💡Okay, so I've got this problem here: Given the expansion of (1 - 2x)^2017, which is equal to a0 + a1x + a2x² + ... + a2017x²⁰¹⁷. I need to find the value of the sum a1/2 + a2/2² + ... + a2017/2²⁰¹⁷. The options are A: 2, B: 0, C: -1, D: -2.Hmm, let's see. I remember that when you have a polynomial expansion like this, you can find the coefficients by plugging in specific values of x. Maybe I can use that here.First, if I set x = 0, the entire expression becomes (1 - 0)^2017, which is just 1. On the right side, plugging in x = 0 gives a0 + 0 + 0 + ... + 0, so a0 must be 1. That's straightforward.Now, the sum I need is a1/2 + a2/2² + ... + a2017/2²⁰¹⁷. This looks like evaluating the polynomial at x = 1/2, but without the a0 term. Because if I plug x = 1/2 into the expansion, I get a0 + a1*(1/2) + a2*(1/2)² + ... + a2017*(1/2)²⁰¹⁷. So, that's exactly the sum I need plus a0.So, if I compute (1 - 2*(1/2))^2017, that should give me the value of the polynomial at x = 1/2. Let's calculate that: 1 - 2*(1/2) is 1 - 1, which is 0. So, 0^2017 is 0. Therefore, the sum a0 + a1/2 + a2/2² + ... + a2017/2²⁰¹⁷ equals 0. But I already found that a0 is 1, so subtracting a0 from both sides gives me a1/2 + a2/2² + ... + a2017/2²⁰¹⁷ = -1.Wait, that seems too straightforward. Let me double-check. If I plug x = 1/2 into (1 - 2x)^2017, I get (1 - 1)^2017 = 0. On the right side, that's a0 + a1*(1/2) + a2*(1/2)² + ... + a2017*(1/2)²⁰¹⁷. Since a0 is 1, the rest of the sum must be -1 to make the total 0. Yeah, that makes sense.So, the answer should be -1, which is option C.I don't think I made any mistakes here. It's a standard technique in binomial expansions to plug in specific x values to find sums of coefficients. Plugging in x = 0 gives a0, and plugging in x = 1/2 gives the sum including a0, so subtracting a0 gives the desired sum. Everything checks out.**Final Answer**boxed{C}

💡Alright, so I have this problem where I need to cut a square into four equal parts, and each part has to have the same size and shape. Plus, each part needs to contain one circle and one star. Hmm, okay, let's see.First, I need to visualize the square. It's divided into smaller cells, and within these cells, there are circles and stars arranged in some pattern. The goal is to make cuts along the cell borders so that each of the four resulting pieces has exactly one circle and one star.I guess the first step is to figure out where the circles and stars are located. If I can map out their positions, I can better plan how to make the cuts. Maybe there's a symmetrical arrangement? That would make things easier because symmetry often helps in dividing shapes equally.Let me think about symmetry. If the square has circles and stars placed symmetrically, then making cuts along the vertical and horizontal midlines could work. But wait, if I just cut it into four equal quadrants, would each quadrant have one circle and one star? I'm not sure. Maybe the symbols are not placed symmetrically.Alternatively, perhaps the symbols are arranged in a grid pattern. If that's the case, I can try to identify rows and columns that have a balanced number of circles and stars. For example, if each row has two circles and two stars, then cutting along the middle of the rows and columns might distribute the symbols evenly.But I need to be careful. If I make a vertical cut, I have to ensure that the number of circles and stars on both sides of the cut is equal. The same goes for a horizontal cut. It's like solving a puzzle where each piece must have exactly one circle and one star.Maybe I can start by counting the total number of circles and stars. If there are four circles and four stars, then each part should have one circle and one star. That makes sense. But if there are more or fewer, I need to adjust my approach.Assuming there