answer：To find the smallest positive perfect square divisible by 6, 14, 22, and 30, I need to determine the least common multiple (LCM) of these numbers and then adjust it to be a perfect square.First, I'll find the prime factorization of each number:- 6 = 2 × 3- 14 = 2 × 7- 22 = 2 × 11- 30 = 2 × 3 × 5Next, I'll calculate the LCM by taking the highest powers of all prime factors:- The highest power of 2 is 2¹- The highest power of 3 is 3¹- The highest power of 5 is 5¹- The highest power of 7 is 7¹- The highest power of 11 is 11¹So, the LCM is 2 × 3 × 5 × 7 × 11 = 2310.Since 2310 is not a perfect square, I need to multiply it by additional factors to make all exponents even:- Multiply by another 2 to get 2²- Multiply by another 3 to get 3²- Multiply by another 5 to get 5²- Multiply by another 7 to get 7²- Multiply by another 11 to get 11²The smallest perfect square is now 2² × 3² × 5² × 7² × 11² = 4 × 9 × 25 × 49 × 121 = 5336100.

answer：Alright, so I have this problem where a traveler starts at the origin and moves in different directions, and I need to find the straight-line distance from where he started to where he ended up. The answer choices are given, and I need to figure out which one is correct. Let me try to break this down step by step.First, the traveler starts at the origin, which is the point (0,0) on a coordinate plane. He walks 4 miles due west. Okay, west is to the left on the coordinate plane, so moving west would decrease the x-coordinate. So, after walking 4 miles west, his new position should be at (-4, 0).Next, he turns 120 degrees to his left and walks another 5 miles. Hmm, turning left from west by 120 degrees. Let me visualize this. If he's facing west, turning left would mean he's now facing south of east. Wait, no, let me think again. If you're facing west and turn left, you're actually turning towards the south direction, but 120 degrees is a pretty big turn. Maybe it's better to represent this using angles in standard position.In standard position, 0 degrees is along the positive x-axis (east), and angles increase counterclockwise. So, if he starts facing west, that's 180 degrees. Turning left by 120 degrees from west would mean adding 120 degrees to his current direction. So, 180 + 120 = 300 degrees. So, he's now facing 300 degrees from the positive x-axis, which is equivalent to -60 degrees. That means he's moving in the direction of 60 degrees below the positive x-axis, or 60 degrees south of east.Now, I need to find the components of this 5-mile walk in the x and y directions. The x-component (east-west direction) would be 5 * cos(300°), and the y-component (north-south direction) would be 5 * sin(300°). Let me calculate these.Cos(300°) is equal to cos(-60°), which is 0.5. So, 5 * 0.5 = 2.5 miles east.Sin(300°) is equal to sin(-60°), which is -√3/2. So, 5 * (-√3/2) = - (5√3)/2 miles. That means he moves 2.5 miles east and (5√3)/2 miles south.So, adding this to his previous position at (-4, 0):x-coordinate: -4 + 2.5 = -1.5y-coordinate: 0 - (5√3)/2 = - (5√3)/2So, his new position is (-1.5, - (5√3)/2).Now, he