🧠:[<think>Okay, so I have this problem here where I need to find the angle α between 0 and π/2. The vectors given are a = (cos α, sin α) and b = (-1/2, sqrt(3)/2). The condition is that the magnitude of 2a + b is equal to the magnitude of a - 2b. Alright, let me try to figure this out step by step.First, I remember that the magnitude of a vector (x, y) is sqrt(x² + y²). So, to find when |2a + b| = |a - 2b|, I can start by computing both vectors 2a + b and a - 2b, then set their magnitudes equal and solve for α.Let me write down the expressions for both vectors.Starting with 2a + b:2a would be 2*(cos α, sin α) = (2 cos α, 2 sin α)Then adding vector b: (2 cos α - 1/2, 2 sin α + sqrt(3)/2)Similarly, a - 2b:First compute 2b: 2*(-1/2, sqrt(3)/2) = (-1, sqrt(3))Then subtract that from a: (cos α - (-1), sin α - sqrt(3)) = (cos α + 1, sin α - sqrt(3))So now, the magnitudes of these two vectors should be equal.So, |2a + b| = sqrt[(2 cos α - 1/2)² + (2 sin α + sqrt(3)/2)²]and |a - 2b| = sqrt[(cos α + 1)² + (sin α - sqrt(3))²]Setting these equal, we can square both sides to eliminate the square roots:(2 cos α - 1/2)² + (2 sin α + sqrt(3)/2)² = (cos α + 1)² + (sin α - sqrt(3))²Now, let's expand both sides.Left side:(2 cos α - 1/2)² = 4 cos² α - 2 cos α + 1/4(2 sin α + sqrt(3)/2)² = 4 sin² α + 2*2 sin α*(sqrt(3)/2) + (sqrt(3)/2)^2Wait, let me compute each term step by step.First term: (2 cos α - 1/2)^2= (2 cos α)^2 + (-1/2)^2 + 2*(2 cos α)*(-1/2)= 4 cos² α + 1/4 - 2 cos αSecond term: (2 sin α + sqrt(3)/2)^2= (2 sin α)^2 + (sqrt(3)/2)^2 + 2*(2 sin α)*(sqrt(3)/2)= 4 sin² α + 3/4 + 2 sin α*sqrt(3)So, left side becomes:4 cos² α + 1/4 - 2 cos α + 4 sin² α + 3/4 + 2 sqrt(3) sin αCombine like terms:4 cos² α + 4 sin² α + (-2 cos α) + 2 sqrt(3) sin α + 1/4 + 3/4Simplify:4(cos² α + sin² α) + (-2 cos α + 2 sqrt(3) sin α) + 1Since cos² α + sin² α = 1, this becomes:4*1 + (-2 cos α + 2 sqrt(3) sin α) + 1So, 4 + 1 - 2 cos α + 2 sqrt(3) sin αWhich simplifies to:5 - 2 cos α + 2 sqrt(3) sin αNow, the right side: (cos α + 1)^2 + (sin α - sqrt(3))^2First term: (cos α + 1)^2= cos² α + 2 cos α + 1Second term: (sin α - sqrt(3))^2= sin² α - 2 sqrt(3) sin α + 3So, right side becomes:cos² α + 2 cos α + 1 + sin² α - 2 sqrt(3) sin α + 3Combine like terms:(cos² α + sin² α) + 2 cos α - 2 sqrt(3) sin α + 1 + 3Again, cos² α + sin² α = 1, so:1 + 2 cos α - 2 sqrt(3) sin α + 4Simplify:5 + 2 cos α - 2 sqrt(3) sin αSo now, we have left side: 5 - 2 cos α + 2 sqrt(3) sin αequals right side: 5 + 2 cos α - 2 sqrt(3) sin αSet them equal:5 - 2 cos α + 2 sqrt(3) sin α = 5 + 2 cos α - 2 sqrt(3) sin αSubtract 5 from both sides:-2 cos α + 2 sqrt(3) sin α = 2 cos α - 2 sqrt(3) sin αBring all terms to left side:-2 cos α + 2 sqrt(3) sin α - 2 cos α + 2 sqrt(3) sin α = 0Combine like terms:(-2 cos α - 2 cos α) + (2 sqrt(3) sin α + 2 sqrt(3) sin α) = 0Which is:-4 cos α + 4 sqrt(3) sin α = 0Divide both sides by 4:- cos α + sqrt(3) sin α = 0Rearranged:sqrt(3) sin α = cos αDivide both sides by cos α (assuming cos α ≠ 0, but since α is between 0 and π/2, cos α is positive, so it's okay):sqrt(3) tan α = 1Then tan α = 1 / sqrt(3)Which gives α = arctangent(1/sqrt(3))I remember that tan(π/6) = 1/sqrt(3), so α = π/6Therefore, the angle α is π/6 radians, which is 30 degrees.Let me double-check the steps to make sure I didn't make any mistakes. Starting from the magnitudes, expanding both sides, simplifying, and leading to the equation sqrt(3) sin α = cos α, which leads to tan α = 1/sqrt(3). Yes, that seems correct. Since arctan(1/sqrt(3)) is indeed π/6, and that's within the interval (0, π/2). So, the answer should be π/6.Final AnswerThe angle ( alpha ) is (boxed{dfrac{pi}{6}}).</think>Given ( alpha in left(0, frac{pi}{2}right) ), vectors ( overrightarrow{a} = (cos alpha, sin alpha) ) and ( overrightarrow{b} = left(-frac{1}{2}, frac{sqrt{3}}{2}right) ), we need to find the angle ( alpha ) when ( |2 overrightarrow{a} + overrightarrow{b}| = | overrightarrow{a} - 2 overrightarrow{b}| ).First, we compute the vectors ( 2 overrightarrow{a} + overrightarrow{b} ) and ( overrightarrow{a} - 2 overrightarrow{b} ):- ( 2 overrightarrow{a} + overrightarrow{b} = left(2 cos alpha - frac{1}{2}, 2 sin alpha + frac{sqrt{3}}{2}right) )- ( overrightarrow{a} - 2 overrightarrow{b} = left(cos alpha + 1, sin alpha - sqrt{3}right) )Next, we calculate their magnitudes and set them equal:[sqrt{left(2 cos alpha - frac{1}{2}right)^2 + left(2 sin alpha + frac{sqrt{3}}{2}right)^2} = sqrt{left(cos alpha + 1right)^2 + left(sin alpha - sqrt{3}right)^2}]Squaring both sides to eliminate the square roots:[left(2 cos alpha - frac{1}{2}right)^2 + left(2 sin alpha + frac{sqrt{3}}{2}right)^2 = left(cos alpha + 1right)^2 + left(sin alpha - sqrt{3}right)^2]Expanding both sides:- Left side: [ 4 cos^2 alpha - 2 cos alpha + frac{1}{4} + 4 sin^2 alpha + 2 sqrt{3} sin alpha + frac{3}{4} = 4 (cos^2 alpha + sin^2 alpha) - 2 cos alpha + 2 sqrt{3} sin alpha + 1 ] Simplifies to: [ 5 - 2 cos alpha + 2 sqrt{3} sin alpha ]- Right side: [ cos^2 alpha + 2 cos alpha + 1 + sin^2 alpha - 2 sqrt{3} sin alpha + 3 = 1 + 2 cos alpha - 2 sqrt{3} sin alpha + 4 ] Simplifies to: [ 5 + 2 cos alpha - 2 sqrt{3} sin alpha ]Equating both sides:[5 - 2 cos alpha + 2 sqrt{3} sin alpha = 5 + 2 cos alpha - 2 sqrt{3} sin alpha]Simplifying:[-4 cos alpha + 4 sqrt{3} sin alpha = 0]Dividing by 4:[-cos alpha + sqrt{3} sin alpha = 0]Solving for ( alpha ):[sqrt{3} sin alpha = cos alpha implies tan alpha = frac{1}{sqrt{3}} implies alpha = frac{pi}{6}]Thus, the angle ( alpha ) is (boxed{dfrac{pi}{6}}).,<think>Okay, so I need to find the angle α between 0 and π/2 where the magnitude of the vector 2a + b is equal to the magnitude of the vector a - 2b. The vectors are given as a = (cos α, sin α) and b = (-1/2, √3/2). Hmm, let me start by recalling how to compute the magnitude of a vector. The magnitude of a vector (x, y) is √(x² + y²). So, if I can find expressions for |2a + b| and |a - 2b|, set them equal, and solve for α, that should work.First, let's compute 2a + b. Since a is (cos α, sin α), multiplying by 2 gives (2cos α, 2sin α). Then adding vector b which is (-1/2, √3/2), so component-wise addition:2a + b = (2cos α - 1/2, 2sin α + √3/2).Similarly, a - 2b would be (cos α - 2*(-1/2), sin α - 2*(√3/2)). Let's compute that:First component: cos α - 2*(-1/2) = cos α + 1.Second component: sin α - 2*(√3/2) = sin α - √3.So, a - 2b = (cos α + 1, sin α - √3).Now, compute the magnitudes of both vectors. Let's start with |2a + b|:|2a + b| = √[(2cos α - 1/2)² + (2sin α + √3/2)²].Similarly, |a - 2b| = √[(cos α + 1)² + (sin α - √3)²].We need to set these equal:√[(2cos α - 1/2)² + (2sin α + √3/2)²] = √[(cos α + 1)² + (sin α - √3)²].To simplify, I can square both sides to eliminate the square roots:(2cos α - 1/2)² + (2sin α + √3/2)² = (cos α + 1)² + (sin α - √3)².Now, expand all the terms.Starting with the left-hand side (LHS):First term: (2cos α - 1/2)² = (2cos α)^2 - 2*(2cos α)*(1/2) + (1/2)^2 = 4cos² α - 2cos α + 1/4.Second term: (2sin α + √3/2)^2 = (2sin α)^2 + 2*(2sin α)*(√3/2) + (√3/2)^2 = 4sin² α + 2√3 sin α + 3/4.So adding both terms:LHS = 4cos² α - 2cos α + 1/4 + 4sin² α + 2√3 sin α + 3/4.Combine like terms:4cos² α + 4sin² α = 4(cos² α + sin² α) = 4*1 = 4.Then, -2cos α + 2√3 sin α.And constants: 1/4 + 3/4 = 1.So overall, LHS = 4 - 2cos α + 2√3 sin α + 1 = 5 - 2cos α + 2√3 sin α.Wait, hold on, 4 + 1 is 5? Wait, 4 from the cos² and sin² terms, then 1/4 + 3/4 is 1, so total 4 + 1 = 5. Then the other terms: -2cos α + 2√3 sin α. Yes, so LHS = 5 - 2cos α + 2√3 sin α.Now the right-hand side (RHS):(cos α + 1)^2 + (sin α - √3)^2.First term: (cos α + 1)^2 = cos² α + 2cos α + 1.Second term: (sin α - √3)^2 = sin² α - 2√3 sin α + 3.Adding them together:cos² α + 2cos α + 1 + sin² α - 2√3 sin α + 3.Combine like terms:cos² α + sin² α = 1.Then 2cos α - 2√3 sin α.Constants: 1 + 3 = 4.So RHS = 1 + 4 + 2cos α - 2√3 sin α = 5 + 2cos α - 2√3 sin α.Wait, let's check that again:cos² α + sin² α = 1.Then 2cos α (from first term) and -2√3 sin α (from second term).Then constants: 1 (from first term) and 3 (from second term). So 1 + 3 = 4.Therefore, RHS = 1 + 2cos α - 2√3 sin α + 4 = 5 + 2cos α - 2√3 sin α.So now, setting LHS = RHS:5 - 2cos α + 2√3 sin α = 5 + 2cos α - 2√3 sin α.Subtract 5 from both sides:-2cos α + 2√3 sin α = 2cos α - 2√3 sin α.Bring all terms to one side:-2cos α + 2√3 sin α - 2cos α + 2√3 sin α = 0.Wait, no. Let's subtract the right side from both sides:-2cos α + 2√3 sin α - 2cos α + 2√3 sin α = 0.Wait, that's combining like terms incorrectly. Wait, actually, subtract (2cos α - 2√3 sin α) from both sides:Left side: (-2cos α + 2√3 sin α) - (2cos α - 2√3 sin α) = -2cos α + 2√3 sin α - 2cos α + 2√3 sin α = (-2cos α - 2cos α) + (2√3 sin α + 2√3 sin α) = -4cos α + 4√3 sin α.Set this equal to zero:-4cos α + 4√3 sin α = 0.Divide both sides by 4:-cos α + √3 sin α = 0.Bring cos α to the other side:√3 sin α = cos α.Divide both sides by cos α (since α is in (0, π/2), cos α ≠ 0):√3 tan α = 1.So tan α = 1/√3.Therefore, α = arctangent of 1/√3. The angle whose tangent is 1/√3 is π/6, since tan(π/6) = 1/√3.So α = π/6.Let me double-check this result. If α is π/6, then cos α is √3/2 and sin α is 1/2.Compute vector a: (√3/2, 1/2).Vector b is (-1/2, √3/2).Compute 2a + b: 2*(√3/2, 1/2) + (-1/2, √3/2) = (√3, 1) + (-1/2, √3/2) = (√3 - 1/2, 1 + √3/2).Compute the magnitude: sqrt[(√3 - 1/2)^2 + (1 + √3/2)^2].Similarly, compute a - 2b: (√3/2 - 2*(-1/2), 1/2 - 2*(√3/2)) = (√3/2 + 1, 1/2 - √3).Compute the magnitude: sqrt[(√3/2 + 1)^2 + (1/2 - √3)^2].Let me compute both magnitudes.First, |2a + b|:First component squared: (√3 - 1/2)^2 = (√3)^2 - 2*(√3)*(1/2) + (1/2)^2 = 3 - √3 + 1/4.Second component squared: (1 + √3/2)^2 = 1 + 2*(√3/2) + (√3/2)^2 = 1 + √3 + 3/4.Total: (3 - √3 + 1/4) + (1 + √3 + 3/4) = 3 + 1/4 + 1 + 3/4 + (-√3 + √3) = 4 + 1 = 5. So sqrt(5).Now |a - 2b|:First component squared: (√3/2 + 1)^2 = (√3/2)^2 + 2*(√3/2)*1 + 1^2 = 3/4 + √3 + 1.Second component squared: (1/2 - √3)^2 = (1/2)^2 - 2*(1/2)*√3 + (√3)^2 = 1/4 - √3 + 3.Total: (3/4 + √3 + 1) + (1/4 - √3 + 3) = (3/4 + 1 + 1/4 + 3) + (√3 - √3) = (4 + 3) + 0 = 7. Wait, sqrt(7)? But sqrt(5) is not equal to sqrt(7). Wait, that's a problem. Hmm, did I make a mistake in the calculation?Wait, hold on. If α = π/6, then let's recompute:Wait, vector a is (cos(π/6), sin(π/6)) = (√3/2, 1/2). Correct.Vector b is (-1/2, √3/2). Correct.Compute 2a + b: 2*(√3/2, 1/2) = (√3, 1). Then add (-1/2, √3/2): (√3 - 1/2, 1 + √3/2). Correct.Compute the magnitude squared:(√3 - 1/2)^2 + (1 + √3/2)^2.Compute each term:First term: (√3 - 1/2)^2 = (√3)^2 - 2*(√3)*(1/2) + (1/2)^2 = 3 - √3 + 1/4. Correct.Second term: (1 + √3/2)^2 = 1^2 + 2*1*(√3/2) + (√3/2)^2 = 1 + √3 + 3/4. Correct.Total sum: (3 - √3 + 1/4) + (1 + √3 + 3/4) = 3 + 1/4 + 1 + 3/4 + (-√3 + √3) = 4 + 1 = 5. So the magnitude is sqrt(5). Correct.Now for |a - 2b|:a - 2b = (√3/2 - 2*(-1/2), 1/2 - 2*(√3/2)).Compute first component: √3/2 - 2*(-1/2) = √3/2 + 1.Second component: 1/2 - 2*(√3/2) = 1/2 - √3.So the vector is (√3/2 + 1, 1/2 - √3).Compute magnitude squared:(√3/2 + 1)^2 + (1/2 - √3)^2.First term: (√3/2 + 1)^2 = (√3/2)^2 + 2*(√3/2)*1 + 1^2 = 3/4 + √3 + 1.Second term: (1/2 - √3)^2 = (1/2)^2 - 2*(1/2)*√3 + (√3)^2 = 1/4 - √3 + 3.Adding both terms: (3/4 + √3 + 1) + (1/4 - √3 + 3) = 3/4 + 1 + 1/4 + 3 + (√3 - √3) = (4/4) + 4 = 1 + 4 = 5. So the magnitude is sqrt(5). Therefore, they are equal. So yes, when α = π/6, both magnitudes are equal to sqrt(5), which matches. Therefore, the answer is correct.But wait, my initial calculation when I thought |a - 2b| squared was 7 must have been a mistake. Let me check again. When I computed the magnitude squared of a - 2b:(√3/2 + 1)^2 + (1/2 - √3)^2.First term: (sqrt(3)/2 + 1)^2:= (sqrt(3)/2)^2 + 2*(sqrt(3)/2)*1 + 1^2= 3/4 + sqrt(3) + 1.Second term: (1/2 - sqrt(3))^2:= (1/2)^2 - 2*(1/2)*sqrt(3) + (sqrt(3))^2= 1/4 - sqrt(3) + 3.Adding both terms:3/4 + sqrt(3) + 1 + 1/4 - sqrt(3) + 3= (3/4 + 1/4) + (sqrt(3) - sqrt(3)) + (1 + 3)= 1 + 0 + 4= 5.Yes, so sqrt(5). So both magnitudes are sqrt(5). So that's correct. Therefore, α = π/6 is indeed the solution. So I think my answer is correct.Final AnswerThe angle ( α ) is boxed{dfrac{pi}{6}}.</think>Given ( alpha in (0, frac{pi}{2}) ), vectors ( overrightarrow{a} = (cos alpha, sin alpha) ) and ( overrightarrow{b} = left( -frac{1}{2}, frac{sqrt{3}}{2} right) ), we need to find the angle ( alpha ) when ( |2overrightarrow{a} + overrightarrow{b}| = |overrightarrow{a} - 2overrightarrow{b}| ).First, we compute the vectors ( 2overrightarrow{a} + overrightarrow{b} ) and ( overrightarrow{a} - 2overrightarrow{b} ):[2overrightarrow{a} + overrightarrow{b} = left( 2cos alpha - frac{1}{2}, 2sin alpha + frac{sqrt{3}}{2} right)][overrightarrow{a} - 2overrightarrow{b} = left( cos alpha + 1, sin alpha - sqrt{3} right)]Next, we calculate the magnitudes squared of these vectors:[|2overrightarrow{a} + overrightarrow{b}|^2 = left( 2cos alpha - frac{1}{2} right)^2 + left( 2sin alpha + frac{sqrt{3}}{2} right)^2][= 4cos^2 alpha - 2cos alpha + frac{1}{4} + 4sin^2 alpha + 2sqrt{3}sin alpha + frac{3}{4}][= 4(cos^2 alpha + sin^2 alpha) - 2cos alpha + 2sqrt{3}sin alpha + 1][= 5 - 2cos alpha + 2sqrt{3}sin alpha][|overrightarrow{a} - 2overrightarrow{b}|^2 = left( cos alpha + 1 right)^2 + left( sin alpha - sqrt{3} right)^2][= cos^2 alpha + 2cos alpha + 1 + sin^2 alpha - 2sqrt{3}sin alpha + 3][= (cos^2 alpha + sin^2 alpha) + 2cos alpha - 2sqrt{3}sin alpha + 4][= 5 + 2cos alpha - 2sqrt{3}sin alpha]Setting the magnitudes equal, we get:[5 - 2cos alpha + 2sqrt{3}sin alpha = 5 + 2cos alpha - 2sqrt{3}sin alpha]Simplifying:[-4cos alpha + 4sqrt{3}sin alpha = 0][-cos alpha + sqrt{3}sin alpha = 0][sqrt{3}sin alpha = cos alpha][tan alpha = frac{1}{sqrt{3}}]Thus, ( alpha = arctan left( frac{1}{sqrt{3}} right) = frac{pi}{6} ).The angle ( alpha ) is boxed{dfrac{pi}{6}}.]

