)Now, I can cross-multiply to eliminate the fractions. That should give me:(2(a + b))^2 = ab * 1Wait, no, that's not quite right. If I have (2(a + b))/ab = 1/(a + b), then cross-multiplying would mean multiplying both sides by ab and by (a + b). So, let me do that properly.Multiply both sides by ab and by (a + b):2(a + b) * (a + b) = ab * 1So that simplifies to:2(a + b)^2 = abAlright, now I have 2(a + b)^2 = ab. Let me expand the left side:2(a^2 + 2ab + b^2) = abWhich becomes:2a^2 + 4ab + 2b^2 = abNow, let's bring all terms to one side to set the equation to zero:2a^2 + 4ab + 2b^2 - ab = 0Combine like terms:2a^2 + 3ab + 2b^2 = 0Hmm, so now I have a quadratic equation in terms of a and b. Let me see if I can solve for a in terms of b or vice versa. Maybe I can treat this as a quadratic in a.So, let's write it as:2a^2 + 3ab + 2b^2 = 0This is a quadratic equation in a, where the coefficients are:A = 2, B = 3b, C = 2b^2I can use the quadratic formula to solve for a:a = [-B ± sqrt(B^2 - 4AC)] / (2A)Plugging in the values:a = [-3b ± sqrt((3b)^2 - 4*2*2b^2)] / (2*2)Let's compute the discriminant:(3b)^2 - 4*2*2b^2 = 9b^2 - 16b^2 = -7b^2Wait, that's negative. So the discriminant is negative, which means there are no real solutions for a. That suggests that there are no real pairs (a, b) that satisfy the original equation.But let me double-check my steps to make sure I didn't make a mistake.Starting from the beginning:2/a + 2/b = 1/(a + b)Combine left side: (2a + 2b)/ab = 1/(a + b)Cross-multiplying: 2(a + b)^2 = abExpanding: 2a^2 + 4ab + 2b^2 = abBringing all terms to one side: 2a^2 + 3ab + 2b^2 = 0Quadratic in a: 2a^2 + 3ab + 2b^2 = 0Discriminant: (3b)^2 - 4*2*2b^2 = 9b^2 - 16b^2 = -7b^2Yes, that seems correct. So the discriminant is negative, meaning no real solutions for a. Therefore, there are no pairs (a, b) of non-zero real numbers that satisfy the equation.So the answer should be none, which is option A.Wait, but let me think again. Maybe I made a mistake in cross-multiplying. Let me check that step.Original equation after combining fractions:(2(a + b))/ab = 1/(a + b)Cross-multiplying: 2(a + b)^2 = abYes, that's correct. So 2(a + b)^2 = ab.Expanding: 2a^2 + 4ab + 2b^2 = abBringing ab to the left: 2a^2 + 3ab + 2b^2 = 0Yes, that's correct. So the quadratic equation is correct.Alternatively, maybe I can try solving for b instead of a. Let me see.Starting from 2a^2 + 3ab + 2b^2 = 0Treat it as a quadratic in b:2b^2 + 3ab + 2a^2 = 0Using quadratic formula for b:b = [-3a ± sqrt(9a^2 - 16a^2)] / 4Which is:b = [-3a ± sqrt(-7a^2)] / 4Again, the discriminant is negative, so no real solutions for b either.Therefore, regardless of whether I solve for a or b, I end up with a negative discriminant, meaning no real solutions.So, I think my conclusion is correct. There are no pairs (a, b) of non-zero real numbers that satisfy the given equation.**Final Answer**boxed{A}

💡Alright, so I have this problem here: there's a line l given by the equation y equals the square root of 3 times x, and then there's a point P(x, y) that's moving around on the circle defined by (x minus 2) squared plus y squared equals 1. I need to find the minimum distance from point P to the line l. Hmm, okay.First, I remember that the distance from a point to a line can be calculated using a specific formula. Let me recall what that formula is. I think it's something like the absolute value of (Ax + By + C) divided by the square root of (A squared plus B squared), right? Yeah, that sounds familiar. So, if I have a line in the form Ax + By + C = 0, then the distance from a point (x0, y0) to that line is |Ax0 + By0 + C| divided by sqrt(A squared plus B squared).So, in this case, the line l is given by y = sqrt(3)x. I need to write this in the standard form Ax + By + C = 0. Let me rearrange it: sqrt(3)x minus y equals 0. So, A is sqrt(3), B is -1, and C is 0.Now, the point P(x, y) is moving on the circle (x - 2)^2 + y^2 = 1. That means the center of the circle is at (2, 0) and the radius is 1. So, the circle is centered at (2, 0) with radius 1.I think the strategy here is to find the distance from the center of the circle to the line l, and then subtract the radius of the circle to get the minimum distance from any point on the circle to the line. Is that right? Let me think. If the distance from the center to the line is greater than the radius, then the minimum distance from the circle to the line would be that distance minus