answer：Alright, so I have these two trigonometric expressions to simplify. Let me take them one at a time. Starting with the first one: (1) sin 420° cos 330° + sin(-690°) cos(-660°)Hmm, okay. I remember that trigonometric functions are periodic, so maybe I can reduce these angles to something within 0° to 360°, or even 0° to 90°, to make it easier. Let me recall the periods:- Sine and cosine have a period of 360°, so adding or subtracting 360° will give the same value.- Tangent has a period of 180°, but that's not directly relevant here.So, let's tackle each term one by one.First term: sin 420°. Since 420° is more than 360°, I can subtract 360° to find an equivalent angle. 420° - 360° = 60°. So sin 420° is the same as sin 60°. I remember that sin 60° is √3/2.Second term: cos 330°. 330° is in the fourth quadrant. Cosine is positive in the fourth quadrant. 330° is 360° - 30°, so it's equivalent to cos(-30°), which is the same as cos 30°. Cos 30° is √3/2. So cos 330° is √3/2.So the first part, sin 420° cos 330°, becomes (√3/2)(√3/2). Multiplying those together: √3 * √3 is 3, and 2*2 is 4, so 3/4.Now the second part: sin(-690°) cos(-660°). Negative angles mean we can add 360° until we get a positive angle within 0° to 360°, or use the fact that sin(-θ) = -sin θ and cos(-θ) = cos θ.Let me apply the identities first:sin(-690°) = -sin 690°cos(-660°) = cos 660°So now, we have -sin 690° * cos 660°. Let's simplify sin 690° and cos 660°.Starting with sin 690°: 690° is more than 360°, so subtract 360°: 690 - 360 = 330°. So sin 690° is sin 330°. Sin 330° is in the fourth quadrant, where sine is negative. 330° is 360° - 30°, so sin 330° = -sin 30° = -1/2.Similarly, cos 660°: 660° is more than 360°, so subtract 360°: 660 - 360 = 300°. Cos 300° is in the fourth quadrant, where cosine is positive. 300° is 360° - 60°, so cos 300° = cos 60° = 1/2.So putting it back into the expression: -sin 690° * cos 660° = -(-1/2) * (1/2). Let's compute that step by step.First, sin(-690°) is -sin 690°, which is -(-1/2) = 1/2.Wait, hold on. Let me clarify:sin(-690°) = -sin 690°, and sin 690° = sin(330°) = -1/2. So sin(-690°) = -(-1/2) = 1/2.Similarly, cos(-660°) = cos 660°, which is cos(300°) = 1/2.So the second term is sin(-690°) * cos(-660°) = (1/2) * (1/2) = 1/4.Wait, hold on, that contradicts my earlier step. Let me check again.Wait, no, the second term is sin(-690°) * cos(-660°). Which is equal to (-sin 690°) * (cos 660°). As sin(-x) = -sin x, and cos(-x) = cos x.So sin(-690°) = -sin 690°, and sin 690° is sin(330°) = -1/2. So sin(-690°) = -(-1/2) = 1/2.Similarly, cos(-660°) = cos 660°, which is cos(300°) = 1/2.So multiplying them: (1/2) * (1/2) = 1/4.Wait, so the second term is 1/4.But hold on, in the original expression, it's sin(-690°) * cos(-660°). So that's (1/2) * (1/2) = 1/4.Therefore, the entire expression is first term + second term: 3/4 + 1/4 = 1.Wait, that seems too straightforward. Let me double-check each step.First term: sin 420° cos 330°.420° - 360° = 60°, so sin 420° = sin 60° = √3/2.330° is 360° - 30°, so cos 330° = cos(-30°) = cos 30° = √3/2.Multiplying: (√3/2)(√3/2) = (3/4). Correct.Second term: sin(-690°) cos(-660°).sin(-690°) = -sin 690°, and 690° - 2*360° = 690 - 720 = -30°, so sin 690° = sin(-30°) = -1/2. Therefore, sin(-690°) = -(-1/2) = 1/2.cos(-660°) = cos 660°, and 660° - 360° = 300°, so cos 660° = cos 300° = 1/2.Multiplying: (1/2)(1/2) = 1/4.So total expression: 3/4 + 1/4 = 1. Hmm, okay, that seems correct.Wait, but 690° - 360° is 330°, not -30°. Wait, 690° divided by 360° is 1.916..., so 690° - 360° = 330°, which is correct. So sin 690° = sin 330° = -1/2. So sin(-690°) = -sin 690° = -(-1/2) = 1/2. That's correct.Similarly, 660° - 360° = 300°, so cos 660° = cos 300° = 1/2. So that's correct.So, yes, the second term is 1/4, so total is 1.Okay, so the first expression simplifies to 1.Moving on to the second expression:(2) [sin(π/2 + α) cos(π/2 - α)] / cos(π + α) + [sin(π - α) cos(π/2 + α)] / sin(π + α)Hmm, okay. Let me recall some trigonometric identities to simplify each term.First, let's look at each part step by step.Starting with the first