turns another 60 degrees to his left. Let me figure out his new direction. He was facing 300 degrees, which is 60 degrees south of east. Turning another 60 degrees to his left would mean adding another 60 degrees to his current direction. So, 300 + 60 = 360 degrees, which brings him back to 0 degrees, or directly east.Wait, that doesn't seem right. If he was facing 300 degrees and turns another 60 degrees left, wouldn't that make him face 360 degrees, which is the same as 0 degrees, meaning directly east? Yes, that seems correct.So, now he's facing directly east and walks another 2 miles. Since he's facing east, this is straightforward. His x-coordinate will increase by 2 miles, and his y-coordinate remains the same.So, adding this to his previous position:x-coordinate: -1.5 + 2 = 0.5y-coordinate: - (5√3)/2 remains the same.So, his final position is (0.5, - (5√3)/2).Now, I need to find the straight-line distance from the origin (0,0) to his final position (0.5, - (5√3)/2). To do this, I can use the distance formula, which is:Distance = √[(x2 - x1)^2 + (y2 - y1)^2]Plugging in the coordinates:Distance = √[(0.5 - 0)^2 + (- (5√3)/2 - 0)^2]Simplify:Distance = √[(0.5)^2 + (- (5√3)/2)^2]Calculate each part:(0.5)^2 = 0.25(- (5√3)/2)^2 = (25 * 3)/4 = 75/4 = 18.75So, Distance = √[0.25 + 18.75] = √[19] ≈ 4.358 milesLooking at the answer choices, √19 is one of them, which is approximately 4.358 miles, and it's less than 5 miles as specified in the problem.Wait, let me double-check my calculations to make sure I didn't make any mistakes.First movement: 4 miles west to (-4, 0) – that seems correct.Second movement: 5 miles at 300 degrees. Calculated x-component as 2.5 and y-component as - (5√3)/2. Adding to (-4, 0) gives (-1.5, - (5√3)/2) – that seems correct.Third movement: 2 miles east, so x increases by 2 to 0.5, y remains the same – correct.Distance calculation: √[(0.5)^2 + (- (5√3)/2)^2] = √[0.25 + 18.75] = √19 – correct.Yes, everything checks out. So, the distance is √19 miles, which is approximately 4.358 miles, and it's less than 5 miles as required.**Final Answer**boxed{C}

answer：Okay, so I have this problem where I'm given that a and b are both greater than 0, and they satisfy the equation 1/a + 3/b = 1. Then, I need to check which of the statements A, B, C, D are correct regarding the minimum values of ab, a + b, a² + b², and 1/(a - 1) + 3/(b - 3). First, I'll try to understand what each statement is claiming and then figure out how to approach each one. Maybe I can use some algebraic manipulation or inequalities like AM-GM to find the minimum values.Starting with Option A: The minimum value of ab is 12.Hmm, so I need to find the minimum value of the product ab given the constraint 1/a + 3/b = 1. Maybe I can express one variable in terms of the other and then find the minimum.Let me try that. Let's solve the equation 1/a + 3/b = 1 for one variable. Let's solve for 1/a:1/a = 1 - 3/bThen, 1/a = (b - 3)/bSo, a = b / (b - 3)Now, ab would be a * b = [b / (b - 3)] * b = b² / (b - 3)So, ab = b² / (b - 3). Now, I need to find the minimum