answer：Okay, so I have this problem about two circles, C and D. Circle C is centered at (3, 5) with a radius of 5 units, and Circle D is centered at (13, 5) with the same radius of 5 units. I need to find the area of the gray region bound by these circles and the x-axis. Hmm, let me try to visualize this.First, I should probably sketch the coordinate system to get a better idea. Circle C is at (3, 5), so that's somewhere to the left, and Circle D is at (13, 5), which is to the right. Both have the same radius, so they should be identical in size. The x-axis is the horizontal line at y=0, so the gray region is probably the area between the two circles and below the x-axis? Or maybe above? Wait, the problem says "bound by the circles and the x-axis," so it's likely the area that is enclosed by both circles and the x-axis.Let me think about how these circles interact with the x-axis. The center of each circle is at y=5, and the radius is 5 units. So, the distance from the center to the x-axis is 5 units, which is exactly the radius. That means each circle will touch the x-axis at exactly one point. For Circle C, the point where it touches the x-axis should be directly below the center, so that's (3, 0). Similarly, Circle D will touch the x-axis at (13, 0).So, the gray region is probably the area between these two points on the x-axis and the arcs of the circles above them. Wait, but both circles are centered at y=5, so above the x-axis. So, the gray region is bound by the two circles and the x-axis, forming a sort of lens-shaped area between the two circles and above the x-axis? Or maybe it's the area below the circles and above the x-axis?Wait, if both circles touch the x-axis at (3,0) and (13,0), then the region bound by the circles and the x-axis would be the area between the two circles and above the x-axis, right? So, it's like a sort of football shape between the two circles, but only the part that's above the x-axis.But actually, since both circles are centered at y=5 and have radius 5, they just touch the x-axis. So, the region bound by the circles and the x-axis is the area between the two circles and above the x-axis, which is actually the area of the rectangle formed by the centers and the x-axis minus the areas of the two semicircles.Wait, let me clarify. The centers are at (3,5) and (13,5), so the distance between the centers is 10 units along the x-axis. Each circle has a radius of 5, so the circles just touch each other at the midpoint between (3,5) and (13,5), which is (8,5). So, the two circles intersect at (8,5). Therefore, the gray region is bound by the two circles and the x-axis, forming a sort of lens shape between the two circles and above the x-axis.But actually, since both circles touch the x-axis at (3,0) and (13,0), and they intersect at (8,5), the gray region is the area between the two circles from (3,0) to (13,0) and up to their intersection point at (8,5). So, it's like a region bounded below by the x-axis and above by the two arcs of the circles.To find the area of this gray region, I can think of it as the area of the rectangle formed by the points (3,0), (13,0), (13,5), and (3,5) minus the areas of the two semicircles that are above the x-axis.Wait, but actually, each circle contributes a semicircle above the x-axis, but since the circles intersect, it's not just two separate semicircles. Instead, the gray region is the area between the two circles and above the x-axis, which is actually the area of the rectangle minus the areas of the two quarter-circles.Wait, let me think again. The rectangle from (3,0) to (13,0) to (13,5) to (3,5) has an area of length times width, which is 10 units (from x=3 to x=13) times 5 units (from y=0 to y=5), so 50 square units.Now, the area above the x-axis and below the circles would be the area of the rectangle minus the areas of the two quarter-circles. Each circle contributes