fraction:sin(π/2 + α) cos(π/2 - α) divided by cos(π + α).I remember that sin(π/2 + α) is equal to cos α, because sin(π/2 + α) = cos α.Similarly, cos(π/2 - α) is equal to sin α, because cos(π/2 - α) = sin α.So the numerator becomes cos α * sin α.The denominator is cos(π + α). I remember that cos(π + α) is equal to -cos α, because cosine is negative in the third quadrant and cos(π + α) = -cos α.So the first fraction becomes [cos α * sin α] / (-cos α). The cos α cancels out, leaving sin α / (-1) = -sin α.Okay, that's the first part.Now the second fraction:sin(π - α) cos(π/2 + α) divided by sin(π + α).Let's simplify each part.sin(π - α) is equal to sin α, because sine is positive in the second quadrant and sin(π - α) = sin α.cos(π/2 + α) is equal to -sin α, because cos(π/2 + α) = -sin α.So the numerator becomes sin α * (-sin α) = -sin² α.The denominator is sin(π + α). I remember that sin(π + α) is equal to -sin α, because sine is negative in the third quadrant and sin(π + α) = -sin α.So the second fraction becomes (-sin² α) / (-sin α). The negatives cancel, and sin α cancels, leaving sin α.So now, putting it all together:First fraction: -sin αSecond fraction: sin αSo the entire expression is (-sin α) + (sin α) = 0.Wait, that's interesting. So the entire expression simplifies to zero.Let me double-check each step to make sure I didn't make a mistake.First fraction:sin(π/2 + α) = cos αcos(π/2 - α) = sin αNumerator: cos α * sin αDenominator: cos(π + α) = -cos αSo [cos α sin α] / (-cos α) = -sin α. Correct.Second fraction:sin(π - α) = sin αcos(π/2 + α) = -sin αNumerator: sin α * (-sin α) = -sin² αDenominator: sin(π + α) = -sin αSo [-sin² α] / [-sin α] = sin α. Correct.Adding them: -sin α + sin α = 0. Perfect.So the second expression simplifies to 0.So, summarizing:(1) The first expression simplifies to 1.(2) The second expression simplifies to 0.**Final Answer**(1) boxed{1}(2) boxed{0}

answer：Okay, so I have this problem about finding the lateral surface area of a regular triangular prism. The height of the prism is given as ( h ), and there's a specific condition about a line passing through the center of the upper base and the midpoint of a side of the lower base being inclined at a 60-degree angle to the plane of the base. Hmm, that sounds a bit complex, but let me try to break it down step by step.First, let me recall what a regular triangular prism is. A regular triangular prism has two congruent equilateral triangles as its bases, and three rectangles as its lateral faces. The height of the prism is the distance between these two bases. So, in this case, the height is given as ( h ).Now, the problem mentions a line passing through the center of the upper base and the midpoint of a side of the lower base. Let me visualize this. The upper base is an equilateral triangle, so its center is the centroid. The centroid of an equilateral triangle is also its center of mass, located at the intersection of its medians. Each median divides the triangle into two smaller triangles of equal area.The lower base is also an equilateral triangle, and we're considering the midpoint of one of its sides. So, if I imagine the prism, the line in question goes from the centroid of the upper base down to the midpoint of a side on the lower base. This line is inclined at 60 degrees to the plane of the base. Since the base is a horizontal plane (assuming the prism is standing upright), the angle between this line and the horizontal is 60 degrees.I think I can model this line as a vector that connects the centroid of the upper base to the midpoint of the lower base's side. Let me denote the centroid of the upper base as point ( G ) and the midpoint of the lower base's side as point ( M ). So, the line ( GM ) makes a 60-degree angle with the base.To find the lateral surface area, I need to find the perimeter of the base multiplied by the height of the prism. Wait, no, actually, for a prism, the lateral surface area is the perimeter of the base times the height. But hold on, in a regular triangular prism, the lateral faces are rectangles, each with one side equal to the side