🔑:Proof: Given vectors overrightarrow{a}=(cos α, sin α), and overrightarrow{b}=(- frac {1}{2}, frac { sqrt {3}}{2}),Since | overrightarrow{a}|=| overrightarrow{b}|, we have overrightarrow{a}cdot overrightarrow{b}=0, which implies -frac {1}{2}cos α + frac { sqrt {3}}{2}sin α=0.Squaring both sides of |2 overrightarrow{a}+ overrightarrow{b}|=| overrightarrow{a}-2 overrightarrow{b}| and simplifying, we get 3(| overrightarrow{a}|^2-| overrightarrow{b}|^2)+8 overrightarrow{a}cdot overrightarrow{b}=0.Since overrightarrow{a}+ overrightarrow{b} is perpendicular to overrightarrow{a}- overrightarrow{b} (...(divide)...),We have sin(α- frac {π}{6})=0. Given that α∈(0, frac {π}{2}), we get α= frac {π}{6}.Therefore, the final answer is boxed{α= frac {π}{6}}.

🧠:[<think>Okay, so I need to figure out the minimum number of cells that must be painted in a 65x65 chessboard so that every unpainted cell is isolated in terms of knight moves. That means, from any unpainted cell, there shouldn't be another unpainted cell a knight's move away. Hmm, this sounds like a graph problem where each cell is a node, and edges connect cells that are a knight's move apart. Then, painting a cell would be like removing that node from the graph, and we want the remaining graph to have no edges—so it's an independent set. Wait, no, actually, the remaining unpainted cells should form an independent set, and the painted cells are the ones not in this set. But since we want the minimum number of painted cells, this is equivalent to finding the maximum independent set and then subtracting its size from the total. But maximum independent set is NP-hard, so maybe there's a pattern or structure here because of the knight's moves on a chessboard?Knight moves alternate between light and dark squares. On a standard chessboard, a knight on a white square can only move to black squares and vice versa. So, the chessboard is bipartite with respect to knight moves? Let me check. If you color the chessboard in the standard black and white pattern, then a knight moves from a black to a white square and vice versa. So yes, the graph is bipartite. Therefore, the maximum independent set would be the larger of the two color classes. In a standard 8x8 chessboard, there are 32 white and 32 black squares, so both color classes are equal. But on an odd-sized chessboard like 65x65, the numbers are unequal. Let's see, 65 is odd, so the total number of squares is 65^2 = 4225. The number of black squares would be ceil(4225/2) = 2113, and white squares would be floor(4225/2) = 2112. Wait, actually, since 65 is odd, each row alternates 33 and 32 cells of alternating colors. So, the total number of black squares would be 33*33 + 32*32? Wait, maybe not. Let me think again.Each row has 65 cells. If the first cell is black, then the colors alternate. So for an odd-length row, the number of black squares per row is (65 + 1)/2 = 33, and white squares are 32. Then, there are 33 rows with 33 black squares and 32 white squares, and 32 rows with 32 black squares and 33 white squares? Wait, no, actually, each row alternates starting with black or white. But actually, in the standard chessboard coloring, adjacent rows start with the opposite color. So in a 65x65 chessboard, the coloring would alternate starting with black and white for each row. Since 65 is odd, each row starts with the opposite color of the previous row. So, for the first row, starting with black, the colors would be black, white, black,..., black (33 blacks and 32 whites). The next row starts with white, so 32 whites, 33 blacks. Then the third row starts with black again, etc. So, over 65 rows, how many rows start with black? Since 65 is odd, there are 33 rows starting with black and 32 starting with white. Therefore, total number of black squares would be 33 rows * 33 blacks + 32 rows * 33 blacks? Wait, no. Wait, rows starting with black have 33 blacks and 32 whites, and rows starting with white have 32 blacks and 33 whites. So, total black squares would be 33 rows * 33 blacks + 32 rows * 32 blacks. Let me calculate that:33*33 = 108932*32 = 1024Total blacks: 1089 + 1024 = 2113Similarly, total white squares: 33 rows *32 whites + 32 rows *33 whites = 33*32 +32*33= 2*(32*33)= 2112Yes, so total blacks are 2113, whites 2112.Since the knight moves from black to white and vice versa, the graph is bipartite with partitions of size 2113 and 2112. The maximum independent set in a bipartite graph is the size of the larger partition. So, the maximum independent set is 2113. Therefore, the minimum number of cells to paint would be the total cells minus the maximum independent set, which is 4225 - 2113 = 2112. Wait, but wait: the problem says that from any unpainted cell, there are no other unpainted cells a knight's move away. So the set of unpainted cells must be an independent set. Therefore, the maximum possible size of such a set is the maximum independent set, which is 2113. Therefore, the minimum number of painted cells is 4225 - 2113 = 2112. So the answer is 2112?But let me verify this reasoning. If we color all the black squares (2113) and leave the white squares (2112) unpainted, then since knights move from black to white, no two white squares are reachable by a knight's move. Therefore, the white squares form an independent set. Similarly, if we color all the white squares, the black squares would form an independent set. But since 2113 is larger, coloring the white squares (2112 cells) would leave the larger independent set (2113) unpainted. Wait, but if we want the unpainted cells to form an independent set, we need to choose the color class that's larger as the unpainted set, and paint the smaller one. Wait, no. Wait: If you paint the smaller color class, then the larger color class remains unpainted, which is an independent set. So to minimize the number of painted cells, we want to paint the smaller color class. The smaller color class is the white squares with 2112 cells. Therefore, painting 2112 cells (all white squares) leaves all black squares (2113) unpainted, which are an independent set. Therefore, from any unpainted cell (black), there's no other unpainted cell a knight's move away, since all unpainted cells are black, and knight moves from black go to white, which are all painted. Therefore, the minimal number of painted cells is 2112.Alternatively, if we paint all the black squares (2113), then the white squares (2112) would be unpainted, which is also an independent set. But since we need the minimal number of painted cells, painting the smaller color class (2112) is better. Therefore, the answer is 2112.But let me double-check. Suppose we leave the black squares (2113) unpainted. Since all knight moves from a black square go to white squares, which are painted. Therefore, no two black squares are adjacent via knight moves, so they form an independent set. Therefore, painting all white squares (2112) is sufficient. Similarly, painting all black squares (2113) would leave white squares (2112) unpainted, but since we want the minimal number, 2112 is better.Therefore, the answer is 2112. But let me confirm this with a smaller chessboard. Let's take a 3x3 chessboard. Total cells 9. How many need to be painted? The color classes would be 5 and 4. So painting the smaller one (4) would leave 5 unpainted, which is an independent set. Indeed, in a 3x3, if you paint the 4 white squares, then the 5 black squares are all isolated. Wait, but in a 3x3 chessboard, can a knight move from one black square to another? Let's see. A knight moves 2 in one direction and 1 in the other. On 3x3, the possible knight moves are limited. For example, the center cell can't have any knight moves because moving 2 in any direction would go off the board. The corner cells: from (1,1), a knight would move to (2,3) or (3,2). Let's see, (1,1) is a corner. Knight moves 2 right and 1 down: (1+2, 1+1)=(3,2). Similarly, 2 down and 1 right: (1+1,1+2)=(2,3). So those are the two moves. Let's color the chessboard. Starting with (1,1) as black. Then (1,1) black, (1,2) white, (1,3) black. (2,1) white, (2,2) black, (2,3) white. (3,1) black, (3,2) white, (3,3) black. So black squares are (1,1), (1,3), (2,2), (3,1), (3,3) – 5 blacks. White squares are 4. Now, if we paint all white squares, then the black squares are: (1,1), (1,3), (2,2), (3,1), (3,3). Are any of these reachable by a knight's move? From (1,1), knight moves to (2,3) or (3,2), which are white (painted). From (1,3), moves to (2,1) or (3,2), which are painted. From (3,1), moves to (1,2) or (2,3), painted. From (3,3), moves to (1,2) or (2,1), painted. The center (2,2) can't move anywhere. So indeed, all black squares are isolated. Therefore, painting the smaller color class (4 cells) works, and the answer is 4. Similarly, if you paint 5 cells (the black squares), you get 4 unpainted, which is also okay but not minimal. Therefore, the logic holds here.So applying the same logic to 65x65: paint the smaller color class (2112 cells), leaving the larger color class (2113) unpainted, which is an independent set. Therefore, the answer is 2112. But let me check another example. Let's take 2x2. Though knight moves on a 2x2 are