value of this expression.To find the minimum, I can take the derivative with respect to b and set it to zero. But since I'm not sure if calculus is the intended method here, maybe I can use the AM-GM inequality instead.Wait, AM-GM applies to sums, but I have a product. Maybe I can manipulate the given equation to apply AM-GM.Let me consider the given equation: 1/a + 3/b = 1.I can write this as (1/a) + (3/b) = 1.If I let x = 1/a and y = 3/b, then x + y = 1, and I need to find the minimum of ab.But ab = (1/x) * (3/y) = 3/(xy). So, I need to minimize 3/(xy) given that x + y = 1.Since x + y = 1, the maximum of xy occurs when x = y = 1/2, by AM-GM. So, the maximum of xy is (1/2)^2 = 1/4.Therefore, the minimum of 3/(xy) would be when xy is maximum, so 3/(1/4) = 12. So, the minimum value of ab is 12. That makes Option A correct.Okay, moving on to Option B: The minimum value of a + b is 4√3.So, I need to find the minimum of a + b given 1/a + 3/b = 1.Again, maybe I can use the Cauchy-Schwarz inequality or AM-GM here.Alternatively, I can express a in terms of b or vice versa and then find the minimum.From earlier, I have a = b / (b - 3). So, a + b = b / (b - 3) + b.Let me write that as a function of b: f(b) = b / (b - 3) + b.Simplify f(b):f(b) = [b + b(b - 3)] / (b - 3) = [b + b² - 3b] / (b - 3) = (b² - 2b) / (b - 3)Hmm, that might not be the easiest way. Maybe I can use calculus here.Let me take the derivative of f(b):f(b) = b / (b - 3) + bFirst, let's differentiate term by term.The derivative of b / (b - 3) is [ (1)(b - 3) - b(1) ] / (b - 3)^2 = (b - 3 - b) / (b - 3)^2 = (-3) / (b - 3)^2.The derivative of b is 1.So, f'(b) = (-3)/(b - 3)^2 + 1.Set f'(b) = 0:(-3)/(b - 3)^2 + 1 = 0=> 1 = 3/(b - 3)^2=> (b - 3)^2 = 3=> b - 3 = sqrt(3) or b - 3 = -sqrt(3)But since b > 0 and from earlier, a = b/(b - 3), we need b > 3 to have a positive a. So, b - 3 = sqrt(3) => b = 3 + sqrt(3)Then, a = (3 + sqrt(3)) / (sqrt(3)) = (3 + sqrt(3)) / sqrt(3) = sqrt(3) + 1So, a + b = (sqrt(3) + 1) + (3 + sqrt(3)) = 4 + 2sqrt(3)Wait, 4 + 2√3 is approximately 4 + 3.464 = 7.464, which is less than 4√3 (which is about 6.928). Wait, no, 4√3 is about 6.928, but 4 + 2√3 is about 7.464, which is actually larger. So, that contradicts the option B which says the minimum is 4√3.Wait, maybe I made a mistake in the calculation.Wait, let me check:If b = 3 + sqrt(3), then a = (3 + sqrt(3)) / sqrt(3) = (3/sqrt(3)) + (sqrt(3)/sqrt(3)) = sqrt(3) + 1.So, a + b = (sqrt(3) + 1) + (3 + sqrt(3)) = 4 + 2sqrt(3).But 4 + 2√3 is approximately 4 + 3.464 = 7.464.But 4√3 is approximately 6.928, which is less than 7.464. So, the minimum value of a + b is 4 + 2√3, not 4√3. So, Option B is incorrect.Wait, maybe I should try using the Cauchy-Schwarz inequality.Given that 1/a + 3/b = 1, and we need to minimize a + b.Let me consider vectors. Let me think of (sqrt(a), sqrt(b)) and (1/sqrt(a), sqrt(3)/sqrt(b)).But maybe that's complicating things.Alternatively, using the method of Lagrange multipliers, but that's calculus again.Alternatively, using AM-GM.Let me consider that a + b is to be minimized, given 1/a + 3/b = 1.Let me write 1/a + 3/b = 1 as (1/a) + (3/b) = 1.Let me consider variables x = 1/a and y = 3/b, so x + y = 1.We need to minimize a + b = 1/x + 3/y.So, minimize 1/x + 3/y with x + y = 1, where x > 0, y > 0.Using Cauchy-Schwarz inequality:(1/x + 3/y)(x + y) ≥ (1 + sqrt(3))^2Since x + y = 1, we have (1/x + 3/y) ≥ (1 + sqrt(3))^2.But (1 + sqrt(3))^2 = 1 + 2sqrt(3) + 3 = 4 + 2sqrt(3).So, 1/x + 3/y ≥ 4 + 2sqrt(3).Thus, the minimum of a + b is 4 + 2sqrt(3), which is approximately 7.464, as before. So, Option B is incorrect because it claims the minimum is 4√3, which is about 6.928.Wait, but 4 + 2√3 is actually greater than 4√3, since 4 + 2√3 ≈ 7.464 and 4√3 ≈ 6.928. So, the minimum is indeed 4 + 2√3, not 4√3. So, Option B is incorrect.Moving on to Option C: The minimum value of a² + b² is 24.Again, using the constraint 1/a + 3/b = 1.I need to find the minimum of a² + b².Perhaps I can use the method of Lagrange multipliers, but since I'm not sure, maybe I can express a in terms of b and then find the minimum.From earlier, a = b / (b - 3).So, a² + b² = (b²)/(b - 3)^2 + b².Let me write this as f(b) = b²/(b - 3)^2 + b².To find the minimum, I can take the derivative of f(b) with respect to b and set it to zero.But this might get complicated. Alternatively, maybe I can use the Cauchy-Schwarz inequality or other inequalities.Alternatively, I can consider that a² + b² is minimized when a and b are as small as possible, but given the constraint.Wait, perhaps I can use the method of substitution.Let me let t = b - 3, so b = t + 3, where t > 0.Then, a = (t + 3)/t = 1 + 3/t.So, a = 1 + 3/t, and b = t + 3.Then, a² + b² = (1 + 3/t)^2 + (t + 3)^2.Expanding:(1 + 6/t + 9/t²) + (t² + 6t + 9) = 1 + 6/t + 9/t² + t² + 6t + 9.Simplify:t² + 6t + 10 + 6/t + 9/t².Now, to find the minimum of this expression, maybe I can consider it as a function of t and take the derivative.Let me denote f(t) = t² + 6t + 10 + 6/t + 9/t².Compute f'(t):f'(t) = 2t + 6 - 6/t² - 18/t³.Set f'(t) = 0:2t + 6 - 6/t² - 18/t³ = 0.Multiply both sides by t³ to eliminate denominators:2t^4 + 6t^3 - 6t - 18 = 0.Hmm, this is a quartic equation, which might be difficult to solve. Maybe I can factor it.Let me factor out 2:2(t^4 + 3t^3 - 3t - 9) = 0.So, t^4 + 3t^3 - 3t - 9 = 0.Let me try to factor this polynomial.Looking for rational roots using Rational Root Theorem: possible roots are ±1, ±3, ±9.Test t = 1: 1 + 3 - 3 - 9 = -8 ≠ 0.t = 3: 81 + 81 - 9 - 9 = 144 ≠ 0.t = -1: 1 - 3 + 3 - 9 = -8 ≠ 0.t = -3: 81 - 81 + 9 - 9 = 0. Oh, t = -3 is a root.So, we can factor out (t + 3):Using polynomial division or synthetic division:Divide t^4 + 3t^3 - 3t - 9 by (t + 3).Using synthetic division:-3 | 1 3 0 -3 -9 -3 0 0 9 1 0 0 -3 0So, the polynomial factors as (t + 3)(t^3 - 3).Thus, t^3 - 3 = 0 => t = cube root of 3 ≈ 1.442.Since t > 0, we have t = ∛3.So, t = ∛3.Now, let's compute a and b:b = t + 3 = ∛3 + 3.a = 1 + 3/t = 1 + 3/∛3 = 1 + ∛9.Now, compute a² + b²:a² = (1 + ∛9)^2 = 1 + 2∛9 + (∛9)^2.b² = (∛3 + 3)^2 = (∛3)^2 + 6∛3 + 9.So, a² + b² = [1 + 2∛9 + (∛9)^2] + [(∛3)^2 + 6∛3 + 9].Simplify:1 + 9 + 2∛9 + 6∛3 + (∛9)^2 + (∛3)^2.Note that ∛9 = (∛3)^2, so (∛9)^2 = (∛3)^4 = (∛3)^3 * ∛3 = 3 * ∛3.Similarly, (∛3)^2 is just ∛9.So, let's substitute:= 10 + 2∛9 + 6∛3 + 3∛3 + ∛9Combine like terms:= 10 + (2∛9 + ∛9) + (6∛3 + 3∛3)= 10 + 3∛9 + 9∛3Hmm, that's approximately:∛9 ≈ 2.080, ∛3 ≈ 1.442.So, 3∛9 ≈ 6.240, 9∛3 ≈ 12.978.Total ≈ 10 + 6.240 + 12.978 ≈ 29.218.But Option C claims the minimum is 24, which is less than this. So, perhaps my approach is wrong.Wait, maybe I made a mistake in the derivative calculation.Let me double-check the derivative of f(t) = t² + 6t + 10 + 6/t + 9/t².f'(t) = 2t + 6 - 6/t² - 18/t³.Yes, that's correct.Setting f'(t) = 0:2t + 6 - 6/t² - 18/t³ = 0.Multiplying by t³:2t^4 + 6t^3 - 6t - 18 = 0.Factoring out 2:2(t^4 + 3t^3 - 3t - 9) = 0.Then, factoring (t + 3)(t^3 - 3) = 0.So, t = -3 or t = ∛3.Since t > 0, t = ∛3.So, the calculation seems correct.Thus, the minimum value of a² + b² is approximately 29.218, which is greater than 24. So, Option C is incorrect.Wait, but maybe I can find a better approach.Alternatively, using Cauchy-Schwarz:Given 1/a + 3/b = 1.We can write this as (1/a) + (3/b) = 1.We need to find the minimum of a² + b².Let me consider vectors (a, b) and (1/a, 3/b).But I'm not sure.Alternatively, using the method of Lagrange multipliers.Let me set up the function to minimize: f(a, b) = a² + b².Subject to the constraint g(a, b) = 1/a + 3/b - 1 = 0.The Lagrangian is L = a² + b² - λ(1/a + 3/b - 1).Taking partial derivatives:∂L/∂a = 2a + λ/a² = 0∂L/∂b = 2b + 3λ/b² = 0∂L/∂λ = -(1/a + 3/b - 1) = 0From the first equation: 2a + λ/a² = 0 => λ = -2a³From the second equation: 2b + 3λ/b² = 0 => 2b + 3(-2a³)/b² = 0 => 2b - 6a³/b² = 0 => 2b = 6a³/b² => b³ = 3a³ => b = a * (3)^(1/3)So, b = a * ∛3.Now, substitute into the constraint: 1/a + 3/b = 1.Since b = a * ∛3, then 3/b = 3/(a * ∛3) = 3/(a * ∛3) = (3)/(a * ∛3) = (3)/(a * ∛3) = (3)/(a * ∛3) = (3)/(a * ∛3) = (3)/(a * ∛3) = (3)/(a * ∛3) = (3)/(a * ∛3).Wait, let me compute 3/b:3/b = 3/(a * ∛3) = 3/(a * ∛3) = (3)/(a * ∛3) = (3)/(a * ∛3) = (3)/(a * ∛3) = (3)/(a * ∛3) = (3)/(a * ∛3).Wait, maybe I can rationalize this:3/(a * ∛3) = 3/(a * ∛3) = (3 * ∛9)/(a * ∛3 * ∛9) = (3 * ∛9)/(a * ∛27) = (3 * ∛9)/(a * 3) = ∛9 / a.So, 3/b = ∛9 / a.Thus, the constraint becomes:1/a + ∛9 / a = 1 => (1 + ∛9)/a = 1 => a = 1 + ∛9.Then, b = a * ∛3 = (1 + ∛9) * ∛3 = ∛3 + ∛27 = ∛3 + 3.Wait, ∛27 is 3, so b = ∛3 + 3.So, a = 1 + ∛9, b = 3 + ∛3.Now, compute a² + b²:a² = (1 + ∛9)^2 = 1 + 2∛9 + (∛9)^2.b² = (3 + ∛3)^2 = 9 + 6∛3 + (∛3)^2.So, a² + b² = 1 + 2∛9 + (∛9)^2 + 9 + 6∛3 + (∛3)^2.Simplify:= 10 + 2∛9 + 6∛3 + (∛9)^2 + (∛3)^2.As before, since ∛9 = (∛3)^2, let me denote x = ∛3.Then, ∛9 = x², and (∛9)^2 = x^4 = (x^3) * x = 3x.Similarly, (∛3)^2 = x².So, substituting:= 10 + 2x² + 6x + 3x + x²= 10 + 3x² + 9x.Now, x = ∛3 ≈ 1.442.So, 3x² ≈ 3*(2.080) ≈ 6.240.9x ≈ 12.978.Thus, total ≈ 10 + 6.240 + 12.978 ≈ 29.218.So, the minimum value of a² + b² is approximately 29.218, which is greater than 24. Therefore, Option C is incorrect.Now, moving on to Option D: The minimum value of 1/(a - 1) + 3/(b - 3) is 2.Hmm, so I need to find