length of the base and the other equal to the height of the prism. So, the lateral surface area should be the perimeter of the base times the height.But before I jump into that, I need to figure out the side length of the base triangle because I don't have it directly. The problem gives me the height ( h ) of the prism, but I need to relate this to the side length using the given angle condition.Let me consider the coordinates to model this. Let's place the prism in a coordinate system where the lower base is on the xy-plane, and the upper base is parallel to it at height ( h ). Let me define the coordinates of the lower base. Since it's an equilateral triangle, I can place one vertex at ( (0, 0, 0) ), another at ( (a, 0, 0) ), and the third at ( (a/2, (asqrt{3})/2, 0) ), where ( a ) is the side length of the base triangle.Similarly, the upper base will have vertices at ( (0, 0, h) ), ( (a, 0, h) ), and ( (a/2, (asqrt{3})/2, h) ).The centroid ( G ) of the upper base is the average of its vertices' coordinates. So, ( G ) will be at ( left( frac{0 + a + a/2}{3}, frac{0 + 0 + (asqrt{3})/2}{3}, h right) ). Calculating that, the x-coordinate is ( frac{3a/2}{3} = a/2 ), the y-coordinate is ( frac{(asqrt{3})/2}{3} = (asqrt{3})/6 ), and the z-coordinate is ( h ). So, ( G = left( a/2, (asqrt{3})/6, h right) ).Now, the midpoint ( M ) of a side of the lower base. Let me choose the side between ( (0, 0, 0) ) and ( (a, 0, 0) ) for simplicity. The midpoint of this side is ( left( a/2, 0, 0 right) ).So, the line ( GM ) connects ( G = left( a/2, (asqrt{3})/6, h right) ) to ( M = left( a/2, 0, 0 right) ). Let me find the vector representation of this line. The vector ( vec{GM} ) is ( M - G ), which is ( left( a/2 - a/2, 0 - (asqrt{3})/6, 0 - h right) = left( 0, - (asqrt{3})/6, -h right) ).The angle between this vector and the plane of the base is 60 degrees. Wait, actually, the angle between the line and the plane is 60 degrees. But when we talk about the angle between a line and a plane, it's defined as the complement of the angle between the line and the normal to the plane. Alternatively, sometimes it's defined as the angle between the line and its projection onto the plane. Hmm, I need to clarify this.I think in this context, the angle between the line and the plane is the angle between the line and its projection onto the plane. So, if the angle is 60 degrees, then the angle between the line and the plane is 60 degrees, which would mean that the angle between the line and the normal to the plane is 30 degrees, since they add up to 90 degrees.But let me verify that. The angle between a line and a plane is the smallest angle between the line and any line in the plane. It's equal to the complement of the angle between the line and the normal to the plane. So, if ( theta ) is the angle between the line and the plane, then ( 90^circ - theta ) is the angle between the line and the normal.Given that the angle is 60 degrees, so ( theta = 60^circ ), which means the angle between the line and the normal is ( 30^circ ).Alternatively, sometimes the angle between the line and the plane is considered as the angle between the line and its projection onto the plane, which would be 60 degrees. So, in that case, the angle between the line and the normal would be ( 90^circ - 60^circ = 30^circ ).Either way, the relationship between the components of the vector can be used to find the angle.Let me consider the vector ( vec{GM} = (0, - (asqrt{3})/6, -h) ). The projection of this vector onto the base plane (the xy-plane) is ( (0, - (asqrt{3})/6, 0) ). The magnitude of this projection is ( | vec{GM}_{proj} | = sqrt{0^2 + ( - (asqrt{3})/6 )^2 + 0^2 } = (asqrt{3})/6 ).The magnitude of the vector ( vec{GM} ) itself is ( | vec{GM} | = sqrt{0^2 + ( - (asqrt{3})/6 )^2 + (-h)^2 } = sqrt{ (a^2 * 3)/36 + h^2 } = sqrt{ a^2 / 12 + h^2 } ).Now, the angle ( theta ) between the line ( GM ) and its projection onto the base plane is given as 60 degrees. So, using trigonometry, the cosine of this angle is equal to the adjacent side over the hypotenuse, which is the projection's magnitude