a quarter-circle because the center is at (3,5) and (13,5), and the radius is 5, so the quarter-circle is the part from the center down to the x-axis.Wait, no, actually, each circle contributes a semicircle above the x-axis, but since the circles intersect, the overlapping area is counted twice if I just subtract two semicircles. Hmm, this is getting a bit confusing.Maybe a better approach is to calculate the area of the region bounded by the two circles and the x-axis by integrating or using geometric formulas. Since both circles have the same radius and their centers are aligned horizontally, the problem has symmetry along the vertical line x=8.So, perhaps I can calculate the area of one side and double it. Let me try to find the area of the region from x=3 to x=8 and then double it to get the total gray area.First, I need the equations of both circles.For Circle C centered at (3,5) with radius 5:[(x - 3)^2 + (y - 5)^2 = 25]For Circle D centered at (13,5) with radius 5:[(x - 13)^2 + (y - 5)^2 = 25]I need to find the area between these two circles from x=3 to x=13 and above the x-axis (y=0). But since both circles touch the x-axis at (3,0) and (13,0), the region is bounded below by y=0 and above by the two circles.To find this area, I can set up an integral from x=3 to x=13 of the upper functions minus the lower functions. However, since the circles intersect at (8,5), the upper boundary changes at x=8.So, from x=3 to x=8, the upper boundary is the upper half of Circle C, and from x=8 to x=13, the upper boundary is the upper half of Circle D. The lower boundary is y=0 for the entire region.Therefore, the area A can be expressed as:[A = int_{3}^{8} sqrt{25 - (x - 3)^2} , dx + int_{8}^{13} sqrt{25 - (x - 13)^2} , dx]These integrals represent the areas under the upper halves of the circles from x=3 to x=8 and from x=8 to x=13, respectively.But integrating might be a bit complicated. Maybe there's a geometric way to find this area without calculus.Since both circles have the same radius and their centers are 10 units apart, the distance between the centers is equal to twice the radius. Wait, no, the radius is 5, so twice the radius is 10, which is exactly the distance between the centers. That means the two circles touch each other at one point, which is (8,5). So, they are externally tangent to each other at that point.Therefore, the region bounded by the two circles and the x-axis is actually the area of the rectangle from (3,0) to (13,0) to (13,5) to (3,5) minus the areas of the two quarter-circles at the top.Wait, but if the circles are externally tangent, then the area above the x-axis and below the circles is just the area of the rectangle minus the areas of the two quarter-circles.Each quarter-circle has an area of (1/4)πr² = (1/4)π(5)² = (25/4)π. Since there are two such quarter-circles, the total area to subtract is 2*(25/4)π = (25/2)π.The area of the rectangle is 10*5 = 50.So, the gray area would be 50 - (25/2)π.Wait, but let me confirm. The rectangle is 10 units wide and 5 units tall, so 50. The two quarter-circles make up half a circle, which is (1/2)πr² = (1/2)π(25) = (25/2)π. So, subtracting that from the rectangle gives the gray area.Yes, that makes sense. So, the area of the gray region is 50 - (25/2)π.But let me double-check. If I consider the area above the x-axis and below the circles, it's the area of the rectangle minus the areas of the two quarter-circles. Since each circle contributes a quarter-circle above the rectangle, subtracting both gives the gray area.Alternatively, I can think of the gray area as the area between the two circles and above the x-axis, which is the area of the rectangle minus the areas of the two quarter-circles.Yes, that seems correct. So, the final answer should be 50 - (25/2)π, which can also be written as 50 - 12.5π.I think that's the correct approach. I don't see any mistakes in the reasoning.