impossible, but just for testing. But 2x2 is even, color classes equal. Paint 2 cells. But knight can't move anywhere, so all cells can be unpainted. Wait, but 2x2: each cell is a separate component. So maximum independent set is 4, but since it's disconnected. Wait, maybe the logic is slightly different for even vs odd. But in our problem, it's 65x65 which is odd. So the logic holds. Therefore, I think the answer is 2112.But wait, the problem states "from any unpainted cell it is impossible to move to another unpainted cell with a knight's move". So we need that the set of unpainted cells is an independent set. The maximum independent set in the knight's graph on chessboard. However, in standard chessboard coloring, the bipartition splits the graph into two color classes, which are independent sets because knights alternate colors. Therefore, the maximum independent set is indeed the larger color class, which has 2113 cells. Therefore, the minimum number of painted cells is 4225 - 2113 = 2112. So yes, 2112 is the answer.But wait, is the knight's graph on a chessboard actually bipartite? Because in chess, a knight alternates colors with each move, so the graph is bipartite with the two color classes as the partitions. Therefore, yes, the maximum independent set is indeed the size of the larger color class, and the rest must be painted. So yes, the answer is 2112.Therefore, I think the answer is boxed{2112}.Final Answerboxed{2112}</think>To solve the problem of determining the minimum number of cells that need to be painted in a 65x65 chessboard such that no unpainted cell can move to another unpainted cell with a knight's move, we can use the following reasoning:1. Knight's Move and Chessboard Coloring: A knight alternates between black and white squares with each move. Therefore, the chessboard can be divided into two color classes (black and white), forming a bipartite graph where each color class is an independent set.2. Color Class Calculation: For a 65x65 chessboard: - Each row alternates between starting with black and white. - There are 33 rows starting with black and 32 rows starting with white. - Each row starting with black has 33 black cells and 32 white cells. - Each row starting with white has 32 black cells and 33 white cells. - Total black cells: (33 times 33 + 32 times 32 = 1089 + 1024 = 2113) - Total white cells: (33 times 32 + 32 times 33 = 2112)3. Independent Set: The maximum independent set in a bipartite graph is the larger of the two color classes. Here, the larger color class has 2113 cells (black), and the smaller has 2112 cells (white).4. Minimum Painted Cells: To ensure all unpainted cells form an independent set, we paint the smaller color class. This minimizes the number of painted cells while leaving the larger color class (2113 cells) unpainted, which is an independent set.Thus, the minimum number of cells that need to be painted is the total number of cells minus the size of the larger color class:[ 65 times 65 - 2113 = 4225 - 2113 = 2112 ]The final answer is (boxed{2112}).,<think>Okay, so I have this problem where I need to figure out the minimum number of cells that need to be painted in a 65x65 chessboard-like grid. The condition is that from any unpainted cell, you can't move to another unpainted cell using a knight's move. A knight's move in chess is that L-shaped move, right? Like two squares in one direction and then one square perpendicular, or vice versa. So the goal is to paint certain cells such that all remaining unpainted cells are isolated in terms of knight moves. That means any two unpainted cells can't be a knight's move apart. And I need to do this with the fewest painted cells possible.First, I need to understand the problem better. Let me paraphrase: we want to paint some cells so that the remaining unpainted cells are all "knight's move isolated." So, no two unpainted cells can be a knight's move away from each other. Therefore, the painted cells must block all possible knight moves between unpainted cells. But how does painting cells block knight moves? Wait, actually, the problem states that from any unpainted cell, you can't move to another unpainted cell with a knight's move. So, if I have an unpainted cell, all cells that are a knight's move away from it must be painted. So, in other words, each unpainted cell must have all its knight's move neighbors painted. Therefore, each unpainted cell is surrounded by painted cells in all possible knight's move positions. So, if two cells are a knight's move apart, at least one of them must be painted. Hence, this is similar to a vertex cover problem in graph theory, where the graph has edges between cells that are a knight's move apart, and we need a vertex cover of this graph. The vertex cover is a set of vertices (cells) such that every edge (knight's move) is incident to at least one vertex in the set. So, in this case, the problem reduces to finding a minimum vertex cover in a graph where each node is a cell on the 65x65 grid, and edges connect cells that are a knight's move apart.Vertex cover problems are generally NP-hard, but maybe there's some structure here due to the grid and knight's moves that can be exploited. Alternatively, maybe the problem can be transformed into a different problem that's easier to solve. Let me think.Alternatively, maybe instead of vertex cover, this is similar to a dominating set problem. But in a dominating set, each node is either in the set or adjacent to a node in the set. Here, it's slightly different: we need that any two nodes not in the set are not adjacent (in terms of knight's moves). That is, the complement of the set is an independent set. So, the problem is equivalent to finding the maximum independent set in the knight's graph, and then the minimum vertex cover would be the total number of cells minus the size of the maximum independent set. But maximum independent set is also NP-hard. However, for certain graphs, especially those with a lot of structure, like bipartite graphs, the problem can be easier. Since knight's moves on a chessboard have a bipartite structure.Wait, a knight alternates between black and white squares with each move. So the knight's graph is bipartite. So, the chessboard is a bipartite graph with the two colorings as partitions. In a bipartite graph, the size of the minimum vertex cover is equal to the size of the maximum matching (Konig's theorem). But I don't know if that helps here. Alternatively, in a bipartite graph, the maximum independent set is equal to the total number of vertices minus the size of the minimum vertex cover. But since Konig's theorem relates minimum vertex cover and maximum matching, perhaps we can use that.But maybe there's a simpler way. If the knight's graph is bipartite, then the maximum independent set would be the larger of the two color classes. Since in a bipartite graph, the two color classes are both independent sets. Therefore, if we color the chessboard in the standard black and white pattern, then all the black squares form one independent set, and all the white squares form another. Since a knight always moves from a black to a white square or vice versa, so the knight's graph is indeed bipartite with the two color classes as black and white squares. Therefore, the maximum independent set in the knight's graph would be the larger of the two color classes.But on a standard chessboard (8x8), the number of black and white squares is equal when the side length is even. But here we have a 65x65 grid, which is odd. So the number of black squares would be (65^2 +1)/2 and white squares would be (65^2 -1)/2. Therefore, the maximum independent set would be the larger color class, which is the black squares with (65^2 +1)/2 cells. Therefore, the minimum vertex cover would be the total number of cells minus the maximum independent set, which is 65^2 - (65^2 +1)/2 = (65^2 -1)/2. Therefore, the minimum number of cells to paint would be (65^2 -1)/2. But wait, that would be 4225 - 2113 = 2112. But let me check the numbers:65 squared is 4225. Then (4225 -1)/2 is 4224/2 = 2112. So, if the maximum independent set is 2113, then the minimum vertex cover is 2112. So, is that correct? But wait, in an odd by odd chessboard, the number of black squares is indeed (n^2 +1)/2 when n is odd, so 65x65 gives (4225 +1)/2 = 2113, and white squares are 2112. Therefore, the maximum independent set is 2113. So the minimum vertex cover is 2112. But wait, vertex cover is the number of cells you need to paint so that every edge (knight's move) has at least one endpoint painted. So in this case, if we paint all the white squares (2112), then every knight's move is from a black to a white square, so the white squares are painted, so the edge is covered. Therefore, the vertex cover is 2112. Conversely, if we paint all the black squares (2113), that would also be a vertex cover, but it's larger. So the minimum is 2112.Therefore, the answer should be 2112. But let me verify this.Wait, but in the problem statement, it's phrased as: "from any unpainted cell it is impossible to move to another unpainted cell with a knight's move." So, the set of unpainted cells must be an