are four circles and four stars, I can proceed. Now, how do I ensure that each part has one of each? Perhaps by making sure that each cut separates one circle and one star from the rest. It's like pairing each circle with a star in each quadrant.I'm also thinking about the shape of the pieces. They need to be the same size and shape. So, if I make straight cuts along the grid lines, the resulting pieces will be rectangles or squares, which are easy to manage. But if the symbols are not aligned in a way that allows for such simple cuts, I might need to make more complex shapes.Wait, maybe the key is to look for patterns or pairs. If I can identify pairs of circles and stars that are adjacent or close to each other, I can design cuts that isolate these pairs into separate regions. That way, each region will have one circle and one star.But how do I ensure the regions are the same size and shape? That's the tricky part. I need to make sure that the cuts are symmetrical and that each resulting piece is congruent. Maybe by using rotational symmetry, where each piece is a rotation of the others.I'm also considering if there's a standard method for such problems. I recall that sometimes dividing a shape into congruent parts with specific elements can be achieved by identifying centers of symmetry or using folding techniques. Maybe I can imagine folding the square in certain ways to see how the symbols align.Another thought: if I number the cells from 1 to 16 (assuming a 4x4 grid), I can assign each cell a position and then map out where the circles and stars are. This might help in systematically planning the cuts.But without seeing the actual image, it's a bit challenging. I have to rely on the description and my imagination. Maybe I can sketch a rough grid and place circles and stars in a way that seems balanced. Then, I can try making cuts and checking if each quadrant meets the requirements.I'm also thinking about the properties of squares and symmetry. A square has four lines of symmetry: two diagonals, the vertical midline, and the horizontal midline. If I use these lines to make cuts, I can ensure that the resulting pieces are congruent.But again, the placement of circles and stars is crucial. If they're not aligned with these symmetry lines, the cuts might not work as intended. I need to make sure that each cut separates one circle and one star into separate regions.Maybe I can approach this step by step. First, identify a circle and a star that can be grouped together in one quadrant. Then, plan the cuts accordingly, ensuring that the remaining symbols are distributed equally among the other quadrants.It's also important to verify that after making the cuts, each piece has exactly one circle and one star. I can do this by checking each quadrant individually, counting the symbols, and ensuring they match the requirement.In summary, my approach would be:1. Visualize or sketch the square with circles and stars.2. Identify the positions of circles and stars.3. Plan cuts along symmetry lines or grid lines to ensure equal distribution.4. Verify that each resulting piece has one circle and one star.5. Adjust the cuts if necessary to meet all requirements.This seems like a logical way to tackle the problem. I just need to be meticulous in planning and checking each step to ensure the final result meets all the given conditions.

