💡First, I need to ensure all numbers are in the same format to simplify the addition process. The given numbers are 85.9, 5.31, and 43/2. I notice that 43/2 is a fraction, so I'll convert it to a decimal.Next, I'll perform the division: 43 divided by 2 equals 21.5. Now, all numbers are in decimal form: 85.9, 5.31, and 21.5.Finally, I'll add the three decimal numbers together: 85.9 plus 5.31 plus 21.5 equals 112.71.

💡Okay, so I have this problem about a cylinder and a plane cutting through it. The plane passes through a generator of the cylinder, which I think means it passes through one of the straight lines that make up the cylinder's surface. This plane forms an acute angle α with the plane of the axial section that contains the same generator. Hmm, axial section probably means the section that includes the axis of the cylinder, right? So, if I imagine a cylinder standing upright, the axial section would be a rectangle when you cut it along its height.Now, the diagonal of the rectangle obtained from this section is equal to l, and this diagonal forms an angle β with the base plane. The base plane is probably the plane that's perpendicular to the cylinder's axis, so it's the circular base of the cylinder. I need to find the volume of the cylinder.Alright, let's start by recalling that the volume of a cylinder is given by V = πR²H, where R is the radius and H is the height. So, if I can find R and H, I can compute the volume.First, let's try to visualize the situation. There's a cylinder, and a plane cuts through it, passing through a generator. The plane makes an angle α with the axial section. The section created by this plane is a rectangle, and the diagonal of this rectangle is l, making an angle β with the base.I think it might help to draw a diagram. Since I can't draw, I'll try to imagine it. The axial section is a rectangle with height H and width 2R (since the diameter is 2R). The plane cutting through the cylinder at an angle α to this axial section will create another rectangle, but this rectangle is inclined at angle α. The diagonal of this inclined rectangle is l, and it makes an angle β with the base.Wait, so the diagonal l is in the inclined plane, and it forms an angle β with the base. That probably means that if we project the diagonal onto the base, the angle between the projection and the base is β. Hmm, not sure if that's the right way to think about it.Alternatively, maybe the diagonal l is in the inclined plane, and it makes an angle β with the base plane. So, if we consider the base plane as the horizontal plane, then the diagonal l is slanting upwards at angle β from the base.Let me think about the relationship between the diagonal l, the angle β, and the dimensions of the rectangle. In the inclined plane, the rectangle has sides equal to the height of the cylinder H and the length of the generator, which is the same as the circumference? Wait, no, the generator is just a straight line on the cylinder's surface, so the length of the generator would be the same as the height H, right?Wait, no, the generator is a line that's part of the cylinder's surface. If the cylinder is circular, the generators are all the straight lines that make up the cylinder. So, each generator has the same length as the height of the cylinder, H.So, the rectangle formed by the plane cutting through the cylinder has sides equal to H and something else. Since the plane is passing through a generator, the length of the generator is H, and the other side of the rectangle is the chord of the circular base.Wait, no, actually, if the plane cuts through the cylinder, it will create an ellipse, but since the plane is passing through a generator, it's actually a rectangle. Because when you cut a cylinder with a plane that contains a generator, the intersection is a rectangle. So, the rectangle has sides equal to H and 2R, but since it's inclined, the diagonal is l.Wait, no, the rectangle isn't necessarily 2R in width because the plane is inclined. So, the width of the rectangle would be the length of the chord in the circular base, but since the plane is passing through a generator, the chord is actually the diameter, which is 2R. Wait, but if the plane is inclined, the chord might not be the diameter. Hmm, I'm getting confused.Let me try to approach this step by step.1. The cylinder has radius R and height H.2. A plane passes through a generator, so it contains one of the straight lines (generators) on the cylinder's surface.3. This plane forms an angle α with the axial section. The axial section is the plane that contains the axis of the cylinder and the generator, so it's a rectangle with sides H and 2R.4. The plane in question is at an angle α to this axial section. So, the dihedral angle between the two planes is α.5. The intersection of the plane with the cylinder is a rectangle, and the diagonal