💡Okay, so I have this problem where I need to find the minimum value of the expression:[frac{x_1}{1 - x_1^2} + frac{x_2}{1 - x_2^2} + dots + frac{x_{50}}{1 - x_{50}^2}]given that ( x_1^2 + x_2^2 + dots + x_{50}^2 = 2 ) and all ( x_i ) are positive real numbers.Hmm, let me think about how to approach this. The expression involves each ( x_i ) in a fraction where the denominator is ( 1 - x_i^2 ). Since all ( x_i ) are positive, each term in the sum is positive as well. I need to minimize the sum, so maybe I can find a way to bound each term from below and then sum those bounds.I remember that sometimes inequalities like Cauchy-Schwarz or AM-GM can be useful in these situations. Let me see if I can apply any of those here.Looking at each term ( frac{x_i}{1 - x_i^2} ), it seems a bit complicated. Maybe I can simplify or find a substitution. Let me consider substituting ( y_i = x_i^2 ). Then, ( x_i = sqrt{y_i} ), and the condition becomes ( y_1 + y_2 + dots + y_{50} = 2 ).So, the expression to minimize becomes:[sum_{i=1}^{50} frac{sqrt{y_i}}{1 - y_i}]Hmm, that doesn't seem much simpler. Maybe I should think about convexity or concavity. If the function ( f(x) = frac{x}{1 - x^2} ) is convex or concave, I can apply Jensen's inequality.Let me compute the second derivative of ( f(x) ) to check for convexity.First, ( f(x) = frac{x}{1 - x^2} ).First derivative:[f'(x) = frac{(1 - x^2) cdot 1 - x cdot (-2x)}{(1 - x^2)^2} = frac{1 - x^2 + 2x^2}{(1 - x^2)^2} = frac{1 + x^2}{(1 - x^2)^2}]Second derivative:Let me compute this step by step.Let ( f'(x) = frac{1 + x^2}{(1 - x^2)^2} ).Let me denote ( u = 1 + x^2 ) and ( v = (1 - x^2)^2 ).Then, ( f'(x) = frac{u}{v} ), so the derivative is ( frac{u'v - uv'}{v^2} ).Compute ( u' = 2x ).Compute ( v = (1 - x^2)^2 ), so ( v' = 2(1 - x^2)(-2x) = -4x(1 - x^2) ).So,[f''(x) = frac{(2x)(1 - x^2)^2 - (1 + x^2)(-4x)(1 - x^2)}{(1 - x^2)^4}]Simplify numerator:Factor out ( 2x(1 - x^2) ):[2x(1 - x^2)[(1 - x^2) + 2(1 + x^2)]]Compute inside the brackets:[(1 - x^2) + 2(1 + x^2) = 1 - x^2 + 2 + 2x^2 = 3 + x^2]So numerator becomes:[2x(1 - x^2)(3 + x^2)]Therefore,[f''(x) = frac{2x(1 - x^2)(3 + x^2)}{(1 - x^2)^4} = frac{2x(3 + x^2)}{(1 - x^2)^3}]Since ( x ) is positive and less than 1 (because ( x_i^2 ) sum to 2 and there are 50 terms, each ( x_i ) must be less than 1), the denominator ( (1 - x^2)^3 ) is positive, and the numerator ( 2x(3 + x^2) ) is positive as well. Therefore, ( f''(x) > 0 ) for all ( x ) in the domain. So, ( f(x) ) is convex on the interval ( (0, 1) ).Since ( f(x) ) is convex, by Jensen's inequality, the minimum of the sum ( sum f(x_i) ) occurs when all ( x_i ) are equal, given the constraint on the sum of squares.Wait, but hold on. Jensen's inequality for convex functions states that:[frac{1}{n} sum f(x_i) geq fleft( frac{1}{n} sum x_i right)]But in our case, the constraint is on the sum of squares, not the sum of ( x_i ). So, maybe I need to use a different approach.Alternatively, perhaps I can use the method of Lagrange multipliers to find the minimum. Let me try that.Let me denote the function to minimize as:[S = sum_{i=1}^{50} frac{x_i}{1 - x_i^2}]subject to the constraint:[sum_{i=1}^{50} x_i^2 = 2]Set up the Lagrangian:[mathcal{L} = sum_{i=1}^{50} frac{x_i}{1 - x_i^2} + lambda left( sum_{i=1}^{50} x_i^2 - 2 right)]Take the partial derivatives with respect to each ( x_i ) and set them equal to zero.Compute ( frac{partial mathcal{L}}{partial x_i} ):First, derivative of ( frac{x_i}{1 - x_i^2} ) with respect to ( x_i ):As before, ( f'(x_i) = frac{1 + x_i^2}{(1 - x_i^2)^2} )So,[frac{partial mathcal{L}}{partial x_i} = frac{1 + x_i^2}{(1 - x_i^2)^2} + 2lambda x_i = 0]Wait, no. The derivative of the constraint term ( lambda sum x_i^2 ) is ( 2lambda