💡Alright, so I have this problem where I need to find the sum of three integers a, b, and c. The problem states that the greatest common divisor (GCD) of two quadratic polynomials, x² + ax + b and x² + bx + c, is x + 1. Additionally, the least common multiple (LCM) of these two polynomials is given as x³ - 4x² + x + 6. My goal is to find a + b + c.First, I remember that for any two polynomials, the product of their GCD and LCM equals the product of the polynomials themselves. So, if I denote the two quadratics as f(x) = x² + ax + b and g(x) = x² + bx + c, then:f(x) * g(x) = GCD(f(x), g(x)) * LCM(f(x), g(x))Given that the GCD is x + 1 and the LCM is x³ - 4x² + x + 6, I can write:(x² + ax + b)(x² + bx + c) = (x + 1)(x³ - 4x² + x + 6)Now, I need to compute the right-hand side. Let me expand (x + 1)(x³ - 4x² + x + 6). Multiplying term by term:First, x * x³ = x⁴x * (-4x²) = -4x³x * x = x²x * 6 = 6xThen, 1 * x³ = x³1 * (-4x²) = -4x²1 * x = x1 * 6 = 6Now, combining all these terms:x⁴ - 4x³ + x² + 6x + x³ - 4x² + x + 6Combine like terms:x⁴ + (-4x³ + x³) + (x² - 4x²) + (6x + x) + 6Simplify:x⁴ - 3x³ - 3x² + 7x + 6So, the right-hand side simplifies to x⁴ - 3x³ - 3x² + 7x + 6.Now, the left-hand side is (x² + ax + b)(x² + bx + c). Let me expand this product:(x² + ax + b)(x² + bx + c) = x²*x² + x²*bx + x²*c + ax*x² + ax*bx + ax*c + b*x² + b*bx + b*cSimplify each term:x⁴ + bx³ + cx² + ax³ + abx² + acx + bx² + b²x + bcCombine like terms:x⁴ + (b + a)x³ + (c + ab + b)x² + (ac + b²)x + bcSo, the expanded form is:x⁴ + (a + b)x³ + (ab + b + c)x² + (ac + b²)x + bcNow, set this equal to the right-hand side which we found earlier:x⁴ - 3x³ - 3x² + 7x + 6Therefore, we can equate the coefficients of corresponding powers of x:1. Coefficient of x⁴: 1 = 1 (which is already satisfied)2. Coefficient of x³: a + b = -33. Coefficient of x²: ab + b + c = -34. Coefficient of x: ac + b² = 75. Constant term: bc = 6So, we have a system of equations:1. a + b = -32. ab + b + c = -33. ac + b² = 74. bc = 6Now, I need to solve this system for integers a, b, c.Let me start with equation 4: bc = 6. Since b and c are integers, possible pairs (b, c) are:(1, 6), (2, 3), (3, 2), (6, 1), (-1, -6), (-2, -3), (-3, -2), (-6, -1)I can test these pairs in the other equations to see which ones satisfy all conditions.Let me also note equation 1: a + b = -3, so a = -3 - b.Let me substitute a = -3 - b into the other equations.Equation 2: ab + b + c = -3Substitute a:(-3 - b)b + b + c = -3Expand:-3b - b² + b + c = -3Simplify:-2b - b² + c = -3So, c = b² + 2b - 3Equation 3: ac + b² = 7Substitute a = -3 - b and c from equation 2:(-3 - b)c + b² = 7But c = b² + 2b - 3, so:(-3 - b)(b² + 2b - 3) + b² = 7Let me expand this:First, multiply (-3 - b) with (b² + 2b - 3):-3*(b² + 2b - 3) = -3b² -6b + 9-b*(b² + 2b - 3) = -b³ -2b² + 3bCombine these:-3b² -6b + 9 - b³ -2b² + 3b = -b³ -5b² -3b + 9Now, add b²:(-b³ -5b² -3b + 9) + b² = -b³ -4b² -3b + 9Set equal to 7:-b³ -4b² -3b + 9 = 7Subtract 7:-b³ -4b² -3b + 2 = 0Multiply both sides by -1:b³ + 4b² + 3b - 2 = 0Now, I need to find integer roots of this cubic equation. Possible rational roots are factors of 2 over factors of 1, so ±1, ±2.Let me test b = 1:1 + 4 + 3 - 2 = 6 ≠ 0b = -1:-1 + 4 - 3 - 2 = -2 ≠ 0b = 2:8 + 16 + 6 - 2 = 28 ≠ 0b = -2:-8 + 16 -6 -2 = 0Yes, b = -2 is a root.So, (b + 2) is a factor. Let's perform polynomial division or factor it out.Divide b³ + 4b² + 3b - 2 by (b + 2):Using synthetic division:-2 | 1 4 3 -2 -2 -4 2 1 2 -1 0So, the cubic factors as (b + 2)(b² + 2b - 1) = 0Set equal to zero:(b + 2)(b² + 2b - 1) = 0So, b = -2 or b² + 2b -1 = 0The quadratic equation b² + 2b -1 = 0 has solutions:b = [-2 ± sqrt(4 + 4)] / 2 = [-2 ± sqrt(8)] / 2 = [-2 ± 2√2]/2 = -1 ± √2But since b must be integer, only b = -2 is valid.So, b = -2Now, from equation 1: a + b = -3 => a = -3 - (-2) = -1From equation 4: bc = 6 => c = 6 / b = 6 / (-2) = -3So, a = -1, b = -2, c = -3Let me verify these values in equation 2 and 3.Equation 2: ab + b + c = (-1)(-2) + (-2) + (-3) = 2 -2 -3 = -3, which matches.Equation 3: ac + b² = (-1)(-3) + (-2)^2 = 3 + 4 = 7, which also matches.So, the values are consistent.Therefore, a = -1, b = -2, c = -3Thus, a + b + c = -1 + (-2) + (-3) = -6**Final Answer**The sum of a, b, and c is boxed{-6}.