divided by the vector's magnitude.So, ( cos(60^circ) = frac{ | vec{GM}_{proj} | }{ | vec{GM} | } ).We know that ( cos(60^circ) = 0.5 ), so:( 0.5 = frac{ (asqrt{3}/6) }{ sqrt{ a^2 / 12 + h^2 } } ).Let me write that equation:( frac{asqrt{3}/6}{sqrt{a^2 / 12 + h^2}} = 0.5 ).Let me solve for ( a ) in terms of ( h ).First, cross-multiplying:( asqrt{3}/6 = 0.5 times sqrt{a^2 / 12 + h^2} ).Multiply both sides by 6:( asqrt{3} = 3 times sqrt{a^2 / 12 + h^2} ).Now, square both sides to eliminate the square root:( (asqrt{3})^2 = (3)^2 times (a^2 / 12 + h^2) ).Calculating left side:( 3a^2 = 9 times (a^2 / 12 + h^2) ).Simplify the right side:( 3a^2 = 9*(a^2 / 12) + 9*h^2 )( 3a^2 = (9a^2)/12 + 9h^2 )Simplify ( (9a^2)/12 ) to ( (3a^2)/4 ):( 3a^2 = (3a^2)/4 + 9h^2 )Subtract ( (3a^2)/4 ) from both sides:( 3a^2 - (3a^2)/4 = 9h^2 )( (12a^2 - 3a^2)/4 = 9h^2 )( (9a^2)/4 = 9h^2 )Multiply both sides by 4:( 9a^2 = 36h^2 )Divide both sides by 9:( a^2 = 4h^2 )Take square roots:( a = 2h )So, the side length ( a ) of the base triangle is ( 2h ).Now, knowing that, let's find the lateral surface area. The lateral surface area of a prism is given by the perimeter of the base times the height of the prism. For a regular triangular prism, the base is an equilateral triangle with side length ( a ), so the perimeter is ( 3a ).Therefore, lateral surface area ( LSA = 3a times h ).Substituting ( a = 2h ):( LSA = 3*(2h)*h = 6h^2 ).Wait, hold on, that seems straightforward, but let me double-check my steps because sometimes when dealing with angles and projections, it's easy to mix up the relationships.So, starting from the vector ( vec{GM} = (0, - (asqrt{3})/6, -h) ). The projection onto the base is ( (0, - (asqrt{3})/6, 0) ), which has magnitude ( (asqrt{3})/6 ). The magnitude of ( vec{GM} ) is ( sqrt{(asqrt{3}/6)^2 + h^2} ). Then, the angle between ( vec{GM} ) and its projection is 60 degrees, so ( cos(60^circ) = ) adjacent / hypotenuse, which is ( (asqrt{3}/6) / | vec{GM} | ). Plugging in the values, we arrived at ( a = 2h ).So, the side length is ( 2h ), so the perimeter is ( 6h ), and multiplying by height ( h ) gives ( 6h^2 ). That seems correct.Alternatively, let me think about another approach to verify. Maybe using the triangle formed by the centroid, midpoint, and the height.In the prism, the line from centroid ( G ) to midpoint ( M ) forms a triangle with the height of the prism. The vertical component is the height ( h ), and the horizontal component is the distance from centroid to midpoint in the base.In an equilateral triangle, the centroid divides the median in a 2:1 ratio. So, the distance from centroid to a vertex is ( frac{2}{3} ) of the median, and the distance from centroid to the midpoint (which is a point on the median) is ( frac{1}{3} ) of the median.Wait, actually, in an equilateral triangle, the centroid is located at a distance of ( frac{sqrt{3}}{3}a ) from each side. Wait, let me clarify.The length of the median in an equilateral triangle is also the height of the triangle, which is ( frac{sqrt{3}}{2}a ). So, the centroid is located at ( frac{1}{3} ) of the median from the base, which is ( frac{sqrt{3}}{6}a ) from the base.So, in our case, the distance from centroid ( G ) to midpoint ( M ) in the base is ( frac{sqrt{3}}{6}a ). So, in the triangle formed by ( G ), ( M ), and the projection of ( G ) onto the base, we have a right triangle with one leg as ( frac{sqrt{3}}{6}a ), the other leg as ( h ), and the hypotenuse as the line ( GM ).Given that the angle between ( GM ) and the base is 60 degrees, which is the angle between the hypotenuse and the base. So, in this right triangle, the angle at ( M ) is 60 degrees.Wait, no. The angle between the line ( GM ) and the base is 60 degrees, which would correspond to the angle between the hypotenuse and the adjacent side (the projection). So, in the right triangle, the angle between the hypotenuse and the adjacent side is 60 degrees, so the opposite side is ( h ), adjacent is ( frac{sqrt{3}}{6}a ), and hypotenuse is ( sqrt{ (frac{sqrt{3}}{6}a)^2 + h^2 } ).So, using trigonometry, ( tan(60^circ) = frac{h}{ frac{sqrt{3}}{6}a } ).Since ( tan(60^circ) = sqrt{3} ), we have:( sqrt{3} = frac{h}{ frac{sqrt{3}}{6}a } ).Solving for ( a ):Multiply both sides by ( frac{sqrt{3}}{6}a ):( sqrt{3} * frac{sqrt{3}}{6}a = h ).Simplify ( sqrt{3} * sqrt{3} = 3 ):( frac{3}{6}a = h ).Simplify ( frac{3}{6} = frac{1}{2} ):( frac{1}{2}a = h ).Multiply both sides by 2:( a = 2h ).So, same result as before, ( a = 2h ). Therefore, the side length is ( 2h ), so the perimeter is ( 3 * 2h = 6h ). The lateral surface area is perimeter times height, so ( 6h * h = 6h^2 ).Therefore, the lateral surface area is ( 6h^2 ).Wait, but just to make sure, let me think about the angle again. The angle between the line and the plane is 60 degrees. So, in the right triangle, the angle between the hypotenuse and the adjacent side is 60 degrees, which is consistent with our calculation.Alternatively, if we consider the angle between the line and the normal, which would be 30 degrees, we can use sine instead of tangent. Let's try that.In the right triangle, the angle between the hypotenuse and the normal (which is along the height) is 30 degrees. So, ( sin(30^circ) = frac{text{opposite}}{text{hypotenuse}} = frac{ frac{sqrt{3}}{6}a }{ | vec{GM} | } ).But ( sin(30^circ) = 0.5 ), so:( 0.5 = frac{ frac{sqrt{3}}{6}a }{ sqrt{ (frac{sqrt{3}}{6}a)^2 + h^2 } } ).Which is the same equation as before, leading to ( a = 2h ). So, consistent result.Therefore, I can be confident that the side length ( a = 2h ). Thus, the lateral surface area is ( 6h^2 ).**Final Answer**The lateral surface area of the regular triangular prism is boxed{6h^2}.

answer：Okay, so I have this problem here: If the complex number ( z ) satisfies ( z^{24} = 1 ) and ( z^6 ) is a real number, how many such complex numbers exist? The options are (A) 0, (B) 4, (C) 6, (D) 12, (E) 24.Hmm, let me think. I remember that complex numbers satisfying ( z^n = 1 ) are called the nth roots of unity. So, ( z^{24} = 1 ) means that ( z ) is a 24th root of unity. These are evenly spaced around the unit circle in the complex plane, right? So, there are 24 distinct 24th roots of unity.Now, the problem also says that ( z^6 ) is a real number. So, I need to find all 24th roots of unity ( z ) such that when you raise them to the 6th power, the result is real. Let me recall that a complex number is real if and only if it lies on the real axis in the complex plane. So, ( z^6 ) must lie on the real axis. That means ( z^6 ) is either 1, -1, or any real number, but since ( z ) is a root of unity, ( z^6 ) must also be a root of unity. Specifically, since ( z^{24} = 1 ), ( z^6 ) must satisfy ( (z^6)^4 = 1 ), so ( z^6 ) is a 4th root of unity. The 4th roots of unity are 1, -1, ( i ), and ( -i ). But wait, ( z^6 ) is supposed to be real, so it can't be ( i ) or ( -i ). Therefore, ( z^6 ) must be either 1 or -1.So, ( z^6 = 1 ) or ( z^6 = -1 ). Let me consider these two cases separately.First, if ( z^6 = 1 ), then ( z ) is a 6th root of unity. But ( z ) is also a 24th root of unity. So, the 6th roots of unity are a subset of the 24th roots of unity. How many 6th roots of unity are there? There are 6, right? They are ( e^{2pi i k/6} ) for ( k = 0, 1, 2, 3, 4, 5 ).Similarly, if ( z^6 = -1 ), then ( z ) is a 6th root of -1. But ( -1 ) is ( e^{pi i} ), so the 6th roots of -1 would be ( e^{pi i /6 + 2pi i k /6} ) for ( k = 0, 1, 2, 3, 4, 5 ). That gives another 6 roots.So, in total, we have 6 roots from ( z^6 = 1 ) and 6 roots from ( z^6 = -1 ), making 12 roots in total. But wait, are these all distinct? Let me check.The 6th roots of unity are ( e^{2pi i k/6} ), and the 6th roots of -1 are ( e^{pi i /6 + 2pi i k /6} ). Let me write them out:For ( z^6 = 1 ):- ( k = 0 ): ( e^{0} = 1 )- ( k = 1 ): ( e^{2pi i /6} = e^{pi i /3} )- ( k = 2 ): ( e^{4pi i /6} = e^{2pi i /3} )- ( k = 3 ): ( e^{6pi i /6} = e^{pi i} = -1 )- ( k = 4 ): ( e^{8pi i /6} = e^{4pi i /3} )- ( k = 5 ): ( e^{10pi i /6} = e^{5pi i /3} )For ( z^6 = -1 ):- ( k = 0 ): ( e^{pi i /6} )- ( k = 1 ): ( e^{pi i /6 + 2pi i /6} = e^{3pi i /6} = e^{pi i /2} = i )- ( k = 2 ): ( e^{pi i /6 + 4pi i /6} = e^{5pi i /6} )- ( k = 3 ): ( e^{pi i /6 + 6pi i /6} = e^{pi i /6 + pi i} = e^{7pi i /6} )- ( k = 4 ): ( e^{pi i /6 + 8pi i /6} = e^{9pi i /6} = e^{3pi i /2} = -i )- ( k = 5 ): ( e^{pi i /6 + 10pi i /6} = e^{11pi i /6} )Looking at these, none of the roots from ( z^6 = 1 ) are the same as those from ( z^6 = -1 ). Because the angles are different. For example, ( e^{pi i /3} ) is 60 degrees, while ( e^{pi i /6} ) is 30 degrees. Similarly, ( e^{2pi i /3} ) is 120 degrees, and ( e^{5pi i /6} ) is 150 degrees, and so on. So, all 12 roots are distinct.Therefore, the total number of such complex numbers ( z ) is 12.Wait, but hold on. Let me think again. Is there a possibility that some of these roots might not satisfy ( z^{24} = 1 )?No, because all the roots we considered are 24th roots of unity. The 6th roots of unity are a subset of the 24th roots of unity, and similarly, the 6th roots of -1 are also 24th roots of unity because ( (-1)^{24} = 1 ). So, all these 12 roots satisfy ( z^{24} = 1 ).Alternatively, another way to approach this is to consider the angle of ( z ). Since ( z ) is a 24th root of unity, it can be written as ( z = e^{2pi i k /24} ) where ( k ) is an integer from 0 to 23.Then, ( z^6 = e^{2pi i k /24 times 6} = e^{pi i k /2} ). For ( z^6 ) to be real, ( e^{pi i k /2} ) must be real. When is ( e^{itheta} ) real? It's when ( theta ) is a multiple of ( pi ), right? Because ( e^{itheta} ) is real if and only if ( theta ) is 0, ( pi ), ( 2pi ), etc., which correspond to 1, -1, 1, etc.So, ( e^{pi i k /2} ) is real when ( pi k /2 ) is a multiple of ( pi ), which happens when ( k /2 ) is an integer, meaning ( k ) is even.Wait, hold on. Let me clarify.( e^{pi i k /2} ) is real when ( pi k /2 ) is congruent to 0 or ( pi ) modulo ( 2pi ). So, ( pi k /2 = npi ) for some integer ( n ). That simplifies to ( k /2 = n ), so ( k = 2n ). So, ( k ) must be even.But ( k ) is an integer from 0 to 23. So, how many even ( k ) are there? From 0 to 23, there are 24 numbers, half of which are even. So, 12 even numbers. Therefore, there are 12 such ( z ).Wait, but hold on. If ( k ) is even, then ( z^6 = e^{pi i k /2} = e^{npi i} = (-1)^n ), which is either 1 or -1, both real numbers. So, that's consistent with our earlier conclusion.Therefore, the number of such complex numbers ( z ) is 12.So, the answer should be (D) 12.But let me double-check to make sure I didn't make a mistake.Another approach: The 24th roots of unity are solutions to ( z^{24} = 1 ). These are ( e^{2pi i k /24} ) for ( k = 0, 1, 2, ..., 23 ).We need ( z^6 ) to be real. So, ( z^6 = e^{2pi i k /24 times 6} = e^{pi i k /2} ). For this to be real, ( e^{pi i k /2} ) must be either 1 or -1.So, ( e^{pi i k /2} = 1 ) when ( pi k /2 = 2pi n ) for some integer ( n ), which implies ( k = 4n ). Similarly, ( e^{pi i k /2} = -1 ) when ( pi k /2 = pi + 2pi n ), which implies ( k = 2 + 4n ).So, ( k ) must be congruent to 0 or 2 modulo 4. That is, ( k ) is even, but more specifically, ( k ) is 0, 2, 4, ..., 22. So, how many such ( k ) are there?From 0 to 23, stepping by 2: that's 12 numbers. So, 12 values of ( k ) satisfy the condition, meaning 12 such complex numbers ( z ).Yes, that confirms it again. So, the answer is 12, which is option (D).**Final Answer**boxed{D}

answer：Okay, so I need to find the value of (2S), where (S) is the sum of these functions (f(m, 2008)) starting from (m=2) up to (m=2008). Each (f(m, n)) is defined as the product from (k = m) to (k = n) of ((1 - frac{1}{k})). First, let me write down what (f(m, n)) actually is. It's a product of terms like ((1 - frac{1}{k})), which simplifies to (frac{k-1}{k}). So, for each (f(m, n)), it's (frac{m-1}{m} times frac{m}{m+1} times frac{m+1}{m+2} times ldots times frac{n-1}{n}).Wait, that looks telescoping! Because when you multiply all these fractions together, most terms should cancel out. Let me test this with a small example. Suppose (m=2) and (n=4). Then (f(2,4)) would be (frac{1}{2} times frac{2}{3} times frac{3}{4}). Multiplying these together, the 2 in the numerator cancels with the 2 in the denominator, the 3 cancels similarly, leaving just (frac{1}{4}). So, in