of this rectangle is l, which makes an angle β with the base plane.So, the rectangle has sides H and something else. Let me denote the sides as H and x. The diagonal is l, so by the Pythagorean theorem, H² + x² = l².Also, the diagonal makes an angle β with the base plane. The base plane is the plane perpendicular to the cylinder's axis, so the angle β is the angle between the diagonal and the horizontal.In the rectangle, the diagonal makes an angle β with the horizontal side. So, tan β = opposite/adjacent = H / x.So, tan β = H / x, which gives x = H / tan β.But we also have H² + x² = l².Substituting x = H / tan β into the equation:H² + (H / tan β)² = l²H² + H² / tan² β = l²H² (1 + 1 / tan² β) = l²H² ( (tan² β + 1) / tan² β ) = l²But tan² β + 1 = sec² β, so:H² ( sec² β / tan² β ) = l²Simplify sec² β / tan² β:sec² β = 1 / cos² βtan² β = sin² β / cos² βSo, sec² β / tan² β = (1 / cos² β) / (sin² β / cos² β) = 1 / sin² βTherefore, H² (1 / sin² β) = l²So, H² = l² sin² βThus, H = l sin βOkay, so we found H in terms of l and β.Now, we need to find R.From the earlier substitution, x = H / tan β = (l sin β) / tan β = l sin β / (sin β / cos β) ) = l cos βSo, x = l cos βBut x is the other side of the rectangle, which is the length of the chord in the circular base. Wait, but since the plane is passing through a generator, the chord should be the diameter, right? Because the plane contains the generator and is inclined, so the intersection with the base is a diameter.Wait, but in that case, x should be equal to 2R, but we found x = l cos β. So, 2R = l cos β => R = (l cos β)/2But wait, is x equal to 2R? Let me think.When the plane cuts through the cylinder, passing through a generator, the intersection with the base is a straight line. Since the plane contains the generator, which is a straight line on the cylinder, the intersection with the base should be a diameter. Therefore, x is the diameter, which is 2R.But earlier, we found x = l cos β. So, 2R = l cos β => R = (l cos β)/2So, R = (l cos β)/2But wait, earlier, I thought that the rectangle has sides H and x, where x is the diameter. But if the plane is inclined at an angle α with respect to the axial section, then maybe x is not the diameter but something else.Wait, perhaps I made a mistake here. Let me reconsider.The plane is passing through a generator, so the intersection with the cylinder is a rectangle. The sides of this rectangle are H and the length of the chord in the base. But since the plane is inclined, the chord is not the diameter unless the plane is perpendicular to the base.Wait, no, if the plane contains a generator, which is a straight line on the cylinder, then the intersection with the base must be a straight line, which is a chord. But since the generator is a straight line on the cylinder, the chord must pass through the point where the generator meets the base. So, depending on the angle α, the chord can be of different lengths.Wait, this is getting complicated. Maybe I need to use some trigonometry here.Since the plane is inclined at an angle α with respect to the axial section, which is the plane containing the axis and the generator. So, the dihedral angle between the two planes is α.In the axial section, the rectangle has sides H and 2R. The other plane, which is inclined at angle α, also cuts the cylinder into a rectangle. The diagonal of this rectangle is l, and it makes an angle β with the base.So, perhaps we can relate the sides of the rectangle in the inclined plane to the sides in the axial section.Let me denote the sides of the inclined rectangle as H and y. Then, the diagonal is l, so H² + y² = l².Also, the angle β is the angle between the diagonal and the base, so tan β = H / y.So, from tan β = H / y, we get y = H / tan β.Substituting into H² + y² = l²:H² + (H / tan β)^2 = l²Which is the same equation as before, leading to H = l sin β and y = l cos β.But in the axial section, the rectangle has sides H and 2R, so the diagonal of the axial section is sqrt(H² + (2R)^2). But in our case, the plane is inclined at angle α, so the diagonal l is related to the diagonal of the axial section.Wait, maybe we can relate the two diagonals using the angle α.The diagonal of the inclined plane is l, and the diagonal of the axial section is sqrt(H² + (2R)^2). The angle between these two planes is α, so perhaps the angle between the diagonals is α.But I'm not sure. Alternatively, since the two planes intersect along the generator, the angle between the planes is α, so the angle between their normals is α.Alternatively, maybe we can consider the relationship between the areas or something else.Wait, perhaps it's better to think in terms of projection.The diagonal l in the inclined plane can be projected onto the axial section. Since the angle between the planes is α, the projection of l onto the axial section would be l cos α.But the