💡Okay, so I have this problem where I'm given a complex number ( z ) with ( |z| = 1 ), which means it's on the unit circle in the complex plane. The argument of ( z ) is ( theta ), so I can write ( z ) as ( costheta + isintheta ). I need to find the locus of the point ( p = z^2 + 2z costheta ) as ( z ) varies around the unit circle. The options are a circle, hyperbola, parabola, or two intersecting lines.First, I should probably express ( z ) in terms of ( theta ). Since ( |z| = 1 ), ( z = e^{itheta} = costheta + isintheta ). That makes sense.Now, let's compute ( z^2 ). Using the formula for squaring a complex number in polar form, ( z^2 = (costheta + isintheta)^2 ). Expanding this, I get ( cos^2theta - sin^2theta + 2icosthetasintheta ). Using double-angle identities, this simplifies to ( cos2theta + isin2theta ). Okay, that's straightforward.Next, I need to compute ( 2z costheta ). Substituting ( z = costheta + isintheta ), this becomes ( 2(costheta + isintheta)costheta ). Multiplying through, I get ( 2cos^2theta + 2icosthetasintheta ). Hmm, that's also correct.Now, adding ( z^2 ) and ( 2zcostheta ) together, I have:( p = z^2 + 2zcostheta = (cos2theta + isin2theta) + (2cos^2theta + 2icosthetasintheta) ).Let me combine the real and imaginary parts separately. The real part is ( cos2theta + 2cos^2theta ) and the imaginary part is ( sin2theta + 2costhetasintheta ).I remember that ( cos2theta = 2cos^2theta - 1 ), so substituting that into the real part gives ( (2cos^2theta - 1) + 2cos^2theta = 4cos^2theta - 1 ). Wait, is that right? Let me check:Wait, ( cos2theta = 2cos^2theta - 1 ), so ( cos2theta + 2cos^2theta = (2cos^2theta - 1) + 2cos^2theta = 4cos^2theta - 1 ). Yes, that seems correct.For the imaginary part, ( sin2theta = 2sinthetacostheta ), so ( sin2theta + 2costhetasintheta = 2sinthetacostheta + 2sinthetacostheta = 4sinthetacostheta ). Alternatively, this is ( 2sin2theta ). Hmm, that's another way to write it.So, putting it all together, ( p = (4cos^2theta - 1) + i(2sin2theta) ). Hmm, maybe I can express this in terms of a single angle. Let me see.Let me denote ( x = 4cos^2theta - 1 ) and ( y = 2sin2theta ). I need to eliminate ( theta ) to find the relation between ( x ) and ( y ).From ( x = 4cos^2theta - 1 ), I can solve for ( cos^2theta ):( cos^2theta = frac{x + 1}{4} ).Also, from ( y = 2sin2theta ), and since ( sin2theta = 2sinthetacostheta ), we have ( y = 4sinthetacostheta ).But ( sinthetacostheta = frac{1}{2}sin2theta ), so ( y = 2sin2theta ), which is consistent.Now, I can use the identity ( sin^2theta + cos^2theta = 1 ). Let me express ( sintheta ) and ( costheta ) in terms of ( x ) and ( y ).From ( cos^2theta = frac{x + 1}{4} ), so ( costheta = sqrt{frac{x + 1}{4}} ) or ( -sqrt{frac{x + 1}{4}} ). Similarly, ( sintheta = frac{y}{4costheta} ) from ( y = 4sinthetacostheta ).But this might get complicated. Maybe another approach. Let me express ( cos2theta ) and ( sin2theta ) in terms of ( x ) and ( y ).Wait, from ( x = 4cos^2theta - 1 ), and since ( cos2theta = 2cos^2theta - 1 ), then ( cos2theta = 2left(frac{x + 1}{4}right) - 1 = frac{x + 1}{2} - 1 = frac{x - 1}{2} ).Similarly, ( sin2theta = frac{y}{2} ) because ( y = 2sin2theta ).Now, using the identity ( cos^22theta + sin^22theta = 1 ), substituting the expressions in terms of ( x ) and ( y ):( left(frac{x - 1}{2}right)^2 + left(frac{y}{2}right)^2 = 1 ).Expanding this, we get:( frac{(x - 1)^2}{4} + frac{y^2}{4} = 1 ).Multiplying both sides by 4:( (x - 1)^2 + y^2 = 4 ).This is the equation of a circle with center at (1, 0) and radius 2.Wait, but earlier I thought the center was (-1, 0). Did I make a mistake? Let me check.From ( x = 4cos^2theta - 1 ), when ( theta = 0 ), ( costheta = 1 ), so ( x = 4(1) - 1 = 3 ), and ( y = 2sin0 = 0 ). So the point is (3, 0). When ( theta = pi/2 ), ( costheta = 0 ), so ( x = 4(0) - 1 = -1 ), and ( y = 2sinpi = 0 ). Wait, that's ( -1, 0). Hmm, but according to the equation, the center is at (1, 0). But when ( theta = 0 ), the point is (3, 0), which is 2 units away from (1, 0). Similarly, when ( theta = pi/2 ), the point is (-1, 0), which is 2 units away from (1, 0). So yes, the center is at (1, 0) and radius 2.Wait, but in my earlier step, I thought the center was (-1, 0). That was a mistake. The equation clearly shows the center is at (1, 0).So, the locus is a circle with center (1, 0) and radius 2.Therefore, the answer should be (A) A circle.