answer：Okay, so I have this problem here about a sample of size 10 that forms an arithmetic sequence. The common difference isn't zero, which means it's not a constant sequence. They tell me that the third term, a₃, is 8. Also, the first term, a₁, the third term, a₃, and the seventh term, a₇, form a geometric sequence. I need to find the mean and median of this sample, and then choose the correct option from A to D.Alright, let's break this down step by step. First, since it's an arithmetic sequence, I know that each term is the previous term plus a common difference, which I'll call d. So, the nth term of an arithmetic sequence can be written as aₙ = a₁ + (n-1)d. Given that a₃ = 8, I can write that as:a₃ = a₁ + 2d = 8.Okay, so that's one equation. Now, they also mention that a₁, a₃, a₇ form a geometric sequence. In a geometric sequence, each term is the previous term multiplied by a common ratio, r. So, if a₁, a₃, a₇ are in geometric progression, then the ratio between a₃ and a₁ should be the same as the ratio between a₇ and a₃. Mathematically, that means:a₃ / a₁ = a₇ / a₃.So, substituting the expressions for a₃ and a₇ from the arithmetic sequence, we have:(a₁ + 2d) / a₁ = (a₁ + 6d) / (a₁ + 2d).Hmm, that looks a bit complicated, but maybe I can simplify it. Let me write it out:(a₁ + 2d) / a₁ = (a₁ + 6d) / (a₁ + 2d).Cross-multiplying to eliminate the fractions:(a₁ + 2d)^2 = a₁(a₁ + 6d).Expanding both sides:(a₁^2 + 4a₁d + 4d²) = a₁² + 6a₁d.Subtracting a₁² from both sides:4a₁d + 4d² = 6a₁d.Bringing all terms to one side:4a₁d + 4d² - 6a₁d = 0-2a₁d + 4d² = 0.Factor out a 2d:2d(-a₁ + 2d) = 0.Since the common difference d is not zero, we can divide both sides by 2d:-a₁ + 2d = 0=> a₁ = 2d.Okay, so now I have a relationship between a₁ and d: a₁ is twice the common difference.Earlier, I had another equation from a₃ = 8:a₁ + 2d = 8.But since a₁ = 2d, substitute that into the equation:2d + 2d = 84d = 8d = 2.So, the common difference d is 2. Then, a₁ = 2d = 4.Great, so now I can write out the entire arithmetic sequence. Since it's a sample of size 10, n goes from 1 to 10.Let me list them out:a₁ = 4a₂ = a₁ + d = 4 + 2 = 6a₃ = 8 (given)a₄ = 8 + 2 = 10a₅ = 10 + 2 = 12a₆ = 12 + 2 = 14a₇ = 14 + 2 = 16a₈ = 16 + 2 = 18a₉ = 18 + 2 = 20a₁₀ = 20 + 2 = 22So, the sequence is: 4, 6, 8, 10, 12, 14, 16, 18, 20, 22.Now, I need to find the mean and median of this sample.Starting with the mean. The mean is the average of all the numbers. Since it's an arithmetic sequence, the mean is equal to the average of the first and last terms. That's a property of arithmetic sequences.So, mean = (a₁ + a₁₀) / 2 = (4 + 22) / 2 = 26 / 2 = 13.Alternatively, I could add all the terms and divide by 10. Let me check that just to be sure.Adding them up:4 + 6 = 1010 + 8 = 1818 + 10 = 2828 + 12 = 4040 + 14 = 5454 + 16 = 7070 + 18 = 8888 + 20 = 108108 + 22 = 130.Total sum is 130. Divide by 10: 130 / 10 = 13. Yep, that's correct.Now, the median. Since there are 10 terms, which is even, the median will be the average of the 5th and 6th terms.Looking at the sequence:1: 42: 63: 84: 105: 126: 147: 168: 189: 2010: 22So, the 5th term is 12 and the 6th term is 14. The median is (12 + 14)/2 = 26/2 = 13.Therefore, both the mean and the median are 13.Looking back at the options:A: 13, 12B: 13, 13C: 12, 13D: 13, 14So, the correct answer is B: 13, 13.I think that makes sense. The arithmetic sequence is symmetric in a way when considering the mean and median, especially since it's evenly spaced. So, for an even number of terms, the median is the average of the two middle terms, which in this case are 12 and 14, averaging to 13. The mean is also 13 because of the symmetry in the arithmetic sequence.Just to double-check, I can think about the properties of arithmetic sequences. The mean is always equal to the median in an arithmetic sequence because of the uniform distribution of the terms. So, regardless of whether the number of terms is odd or even, the mean and median should be the same. In this case, even though it's 10 terms, which is even, the median still equals the mean because of the nature of the arithmetic progression.Therefore, I'm confident that the answer is B: 13, 13.**Final Answer**boxed{B}

answer：First, I need to calculate the square root of 4, which is straightforward. The square root of 4 is 2.Next, I need to evaluate the absolute value expression |√3 - 2|. Since √3 is approximately 1.732, subtracting 2 from it gives a negative number. Taking the absolute value of a negative number makes it positive, so |√3 - 2| becomes 2 - √3.Then, I need to compute (-1) raised to the power of 2023. Since 2023 is an odd number, (-1) raised to an odd power remains -1.Now, I can substitute these values back into the original expression:√4 - |√3 - 2| + (-1)^2023 = 2 - (2 - √3) - 1.Simplifying this expression step by step:2 - 2 + √3 - 1 = √3 - 1.So, the final result of the calculation is √3 - 1.