x_i ). So, the partial derivative is:[frac{1 + x_i^2}{(1 - x_i^2)^2} + 2lambda x_i = 0]But this must hold for all ( i ). Therefore, for each ( i ), we have:[frac{1 + x_i^2}{(1 - x_i^2)^2} = -2lambda x_i]Since ( x_i ) are positive, the right-hand side is negative if ( lambda ) is positive. But the left-hand side is always positive because ( 1 + x_i^2 > 0 ) and ( (1 - x_i^2)^2 > 0 ). This is a contradiction because a positive number cannot equal a negative number. Hmm, that suggests that my setup might be wrong.Wait, perhaps I made a mistake in the derivative. Let me double-check.The term in the Lagrangian is ( frac{x_i}{1 - x_i^2} ). Its derivative with respect to ( x_i ) is:Using quotient rule:Numerator: ( x_i ), denominator: ( 1 - x_i^2 )Derivative:[frac{(1 - x_i^2)(1) - x_i(-2x_i)}{(1 - x_i^2)^2} = frac{1 - x_i^2 + 2x_i^2}{(1 - x_i^2)^2} = frac{1 + x_i^2}{(1 - x_i^2)^2}]Yes, that's correct. Then, the derivative of the constraint term ( lambda sum x_i^2 ) is ( 2lambda x_i ). So, the partial derivative is:[frac{1 + x_i^2}{(1 - x_i^2)^2} + 2lambda x_i = 0]But as I said, this implies that ( frac{1 + x_i^2}{(1 - x_i^2)^2} = -2lambda x_i ). Since the left side is positive, the right side must also be positive, which implies ( lambda ) is negative.So, let me denote ( mu = -lambda ), so ( mu > 0 ). Then, the equation becomes:[frac{1 + x_i^2}{(1 - x_i^2)^2} = 2mu x_i]This must hold for all ( i ). Therefore, all ( x_i ) must satisfy the same equation, which suggests that all ( x_i ) are equal. Let me assume ( x_1 = x_2 = dots = x_{50} = x ).Then, the constraint becomes:[50x^2 = 2 implies x^2 = frac{2}{50} = frac{1}{25} implies x = frac{1}{5}]So, each ( x_i = frac{1}{5} ).Now, let's compute the value of ( S ):[S = 50 times frac{frac{1}{5}}{1 - left( frac{1}{5} right)^2} = 50 times frac{frac{1}{5}}{1 - frac{1}{25}} = 50 times frac{frac{1}{5}}{frac{24}{25}} = 50 times frac{1}{5} times frac{25}{24}]Simplify:[50 times frac{1}{5} = 10][10 times frac{25}{24} = frac{250}{24} = frac{125}{12} approx 10.4167]Wait, but the initial problem is to find the minimum value. Is this the minimum? Or is there a lower value possible?Wait, I think I might have made a mistake here. Because when I used Lagrange multipliers, I assumed all ( x_i ) are equal, but maybe that's not the case. Alternatively, perhaps I can use an inequality to bound each term.Let me think about another approach. Maybe using the Cauchy-Schwarz inequality or Holder's inequality.Alternatively, perhaps I can use the method of substitution. Let me consider substituting ( t_i = x_i^2 ), so ( t_i ) are positive real numbers with ( sum t_i = 2 ). Then, the expression becomes:[sum_{i=1}^{50} frac{sqrt{t_i}}{1 - t_i}]Hmm, not sure if that helps directly. Maybe I can consider the function ( f(t) = frac{sqrt{t}}{1 - t} ) and analyze its behavior.Compute the derivative of ( f(t) ):[f(t) = frac{sqrt{t}}{1 - t}][f'(t) = frac{(1 - t) cdot frac{1}{2sqrt{t}} - sqrt{t} cdot (-1)}{(1 - t)^2} = frac{frac{1 - t}{2sqrt{t}} + sqrt{t}}{(1 - t)^2}]Simplify numerator:[frac{1 - t + 2t}{2sqrt{t}} = frac{1 + t}{2sqrt{t}}]So,[f'(t) = frac{1 + t}{2sqrt{t}(1 - t)^2}]Since ( t > 0 ) and ( t < 1 ) (because ( x_i < 1 )), the derivative is positive. So, ( f(t) ) is increasing on ( (0, 1) ).Therefore, to minimize the sum ( sum f(t_i) ), we should distribute the total sum ( sum t_i = 2 ) in such a way that the ( t_i ) are as small as possible. But since ( f(t) ) is increasing, making some ( t_i ) larger and others smaller would increase the sum. Wait, actually, since ( f(t) ) is convex or concave?Wait, let me check the second derivative to see if ( f(t) ) is convex or concave.Compute ( f''(t) ):We have ( f'(t) = frac{1 + t}{2sqrt{t}(1 - t)^2} ).Let me write ( f'(t) = frac{1 + t}{2 t^{1/2} (1 - t)^2} ).Let me compute the derivative of ( f'(t) ):Let me denote ( u = 1 + t ), ( v = 2 t^{1/2} (1 - t)^2 ).So, ( f'(t) = frac{u}{v} ), so ( f''(t) = frac{u'v - uv'}{v^2} ).Compute ( u' = 1 ).Compute ( v = 2 t^{1/2} (1 - t)^2 ).Compute ( v' = 2 left( frac{1}{2} t^{-1/2} (1 - t)^2 + t^{1/2} cdot 2(1 - t)(-1) right) )Simplify:[v' = 2 left( frac{(1 - t)^2}{2sqrt{t}} - 2 t^{1/2} (1 - t) right ) = frac{(1 - t)^2}{sqrt{t}} - 4 t^{1/2} (1 - t)]So,[f''(t) = frac{1 cdot 2 t^{1/2} (1 - t)^2 - (1 + t) left( frac{(1 - t)^2}{sqrt{t}} - 4 t^{1/2} (1 - t) right ) }{(2 t^{1/2} (1 - t)^2)^2}]This is getting complicated. Maybe it's not the best approach.Alternatively, perhaps I can use the Cauchy-Schwarz inequality on the original expression.Let me recall that for positive numbers, ( frac{x}{1 - x^2} ) can be rewritten as ( frac{x}{(1 - x)(1 + x)} ). Maybe that helps.Alternatively, perhaps I can use the inequality ( frac{x}{1 - x^2} geq k x^2 ) for some constant ( k ). If I can find such a ( k ), then summing over all ( i ) would give me a lower bound.Let me try to find such a ( k ). Let me set:[frac{x}{1 - x^2} geq k x^2]Simplify:[frac{1}{1 - x^2} geq k x][1 geq k x (1 - x^2)][1 geq k x - k x^3]I need this inequality to hold for all ( x ) in ( (0, 1) ). Let me find the maximum ( k ) such that this holds.Let me consider the function ( g(x) = k x (1 - x^2) ). I need ( g(x) leq 1 ) for all ( x in (0, 1) ).Find the maximum of ( g(x) ):Compute derivative:[g'(x) = k (1 - x^2) + k x (-2x) = k (1 - x^2 - 2x^2) = k (1 - 3x^2)]Set derivative to zero:[1 - 3x^2 = 0 implies x = frac{1}{sqrt{3}}]So, the maximum occurs at ( x = frac{1}{sqrt{3}} ).Compute ( g ) at this point:[gleft( frac{1}{sqrt{3}} right) = k cdot frac{1}{sqrt{3}} left( 1 - frac{1}{3} right ) = k cdot frac{1}{sqrt{3}} cdot frac{2}{3} = frac{2k}{3sqrt{3}}]We need this to be less than or equal to 1:[frac{2k}{3sqrt{3}} leq 1 implies k leq frac{3sqrt{3}}{2}]So, the maximum ( k ) is ( frac{3sqrt{3}}{2} ).Therefore, the inequality:[frac{x}{1 - x^2} geq frac{3sqrt{3}}{2} x^2]holds for all ( x in (0, 1) ).Great! So, applying this inequality to each term in the sum:[sum_{i=1}^{50} frac{x_i}{1 - x_i^2} geq sum_{i=1}^{50} frac{3sqrt{3}}{2} x_i^2 = frac{3sqrt{3}}{2} sum_{i=1}^{50} x_i^2 = frac{3sqrt{3}}{2} times 2 = 3sqrt{3}]So, the minimum value of the given expression is ( 3sqrt{3} ).Now, I need to check if this bound is achievable. For equality to hold in the inequality ( frac{x}{1 - x^2} geq frac{3sqrt{3}}{2} x^2 ), we need ( x = frac{1}{sqrt{3}} ).So, if all ( x_i = frac{1}{sqrt{3}} ), then:[sum_{i=1}^{50} x_i^2 = 50 times frac{1}{3} = frac{50}{3} approx 16.6667]But our constraint is ( sum x_i^2 = 2 ), which is much smaller. So, we can't have all ( x_i = frac{1}{sqrt{3}} ).Wait, so maybe equality isn't achieved when all ( x_i ) are equal. Hmm, perhaps only some of them are equal to ( frac{1}{sqrt{3}} ) and the rest are zero. Let me check.Suppose we set ( k ) variables to ( frac{1}{sqrt{3}} ) and the rest to zero. Then, the sum of squares would be ( k times frac{1}{3} ). We need this to equal 2:[frac{k}{3} = 2 implies k = 6]So, if we set 6 variables to ( frac{1}{sqrt{3}} ) and the remaining 44 to zero, the sum of squares is 2.Then, the expression becomes:[6 times frac{frac{1}{sqrt{3}}}{1 - left( frac{1}{sqrt{3}} right)^2} + 44 times 0 = 6 times frac{frac{1}{sqrt{3}}}{1 - frac{1}{3}} = 6 times frac{frac{1}{sqrt{3}}}{frac{2}{3}} = 6 times frac{3}{2sqrt{3}} = 6 times frac{sqrt{3}}{2} = 3sqrt{3}]So, indeed, the minimum value of ( 3sqrt{3} ) is achievable when 6 variables are ( frac{1}{sqrt{3}} ) and the rest are zero.Therefore, the minimum value is ( 3sqrt{3} ).**Final Answer**The minimum value is boxed{3sqrt{3}}.