💡Okay, so I have this problem here: A pizza parlor offers six different toppings. How many different one-, two-, and three-topping pizzas can you order? Hmm, let me try to figure this out step by step.First, I need to understand what the question is asking. It wants to know the total number of different pizzas you can make by choosing one topping, two toppings, or three toppings from the six available. So, I need to calculate the number of possible combinations for each case and then add them all together.Starting with one-topping pizzas. If there are six different toppings, then for one-topping pizzas, it's straightforward. Each topping can be a pizza on its own. So, that should be 6 different one-topping pizzas. Okay, that part seems easy.Now, moving on to two-topping pizzas. This is where it gets a bit more complicated because now I have to consider combinations of two different toppings. I remember that combinations are used when the order doesn't matter, which is the case here because a pizza with pepperoni and mushrooms is the same as a pizza with mushrooms and pepperoni. So, I need to calculate the number of combinations of six toppings taken two at a time.The formula for combinations is C(n, k) = n! / (k!(n - k)!), where n is the total number of items, and k is the number of items to choose. Plugging in the numbers, C(6, 2) = 6! / (2!(6 - 2)!) = (6 × 5 × 4!) / (2 × 1 × 4!) = (6 × 5) / 2 = 30 / 2 = 15. So, there are 15 different two-topping pizzas.Next, for three-topping pizzas. Using the same combination formula, I need to find C(6, 3). So, C(6, 3) = 6! / (3!(6 - 3)!) = (6 × 5 × 4 × 3!) / (3 × 2 × 1 × 3!) = (6 × 5 × 4) / (3 × 2 × 1) = 120 / 6 = 20. Therefore, there are 20 different three-topping pizzas.Now, to find the total number of different pizzas, I need to add up the one-topping, two-topping, and three-topping pizzas. That would be 6 + 15 + 20. Let me do the addition: 6 + 15 is 21, and 21 + 20 is 41. So, the total number of different pizzas you can order is 41.Wait, let me double-check my calculations to make sure I didn't make any mistakes. For one-topping pizzas, it's definitely 6 because each topping is unique. For two-topping pizzas, using the combination formula, I got 15, which seems right because 6 choose 2 is a standard combination problem, and I remember that 6 choose 2 is 15. For three-topping pizzas, 6 choose 3 is 20, which also checks out because 6 choose 3 is a common combination and equals 20. Adding them all together, 6 + 15 + 20 equals 41. Yeah, that seems correct.I think I've covered all the possibilities here. I considered each case separately and used the combination formula appropriately since the order of toppings doesn't matter on a pizza. There's no overlap between the different topping counts, so adding them together gives the correct total. I don't think I missed any cases or made any calculation errors. So, I'm confident that the total number of different one-, two-, and three-topping pizzas you can order is 41.

💡First, I need to determine how long Beth's trip took. Since Beth took 20 minutes more than Jerry, and Jerry's trip took 30 minutes, Beth's total travel time is 30 minutes + 20 minutes = 50 minutes. Converting this to hours, I divide by 60: 50 minutes ÷ 60 minutes/hour = 5/6 hours.Next, I'll calculate the distance Beth traveled using her average speed. Beth's average speed was 30 miles per hour, so the distance she covered is:Distance = Speed × TimeDistance = 30 miles/hour × 5/6 hours = 25 miles.Since Beth's route was 5 miles longer than Jerry's, I can find Jerry's distance by subtracting 5 miles from Beth's distance:Jerry's distance = 25 miles - 5 miles = 20 miles.Now, I know Jerry's trip was 20 miles and it took him 30 minutes. Converting his travel time to hours:30 minutes ÷ 60 minutes/hour = 1/2 hour.Finally, I'll calculate Jerry's average speed:Average speed = Distance / TimeAverage speed = 20 miles / (1/2 hour) = 40 miles per hour.So, Jerry's average speed was 40 miles per hour.