independent set. Therefore, the maximum independent set is the largest possible set of unpainted cells, so the minimum number of painted cells is total cells minus maximum independent set. Therefore, if maximum independent set is 2113, then minimum painted cells is 4225 - 2113 = 2112. So that's consistent with vertex cover.Therefore, the answer is 2112. However, let me confirm with a smaller chessboard. Let's take a 3x3 chessboard. Then, the number of black squares is 5, white squares is 4. So maximum independent set is 5. Then, the minimum vertex cover would be 4. Let's see: if we paint all 4 white squares, then the 5 black squares are unpainted. Since a knight on a 3x3 board can move from the center to the corners, but the corners are all black (if the center is white). Wait, no. Wait, in a 3x3 chessboard, the knight's graph is actually disconnected. Knights on a 3x3 can only move from the center to the corners, but the corners can't move anywhere else. Wait, in a 3x3 grid, the knight in the center can move to all four corners, but the corners can only move back to the center. The edges (the squares between the corners) can't be reached by a knight's move. So actually, the knight's graph on 3x3 is a star graph with the center connected to four corners. So, in this case, the maximum independent set would be the four corners plus one of the edge squares (which aren't connected to anything), so maybe larger? Wait, no. Wait, the edges (squares that are on the sides but not corners) can't be reached by any knight's move, right? Because a knight moves two in one direction and one in the other, so from a corner (say, (1,1)), you can move to (2,3) or (3,2), but in a 3x3 grid, those squares don't exist. Similarly, from the edge square (1,2), a knight would move to (2,4) or (3,1), which again are outside the grid. So actually, in the 3x3 grid, the only possible knight moves are between the center and the four corners. Therefore, the knight's graph has four edges: center connected to each corner. The rest of the squares (edges) are isolated vertices. Therefore, the maximum independent set would be all the edge squares (four of them) plus the center? Wait, no, because the center is connected to the corners. If we include the center in the independent set, we can't include any of the corners. If we include the corners, we can't include the center. So the maximum independent set would be either the four corners plus the four edges (since edges are not connected to anything), but wait, edges aren't connected to anything. Wait, actually, the edges are separate. So the knight's graph on 3x3 consists of the center connected to four corners, and the four edges are isolated. Therefore, the independent set can include all four edges plus either the center or all four corners. So, maximum independent set would be 4 (edges) + max(1 (center) or 4 (corners)). Therefore, 4 + 4 = 8. But the total cells are 9. Wait, but in reality, the four edges and the four corners and the center. Wait, the 3x3 grid has 9 cells. The four corners, four edges, and the center. So, if we take the four edges (which are not connected to anything) plus the four corners (but the corners are connected to the center). Wait, if we take the edges (4) and the center (1), that's 5, which is an independent set because the center is connected to the corners, not to the edges. Alternatively, edges (4) plus corners (4) would be 8, but the corners are connected to the center, which is not in the set, so maybe that's okay. Wait, but in an independent set, no two vertices can be adjacent. So, the corners are connected to the center, but if the center is not in the set, then having the corners in the set is okay, as long as the corners themselves are not connected. But the corners are not connected to each other. Because a knight can't move from one corner to another. So, in the 3x3 grid, each corner is only connected to the center. So, the four corners are not connected to each other, so they can all be in the independent set. Similarly, the edges are not connected to anything, so they can be included. Therefore, maximum independent set would be 4 corners + 4 edges = 8. Then, the minimum vertex cover would be 1 (the center). Because if you paint the center, all edges (center to corners) are covered. The edges themselves are not connected to anything, so they don't need to be covered. Wait, but in vertex cover, we need to cover all edges. The edges (i.e., the connections) between center and corners are covered by painting the center. The edges (as in the edge squares) are not part of any knight's move, so they don't need to be covered. Therefore, in this case, the minimum vertex cover is indeed 1 (the center), and the maximum independent set is 8. But total cells are 9, so 9 - 1 = 8, which matches. Therefore, in the 3x3 case, painting just the center cell would suffice to block all possible knight moves between unpainted cells. Because if you leave the four corners and four edges unpainted, the only possible knight moves are from the center (which is painted) to the corners. Since the center is painted, you can't move from an unpainted corner to the center (which is painted). And the edges can't move anywhere. Therefore, the maximum independent set is indeed 8, and the minimum vertex cover is 1.But in this case, Konig's theorem applies because the graph is bipartite. The knight's graph is bipartite as usual. In the 3x3 case, the two color classes would be: let's say black and white. The corners and center might alternate colors. Let's check. In a standard chessboard coloring, (1,1) is black, (1,2) is white, (1,3) is black, (2,1) is white, (2,2) is black, (2,3) is white, (3,1) is black, (3,2) is white, (3,3) is black. So the corners (1,1), (1,3), (3,1), (3,3) are black; center (2,2) is black. The edges (1,2), (2,1), (2,3), (3,2) are white. So, the two color classes are black (5 cells) and white (4 cells). Therefore, the maximum independent set should be the larger color class, which is black with 5 cells. But earlier, we thought the maximum independent set is 8. That seems contradictory. Wait, maybe my analysis was wrong. Wait, in the standard chessboard coloring, the independent sets are the color classes. So in the knight's graph, which is bipartite, the maximum independent set is indeed the larger color class. But in the 3x3 case, the larger color class is black with 5 cells, but earlier, we thought we could have 8 cells in the independent set. That must be incorrect.Wait, so maybe my mistake was in thinking that the edges are not connected. Wait, but in the knight's graph, edges (the edge squares) can't make any moves, so they are isolated nodes. So the knight's graph is bipartite between black and white, but the white squares include the edges and the center is black. Wait, no, the center is black. Wait, (2,2) is black. The white squares are (1,2), (2,1), (2,3), (3,2). So, in the knight's graph, the white squares can't move to any other white squares, because knight moves alternate colors. Similarly, the black squares include the corners and the center. From the center, you can move to the corners (which are black?), Wait, no, knight moves alternate colors. Wait, if the center is black, then knight moves from the center would go to white squares, but in the 3x3 grid, the knight moves from the center (2,2) would be to (1,3), (3,1), (1,1), (3,3), but those are all corners, which are black. Wait, that can't be. Wait, in reality, in standard chessboard coloring, a knight moves from black to white and vice versa. But in the 3x3 grid, the center is black, and the corners are black, so how does that work? Wait, perhaps the coloring is different? Let me check again.If we color the grid starting with (1,1) as black, then:Row 1: B, W, BRow 2: W, B, WRow 3: B, W, BSo, (1,1)=B, (1,2)=W, (1,3)=B(2,1)=W, (2,2)=B, (2,3)=W(3,1)=B, (3,2)=W, (3,3)=BTherefore, the corners are all black, the edges (squares adjacent to corners) are white, and the center is black. So, a knight move from the center (2,2) which is black would go to (1,3), (3,1), (1,1), (3,3), which are all black. But that's impossible, because knight moves should switch colors. So this suggests that in the 3x3 grid, the knight's graph isn't actually bipartite? Wait, that can't be. Wait, maybe because the 3x3 grid is too small, the knight moves are only between black squares? That seems contradictory. Wait, actually, in reality, a knight on a chessboard always alternates colors. So from a black square, it must move to a white square. But in the 3x3 grid, from the center (2,2) which is black, the possible knight moves would be:(2+1, 2+2)=(3,4) which is outside,(2+2, 2+1)=(4,3) outside,Similarly negative directions: (2-1, 2-2)=(1,0) outside,(2-2, 2-1)=(0,1) outside. Wait, actually, in the 3x3 grid, the center can't make any knight moves? Wait, but earlier we thought the center can move to the corners. Wait, no. Wait, a knight moves two in one direction and one in the other. So from (2,2), moving two up and one left would be (4,1), which is outside. Similarly, two left and one up would be (0,3), outside. Wait, actually, in a 3x3 grid, the center can't move