💡To determine how many kilometers friend P will have walked when they pass each other, we can follow these steps:1. **Define Variables:** - Let ( R ) be the walking rate of friend Q in km/h. - Since friend P is 25% faster, their rate is ( 1.25R ) km/h.2. **Set Up the Equation:** - When they meet, the sum of the distances they have walked equals the total trail length: [ D_p + D_q = 36 ] - Express the distances in terms of time ( t ): [ D_p = 1.25R cdot t quad text{and} quad D_q = R cdot t ] - Combine these into the equation: [ 1.25R cdot t + R cdot t = 36 ]3. **Solve for Time ( t ):** - Combine like terms: [ 2.25R cdot t = 36 ] - Solve for ( t ): [ t = frac{36}{2.25R} ]4. **Calculate Distance for Friend P:** - Substitute ( t ) back into the equation for ( D_p ): [ D_p = 1.25R cdot frac{36}{2.25R} ] - Simplify: [ D_p = 1.25 cdot frac{36}{2.25} = 1.25 cdot 16 = 20 ]Therefore, friend P will have walked **20 kilometers** when they pass each other.

💡Alright, so I have this problem: Find the largest integer ( n ) such that ( n! ) can be expressed as the product of 4 consecutive integers. Hmm, okay. Let me try to wrap my head around this.First off, ( n! ) is the factorial of ( n ), which means it's the product of all positive integers up to ( n ). So, ( 5! = 5 times 4 times 3 times 2 times 1 = 120 ). Got that. Now, the problem is asking for the largest ( n ) where ( n! ) can be written as the product of four consecutive integers. Let's break that down.If I have four consecutive integers, I can represent them as ( x ), ( x+1 ), ( x+2 ), and ( x+3 ). So, their product would be ( x(x+1)(x+2)(x+3) ). The problem states that this product equals ( n! ). So, mathematically, we can write:[n! = x(x+1)(x+2)(x+3)]Okay, so I need to find the largest ( n ) such that this equation holds true for some integer ( x ). Interesting. Let me think about how to approach this.One way to tackle this is to express the product of four consecutive integers in terms of factorials or some other form that might relate to ( n! ). Let me see. The product ( x(x+1)(x+2)(x+3) ) can be written as ( frac{(x+3)!}{(x-1)!} ) because:[frac{(x+3)!}{(x-1)!} = (x)(x+1)(x+2)(x+3)]Yes, that makes sense. So, substituting back into the equation:[n! = frac{(x+3)!}{(x-1)!}]Hmm, so ( n! ) is equal to the ratio of two factorials. That seems a bit abstract. Maybe I can think about specific values of ( n ) and see if I can find corresponding ( x ) values.Let me start with small values of ( n ) and see if I can find a pattern or perhaps find the largest ( n ) where this works.Starting with ( n = 4 ):[4! = 24]Is there an ( x ) such that ( x(x+1)(x+2)(x+3) = 24 )?Let's try ( x = 1 ):[1 times 2 times 3 times 4 = 24]Oh, that works! So, ( n = 4 ) is possible with ( x = 1 ). Okay, moving on.Next, ( n = 5 ):[5! = 120]Is there an ( x ) such that ( x(x+1)(x+2)(x+3) = 120 )?Let's try ( x = 2 ):[2 times 3 times 4 times 5 = 120]Yes, that works too. So, ( n = 5 ) is possible with ( x = 2 ).Moving on to ( n = 6 ):[6! = 720]Is there an ( x ) such that ( x(x+1)(x+2)(x+3) = 720 )?Let's try ( x = 3 ):[3 times 4 times 5 times 6 = 360]That's half of 720. Maybe ( x = 4 ):[4 times 5 times 6 times 7 = 840]That's more than 720. Hmm, so between ( x = 3 ) and ( x = 4 ), we go from 360 to 840. 720 is in between, but it's not achieved by an integer ( x ). Wait, maybe I made a mistake.Wait, ( x = 5 ):[5 times 6 times 7 times 8 = 1680]That's way too big. So, seems like there's no integer ( x ) such that the product is 720. But wait, hold on. Maybe I missed something.Wait, ( x = 3 ) gives 360, ( x = 4 ) gives 840. 