projection of l onto the axial section should be equal to the diagonal of the axial section, which is sqrt(H² + (2R)^2).Wait, is that correct? Let me think.If we have two planes intersecting at an angle α, and a line segment l lies in one plane, then the projection of l onto the other plane is l cos α.So, the diagonal l in the inclined plane, when projected onto the axial section, becomes l cos α.But the diagonal in the axial section is sqrt(H² + (2R)^2). Therefore:sqrt(H² + (2R)^2) = l cos αWe already have H = l sin β, so substituting:sqrt( (l sin β)^2 + (2R)^2 ) = l cos αSquaring both sides:(l sin β)^2 + (2R)^2 = l² cos² αSo,l² sin² β + 4R² = l² cos² αSolving for R²:4R² = l² cos² α - l² sin² β4R² = l² (cos² α - sin² β)Therefore,R² = (l² / 4)(cos² α - sin² β)But wait, this seems a bit off because if α and β are both acute angles, cos² α - sin² β could be positive or negative, which doesn't make sense for R².Wait, maybe I made a mistake in the projection. Let me think again.The projection of the diagonal l onto the axial section is l cos α, but the diagonal in the axial section is sqrt(H² + (2R)^2). So, we have:sqrt(H² + (2R)^2) = l cos αBut we already found that H = l sin β, so substituting:sqrt( (l sin β)^2 + (2R)^2 ) = l cos αSquaring both sides:l² sin² β + 4R² = l² cos² αSo,4R² = l² cos² α - l² sin² β4R² = l² (cos² α - sin² β)Therefore,R² = (l² / 4)(cos² α - sin² β)Hmm, but R² must be positive, so cos² α - sin² β must be positive. That means cos² α > sin² β.Is that necessarily true? Since α and β are both acute angles, their squares are less than 1, but it's not guaranteed that cos² α > sin² β. Maybe I need to reconsider the projection.Alternatively, perhaps the angle between the diagonals is α, not the angle between the planes. Wait, the problem says the plane forms an angle α with the plane of the axial section. So, the dihedral angle between the two planes is α.In that case, the angle between the normals of the two planes is α. But how does that relate to the diagonals?Alternatively, maybe the angle between the diagonals is α. If the two planes intersect along the generator, then the angle between the diagonals (which lie in each plane) is α.So, if we have two lines (the diagonals) intersecting at the generator, forming an angle α between them.In that case, we can use the formula for the angle between two lines in terms of their direction vectors.But I'm not sure. Maybe it's better to use vector analysis.Let me consider the axial section plane as the x-z plane, with the cylinder's axis along the z-axis. The generator is along the z-axis. The inclined plane makes an angle α with the x-z plane.In the axial section, the diagonal is from (0,0,0) to (2R,0,H). The diagonal in the inclined plane would be from (0,0,0) to some point (x,y,H), forming an angle α with the x-z plane.Wait, this is getting too abstract. Maybe I need to use some trigonometry.Let me denote the diagonal in the inclined plane as l, making an angle β with the base. So, in the inclined plane, the diagonal has a vertical component H and a horizontal component y, such that tan β = H / y.From this, we have y = H / tan β.Also, the length of the diagonal is l, so H² + y² = l².Substituting y = H / tan β:H² + (H² / tan² β) = l²H² (1 + 1 / tan² β) = l²H² ( (tan² β + 1) / tan² β ) = l²Since tan² β + 1 = sec² β, we have:H² ( sec² β / tan² β ) = l²Simplify sec² β / tan² β:sec² β = 1 / cos² βtan² β = sin² β / cos² βSo,(1 / cos² β) / (sin² β / cos² β) = 1 / sin² βTherefore,H² / sin² β = l²So,H² = l² sin² βThus,H = l sin βOkay, so H is l sin β.Now, from y = H / tan β = (l sin β) / (sin β / cos β) ) = l cos βSo, y = l cos βBut y is the horizontal component in the inclined plane, which is the length of the chord in the base of the cylinder.Wait, but in the axial section, the horizontal component is 2R, but in the inclined plane, the horizontal component is y = l cos β.But how are these related?Since the plane is inclined at angle α with respect to the axial section, the horizontal component y in the inclined plane is related to the horizontal component in the axial section (which is 2R) by the angle α.Wait, maybe the horizontal component y is equal to 2R cos α, because when you project 2R onto the inclined plane, it gets scaled by cos α.So, y = 2R cos αBut we have y = l cos βTherefore,2R cos α = l cos βSo,R = (l cos β) / (2 cos α)Okay, so R is (l cos β) / (2 cos α)Now, we have both H and R in terms of l, α, and β.So, H = l sin βR = (l cos β) / (2 cos α)Now, the volume of the cylinder is V = π R² HSubstituting the values:V = π ( (l cos β) / (2 cos α) )² * (l sin β)Simplify:V = π ( l² cos² β / (4 cos² α) ) * l sin βCombine terms:V = π ( l³ cos² β sin β ) / (4 cos² α )So, the volume is (π l³ cos² β sin β) / (4 cos² α)I think that's the answer.**Final Answer**The volume of the cylinder is boxed{dfrac{pi l^3 cos^2 beta sin beta}{4 cos^2 alpha}}.