anywhere. Wait, that's different. Then, in the 3x3 grid, the knight's graph is just the four corners connected to the center? Wait, no. Wait, from the corners:From (1,1): two possible knight moves: (1+2,1+1)=(3,2) and (1+1,1+2)=(2,3). Both are within the grid. (3,2) is white, (2,3) is white. So, (1,1) is black, connected to two white squares. Similarly, from (1,3), knight moves to (2,1) and (3,2). From (3,1), knight moves to (1,2) and (2,3). From (3,3), knight moves to (1,2) and (2,1). So, in reality, the corners (black squares) are connected to the edge white squares. The center (2,2) is black, but cannot move anywhere. Therefore, the knight's graph in 3x3 is composed of the four corners connected to the edges, and the center is an isolated node. So the graph is bipartite between black and white, but the edges are from black corners to white edges. The center is a black square with no connections. Therefore, the two color classes are:Black: (1,1), (1,3), (3,1), (3,3), (2,2) – 5 cells.White: (1,2), (2,1), (2,3), (3,2) – 4 cells.Therefore, the maximum independent set is the larger color class, which is black with 5 cells. However, if we take the black squares, which include the center and the four corners. However, the four corners are connected to the white edges, but since the white edges are in the other partition. Wait, in the independent set, which is a set of vertices with no edges between them. If we take all black squares, which include the corners and the center. But the corners are connected to the white edges, not to each other or the center. The center is not connected to anything. So, all black squares form an independent set. Similarly, all white squares form an independent set. Therefore, the maximum independent set is indeed the black squares with 5 cells. Therefore, the minimum vertex cover would be the white squares, 4 cells. Therefore, if we paint all white squares (4), then all knight moves are from black to white, so the white squares are painted, hence covering all edges. Then, the unpainted cells are the 5 black cells, which are an independent set. Therefore, from any unpainted cell (black), all their knight moves go to white cells, which are painted. Therefore, the condition is satisfied. So, in this case, the minimum number of cells to paint is 4. However, earlier, when I considered the maximum independent set as 8, that was incorrect because I didn't consider the actual connections. Therefore, my initial analysis was wrong because I didn't correctly model the knight's moves in the 3x3 grid.Therefore, returning to the original problem. For a 65x65 grid, which is an odd-sized chessboard, the two color classes will be of size (65^2 +1)/2 = 2113 (black) and 2112 (white). Since the knight's graph is bipartite between these two color classes, the maximum independent set is 2113, and the minimum vertex cover is 2112. Therefore, painting all the white squares (2112) ensures that all knight moves from black squares (unpainted) go to white squares (painted), so no two unpainted cells can be a knight's move apart. Therefore, the answer is 2112.But let me confirm with another example. Take a 2x2 grid. But knights can't move on a 2x2 grid, so all cells are isolated. Therefore, maximum independent set is 4, minimum vertex cover is 0. But in the problem statement, it's 65x65, which is much larger. Another example: 4x4 grid. Then, the color classes are 8 and 8. So, maximum independent set is 8, minimum vertex cover is 8. So you need to paint 8 cells. But if you paint all the white cells, then the black cells are unpainted, and since every knight move from a black cell goes to a white cell, which is painted. Therefore, it works.Therefore, in general, for an n x n chessboard, the minimum number of cells to paint is floor(n^2 / 2). But when n is odd, it's (n^2 -1)/2. For even n, it's n^2 /2. Therefore, for 65x65, which is odd, (65^2 -1)/2 = (4225 -1)/2 = 4224/2 = 2112. So, that seems to hold.But wait, is there a way to do better than this? Because sometimes, vertex covers can be smaller than the size of one partition in a bipartite graph if there are certain structures. But in the case of a complete bipartite graph, the minimum vertex cover is indeed the size of the smaller partition. However, in the knight's graph, the graph is not complete bipartite. It's a more sparse graph. However, Konig's theorem says that in bipartite graphs, the size of the maximum matching equals the size of the minimum vertex cover. But since we might not have a perfect matching, but in our case, since the graph is bipartite and we can have a perfect matching?Wait, perhaps in the knight's graph on a chessboard, there exists a perfect matching when it's even-sized, but not when it's odd-sized. However, in our case, since the chessboard is odd-sized, the two color classes are unequal, so there cannot be a perfect matching. Therefore, the maximum matching would be limited by the smaller partition. Therefore, the minimum vertex cover would be equal to the size of the maximum matching. But since Konig's theorem states that in bipartite graphs, the size of the minimum vertex cover equals the size of the maximum matching. If the maximum matching is equal to the smaller color class, then the minimum vertex cover would be equal to the smaller color class. But in our case, the two color classes are of size 2113 and 2112. Therefore, the maximum matching can't be larger than 2112, the size of the smaller partition. Therefore, if there exists a matching that covers all the smaller partition, then the minimum vertex cover would be 2112. But in a knight's graph on an odd-sized chessboard, can we have a matching that covers all the white squares? Each white square can be matched to a black square via a knight's move. But does such a matching exist? It might not be possible because of the irregular knight moves.Alternatively, maybe it's possible that the minimum vertex cover is indeed the size of the smaller partition, 2112, as we thought before. Because in a bipartite graph, the minimum vertex cover is at least the size of the smaller partition. Wait, no. Konig's theorem says that in bipartite graphs, the minimum vertex cover is equal to the maximum matching. So, if the maximum matching is equal to the size of the smaller partition, then the minimum vertex cover is equal to the size of the smaller partition. However, if the maximum matching is less than that, then the minimum vertex cover would be less. But in our case, it's possible that there is a perfect matching for the smaller partition.Wait, perhaps the knight's graph on a chessboard is such that it has a perfect matching when the partitions are equal, but not otherwise. But since in our case, the partitions are unequal, the maximum matching would be 2112, which is the size of the smaller partition. Therefore, by Konig's theorem, the minimum vertex cover is 2112. Therefore, the answer is 2112.Therefore, after going through these examples and considering the properties of bipartite graphs, Konig's theorem, and the structure of the knight's moves, I believe that the minimum number of cells that need to be painted is 2112.Final AnswerThe minimum number of cells that need to be painted is boxed{2112}.</think>To solve the problem of determining the minimum number of cells that need to be painted in a 65x65 grid such that no two unpainted cells are a knight's move apart, we can use concepts from graph theory. The problem is equivalent to finding a minimum vertex cover in a bipartite graph where each node represents a cell and edges represent knight's moves.1. Bipartite Graph Structure: The chessboard can be colored in a standard alternating black and white pattern. A knight's move always alternates between black and white cells, making the knight's graph bipartite. For an odd-sized grid (65x65), the number of black cells is ((65^2 + 1)/2 = 2113) and the number of white cells is ((65^2 - 1)/2 = 2112).2. Maximum Independent Set: In a bipartite graph, the maximum independent set is the larger of the two color classes. For the 65x65 grid, the maximum independent set is the number of black cells, which is 2113.3. Minimum Vertex Cover: The minimum vertex cover is the complement of the maximum independent set. This means the minimum number of cells to paint (vertex cover) is the total number of cells minus the size of the maximum independent set. [ 65^2 - 2113 = 4225 - 2113 = 2112 ]4. Verification: By painting all white cells (2112 cells), we ensure that every knight's move from an unpainted black cell lands on a painted white cell. This satisfies the condition that no two unpainted cells can be a knight's move apart.Thus, the minimum number of cells that need to be painted is (boxed{2112}).]