the minimum of 1/(a - 1) + 3/(b - 3).Given that a > 0, b > 0, and 1/a + 3/b = 1.First, let's note that since a > 0 and 1/a is part of the equation, a must be greater than 1, because if a were 1, 1/a would be 1, and 3/b would have to be 0, which is impossible since b > 0. Similarly, b must be greater than 3, because 3/b is part of the sum equal to 1, so if b were 3, 3/b would be 1, and 1/a would have to be 0, which is impossible.So, a > 1 and b > 3.Now, let me define new variables to simplify the expression:Let x = a - 1, so a = x + 1, where x > 0.Let y = b - 3, so b = y + 3, where y > 0.Now, the expression to minimize becomes 1/x + 3/y.The constraint is 1/a + 3/b = 1, which becomes 1/(x + 1) + 3/(y + 3) = 1.So, we have:1/(x + 1) + 3/(y + 3) = 1.We need to minimize 1/x + 3/y.This looks similar to the original constraint, but shifted.Let me consider using the AM-GM inequality.Let me denote u = 1/x and v = 3/y.Then, the expression to minimize is u + v.The constraint is 1/(x + 1) + 3/(y + 3) = 1.But 1/(x + 1) = 1/(1 + x) and 3/(y + 3) = 3/(3 + y).Let me express the constraint in terms of u and v.Since u = 1/x, x = 1/u.Similarly, v = 3/y, so y = 3/v.Thus, the constraint becomes:1/(1 + 1/u) + 3/(3 + 3/v) = 1.Simplify:1/( (u + 1)/u ) + 3/( 3(1 + 1/v) ) = 1.Which is:u/(u + 1) + (3)/(3(1 + 1/v)) = 1.Simplify further:u/(u + 1) + 1/(1 + 1/v) = 1.But 1/(1 + 1/v) = v/(v + 1).So, the constraint becomes:u/(u + 1) + v/(v + 1) = 1.Now, we have to minimize u + v, given that u/(u + 1) + v/(v + 1) = 1.Let me denote s = u + v.We need to find the minimum s such that u/(u + 1) + v/(v + 1) = 1.Let me consider the function f(t) = t/(t + 1), which is increasing for t > 0.So, if u and v are positive, then f(u) + f(v) = 1.We need to minimize s = u + v.Let me consider that f(u) + f(v) = 1.Let me set f(u) = a, f(v) = 1 - a, where 0 < a < 1.Then, u = a/(1 - a), v = (1 - a)/a.So, s = u + v = a/(1 - a) + (1 - a)/a.To minimize s, we can take the derivative with respect to a and set it to zero.But maybe there's a better way.Alternatively, using AM-GM.Let me consider that u/(u + 1) + v/(v + 1) = 1.Let me write this as:u/(u + 1) = 1 - v/(v + 1) = (v + 1 - v)/(v + 1) = 1/(v + 1).So, u/(u + 1) = 1/(v + 1).Cross-multiplying:u(v + 1) = u + 1.So, uv + u = u + 1.Thus, uv = 1.So, u * v = 1.Therefore, v = 1/u.Now, we need to minimize s = u + v = u + 1/u.The minimum of u + 1/u occurs when u = 1, by AM-GM, since u + 1/u ≥ 2√(u * 1/u) = 2.Thus, the minimum value of s = u + v is 2, achieved when u = v = 1.Therefore, the minimum value of 1/(a - 1) + 3/(b - 3) is 2, making Option D correct.Wait, let me verify this.If u = v = 1, then u = 1, so x = 1/u = 1, so a = x + 1 = 2.Similarly, v = 1, so y = 3/v = 3, so b = y + 3 = 6.Now, check the constraint: 1/a + 3/b = 1/2 + 3/6 = 1/2 + 1/2 = 1, which satisfies the condition.Thus, when a = 2 and b = 6, the expression 1/(a - 1) + 3/(b - 3) = 1/1 + 3/3 = 1 + 1 = 2, which is indeed the minimum.So, Option D is correct.In summary:Option A: Correct.Option B: Incorrect.Option C: Incorrect.Option D: Correct.Therefore, the correct options are A and D.