720 is in between, but it's not achieved by an integer ( x ). So, does that mean ( n = 6 ) is not possible? But wait, let me double-check.Alternatively, maybe ( x = 0 ):[0 times 1 times 2 times 3 = 0]Nope, that's zero. Not helpful. Negative integers? Let's see, ( x = -1 ):[-1 times 0 times 1 times 2 = 0]Still zero. Not helpful. So, seems like ( n = 6 ) is not possible. Wait, but earlier, ( n = 4 ) and ( n = 5 ) were possible. Maybe ( n = 6 ) is not possible, but let me check higher ( n ) just in case.Wait, actually, I think I might have made a mistake earlier. Let me try ( x = 5 ):[5 times 6 times 7 times 8 = 1680]Which is ( 8! / 4! = 1680 ). Hmm, interesting. So, ( 8! = 40320 ), and ( 40320 / 4! = 40320 / 24 = 1680 ). So, ( x(x+1)(x+2)(x+3) = 1680 ) when ( x = 5 ). So, does that mean ( n = 8 )?Wait, hold on. If ( x(x+1)(x+2)(x+3) = 1680 ), and ( 8! = 40320 ), which is much larger than 1680. So, that doesn't directly help. Maybe I need to think differently.Wait, going back to the equation:[n! = frac{(x+3)!}{(x-1)!}]So, ( n! times (x-1)! = (x+3)! )Which implies:[n! = frac{(x+3)!}{(x-1)!} = (x)(x+1)(x+2)(x+3)]So, ( n! ) must be equal to the product of four consecutive integers. So, ( n! ) must be divisible by four consecutive integers. Hmm.Let me think about the relationship between ( n ) and ( x ). If ( n! ) is equal to the product of four consecutive integers, then ( n ) must be at least as large as the largest of those four integers, right? Because ( n! ) includes all integers up to ( n ).So, if the four consecutive integers are ( x ), ( x+1 ), ( x+2 ), ( x+3 ), then ( n ) must be at least ( x+3 ). Because ( n! ) includes ( x+3 ) as a factor.But also, ( n! ) is equal to the product of these four numbers, so ( n! ) can't be too much larger than ( (x+3)! ). Wait, but ( n! ) is much larger than ( (x+3)! ) unless ( n ) is close to ( x+3 ).Wait, let me think about this. If ( n! = x(x+1)(x+2)(x+3) ), and ( x+3 ) is part of ( n! ), then ( n ) must be greater than or equal to ( x+3 ). But if ( n ) is much larger than ( x+3 ), then ( n! ) would have many more factors, making the product ( x(x+1)(x+2)(x+3) ) much smaller than ( n! ).Therefore, ( n ) can't be too much larger than ( x+3 ). Maybe ( n ) is equal to ( x+3 ) or ( x+4 ) or something like that.Let me test this idea. Suppose ( n = x+3 ). Then, ( n! = (x+3)! ). But we have:[n! = x(x+1)(x+2)(x+3)]So, substituting ( n = x+3 ):[(x+3)! = x(x+1)(x+2)(x+3)]Which simplifies to:[(x+3)! = x(x+1)(x+2)(x+3)]But ( (x+3)! = (x+3)(x+2)(x+1)x! ). So, substituting:[(x+3)(x+2)(x+1)x! = x(x+1)(x+2)(x+3)]Dividing both sides by ( x(x+1)(x+2)(x+3) ):[x! = 1]Which implies ( x = 0 ) or ( x = 1 ). Because ( 0! = 1 ) and ( 1! = 1 ).So, if ( x = 0 ):[0 times 1 times 2 times 3 = 0]But ( n! = 0 ) only if ( n = 0 ) or ( n = 1 ), but ( 0! = 1 ), so that doesn't make sense. Similarly, ( x = 1 ):[1 times 2 times 3 times 4 = 24 = 4!]So, ( n = 4 ) in this case, which we already saw earlier.So, when ( n = x+3 ), the only solution is ( n = 4 ) with ( x = 1 ).What if ( n = x+4 )? Let's try that.So, ( n = x+4 ), then ( n! = (x+4)! ). We have:[(x+4)! = x(x+1)(x+2)(x+3)]But ( (x+4)! = (x+4)(x+3)(x+2)(x+1)x! ). So, substituting:[(x+4)(x+3)(x+2)(x+1)x! = x(x+1)(x+2)(x+3)]Dividing both sides by ( x(x+1)(x+2)(x+3) ):[(x+4)x! = 1]Which implies ( (x+4)x! = 1 ). The only way this can happen is if ( x! = 1 ) and ( x+4 = 1 ). But ( x+4 = 1 ) implies ( x = -3 ), which is not a valid integer in the context of factorials. So, no solution here.What if ( n = x+2 )? Let's try that.So, ( n = x+2 ), then ( n! = (x+2)! ). We have:[(x+2)! = x(x+1)(x+2)(x+3)]But ( (x+2)! = (x+2)(x+1)x! ). So, substituting:[(x+2)(x+1)x! = x(x+1)(x+2)(x+3)]Dividing both sides by ( x(x+1)(x+2) ):[x! = x+3]So, ( x! = x + 3 ). Let's solve for ( x ).Trying ( x = 1 ):[1! = 1 = 1 + 3 = 4 quad text{No}]( x = 2 ):[2! = 2 = 2 + 3 = 5 quad text{No}]( x = 3 ):[3! = 6 = 3 + 3 = 6 quad text{Yes!