general, (f(m, n) = frac{m-1}{n}).Let me verify that. If (f(m, n) = prod_{k=m}^{n} left(1 - frac{1}{k}right)), then each term is (frac{k-1}{k}), so the product is (frac{m-1}{m} times frac{m}{m+1} times ldots times frac{n-1}{n}). The numerator of each fraction cancels with the denominator of the next one, so everything cancels except the first numerator and the last denominator. So, yes, (f(m, n) = frac{m-1}{n}).So, that simplifies things a lot. Therefore, each term (f(m, 2008)) is equal to (frac{m-1}{2008}). Therefore, the sum (S) is the sum from (m=2) to (m=2008) of (frac{m-1}{2008}).So, (S = sum_{m=2}^{2008} frac{m - 1}{2008}). Let me make a substitution to make this sum easier. Let (k = m - 1). Then when (m=2), (k=1), and when (m=2008), (k=2007). So, the sum becomes (S = sum_{k=1}^{2007} frac{k}{2008}).Therefore, (S = frac{1}{2008} sum_{k=1}^{2007} k). The sum of the first (n) integers is given by the formula (frac{n(n+1)}{2}). So, in this case, (n = 2007), so the sum is (frac{2007 times 2008}{2}).Therefore, substituting back, (S = frac{1}{2008} times frac{2007 times 2008}{2}). The 2008 in the numerator and denominator cancels out, leaving (S = frac{2007}{2}).Therefore, (2S = 2 times frac{2007}{2} = 2007).Wait, let me double-check that. So, (f(m, n) = frac{m-1}{n}). So, (f(2,2008) = frac{1}{2008}), (f(3,2008) = frac{2}{2008}), and so on up to (f(2008,2008) = frac{2007}{2008}). So, the sum (S) is (sum_{k=1}^{2007} frac{k}{2008}), which is (frac{1}{2008} times frac{2007 times 2008}{2}), which simplifies to (frac{2007}{2}). So, (2S = 2007). That seems correct.Alternatively, another way to think about it is recognizing that the sum (S) is the average of the first 2007 integers. Since the average of the first (n) integers is (frac{n+1}{2}), but wait, in this case, the sum is (frac{2007 times 2008}{2}), so when divided by 2008, it's (frac{2007}{2}), so yeah, that's consistent.Just to make sure, let me compute a smaller case. Let’s say (n=4), so (S = f(2,4) + f(3,4) + f(4,4)). From earlier, (f(2,4) = frac{1}{4}), (f(3,4) = frac{2}{4} = frac{1}{2}), (f(4,4) = frac{3}{4}). So, (S = frac{1}{4} + frac{1}{2} + frac{3}{4} = frac{1 + 2 + 3}{4} = frac{6}{4} = frac{3}{2}). Then, (2S = 3), which is equal to (n - 1 = 4 - 1 = 3). So, in this case, (2S = n - 1). So, in the original problem, (2S = 2008 - 1 = 2007). That seems to confirm the result.Therefore, I think the answer is 2007.**Final Answer**boxed{2007}

answer：First, I recognize that there are 40 students in total, and I need to find out how many of them do not play either football or volleyball.I know that 20 students play football and 19 play volleyball. The problem also states that 15 students play exactly one of these two sports. This means that 15 students play only football or only volleyball.To find out how many students play both sports, I can subtract the number of students who play exactly one sport from the total number of students who play either sport. So, 20 (football) + 19 (volleyball) - Both = Total playing either sport. Since 15 students play exactly one sport, the number of students who play both sports would be 20 + 19 - 15 = 24.However, this calculation doesn't seem right because the number of students playing both sports should not exceed the number of students playing either sport individually. I realize I made a mistake in my initial approach.Instead, I should use the principle of inclusion-exclusion. The total number of students playing either football or volleyball is equal to the number of football players plus the number of volleyball players minus the number of students who play both. So, 20 + 19 - Both = Total playing either sport.Given that 15 students play exactly one sport, the number of students who play both sports would be 20 + 19 - 15 = 24. This still doesn't make sense because it exceeds the number of volleyball players.I recognize that there's an inconsistency in the problem statement. The number of students who play exactly one sport cannot be greater than the total number of students playing either sport. Therefore, there must be an error in the problem's data.