answer：Okay, so I have this problem where I need to prove that a certain product is divisible by another product. Let me try to understand what's given and what I need to show.First, the problem states that ( k, m, ) and ( n ) are positive integers, and ( m + k + 1 ) is a prime number greater than ( n + 1 ). I need to work with something called ( c ), which is defined as ( s(s + 1) ). Hmm, I think ( c_r ) probably means ( r(r + 1) ), right? So, ( c_r = r(r + 1) ).The main goal is to show that the product ( (c_{m+1} - c_k)(c_{m+2} - c_k) cdots (c_{m+n} - c_k) ) is divisible by ( c_1 c_2 cdots c_n ). So, I need to look at the difference ( c_{m+i} - c_k ) for ( i ) from 1 to ( n ), multiply them all together, and show that this big product is divisible by the product of ( c_1 ) through ( c_n ).Let me start by computing ( c_r - c_x ) for some ( r ) and ( x ). Since ( c_r = r(r + 1) ), then:( c_r - c_x = r(r + 1) - x(x + 1) )Let me compute this:( c_r - c_x = r^2 + r - x^2 - x )I can factor this expression. Let's see:( r^2 - x^2 = (r - x)(r + x) ), so:( c_r - c_x = (r - x)(r + x) + (r - x) )Factor out ( (r - x) ):( c_r - c_x = (r - x)(r + x + 1) )Okay, so that's a useful factorization. So, ( c_r - c_x = (r - x)(r + x + 1) ).Now, applying this to our problem, where ( r = m + i ) and ( x = k ), so:( c_{m+i} - c_k = (m + i - k)(m + i + k + 1) )Therefore, the product ( (c_{m+1} - c_k)(c_{m+2} - c_k) cdots (c_{m+n} - c_k) ) becomes:( prod_{i=1}^{n} (m + i - k)(m + i + k + 1) )Which can be written as two separate products:( left( prod_{i=1}^{n} (m + i - k) right) times left( prod_{i=1}^{n} (m + i + k + 1) right) )Let me simplify each product separately.First, the product ( prod_{i=1}^{n} (m + i - k) ). Let me denote ( a = m - k + 1 ), so the terms become ( a, a + 1, a + 2, ldots, a + n - 1 ). So, this is ( frac{(a + n - 1)!}{(a - 1)!} ).Similarly, the second product ( prod_{i=1}^{n} (m + i + k + 1) ). Let me denote ( b = m + k + 2 ), so the terms become ( b, b + 1, b + 2, ldots, b + n - 1 ). So, this is ( frac{(b + n - 1)!}{(b - 1)!} ).Therefore, the entire product is:( frac{(a + n - 1)!}{(a - 1)!} times frac{(b + n - 1)!}{(b - 1)!} )Now, let's recall that ( c_1 c_2 cdots c_n = prod_{r=1}^{n} r(r + 1) ). Let me compute this product:( c_1 = 1 times 2 = 2 )( c_2 = 2 times 3 = 6 )( c_3 = 3 times 4 = 12 )And so on, up to ( c_n = n(n + 1) ).So, the product ( c_1 c_2 cdots c_n = 2 times 6 times 12 times cdots times n(n + 1) ).I can write this as:( prod_{r=1}^{n} r(r + 1) = prod_{r=1}^{n} r times prod_{r=1}^{n} (r + 1) = (n!) times (n + 1)! )Wait, is that correct? Let me check:( prod_{r=1}^{n} r = n! )( prod_{r=1}^{n} (r + 1) = 2 times 3 times cdots times (n + 1) = (n + 1)! / 1! = (n + 1)! )So, yes, ( c_1 c_2 cdots c_n = n! times (n + 1)! ).Now, going back to our earlier expression for the product ( (c_{m+1} - c_k) cdots (c_{m+n} - c_k) ), which is:( frac{(a + n - 1)!}{(a - 1)!} times frac{(b + n - 1)!}{(b - 1)!} )Where ( a = m - k + 1 ) and ( b = m + k + 2 ).So, substituting back:( frac{(m - k + n)!