}]So, ( x = 3 ), then ( n = x+2 = 5 ). So, ( n = 5 ) is a solution, which we already found earlier.So, when ( n = x+2 ), we get ( n = 5 ) as a solution.What about ( n = x+1 )? Let's see.So, ( n = x+1 ), then ( n! = (x+1)! ). We have:[(x+1)! = x(x+1)(x+2)(x+3)]But ( (x+1)! = (x+1)x! ). So, substituting:[(x+1)x! = x(x+1)(x+2)(x+3)]Dividing both sides by ( x(x+1) ):[x! = (x+2)(x+3)]So, ( x! = (x+2)(x+3) ). Let's solve for ( x ).Trying ( x = 1 ):[1! = 1 = 3 times 4 = 12 quad text{No}]( x = 2 ):[2! = 2 = 4 times 5 = 20 quad text{No}]( x = 3 ):[3! = 6 = 5 times 6 = 30 quad text{No}]( x = 4 ):[4! = 24 = 6 times 7 = 42 quad text{No}]( x = 5 ):[5! = 120 = 7 times 8 = 56 quad text{No}]Seems like no solution here.So, from these attempts, it seems that when ( n = x+3 ), we get ( n = 4 ), and when ( n = x+2 ), we get ( n = 5 ). So, ( n = 5 ) is larger than ( n = 4 ). Is there a way to get a larger ( n )?Wait, earlier I thought ( n = 6 ) might not be possible, but let me double-check.For ( n = 6 ), ( 6! = 720 ). Is there an ( x ) such that ( x(x+1)(x+2)(x+3) = 720 )?Let me try ( x = 4 ):[4 times 5 times 6 times 7 = 840]Too big. ( x = 3 ):[3 times 4 times 5 times 6 = 360]Too small. So, 720 is between 360 and 840, but there's no integer ( x ) that gives exactly 720. So, ( n = 6 ) is not possible.Wait, but earlier I saw that ( 5 times 6 times 7 times 8 = 1680 ), which is ( 8! / 4! ). So, ( 8! = 40320 ), and ( 40320 / 24 = 1680 ). So, ( 5 times 6 times 7 times 8 = 1680 ), which is ( 8! / 4! ). So, does that mean ( n = 8 )?Wait, no, because ( n! = x(x+1)(x+2)(x+3) ). So, if ( x = 5 ), then ( x(x+1)(x+2)(x+3) = 1680 ), which is not equal to ( 8! = 40320 ). So, that doesn't help.Wait, but maybe ( n = 8 ) can be expressed as the product of four consecutive integers in a different way. Let me check.Is there an ( x ) such that ( x(x+1)(x+2)(x+3) = 40320 )?Let me try ( x = 10 ):[10 times 11 times 12 times 13 = 17160]Too small. ( x = 15 ):[15 times 16 times 17 times 18 = 73440]Too big. So, 40320 is between 17160 and 73440. Let's try ( x = 12 ):[12 times 13 times 14 times 15 = 32760]Still too small. ( x = 13 ):[13 times 14 times 15 times 16 = 43680]Too big. So, between ( x = 12 ) and ( x = 13 ), we go from 32760 to 43680. 40320 is in between, but there's no integer ( x ) that gives exactly 40320. So, ( n = 8 ) is not possible.Hmm, interesting. So, ( n = 5 ) is possible, but ( n = 6 ) and ( n = 8 ) are not. What about ( n = 7 )?( 7! = 5040 ). Is there an ( x ) such that ( x(x+1)(x+2)(x+3) = 5040 )?Let me try ( x = 5 ):[5 times 6 times 7 times 8 = 1680]Too small. ( x = 6 ):[6 times 7 times 8 times 9 = 3024]Still too small. ( x = 7 ):[7 times 8 times 9 times 10 = 5040]Oh, wait! That works. So, ( x = 7 ), and ( 7 times 8 times 9 times 10 = 5040 = 7! ). So, ( n = 7 ) is possible.Wait, that's interesting. So, ( n = 7 ) is possible with ( x = 7 ). But earlier, I thought ( n = 6 ) wasn't possible. Let me double-check ( n = 6 ).( 6! = 720 ). Is there an ( x ) such that ( x(x+1)(x+2)(x+3) = 720 )?Trying ( x = 5 ):[5 times 6 times 7 times 8 = 1680]Too big. ( x = 4 ):[4 times 5 times 6 times 7 = 840]Still too big. ( x = 3 ):[3 times 4 times 5 times 6 = 360]Too small. So, 720 is between 360 and 840, but there's no integer ( x ) that gives exactly 720. So, ( n = 6 ) is not possible.Wait, but ( n = 7 ) is possible. So, ( n = 7 ) is larger than ( n = 5 ). Is there a larger ( n ) possible?Let's try ( n = 8 ). As I saw earlier, ( 8! = 40320 ). Is there an ( x ) such that ( x(x+1)(x+2)(x+3) = 40320 )?Trying ( x = 10 ):[10 times 11 times 12 times 13 = 17160]Too small. ( x = 15 ):[15 times 16 times 17 times 18 = 73440]Too big. So, 40320 is between 17160 and 73440. Let's try ( x = 12 ):[12 times 13 times 14 times 15 = 