🔑:# SolutionCells should be colored in a checkerboard pattern. Thus, left[frac{N^{2}}{2}right] cells will be colored. Since any "knight's move" lands on a cell of a different color, there is no move to a cell of the same color. A "knight's move" can traverse any square table larger than 4x4 such that the knight visits each cell exactly once (see, for example, a 5x5 table). If these moves are numbered, it is clear that fewer than left[frac{N^{2}}{2}right] cells cannot be colored because then there would necessarily be two consecutive uncolored cells in this sequence, i.e., a move from one to the other would be possible. The 35 times 35 table needs to be divided into 169 5 times 5 tables, and they should be numbered sequentially, starting with the first 5 times 5 table. left[frac{65^{2}}{2}right]=2112.| 21 | 16 | 5 | 10 | 23 | || :--- | :--- | :--- | :--- | :--- | :--- || 6 | 11 | 22 | 17 | 4 | || 1 | 20 | 15 | 24 | 9 | 26 || 12 | 7 | 18 | 3 | 14 | || 19 | 2 | 13 | 8 | 25 | || | | | | | |Answer: 2112.

🧠:[<think>Okay, so I need to find the value of 5 cos A - 4 cos A cos C + 5 cos C in triangle ABC where the sides a, b, c satisfy 2b = a + c. Hmm, let's start by recalling some trigonometric identities or theorems related to triangles. Since this is a triangle with sides given, maybe the Law of Cosines will be helpful here. The Law of Cosines relates the sides of a triangle to the cosine of one of its angles. The formulas are:For angle A: a² = b² + c² - 2bc cos AFor angle B: b² = a² + c² - 2ac cos BFor angle C: c² = a² + b² - 2ab cos CGiven that 2b = a + c, perhaps I can express one side in terms of the others. Let's solve for b: b = (a + c)/2. So maybe substituting this into the Law of Cosines equations could help?But the expression we need to find involves cos A and cos C. So maybe I need expressions for cos A and cos C. Let's try to use the Law of Cosines for angles A and C.Starting with angle A:a² = b² + c² - 2bc cos ASimilarly, for angle C:c² = a² + b² - 2ab cos CIf I can express cos A and cos C in terms of a, b, c, then maybe substitute using the relation 2b = a + c.Alternatively, perhaps it's better to express a and c in terms of b. Let's let a = 2b - c. Wait, but since 2b = a + c, then a = 2b - c and c = 2b - a. Hmm, but that might complicate things. Alternatively, maybe introducing a parameter or expressing in terms of ratios. Let me think.Alternatively, maybe using the Law of Sines. The Law of Sines says that a/sin A = b/sin B = c/sin C = 2R, where R is the radius of the circumscribed circle. But with the given condition 2b = a + c, perhaps we can relate the sines of the angles.But combining Law of Sines and Law of Cosines might complicate. Let me try first with Law of Cosines.First, express cos A and cos C using the Law of Cosines.cos A = (b² + c² - a²)/(2bc)cos C = (a² + b² - c²)/(2ab)So we can substitute these into the expression 5 cos A - 4 cos A cos C + 5 cos C.But this might get messy. Let's see:First, compute cos A and cos C:cos A = (b² + c² - a²)/(2bc)cos C = (a² + b² - c²)/(2ab)Then, substituting into the expression:5[(b² + c² - a²)/(2bc)] - 4[(b² + c² - a²)/(2bc) * (a² + b² - c²)/(2ab)] + 5[(a² + b² - c²)/(2ab)]This looks quite complicated. Maybe there's a better way.Alternatively, since we know that 2b = a + c, perhaps we can consider substituting a = 2b - c into the expressions. Let's try that.Given a = 2b - c, let's substitute into cos A:cos A = (b² + c² - a²)/(2bc) = (b² + c² - (2b - c)²)/(2bc)Compute (2b - c)²: 4b² -4bc + c²So, numerator: b² + c² - (4b² -4bc + c²) = b² + c² -4b² +4bc -c² = (-3b² +4bc)Therefore, cos A = (-3b² +4bc)/(2bc) = (-3b +4c)/2c = (-3b)/(2c) + 2Similarly, compute cos C. Let's express c as 2b -a, but wait, since we have a = 2b -c, perhaps it's symmetric. Alternatively, let's substitute c = 2b - a into cos C.cos C = (a² + b² - c²)/(2ab) = (a² + b² - (2b -a)^2)/(2ab)Compute (2b -a)^2: 4b² -4ab +a²Numerator: a² + b² - (4b² -4ab +a²) = a² + b² -4b² +4ab -a² = (-3b² +4ab)Therefore, cos C = (-3b² +4ab)/(2ab) = (-3b)/(2a) + 2So, cos A = (-3b)/(2c) + 2 and cos C = (-3b)/(2a) + 2.Now, substitute these into 5 cos A -4 cos A cos C +5 cos C.First, compute 5 cos A +5 cos C:5[ (-3b)/(2c) +2 ] +5[ (-3b)/(2a) +2 ] =5*(-3b)/(2c) +10 +5*(-3b)/(2a) +10 = -15b/(2c) -15b/(2a) +20Then, compute -4 cos A cos C:First, compute cos A cos C:[ (-3b)/(2c) +2 ] * [ (-3b)/(2a) +2 ] = [ (-3b)/(2c) * (-3b)/(2a) ) ] + [ (-3b)/(2c)*2 + (-3b)/(2a)*2 ] + [2*2]Compute each term:First term: (9b²)/(4ac)Second terms: (-3b)/(c) + (-3b)/(a)Third term: 4So cos A cos C = (9b²)/(4ac) - 3b(1/c +1/a) +4Therefore, -4 cos A cos C = -4*(9b²)/(4ac) +12b(1/c +1/a) -16 = -9b²/ac +12b(1/a +1/c) -16Putting it all together, the entire expression is:-15b/(2c) -15b/(2a) +20 -9b²/ac +12b(1/a +1/c) -16Simplify term by term:First, -15b/(2c) -15b/(2a) +12b(1/a +1/c):Let's factor out b:b[ -15/(2c) -15/(2a) +12/a +12/c ] = b[ (-15/2 +12)/a + (-15/2 +12)/c ] = b[ ( (-15/2 +24/2 )/a + (-15/2 +24/2 )/c ) ] = b[ (9/2)/a +9/2/c ] = (9b/2)(1/a +1/c)Then, remaining terms: 20 -16 -9b²/ac =4 -9b²/acSo the entire expression becomes:(9b/2)(1/a +1/c) +4 -9b²/acHmm, let's compute (1/a +1/c) = (a + c)/(ac). Given that a +c =2b, so (1/a +1/c) =2b/(ac). Therefore:(9b/2)(2b/(ac)) +4 -9b²/ac = (9b/2)*(2b/(ac)) +4 -9b²/ac = (9b²/ac) +4 -9b²/ac = 4Wait, that simplifies to 4? So all the terms involving b²/ac cancel out?Yes, let's check again:First part: (9b/2)(1/a +1/c) = (9b/2)( (a +c)/ac ) = (9b/2)(2b/ac) because a + c =2b. So 9b/2 *2b/ac = 9b²/acThen, the rest is +4 -9b²/ac. Therefore, 9b²/ac -9b²/ac +4 =4So the entire expression simplifies to 4. Therefore, the value of 5 cos A -4 cos A cos C +5 cos C is 4.Wait, so is the answer 4? Let me verify this again step by step because that seems straightforward, but maybe I made a miscalculation.So, starting with 5 cos A -4 cos A cos C +5 cos C, we expressed cos A and cos C in terms of a, b, c, substituted the given relation 2b =a +c, and after simplifying all the terms, everything except 4 canceled out. Therefore, the result is 4. Hmm, that seems correct. Let me check with a specific example.Suppose we take a triangle where 2b =a +c. Let's choose specific values. For simplicity, let’s take b=2, so a +c=4. Let's choose a=1, c=3. Then sides are a=1, b=2, c=3. But wait, in a triangle, the sum of any two sides must be greater than the third side. However, 1 +2=3, which is not greater than 3, so that's a degenerate triangle. Bad example. Let's pick a=2, c=2, so a +c=4, so b=2. Then the triangle is equilateral? Wait, a=2, b=2, c=2. Then all angles are 60 degrees. Let's compute 5 cos A -4 cos A cos C +5 cos C. Since all angles are 60°, cos 60°=0.5. Then expression is 5*(0.5) -4*(0.5*0.5) +5*(0.5)= 2.5 -1 +2.5=4. Which matches the result. So that's a good check.Another example: Let's take a triangle with sides a=3, c=5, then 2b=8, so b=4. Check if triangle with sides 3,4,5 is valid. 3+4>5, 4+5>3, 3+5>4: yes, it's a right triangle with sides 3,4,5. Then angle A is opposite side a=3, angle C opposite c=5 (which is the right angle). So compute cos A and cos C.In 3-4-5 triangle, angle opposite 5 is right angle, so angle C is 90°, cos C =0. Angle A: adjacent over hypotenuse for angle A is 4/5 (since in a right triangle, cos A = adjacent/hypotenuse = 4/5). Wait, but in the Law of Cosines terms:cos A = (b² + c² -a²)/(2bc) = (16 +25 -9)/(2*4*5)= (32)/(40)= 0.8=4/5.cos C = (a² +b² -c²)/(2ab)= (9 +16 -25)/(2*3*4)=0/(24)=0.Then compute 5 cos A -4 cos A cos C +5 cos C =5*(4/5) -4*(4/5*0)+5*0=4 -0 +0=4. Which again gives 4. So this checks out.Another example: Let's take a triangle with sides a=5, c=7, so 2b=12, so b=6. Check if 5,6,7 is a valid triangle. 5+6>7, 6+7>5, 5+7>6: yes. Compute cos A and cos C.cos A = (b² +c² -a²)/(2bc)= (36 +49 -25)/(2*6*7)= (60)/(84)= 5/7≈0.714.cos C= (a² +b² -c²)/(2ab)= (25 +36 -49)/(2*5*6)= (12)/(60)=1/5=0.2.Compute 5 cos A -4 cos A cos C +5 cos C=5*(5/7) -4*(5/7*1/5) +5*(1/5)=25/7 -4*(1/7) +1=25/7 -4/7 +7/7= (25 -4 +7)/7=28/7=4. Again, it works.Therefore, regardless of the triangle, as long as 2b =a +c, this expression equals 4. Therefore, the answer is 4.Final Answerboxed{4}</think>Given a triangle ( triangle ABC ) with sides ( a ), ( b ), and ( c ) satisfying ( 2b = a + c ), we need to find the value of ( 5 cos A - 4 cos A cos C + 5 cos C ).First, we use the Law of Cosines to express ( cos A ) and ( cos C ):[cos A = frac{b^2 + c^2 - a^2}{2bc}][cos C = frac{a^2 + b^2 - c^2}{2ab}]Given ( 2b = a + c ), we substitute ( a = 2b - c ) into the expression for ( cos A ):[cos A = frac{b^2 + c^2 - (2b - c)^2}{2bc} = frac{b^2 + c^2 - (4b^2 - 4bc + c^2)}{2bc} = frac{-3b^2 + 4bc}{2bc} = frac{-3b + 4c}{2c} = -frac{3b}{2c} + 2]Similarly, substituting ( c = 2b - a ) into the expression for ( cos C ):[cos C = frac{a^2 + b^2 - (2b - a)^2}{2ab} = frac{a^2 + b^2 - (4b^2 - 4ab + a^2)}{2ab} = frac{-3b^2 + 4ab}{2ab} = frac{-3b + 4a}{2a} = -frac{3b}{2a} + 2]Next, we substitute these expressions into ( 5 cos A - 4 cos A cos C + 5 cos C ):[5 left( -frac{3b}{2c} + 2 right) + 5 left( -frac{3b}{2a} + 2 right) - 4 left( -frac{3b}{2c} + 2 right) left( -frac{3b}{2a} + 2 right)]Simplifying each term:1. ( 5 cos A + 5 cos C ):[5 left( -frac{3b}{2c} + 2 right) + 5 left( -frac{3b}{2a} + 2 right) = -frac{15b}{2c} - frac{15b}{2a} + 10 + 10 = -frac{15b}{2c} - frac{15b}{2a} + 20]2. ( -4 cos A cos C ):[-4 left( left( -frac{3b}{2c} + 2 right) left( -frac{3b}{2a} + 2 right) right)]Expanding the product inside:[left( frac{9b^2}{4ac} - frac{3b}{c} - frac{3b}{a} + 4 right)]Multiplying by -4:[-4 left( frac{9b^2}{4ac} - frac{3b}{c} - frac{3b}{a} + 4 right) = -frac{9b^2}{ac} + frac{12b}{c} + frac{12b}{a} - 