}{(m - k)!} times frac{(m + k + n + 1)!}{(m + k + 1)!} )Now, we need to show that this is divisible by ( n! times (n + 1)! ).Given that ( m + k + 1 ) is a prime number greater than ( n + 1 ), this might be useful in ensuring that certain factorial terms are coprime or have specific divisibility properties.Let me think about the factors involved. The numerator of the first fraction is ( (m - k + n)! ) and the denominator is ( (m - k)! ). Similarly, the second fraction has numerator ( (m + k + n + 1)! ) and denominator ( (m + k + 1)! ).So, the first fraction simplifies to ( binom{m - k + n}{n} times n! ), because ( frac{(m - k + n)!}{(m - k)!} = binom{m - k + n}{n} times n! ).Similarly, the second fraction simplifies to ( binom{m + k + n + 1}{n + 1} times (n + 1)! ), because ( frac{(m + k + n + 1)!}{(m + k + 1)!} = binom{m + k + n + 1}{n + 1} times (n + 1)! ).Therefore, the entire product becomes:( binom{m - k + n}{n} times n! times binom{m + k + n + 1}{n + 1} times (n + 1)! )Which is:( binom{m - k + n}{n} times binom{m + k + n + 1}{n + 1} times n! times (n + 1)! )Now, since ( m + k + 1 ) is a prime number greater than ( n + 1 ), this implies that ( m + k + 1 ) does not divide any of the terms in ( (n + 1)! ), because ( m + k + 1 > n + 1 ). Therefore, the binomial coefficient ( binom{m + k + n + 1}{n + 1} ) is an integer, and since ( m + k + 1 ) is prime, it doesn't introduce any factors that would interfere with the divisibility by ( n! times (n + 1)! ).Similarly, ( binom{m - k + n}{n} ) is also an integer. Therefore, the entire product is divisible by ( n! times (n + 1)! ), which is exactly ( c_1 c_2 cdots c_n ).So, putting it all together, the product ( (c_{m+1} - c_k)(c_{m+2} - c_k) cdots (c_{m+n} - c_k) ) is divisible by ( c_1 c_2 cdots c_n ).

answer：First, I need to determine the initial number of dolphins in the ocean. Let's denote this number as ( D ).According to the problem, three times the initial number of dolphins join from the river. This means ( 3D ) dolphins come from the river.The total number of dolphins in the ocean after the river dolphins join is given as 260. Therefore, the equation can be set up as:[ D + 3D = 260 ]Combining like terms, the equation simplifies to:[ 4D = 260 ]To find the value of ( D ), I divide both sides of the equation by 4:[ D = frac{260}{4} ][ D = 65 ]Thus, the initial number of dolphins in the ocean was 65.

answer：First, I need to determine the total number of ways to select 5 books from the 10 available without any restrictions. This can be calculated using the combination formula, which is (binom{10}{5}).Next, since 2 specific books must always be included in the selection, I need to adjust the selection process. By including these 2 books, I reduce the problem to selecting the remaining 3 books from the 8 that are not the specific ones. This is calculated using (binom{8}{3}).Finally, I compute the value of (binom{8}{3}) to find the total number of ways to choose the 5 books while ensuring the 2 specific books are always included.