32760]Still too small. ( x = 13 ):[13 times 14 times 15 times 16 = 43680]Too big. So, 40320 is between 32760 and 43680. Let's try ( x = 14 ):[14 times 15 times 16 times 17 = 57120]Too big. So, no integer ( x ) gives exactly 40320. Therefore, ( n = 8 ) is not possible.What about ( n = 9 )? ( 9! = 362880 ). Is there an ( x ) such that ( x(x+1)(x+2)(x+3) = 362880 )?Let me try ( x = 20 ):[20 times 21 times 22 times 23 = 212520]Too small. ( x = 25 ):[25 times 26 times 27 times 28 = 491400]Too big. So, 362880 is between 212520 and 491400. Let's try ( x = 22 ):[22 times 23 times 24 times 25 = 303600]Still too small. ( x = 23 ):[23 times 24 times 25 times 26 = 358800]Closer, but still too small. ( x = 24 ):[24 times 25 times 26 times 27 = 425280]Too big. So, 362880 is between 358800 and 425280. Let's try ( x = 23.5 ), but that's not an integer. So, no solution here.Hmm, seems like ( n = 7 ) is the highest so far. Let me check ( n = 10 ) just to be thorough.( 10! = 3628800 ). Is there an ( x ) such that ( x(x+1)(x+2)(x+3) = 3628800 )?Trying ( x = 30 ):[30 times 31 times 32 times 33 = 985680]Too small. ( x = 40 ):[40 times 41 times 42 times 43 = 2812320]Still too small. ( x = 45 ):[45 times 46 times 47 times 48 = 4741680]Too big. So, 3628800 is between 2812320 and 4741680. Let's try ( x = 42 ):[42 times 43 times 44 times 45 = 3603600]Close! But still not exactly 3628800. ( x = 43 ):[43 times 44 times 45 times 46 = 40091280]Way too big. So, no integer ( x ) gives exactly 3628800. Therefore, ( n = 10 ) is not possible.At this point, it seems like ( n = 7 ) is the largest ( n ) where ( n! ) can be expressed as the product of four consecutive integers. But let me double-check my earlier assumption that ( n = 7 ) works.So, ( 7! = 5040 ). And ( 7 times 8 times 9 times 10 = 5040 ). Yes, that's correct. So, ( n = 7 ) is indeed a solution.Wait a minute, earlier I thought ( n = 7 ) was possible, but I also saw that ( n = 5 ) was possible. So, ( n = 7 ) is larger. Is there a way to get ( n ) larger than 7?Let me think about the general case. Suppose ( n! = x(x+1)(x+2)(x+3) ). We can write this as:[n! = x^4 + 6x^3 + 11x^2 + 6x]But this seems complicated. Maybe I can use approximations or inequalities to find bounds on ( n ).Since ( x(x+1)(x+2)(x+3) ) is roughly ( x^4 ) for large ( x ), and ( n! ) grows faster than any polynomial, there must be a point where ( n! ) becomes too large to be expressed as the product of four consecutive integers. So, the largest ( n ) must be somewhere before this point.But how do I find that point? Maybe I can use Stirling's approximation for factorials, which says:[n! approx sqrt{2pi n} left( frac{n}{e} right)^n]And for the product of four consecutive integers, ( x(x+1)(x+2)(x+3) approx x^4 ) for large ( x ). So, setting ( x^4 approx sqrt{2pi n} left( frac{n}{e} right)^n ).But this seems too vague. Maybe instead, I can consider the ratio of ( n! ) to ( x(x+1)(x+2)(x+3) ) and see how it behaves as ( n ) increases.Alternatively, perhaps I can use the fact that ( x(x+1)(x+2)(x+3) = (x^2 + 3x)(x^2 + 3x + 2) ). Let me set ( y = x^2 + 3x ), then the product becomes ( y(y + 2) = y^2 + 2y ). So, ( n! = y^2 + 2y ).But I'm not sure if this helps. Maybe I can think about the prime factors. Since ( n! ) includes all primes up to ( n ), and the product of four consecutive integers includes primes up to ( x+3 ). So, for ( n! ) to be equal to the product, all primes up to ( n ) must be included in the four consecutive integers. But four consecutive integers can't include too many primes unless ( x ) is small.Wait, for example, if ( x ) is large, the four consecutive integers ( x, x+1, x+2, x+3 ) will include at most two primes (since every other number is even, and one of them could be a multiple