answer：First, I'll denote the number Wilson has in mind as ( x ).According to the problem, if Wilson takes away one-third of the number from it, the result is ( frac{16}{3} ). This can be written as the equation:[ x - frac{1}{3}x = frac{16}{3} ]Next, I'll combine like terms on the left side of the equation:[ frac{3}{3}x - frac{1}{3}x = frac{16}{3} ]This simplifies to:[ frac{2}{3}x = frac{16}{3} ]To solve for ( x ), I'll multiply both sides of the equation by the reciprocal of ( frac{2}{3} ), which is ( frac{3}{2} ):[ frac{3}{2} cdot frac{2}{3}x = frac{16}{3} cdot frac{3}{2} ]This simplifies to:[ x = frac{16}{3} cdot frac{3}{2} ]Multiplying the numerators and denominators separately:[ x = frac{16 cdot 3}{3 cdot 2} ]The 3s in the numerator and denominator cancel out:[ x = frac{16}{2} ]Finally, dividing 16 by 2 gives:[ x = 8 ]So, the number Wilson has in mind is (boxed{8}).

answer：Okay, so I have this problem where I need to prove that there's a unique point X in the plane of triangle ABC such that three expressions involving distances from X to A, B, and C are all equal. The expressions are:1. ( XA^2 + XB^2 + AB^2 )2. ( XB^2 + XC^2 + BC^2 )3. ( XC^2 + XA^2 + CA^2 )All three of these should be equal. Hmm, that's interesting. Let me try to unpack this step by step.First, since A, B, and C are noncollinear, they form a triangle. So, we're working within the plane of this triangle. The point X is somewhere in this plane, and we need to find its location such that the given equations hold.Let me write down the equations more clearly:1. ( XA^2 + XB^2 + AB^2 = XB^2 + XC^2 + BC^2 )2. ( XB^2 + XC^2 + BC^2 = XC^2 + XA^2 + CA^2 )Wait, actually, the problem states that all three expressions are equal, so it's like:( XA^2 + XB^2 + AB^2 = XB^2 + XC^2 + BC^2 = XC^2 + XA^2 + CA^2 )So, all three are equal to each other. Let me denote this common value as K. So,( XA^2 + XB^2 + AB^2 = K )( XB^2 + XC^2 + BC^2 = K )( XC^2 + XA^2 + CA^2 = K )So, all three expressions equal K. Therefore, we can set them equal to each other:1. ( XA^2 + XB^2 + AB^2 = XB^2 + XC^2 + BC^2 )2. ( XB^2 + XC^2 + BC^2 = XC^2 + XA^2 + CA^2 )Let me simplify the first equation:( XA^2 + XB^2 + AB^2 = XB^2 + XC^2 + BC^2 )Subtract ( XB^2 ) from both sides:( XA^2 + AB^2 = XC^2 + BC^2 )So,( XA^2 - XC^2 = BC^2 - AB^2 )Similarly, let's simplify the second equation:( XB^2 + XC^2 + BC^2 = XC^2 + XA^2 + CA^2 )Subtract ( XC^2 ) from both sides:( XB^2 + BC^2 = XA^2 + CA^2 )So,( XB^2 - XA^2 = CA^2 - BC^2 )Hmm, so now I have two equations:1. ( XA^2 - XC^2 = BC^2 - AB^2 )2. ( XB^2 - XA^2 = CA^2 - BC^2 )Let me denote these as Equation (1) and Equation (2).Equation (1): ( XA^2 - XC^2 = BC^2 - AB^2 )Equation (2): ( XB^2 - XA^2 = CA^2 - BC^2 )Let me rearrange