💡Okay, so I have this problem about a trapezoid (ABCD) with bases (AB) and (CD). They tell me that (overline{AM} = overline{MD}) and (overline{BK} = overline{KC}), and I need to prove that lines (AK) and (CM) are not parallel. Hmm, let me try to visualize this first.Alright, so trapezoid (ABCD) has (AB) and (CD) as the two bases, which means they are parallel. Points (M) and (K) are midpoints of sides (AD) and (BC), respectively. So, (M) divides (AD) into two equal parts, and (K) divides (BC) into two equal parts.I think drawing a diagram might help. Let me sketch trapezoid (ABCD) with (AB) at the top and (CD) at the bottom. Points (A) and (B) are on the top base, (C) and (D) on the bottom. Then, (M) is the midpoint of (AD), so it's halfway between (A) and (D). Similarly, (K) is the midpoint of (BC), so it's halfway between (B) and (C).Now, I need to show that lines (AK) and (CM) are not parallel. To do this, maybe I can use coordinate geometry. Assign coordinates to the points and then find the slopes of (AK) and (CM). If the slopes are different, then the lines aren't parallel.Let me assign coordinates. Let's place point (A) at ((0, 0)). Since (AB) is the top base, let me put point (B) at ((b, 0)). Now, since (ABCD) is a trapezoid with bases (AB) and (CD), the sides (AD) and (BC) are the non-parallel sides. Let me assign coordinates to (D) as ((d, h)) and (C) as ((c, h)), where (h) is the height of the trapezoid.Wait, but since (AB) and (CD) are parallel, their slopes must be equal. The slope of (AB) is ((0 - 0)/(b - 0) = 0), so it's horizontal. Therefore, (CD) must also be horizontal, meaning (C) and (D) must have the same y-coordinate, which I've already assigned as (h). That makes sense.Now, point (M) is the midpoint of (AD). So, coordinates of (M) would be the average of coordinates of (A) and (D). So, (M_x = (0 + d)/2 = d/2), and (M_y = (0 + h)/2 = h/2). So, (M) is at ((d/2, h/2)).Similarly, point (K) is the midpoint of (BC). Coordinates of (B) are ((b, 0)) and (C) are ((c, h)). So, (K_x = (b + c)/2), (K_y = (0 + h)/2 = h/2). So, (K) is at (((b + c)/2, h/2)).Now, I need to find the slopes of lines (AK) and (CM). Let's start with line (AK). Point (A) is at ((0, 0)) and point (K) is at (((b + c)/2, h/2)). The slope of (AK) is ((h/2 - 0)/[(b + c)/2 - 0] = (h/2) / [(b + c)/2] = h / (b + c)).Next, let's find the slope of line (CM). Point (C) is at ((c, h)) and point (M) is at ((d/2, h/2)). The slope of (CM) is ((h/2 - h)/[d/2 - c] = (-h/2) / (d/2 - c)). Let's simplify that: numerator is (-h/2), denominator is ((d - 2c)/2), so the slope is ((-h/2) / [(d - 2c)/2] = (-h)/ (d - 2c)).So, the slope of (AK) is (h / (b + c)), and the slope of (CM) is (-h / (d - 2c)). For lines (AK) and (CM) to be parallel, their slopes must be equal. So, we set (h / (b + c) = -h / (d - 2c)).Let me write that equation:[frac{h}{b + c} = frac{-h}{d - 2c}]We can cancel (h) from both sides (assuming (h neq 0), which it is because it's the height of the trapezoid):[frac{1}{b + c} = frac{-1}{d - 2c}]Cross-multiplying:[d - 2c = - (b + c)]Simplify the right side:[d - 2c = -b - c]Bring all terms to one side:[d - 2c + b + c = 0]Simplify:[d + b - c = 0]So, (d + b = c).Hmm, does this equation hold in a general trapezoid? Let me think. In a trapezoid, the lengths of the bases (AB) and (CD) can vary. The coordinates I assigned were arbitrary, but in reality, in a trapezoid, the sides (AD) and (BC) are not necessarily equal or related in a specific way unless it's an isosceles trapezoid.Wait, but in my coordinate system, (AB) is from ((0,0)) to ((b,0)), so its length is (b). Similarly, (CD) is from ((d,h)) to ((c,h)), so its length is (|c - d|). Since (AB) and (CD) are the two bases, they can have different lengths.But in my earlier equation, I got (d + b = c). So, unless (c = d + b), the slopes won't be equal. But in a general trapezoid, there's no reason why (c) should equal (d + b). That would only happen in specific cases, not in general.Therefore, unless the trapezoid is constructed in such a way that (c = d + b), which is not necessarily the case, the slopes of (AK) and (CM) won't be equal. Hence, lines (AK) and (CM) are not parallel in a general trapezoid.Wait, but is there a possibility that even if (c neq d + b), the slopes could still be equal? Let me check my calculations again.Slope of (AK): (h / (b + c)).Slope of (CM): (-h / (d - 2c)).Setting them equal: (h / (b + c) = -h / (d - 2c)).Cancel (h): (1 / (b + c) = -1 / (d - 2c)).Cross-multiplying: (d - 2c = - (b + c)).Which simplifies to (d - 2c = -b - c), then (d = -b + c).So, (d = c - b).So, unless (d = c - b), the slopes won't be equal. But in a general trapezoid, (d) and (c) are independent of (b). So, unless the trapezoid is specifically constructed with (d = c - b), which isn't a standard property, the lines won't be parallel.Therefore, in a general trapezoid, lines (AK) and (CM) are not parallel.Alternatively, maybe I can approach this using vectors or coordinate geometry differently. Let me try another method to confirm.Let me consider vectors. Let me assign vectors to the points. Let me take point (A) as the origin, so (A = (0, 0)). Let me denote vector (overline{AB}) as vector (b), and vector (overline{AD}) as vector (d). Since (AB) and (CD) are parallel, vector (overline{DC}) is a scalar multiple of (overline{AB}). Let me denote (overline{DC} = k overline{AB}), where (k) is some scalar.So, point (D) is at vector (d), and point (C) is at vector (d + k b), because from (D), moving along (overline{DC}) which is (k) times (overline{AB}).Now, point (M) is the midpoint of (AD), so its position vector is ((0 + d)/2 = d/2).Point (K) is the midpoint of (BC). Point (B) is at vector (b), and point (C) is at vector (d + k b). So, the position vector of (K) is ((b + d + k b)/2 = ( (1 + k) b + d ) / 2.Now, vector (AK) is from (A) to (K), so it's the position vector of (K), which is (( (1 + k) b + d ) / 2.Vector (CM) is from (C) to (M). Position vector of (C) is (d + k b), position vector of (M) is (d/2). So, vector (CM = M - C = d/2 - (d + k b) = -d/2 - k b.Now, to check if (AK) and (CM) are parallel, we need to see if one is a scalar multiple of the other.So, suppose (AK = lambda CM), where (lambda) is a scalar.So,[frac{(1 + k) b + d}{2} = lambda ( - frac{d}{2} - k b )]Multiply both sides by 2 to eliminate denominators:[(1 + k) b + d = lambda ( -d - 2 k b )]Rearranging:[(1 + k) b + d = - lambda d - 2 lambda k b]Now, let's collect like terms:For vector (b):[(1 + k) = -2 lambda k]For vector (d):[1 = - lambda]So, from the (d) component, we have (1 = - lambda), which gives (lambda = -1).Now, substitute (lambda = -1) into the equation from the (b) component:[1 + k = -2 (-1) k = 2k]So,[1 + k = 2k]Subtract (k) from both sides:[1 = k]So, (k = 1). Therefore, unless (k = 1), which would mean that (overline{DC} = overline{AB}), making the trapezoid a parallelogram, the vectors (AK) and (CM) are not scalar multiples of each other.But in a general trapezoid, (k) is not necessarily 1. A trapezoid only requires that the two bases are parallel, not that they are equal in length. So, unless it's a parallelogram, (k neq 1), and thus (AK) and (CM) are not parallel.Therefore, in a trapezoid that is not a parallelogram, lines (AK) and (CM) are not parallel.Wait, but if (k = 1), then it's a parallelogram, and in that case, (AK) and (CM) would be parallel? Let me check.If (k = 1), then vector (overline{DC} = overline{AB}), so (C = D + AB = d + b). Then, point (K) is the midpoint of (BC), which is from (B = b) to (C = d + b). So, midpoint (K) is at ((b + d + b)/2 = (2b + d)/2 = b + d/2.Vector (AK) is from (A = 0) to (K = b + d/2), so it's (b + d/2).Vector (CM) is from (C = d + b) to (M = d/2), so it's (d/2 - (d + b) = -d/2 - b).So, vector (CM = - (d/2 + b)). Vector (AK = b + d/2). So, (AK = - CM). Therefore, they are scalar multiples (with (lambda = -1)), so they are parallel.But in a general trapezoid, unless it's a parallelogram ((k = 1)), (AK) and (CM) are not parallel. Therefore, in a trapezoid that is not a parallelogram, (AK) and (CM) are not parallel.But the problem statement just says "trapezoid", not necessarily a parallelogram. So, in general, they are not parallel.Alternatively, maybe I can use affine geometry or another approach, but I think the coordinate and vector methods I've used are sufficient.Another way is to use the concept of midlines in trapezoids. The midline (or median) of a trapezoid is the segment connecting the midpoints of the non-parallel sides, and its length is the average of the lengths of the two bases.But in this case, points (M) and (K) are midpoints of the legs (AD) and (BC), so the midline would be (MK). The length of (MK) is ((AB + CD)/2).But I'm not sure if that directly helps with proving (AK) and (CM) are not parallel. Maybe I can use properties of the midline.Alternatively, maybe using similar triangles. Let me see.If I can show that triangles formed by these lines are not similar, that might help. But I think the coordinate/vector approach was more straightforward.Wait, another thought: in a trapezoid, the line connecting the midpoints of the legs is parallel to the bases. So, (MK) is parallel to (AB) and (CD). But (AK) and (CM) are not necessarily parallel to the bases.But how does that help? Maybe not directly.Alternatively, maybe using the concept of the centroid or something else. Hmm.But I think my earlier approaches are solid. Using coordinates, I showed that unless (k = 1), the slopes are different, so lines aren't parallel. Using vectors, I showed that unless (k = 1), vectors aren't scalar multiples, so lines aren't parallel.Therefore, in a general trapezoid, lines (AK) and (CM) are not parallel.I think I've covered different methods and each time, unless the trapezoid is a parallelogram, the lines aren't parallel. Since the problem doesn't specify it's a parallelogram, just a trapezoid, the conclusion holds.So, I think I've convinced myself that (AK) and (CM) are not parallel in a general trapezoid.

💡First, I need to simplify the expression (47^{1235} - 22^{1235}) modulo 8. To do this, I'll start by finding the remainders of 47 and 22 when divided by 8.Calculating (47 mod 8), I find that (47 equiv 7 pmod{8}). Similarly, (22 mod 8) gives (22 equiv 6 pmod{8}).Now, the expression becomes (7^{1235} - 6^{1235} pmod{8}).Next, I'll simplify each term separately. For (7^{1235}), since (7 equiv -1 pmod{8}), raising both sides to the 1235th power gives (7^{1235} equiv (-1)^{1235} pmod{8}). Because 1235 is odd, ((-1)^{1235} = -1), so (7^{1235} equiv -1 pmod{8}).For (6^{1235}), I'll look for a pattern in the powers of 6 modulo 8. Calculating the first few powers:[6^1 equiv 6 pmod{8}, quad 6^2 equiv 36 equiv 4 pmod{8}, quad 6^3 equiv 6 times 4 = 24 equiv 0 pmod{8}]I notice that starting from (6^3), the powers of 6 modulo 8 are 0. Since 1235 is greater than 3, (6^{1235} equiv 0 pmod{8}).Finally, subtracting the two results:[7^{1235} - 6^{1235} equiv -1 - 0 equiv -1 equiv 7 pmod{8}]

💡To determine how many different two-digit endings Lara likes, we need to find all two-digit numbers divisible by 3. A number is divisible by 3 if the sum of its digits is divisible by 3.1. **Identify the range of two-digit numbers:** - The smallest two-digit number is 10. - The largest two-digit number is 99.2. **Check each two-digit number for divisibility by 3:** - For each number from 10 to 99, calculate the sum of its digits. - If the sum is divisible by 3, the number is included in the count.3. **Count the valid two-digit numbers:** - After checking all numbers, we find that there are 30 two-digit numbers divisible by 3.**Final Answer:**There are boxed{30} different two-digit endings in numbers that Lara likes.