answer：Alright, so I have this problem about a three-digit number with some specific properties. Let me try to break it down step by step. I need to find a number where:1. The sum of the digits in the units and hundreds places equals the digit in the tens place.2. Seven times the digit in the hundreds place is greater by 2 than the sum of the digits in the units and tens places.3. The sum of all three digits is 14.Hmm, okay. Let me assign variables to each digit to make it easier. Let's say the hundreds digit is ( H ), the tens digit is ( T ), and the units digit is ( U ). So the number is ( HTU ) in digits, which is actually ( 100H + 10T + U ) in value.Now, let's translate the given conditions into equations.First condition: The sum of the units and hundreds digits equals the tens digit. So that would be:[ H + U = T ]Got that.Second condition: Sevenfold of the hundreds digit is greater by 2 than the sum of the units and tens digits. So, seven times ( H ) is equal to the sum of ( U ) and ( T ) plus 2. So:[ 7H = U + T + 2 ]Alright.Third condition: The sum of all three digits is 14. So:[ H + T + U = 14 ]Perfect.So now I have three equations:1. ( H + U = T ) -- Equation (1)2. ( 7H = U + T + 2 ) -- Equation (2)3. ( H + T + U = 14 ) -- Equation (3)Let me see how I can solve these equations. Maybe substitution would work here.From Equation (1), I can express ( T ) in terms of ( H ) and ( U ):[ T = H + U ]So I can substitute this into Equations (2) and (3).Substituting into Equation (2):[ 7H = U + (H + U) + 2 ]Simplify that:[ 7H = U + H + U + 2 ]Combine like terms:[ 7H = 2U + H + 2 ]Subtract ( H ) from both sides:[ 6H = 2U + 2 ]Divide both sides by 2:[ 3H = U + 1 ]So, ( U = 3H - 1 ) -- Let's call this Equation (4)Now, substitute ( T = H + U ) into Equation (3):[ H + (H + U) + U = 14 ]Simplify:[ H + H + U + U = 14 ]Combine like terms:[ 2H + 2U = 14 ]Divide both sides by 2:[ H + U = 7 ]Wait, but from Equation (1), ( H + U = T ). So, ( T = 7 ). Interesting.So, now we know that ( T = 7 ). Let me note that down.So, ( T = 7 ). Now, from Equation (1):[ H + U = 7 ]And from Equation (4):[ U = 3H - 1 ]So, substitute ( U ) into ( H + U = 7 ):[ H + (3H - 1) = 7 ]Simplify:[ 4H - 1 = 7 ]Add 1 to both sides:[ 4H = 8 ]Divide by 4:[ H = 2 ]Okay, so the hundreds digit is 2.Now, find ( U ) using Equation (4):[ U = 3H - 1 = 3*2 - 1 = 6 - 1 = 5 ]So, ( U = 5 ).Let me recap:- ( H = 2 )- ( U = 5 )- ( T = 7 )So the number is 275. Let me verify if this satisfies all the given conditions.First condition: ( H + U = 2 + 5 = 7 ), which equals ( T = 7 ). Good.Second condition: ( 7H = 7*2 = 14 ). The sum of ( U + T = 5 + 7 = 12 ). So, 14 is greater than 12 by 2. Perfect.Third condition: ( H + T + U = 2 + 7 + 5 = 14 ). Yes, that's correct.So, all conditions are satisfied. Therefore, the three-digit number is 275.**Final Answer**The three-digit number is boxed{275}.