the radius. If it's less, then the minimum distance would be zero because the line would intersect the circle. But in this case, I think the distance from the center to the line is more than the radius, so subtracting the radius will give the minimum distance.Okay, so let's calculate the distance from the center (2, 0) to the line l. Using the formula I mentioned earlier, it's |A*2 + B*0 + C| divided by sqrt(A squared plus B squared). Plugging in the values, A is sqrt(3), B is -1, and C is 0.So, the numerator becomes |sqrt(3)*2 + (-1)*0 + 0|, which simplifies to |2*sqrt(3)|. The denominator is sqrt((sqrt(3))^2 + (-1)^2). Calculating that, (sqrt(3))^2 is 3, and (-1)^2 is 1, so the denominator is sqrt(3 + 1) which is sqrt(4), and that's 2.So, the distance from the center to the line is |2*sqrt(3)| divided by 2, which is sqrt(3). That makes sense because sqrt(3) is approximately 1.732, which is greater than the radius of 1. So, the minimum distance from the circle to the line should be sqrt(3) minus 1.Wait, let me make sure I'm not making a mistake here. Is it always just the distance from the center minus the radius? I think so, as long as the line doesn't intersect the circle. Since sqrt(3) is about 1.732, and the radius is 1, the distance from the center to the line is greater than the radius, so the line doesn't intersect the circle. Therefore, the closest point on the circle to the line would be along the line connecting the center to the given line, and that distance would be sqrt(3) minus 1.Let me visualize this. Imagine the line l: y = sqrt(3)x, which has a slope of sqrt(3), meaning it's a fairly steep line. The circle is centered at (2, 0) with a radius of 1. So, the center is to the right of the origin, and the circle is relatively small. The distance from the center to the line is sqrt(3), which is about 1.732, so the closest point on the circle to the line would be sqrt(3) minus 1, which is approximately 0.732.Just to double-check, if I were to parametrize the circle, I could express any point P on the circle as (2 + cos(theta), sin(theta)), where theta is the angle parameter. Then, I could plug that into the distance formula from a point to the line and find the minimum distance. But that might be more complicated, and since the distance from the center is straightforward, subtracting the radius should suffice.Alternatively, I can think about the formula for the distance from a point to a line and how it relates to the circle. The distance from any point on the circle to the line will vary, but the minimum distance should occur along the line connecting the center to the given line. So, moving from the center towards the line by the radius length would give the closest point on the circle to the line.Therefore, I'm pretty confident that the minimum distance is sqrt(3) minus 1.**Final Answer**The minimum distance from point ( P ) to line ( l ) is (boxed{sqrt{3} - 1}).

💡Okay, so I have this problem here: When (x^{9} + 3) is divided by (x - 2), what is the remainder? The options are A) 205, B) 515, C) 1024, D) 305, and E) None of these answers. Hmm, I need to figure out the remainder when this polynomial is divided by (x - 2). I remember something called the Remainder Theorem. Let me think... Oh yeah, the Remainder Theorem says that if you divide a polynomial (f(x)) by (x - a), the remainder is just (f(a)). So in this case, since we're dividing by (x - 2), that means (a) is 2. So I just need to plug in 2 into the polynomial (x^{9} + 3) and see what I get.Alright, let's calculate that. (2^{9}) is... Let me break it down. (2^1 = 2), (2^2 = 4), (2^3 = 8), (2^4 = 16), (2^5 = 32), (2^6 = 64), (2^7 = 128), (2^8 = 256), and (2^9 = 512). So, (2^{9} = 512). Then I add 3 to that, so 512 + 3 equals 515. So, according to the Remainder Theorem, the remainder should be 515. Let me check the options: A is 205, B is 515, C is 1024, D is 305, and E is None of these. So 515 is option B. Wait, just to make sure I didn't make a mistake, let me go through it again. (2^9) is 512, right? Yeah, because (2^10) is 1024, so (2^9) is half of that, which is 512. Adding 3 gives 515. Yeah, that seems right. I guess another way to think about it is by polynomial long division, but that would take a lot longer. The Remainder Theorem is much quicker. So I think I'm confident that the remainder is 515, which is option B. I don't see any mistakes in my calculations, so I think that's the correct answer.