💡Okay, so I have this problem here with two parts. First, I need to figure out what shape the curve C is, given its polar equation, and then show that the line l intersects with it. Second, I need to find the dot product of vectors PA and PB, where P is the point (0,1) and A and B are the intersection points of line l and curve C.Let me start with part 1. The polar equation is given as ρ = 2√2 sin(θ + π/4). Hmm, polar equations can sometimes be tricky, but I remember that equations of the form ρ = a sin θ or ρ = a cos θ represent circles. Maybe this one is also a circle? Let me try converting it to Cartesian coordinates to see.I know that in polar coordinates, ρ = 2√2 sin(θ + π/4). To convert this, I can use the identity sin(θ + π/4) = sinθ cos(π/4) + cosθ sin(π/4). Since cos(π/4) and sin(π/4) are both √2/2, this becomes:ρ = 2√2 [sinθ (√2/2) + cosθ (√2/2)] Simplify that: ρ = 2√2*(√2/2)(sinθ + cosθ) The 2√2 and √2/2 multiply to 2√2*(√2)/2 = (2*2)/2 = 2. So, ρ = 2(sinθ + cosθ).Now, in Cartesian coordinates, ρ sinθ = y and ρ cosθ = x. So, substituting these in:ρ = 2(sinθ + cosθ) Multiply both sides by ρ: ρ² = 2ρ sinθ + 2ρ cosθ Which translates to: x² + y² = 2y + 2xLet me rearrange this equation: x² - 2x + y² - 2y = 0 To complete the squares for x and y:For x: x² - 2x = (x - 1)² - 1 For y: y² - 2y = (y - 1)² - 1 So, substituting back:(x - 1)² - 1 + (y - 1)² - 1 = 0 Simplify: (x - 1)² + (y - 1)² - 2 = 0 Which gives: (x - 1)² + (y - 1)² = 2Ah, so this is a circle centered at (1,1) with radius √2. Got it. So, curve C is a circle.Now, I need to show that line l intersects with this circle. The parametric equations for line l are x = t and y = 1 + 2t. Let me write this in Cartesian form to make it easier.From x = t, we can substitute t into y: y = 1 + 2xSo, the equation of line l is y = 2x + 1. To find if this line intersects the circle, I can substitute y from the line into the circle's equation and solve for x.Substituting y = 2x + 1 into (x - 1)² + (y - 1)² = 2:(x - 1)² + (2x + 1 - 1)² = 2 Simplify: (x - 1)² + (2x)² = 2 Expand both terms: (x² - 2x + 1) + 4x² = 2 Combine like terms: 5x² - 2x + 1 = 2 Subtract 2 from both sides: 5x² - 2x - 1 = 0This is a quadratic equation in x. Let's compute the discriminant to see if there are real solutions:Discriminant D = b² - 4ac = (-2)² - 4*5*(-1) = 4 + 20 = 24Since D is positive, there are two real solutions, which means the line intersects the circle at two points. So, part 1 is done.Moving on to part 2. We need to find the dot product of vectors PA and PB, where P is (0,1), and A and B are the intersection points.First, let me find the coordinates of points A and B. From the quadratic equation 5x² - 2x - 1 = 0, the solutions are:x = [2 ± √(4 + 20)] / (2*5) = [2 ± √24]/10 = [2 ± 2√6]/10 = [1 ± √6]/5So, x-coordinates of A and B are (1 + √6)/5 and (1 - √6)/5.Corresponding y-coordinates are y = 2x + 1:For x = (1 + √6)/5: y = 2*(1 + √6)/5 + 1 = (2 + 2√6)/5 + 5/5 = (7 + 2√6)/5For x = (1 - √6)/5: y = 2*(1 - √6)/5 + 1 = (2 - 2√6)/5 + 5/5 = (7 - 2√6)/5So, points A and B are:A: ((1 + √6)/5, (7 + 2√6)/5) B: ((1 - √6)/5, (7 - 2√6)/5)Now, point P is (0,1). Let me find vectors PA and PB.Vector PA is A - P: x-component: (1 + √6)/5 - 0 = (1 + √6)/5 y-component: (7 + 2√6)/5 - 1 = (7 + 2√6 - 5)/5 = (2 + 2√6)/5 So, PA = [(1 + √6)/5, (2 + 2√6)/5]Similarly, vector PB is B - P: x-component: (1 - √6)/5 - 0 = (1 - √6)/5 y-component: (7 - 2√6)/5 - 1 = (7 - 2√6 - 5)/5 = (2 - 2√6)/5 So, PB = [(1 - √6)/5, (2 - 2√6)/5]Now, the dot product PA · PB is:[(1 + √6)/5 * (1 - √6)/5] + [(2 + 2√6)/5 * (2 - 2√6)/5]Let me compute each part separately.First term: [(1 + √6)(1 - √6)] / 25 This is a difference of squares: 1² - (√6)² = 1 - 6 = -5 So, first term: -5/25 = -1/5Second term: [(2 + 2√6)(2 - 2√6)] / 25 Again, difference of squares: (2)^2 - (2√6)^2 = 4 - 4*6 = 4 - 24 = -20 So, second term: -20/25 = -4/5Adding both terms: (-1/5) + (-4/5) = -5/5 = -1Wait, that's straightforward. So, the dot product is -1.But let me double-check my calculations to make sure I didn't make a mistake.First term: (1 + √6)(1 - √6) = 1 - 6 = -5. Divided by 25: -1/5. Correct.Second term: (2 + 2√6)(2 - 2√6) = 4 - (2√6)^2 = 4 - 24 = -20. Divided by 25: -4/5. Correct.Adding them: -1/5 - 4/5 = -5/5 = -1. Yep, that seems right.Alternatively, maybe there's another way to compute this without finding the exact points. Maybe using parametric equations or some geometric properties?Let me think. The parametric equation of line l is x = t, y = 1 + 2t. So, any point on l can be written as (t, 1 + 2t). Let me substitute this into the circle equation:(x - 1)^2 + (y - 1)^2 = 2 Substitute x = t, y = 1 + 2t: (t - 1)^2 + (1 + 2t - 1)^2 = 2 Simplify: (t - 1)^2 + (2t)^2 = 2 Which is: t² - 2t + 1 + 4t² = 2 Combine like terms: 5t² - 2t + 1 = 2 So, 5t² - 2t - 1 = 0, which is the same quadratic as before. So, the solutions are t = [2 ± √(4 + 20)] / 10 = [1 ± √6]/5.So, points A and B correspond to t = (1 + √6)/5 and t = (1 - √6)/5.Now, vector PA is from P(0,1) to A(t, 1 + 2t). So, PA = (t - 0, (1 + 2t) - 1) = (t, 2t). Similarly, PB = (t', 2t'), where t' is the other solution.So, PA · PB = (t)(t') + (2t)(2t') = tt' + 4tt' = 5tt'From the quadratic equation 5t² - 2t - 1 = 0, the product of the roots t1*t2 = c/a = (-1)/5.So, tt' = -1/5. Therefore, PA · PB = 5*(-1/5) = -1.Oh, that's a smarter way. Instead of computing the coordinates and then the vectors, I could have used Vieta's formula to find the product of the roots and then compute the dot product directly. That's much quicker and avoids dealing with radicals.So, either way, whether by computing the coordinates or by using the parametric form and Vieta's theorem, I get the same result: the dot product is -1.I think that's solid. I don't see any mistakes in my reasoning, and both methods lead to the same answer. So, I'm confident that the answer is -1.**Final Answer**The value of overrightarrow{PA} cdot overrightarrow{PB} is boxed{-1}.