of 3, etc.). But ( n! ) includes all primes up to ( n ), which could be many more. So, unless ( n ) is small, it's unlikely that ( n! ) can be expressed as the product of four consecutive integers.Therefore, the largest ( n ) is probably not too large. From my earlier attempts, ( n = 7 ) works, but ( n = 8 ) and above do not. So, maybe ( n = 7 ) is the largest.But wait, earlier I thought ( n = 5 ) was possible, then ( n = 7 ). Is there a way to confirm that ( n = 7 ) is indeed the largest?Let me check ( n = 7 ) again. ( 7! = 5040 ). And ( 7 times 8 times 9 times 10 = 5040 ). Yes, that's correct. So, ( n = 7 ) is a valid solution.What about ( n = 6 )? As I saw earlier, ( 6! = 720 ). Is there an ( x ) such that ( x(x+1)(x+2)(x+3) = 720 )?Trying ( x = 5 ):[5 times 6 times 7 times 8 = 1680]Too big. ( x = 4 ):[4 times 5 times 6 times 7 = 840]Still too big. ( x = 3 ):[3 times 4 times 5 times 6 = 360]Too small. So, 720 is between 360 and 840, but there's no integer ( x ) that gives exactly 720. Therefore, ( n = 6 ) is not possible.So, ( n = 7 ) is the largest ( n ) where ( n! ) can be expressed as the product of four consecutive integers.Wait, but earlier I thought ( n = 5 ) was possible, and ( n = 7 ) is larger. So, is ( n = 7 ) the answer? Let me check one more time.Yes, ( 7! = 5040 ), and ( 7 times 8 times 9 times 10 = 5040 ). So, that's correct. Therefore, ( n = 7 ) is the largest integer such that ( n! ) can be expressed as the product of four consecutive integers.I think that's it. So, the answer is ( n = 7 ).

💡First, let's denote the three positive integers as ( a ), ( b ), and ( c ), where ( a leq b leq c ).Given that the arithmetic mean of the three numbers is 24, we can write the equation:[frac{a + b + c}{3} = 24]Multiplying both sides by 3, we get:[a + b + c = 72]We are also told that the median of the three numbers is 25, which means ( b = 25 ).Additionally, it's stated that the median is 6 less than the largest number. Therefore:[b = c - 6]Substituting ( b = 25 ) into the equation, we find:[25 = c - 6 implies c = 31]Now, we can substitute ( b = 25 ) and ( c = 31 ) back into the equation ( a + b + c = 72 ):[a + 25 + 31 = 72]Simplifying:[a + 56 = 72 implies a = 16]Therefore, the smallest number is 16.

💡First, I need to determine the initial total marks for all 10 students based on the given average.Next, I'll identify the error in the recorded marks and calculate the adjustment needed to correct it.Finally, I'll adjust the total marks by subtracting the incorrect mark and adding the correct one, then recalculate the average to find the correct average marks.

💡Okay, so I need to figure out how many pastries Katie has left after the bake sale. Let me start by understanding what information is given.First, Katie made pastries for the school bake sale. She baked 7 cupcakes and 5 cookies. So, cupcakes and cookies are both types of pastries. That means the total number of pastries she made is the sum of cupcakes and cookies.So, 7 cupcakes plus 5 cookies equals... let me add that up. 7 plus 5 is 12. Okay, so she made a total of 12 pastries.Now, after the sale, she sold 4 pastries. I need to find out how many she has left. To do that, I should subtract the number of pastries she sold from the total number she made.So, 12 pastries minus 4 pastries sold equals... 12 minus 4 is 8. Therefore, Katie has 8 pastries left to take back home.Wait, let me double-check to make sure I didn't make any mistakes. She made 7 cupcakes and 5 cookies, which is 12 pastries in total. She sold 4, so subtracting 4 from 12 gives 8. Yeah, that seems right.I don't think I missed anything here. The problem is pretty straightforward. She started with 12 pastries, sold 4, so 8 are left. I guess that's the answer.