Equation (2):( XB^2 - XA^2 = CA^2 - BC^2 )Which can be written as:( XB^2 = XA^2 + CA^2 - BC^2 )Similarly, from Equation (1):( XA^2 = XC^2 + BC^2 - AB^2 )So, plugging this into Equation (2):( XB^2 = (XC^2 + BC^2 - AB^2) + CA^2 - BC^2 )Simplify:( XB^2 = XC^2 + BC^2 - AB^2 + CA^2 - BC^2 )The ( BC^2 ) terms cancel out:( XB^2 = XC^2 - AB^2 + CA^2 )So,( XB^2 - XC^2 = CA^2 - AB^2 )Wait, that's similar to Equation (1). Let me check.Equation (1): ( XA^2 - XC^2 = BC^2 - AB^2 )Equation (2): ( XB^2 - XA^2 = CA^2 - BC^2 )So, if I add Equation (1) and Equation (2):( (XA^2 - XC^2) + (XB^2 - XA^2) = (BC^2 - AB^2) + (CA^2 - BC^2) )Simplify left side:( XB^2 - XC^2 )Right side:( CA^2 - AB^2 )So, we get:( XB^2 - XC^2 = CA^2 - AB^2 )Which is consistent with what we had earlier.Hmm, so perhaps I can think of these as equations representing lines or something in the plane.I recall that the set of points X such that ( XA^2 - XC^2 = text{constant} ) is a line perpendicular to the line AC. Similarly for other pairs.Wait, yes, that's a property from coordinate geometry. The locus of points X such that the difference of squares of distances from two fixed points is constant is a straight line perpendicular to the line joining those two points.So, in this case, ( XA^2 - XC^2 = BC^2 - AB^2 ) is a line perpendicular to AC.Similarly, ( XB^2 - XA^2 = CA^2 - BC^2 ) is a line perpendicular to AB.So, if I can find the intersection of these two lines, that should give me the point X.Since these two lines are perpendicular to AC and AB respectively, and since AB and AC are sides of the triangle, these two lines should intersect at a unique point.Therefore, X is uniquely determined as the intersection of these two perpendiculars.Wait, but the problem mentions all three expressions being equal, so perhaps I need to ensure that the third condition is automatically satisfied once the first two are met.Let me check that.Suppose I have found X such that:1. ( XA^2 + XB^2 + AB^2 = XB^2 + XC^2 + BC^2 )2. ( XB^2 + XC^2 + BC^2 = XC^2 + XA^2 + CA^2 )Then, does it necessarily follow that ( XA^2 + XB^2 + AB^2 = XC^2 + XA^2 + CA^2 )?Yes, because if the first two are equal, and the second and third are equal, then all three are equal.So, it's sufficient to solve the first two equations, which gives a unique point X, and then the third equation will hold automatically.Therefore, the point X is uniquely determined as the intersection of the two perpendiculars defined by the first two equations.But let me try to formalize this a bit more.Let me assign coordinates to the points to make it easier. Maybe place the triangle ABC in a coordinate system.Let me set point A at (0, 0), point B at (c, 0), and point C at (d, e). Since the points are noncollinear, e ≠ 0.Let X be a point (x, y).Then, ( XA^2 = x^2 + y^2 )( XB^2 = (x - c)^2 + y^2 )( XC^2 = (x - d)^2 + (y - e)^2 )Also, AB^2 = c^2, BC^2 = (d - c)^2 + e^2, CA^2 = d^2 + e^2.So, plugging into the first equation:( XA^2 + XB^2 + AB^2 = XB^2 + XC^2 + BC^2 )Simplify:( XA^2 + AB^2 = XC^2 + BC^2 )Substitute the expressions:( (x^2 + y^2) + c^2 = [(x - d)^2 + (y - e)^2] + [(d - c)^2 + e^2] )Let me expand the right-hand side:( (x^2 - 2dx + d^2 + y^2 - 2ey + e^2) + (d^2 - 2cd + c^2 + e^2) )Simplify:( x^2 - 2dx + d^2 + y^2 - 2ey + e^2 + d^2 - 2cd + c^2 + e^2 )Combine like terms:( x^2 + y^2 - 2dx - 2ey + 2d^2 + 2e^2 - 2cd + c^2 )So, the equation becomes:Left-hand side: ( x^2 + y^2 + c^2 )Right-hand side: ( x^2 + y^2 - 2dx - 2ey + 2d^2 + 2e^2 - 2cd + c^2 )Subtract left-hand side from both sides:0 = -2dx - 2ey + 2d^2 + 2e^2 - 2cdSimplify:2dx + 2ey = 2d^2 + 2e^2 - 2cdDivide both sides by 2:dx + ey = d^2 + e^2 - cdSo, that's the first equation.Similarly, let's take the second equation:( XB^2 + XC^2 + BC^2 = XC^2 + XA^2 + CA^2 )Simplify:( XB^2 + BC^2 = XA^2 + CA^2 )Substitute the expressions:( [(x - c)^2 + y^2] + [(d - c)^2 + e^2] = [x^2 + y^2] + [d^2 + e^2] )Expand the left-hand side:( (x^2 - 2cx + c^2 + y^2) + (d^2 - 2cd + c^2 + e^2) )Simplify:( x^2 - 2cx + c^2 + y^2 + d^2 - 2cd + c^2 + e^2 )Combine like terms:( x^2 + y^2 - 2cx + 2c^2 + d^2 - 2cd + e^2 )The right-hand side is:( x^2 + y^2 + d^2 + e^2 )Subtract right-hand side from both sides:0 = -2cx + 2c^2 - 2cdSimplify:2cx = 2c^2 - 2cdDivide both sides by 2c (assuming c ≠ 0, which it is since AB is a side of the triangle):x = c - dSo, from the second equation, we get x = c - d.Now, plug this into the first equation:dx + ey = d^2 + e^2 - cdSubstitute x = c - d:d(c - d) + ey = d^2 + e^2 - cdExpand:dc - d^2 + ey = d^2 + e^2 - cdBring all terms to one side:dc - d^2 + ey - d^2 - e^2 + cd = 0Combine like terms:( dc + cd ) + (-d^2 - d^2) + ey - e^2 = 0Which simplifies to:2cd - 2d^2 + ey - e^2 = 0So,ey = 2d^2 - 2cd + e^2Thus,y = (2d^2 - 2cd + e^2)/eSimplify:y = (2d(d - c) + e^2)/eSo, we have x = c - d and y = (2d(d - c) + e^2)/eTherefore, the coordinates of X are:x = c - dy = [2d(d - c) + e^2]/eSo, this gives a unique solution for X in terms of the coordinates of A, B, and C.Therefore, such a point X exists and is unique.Alternatively, without coordinates, we can think geometrically.The equations ( XA^2 - XC^2 = text{constant} ) and ( XB^2 - XA^2 = text{constant} ) represent lines perpendicular to AC and AB respectively.Since AB and AC are not parallel (as ABC is a triangle), these two lines must intersect at exactly one point, which is X.Hence, the point X is unique.Therefore, we have proven that there exists a unique point X satisfying the given conditions.

answer：Okay, so I have this problem where Mike is climbing stairs, and depending on how he climbs them, there are some steps left at the top. I need to find the smallest number of steps, n, that's greater than 20. Let me try to understand the problem step by step.First, when Mike climbs 6 steps at a time, there are 5 steps left. That means if he takes as many 6-step climbs as possible, he'll still have 5 steps remaining. So, mathematically, this can be written as n ≡ 5 mod 6. In other words, when n is divided by 6, the remainder is 5.Similarly, when he climbs 7 steps at a time, there are 3 steps left. So, this translates to n ≡ 3 mod 7. That means when n is divided by 7, the remainder is 3.So, I have two congruences:1. n ≡ 5 mod 62. n ≡ 3 mod 7I need to find the smallest n greater than 20 that satisfies both of these conditions. This seems like a problem that can be solved using the Chinese Remainder Theorem, which is used to solve systems of simultaneous congruences.Let me recall how the Chinese Remainder Theorem works. If the moduli are coprime, then there's a unique solution modulo the product of the moduli. In this case, 6 and 7 are coprime since their greatest common divisor is 1. So, there should be a unique solution modulo 42 (since 6*7=42).So, I can express n as:n = 6k + 5, where k is some integer.Now, substituting this into the second congruence:6k + 5 ≡ 3 mod 7Let me simplify this:6k + 5 ≡ 3 mod 7Subtract 5 from both sides:6k ≡ -2 mod 7But -2 mod 7 is the same as 5 mod 7, so:6k ≡ 5 mod 7Now, I need to solve for k. To do this, I need the modular inverse of 6 mod 7. The modular inverse of a number a modulo m is a number x such that a*x ≡ 1 mod m.Let me find the inverse of 6 mod 7. I can test numbers from 1 to 6:6*1 = 6 ≡ 6 mod 76*2 = 12 ≡ 5 mod 76*3 = 18 ≡ 4 mod 76*4 = 24 ≡ 3 mod 76*5 = 30 ≡ 2 mod 76*6 = 36 ≡ 1 mod 7Ah, so 6*6 ≡ 1 mod 7. Therefore, the inverse of 6 mod 7 is 6.Now, multiply both sides of the congruence 6k ≡ 5 mod 7 by 6:6*6k ≡ 6*5 mod 736k ≡ 30 mod 7Simplify 36 mod 7: 36 ÷ 7 is 5 with a remainder of 1, so 36 ≡ 1 mod 7Similarly, 30 ÷ 7 is 4 with a remainder of 2, so 30 ≡ 2 mod 7So, the equation becomes:1k ≡ 2 mod 7Which simplifies to:k ≡ 2 mod 7This means that k can be written as:k = 7m + 2, where m is some integer.Now, substitute this back into the expression for n:n = 6k + 5 = 6*(7m + 2) + 5 = 42m + 12 + 5 = 42m + 17So, n can be expressed as:n = 42m + 17Now, I need the smallest n greater than 20. Let's find the smallest integer m such that n > 20.Let's plug in m = 0:n = 42*0 + 17 = 17, which is less than 20.m = 1:n = 42*1 + 17 = 59, which is greater than 20.m = 2:n = 42*2 + 17 = 84 + 17 = 101, which is also greater than 20, but 59 is smaller.So, the smallest n greater than 20 is 59.Let me verify this solution to make sure it satisfies both original conditions.First condition: n ≡ 5 mod 659 ÷ 6 = 9 with a remainder of 5. So, 59 ≡ 5 mod 6. That checks out.Second condition: n ≡ 3 mod 759 ÷ 7 = 8 with a remainder of 3. So, 59 ≡ 3 mod 7. That also checks out.Therefore, 59 is indeed the smallest number greater than 20 that satisfies both conditions.I think I've covered all the steps and verified the solution. It seems correct.**Final Answer**The smallest possible value of ( n ) that is greater than 20 is boxed{59}.