16]Combining all terms:[-frac{15b}{2c} - frac{15b}{2a} + 20 - frac{9b^2}{ac} + frac{12b}{c} + frac{12b}{a} - 16]Simplifying:[left( -frac{15b}{2c} + frac{12b}{c} right) + left( -frac{15b}{2a} + frac{12b}{a} right) + (20 - 16) - frac{9b^2}{ac}][frac{9b}{2c} + frac{9b}{2a} + 4 - frac{9b^2}{ac}]Using ( a + c = 2b ):[frac{9b}{2} left( frac{1}{a} + frac{1}{c} right) + 4 - frac{9b^2}{ac} = frac{9b}{2} cdot frac{a + c}{ac} + 4 - frac{9b^2}{ac} = frac{9b}{2} cdot frac{2b}{ac} + 4 - frac{9b^2}{ac} = frac{9b^2}{ac} + 4 - frac{9b^2}{ac} = 4]Thus, the value is (boxed{4}).,<think>Okay, let me try to solve this problem. So, we have triangle ABC with sides a, b, c opposite angles A, B, C respectively. The given condition is 2b = a + c. We need to find the value of 5 cos A - 4 cos A cos C + 5 cos C. Hmm, alright. Let's start by recalling some trigonometric identities or theorems related to triangles. The Law of Cosines comes to mind. It relates the sides of a triangle to the cosine of one of its angles. The formula is:For any triangle,a² = b² + c² - 2bc cos Ab² = a² + c² - 2ac cos Bc² = a² + b² - 2ab cos CSince we have expressions involving cos A and cos C, maybe we can express those using the Law of Cosines. Also, we have the given condition 2b = a + c. Maybe we can use this to relate the sides and substitute into the expressions for the cosines.Alternatively, there's the Law of Sines, which states that a/sin A = b/sin B = c/sin C = 2R, where R is the radius of the circumscribed circle. Maybe that could be helpful as well. Let's see.First, let's note that 2b = a + c. Maybe we can express a or c in terms of b. Let's say a = 2b - c. But I don't know if that's immediately helpful. Alternatively, since all sides are related, perhaps we can parameterize the sides. For example, let me consider that since 2b = a + c, maybe we can set a = b - d and c = b + d for some d? That way, a + c = (b - d) + (b + d) = 2b, which satisfies the given condition. That might be a good approach. Let's try that.Let’s denote a = b - d and c = b + d. Then, substituting into the Law of Cosines expressions for angles A and C.First, for angle A:Using Law of Cosines:a² = b² + c² - 2bc cos ASubstituting a = b - d and c = b + d:(b - d)² = b² + (b + d)² - 2b(b + d) cos ALet’s expand both sides:Left side: (b - d)² = b² - 2bd + d²Right side: b² + (b + d)² - 2b(b + d) cos AFirst compute (b + d)² = b² + 2bd + d²So right side becomes: b² + b² + 2bd + d² - 2b(b + d) cos A = 2b² + 2bd + d² - 2b(b + d) cos ASet left side equal to right side:b² - 2bd + d² = 2b² + 2bd + d² - 2b(b + d) cos ASubtract left side from both sides:0 = 2b² + 2bd + d² - 2b(b + d) cos A - (b² - 2bd + d²)Simplify:0 = 2b² + 2bd + d² - 2b(b + d) cos A - b² + 2bd - d²Simplify term by term:2b² - b² = b²2bd + 2bd = 4bdd² - d² = 0So,0 = b² + 4bd - 2b(b + d) cos ALet’s factor out b from the terms involving b:0 = b(b + 4d) - 2b(b + d) cos ADivide both sides by b (assuming b ≠ 0, which it can't be in a triangle):0 = (b + 4d) - 2(b + d) cos AThen,2(b + d) cos A = b + 4dTherefore,cos A = (b + 4d)/(2(b + d))Similarly, let's compute cos C using the Law of Cosines.For angle C:c² = a² + b² - 2ab cos CSubstituting c = b + d and a = b - d:(b + d)² = (b - d)² + b² - 2b(b - d) cos CExpand the left side: b² + 2bd + d²Right side: (b² - 2bd + d²) + b² - 2b(b - d) cos CSimplify: b² - 2bd + d² + b² - 2b(b - d) cos C = 2b² - 2bd + d² - 2b(b - d) cos CSet left side equal to right side:b² + 2bd + d² = 2b² - 2bd + d² - 2b(b - d) cos CSubtract left side from both sides:0 = 2b² - 2bd + d² - 2b(b - d) cos C - (b² + 2bd + d²)Simplify:0 = 2b² - 2bd + d² - 2b(b - d) cos C - b² - 2bd - d²Simplify term by term:2b² - b² = b²-2bd - 2bd = -4bdd² - d² = 0So,0 = b² - 4bd - 2b(b - d) cos CFactor out b from the first two terms:0 = b(b - 4d) - 2b(b - d) cos CDivide both sides by b (again, assuming b ≠ 0):0 = (b - 4d) - 2(b - d) cos CTherefore,2(b - d) cos C = b - 4dHence,cos C = (b - 4d)/(2(b - d))Okay, so we have expressions for cos A and cos C in terms of b and d. Let's note that our goal is to compute 5 cos A - 4 cos A cos C + 5 cos C.So, let's first compute cos A and cos C:cos A = (b + 4d)/(2(b + d))cos C = (b - 4d)/(2(b - d))Now, let's compute each term:First, 5 cos A = 5*(b + 4d)/(2(b + d)) = [5(b + 4d)]/[2(b + d)]Similarly, 5 cos C = 5*(b - 4d)/(2(b - d)) = [5(b - 4d)]/[2(b - d)]Then, the middle term is -4 cos A cos C. Let's compute cos A cos C:cos A cos C = [(b + 4d)/(2(b + d))] * [(b - 4d)/(2(b - d))] = [(b + 4d)(b - 4d)]/[4(b + d)(b - d)] = [b² - (4d)^2]/[4(b² - d²)] = [b² - 16d²]/[4(b² - d²)]Therefore, -4 cos A cos C = -4 * [b² - 16d²]/[4(b² - d²)] = - [b² - 16d²]/[b² - d²] = (-b² + 16d²)/(b² - d²) = (16d² - b²)/(b² - d²)So, combining the three terms:5 cos A -4 cos A cos C +5 cos C = [5(b + 4d)]/[2(b + d)] + [5(b - 4d)]/[2(b - d)] + (16d² - b²)/(b² - d²)Hmm, this looks a bit complicated. Maybe we can combine the first two terms and then add the third term.Let’s compute the first two fractions:[5(b + 4d)]/[2(b + d)] + [5(b - 4d)]/[2(b - d)] = (5/2) [ (b + 4d)/(b + d) + (b - 4d)/(b - d) ]Let me compute the sum inside the brackets:[(b + 4d)/(b + d) + (b - 4d)/(b - d)]To add these two fractions, we need a common denominator, which is (b + d)(b - d)So:[(b + 4d)(b - d) + (b - 4d)(b + d)] / [(b + d)(b - d)]Let’s compute the numerator:First term: (b + 4d)(b - d) = b² - bd + 4bd -4d² = b² + 3bd -4d²Second term: (b - 4d)(b + d) = b² + bd -4bd -4d² = b² -3bd -4d²Adding them together:(b² + 3bd -4d²) + (b² -3bd -4d²) = 2b² -8d²Therefore, the sum becomes (2b² -8d²)/[(b + d)(b - d)] = 2(b² -4d²)/(b² -d²)Therefore, the first two terms combined:(5/2) * [2(b² -4d²)/(b² -d²)] = 5*(b² -4d²)/(b² -d²)So now, the entire expression 5 cos A -4 cos A cos C +5 cos C becomes:5*(b² -4d²)/(b² -d²) + (16d² -b²)/(b² -d²) = [5(b² -4d²) +16d² -b²]/(b² -d²)Simplify the numerator:5b² -20d² +16d² -b² = (5b² -b²) + (-20d² +16d²) = 4b² -4d²Thus, the entire expression becomes:(4b² -4d²)/(b² -d²) = 4(b² -d²)/(b² -d²) = 4Wait, that's 4? So, the value is 4? Let me check the algebra again to be sure.Starting with the numerator:5(b² -4d²) +16d² -b²=5b² -20d² +16d² -b²= (5b² -b²) + (-20d² +16d²)=4b² -4d²Yes, that's correct. Then,4b² -4d² divided by (b² -d²) is 4(b² -d²)/(b² -d²) which is 4. So the entire expression simplifies to 4. Therefore, the answer is 4.But let me verify this with another approach just to be thorough. Because sometimes when you parameterize variables, you might have introduced some constraints or made an assumption that could affect the generality. Let me try using another method to confirm.Alternative approach:Given 2b = a + c. Let's use the Law of Cosines to express cos A and cos C in terms of sides.First, cos A = (b² + c² - a²)/(2bc)Similarly, cos C = (a² + b² - c²)/(2ab)But given 2b = a + c, maybe we can express a and c in terms of b. Let me assume that a = b - x and c = b + x, similar to the previous parameterization but using x instead of d. Then, 2b = (b - x) + (b + x) = 2b, which satisfies the condition.So, sides: a = b - x, c = b + x, so the triangle is parameterized in terms of b and x. Then, substituting into expressions for cos A and cos C.Compute cos A:cos A = (b² + c² - a²)/(2bc) = (b² + (b + x)^2 - (b - x)^2)/(2b(b + x))Compute numerator:b² + (b² + 2bx + x²) - (b² - 2bx + x²) = b² + b² + 2bx + x² - b² + 2bx - x² = b² + 4bxTherefore, cos A = (b² + 4bx)/(2b(b + x)) = (b + 4x)/(2(b + x)) which matches the previous result if x = d.Similarly, compute cos C:cos C = (a² + b² - c²)/(2ab) = ((b - x)^2 + b² - (b + x)^2)/(2b(b - x))Compute numerator:(b² - 2bx + x²) + b² - (b² + 2bx + x²) = b² - 2bx + x² + b² - b² - 2bx - x² = b² -4bxThus, cos C = (b² -4bx)/(2b(b -x)) = (b -4x)/(2(b -x)) which again matches the previous result if x = d.Therefore, proceeding as before, we can compute the expression 5 cos A -4 cos A cos C +5 cos C.Alternatively, perhaps using the parameterization x = d, then expressions are same. So, the calculation would result in 4 as before. Therefore, this confirms the result.Alternatively, maybe there's a way without parameterizing the sides. Let's try another method.Given 2b = a + c. Let's consider using the Law of Cosines for angles A and C.We need expressions for cos A and cos C.From Law of Cosines:cos A = (b² + c² - a²)/(2bc)cos C = (a² + b² - c²)/(2ab)Given that 2b = a + c. Let's denote a = 2b - c. Then, substituting into the expression for cos C:cos C = ((2b - c)^2 + b² - c²)/(2b(2b - c))Let’s compute numerator:(4b² -4bc + c²) + b² - c² = 5b² -4bcThus, cos C = (5b² -4bc)/(2b(2b - c)) = (5b² -4bc)/(4b² -2bc) = [b(5b -4c)]/[2b(2b - c)] = (5b -4c)/(2(2b - c))But since a + c = 2b, then a = 2b - c. So, the denominator is 2b - c + c = 2b, but maybe this isn't helpful.Alternatively, let's express both cos A and cos C in terms of b and then relate them.Alternatively, maybe we can use the Law of Sines. Since a/sin A = b/sin B = c/sin C = 2R. Let’s denote 2R as a constant.Given that 2b = a + c, so 2b = 2R sin A + 2R sin C => 2b = 2R (sin A + sin C) => b = R (sin A + sin C)But from the Law of Sines, b = 2R sin B. Therefore,2R sin B = R (sin A + sin C) => 2 sin B = sin A + sin CAlso, since in a triangle, A + B + C = π, so B = π - A - CTherefore, 2 sin(π - A - C) = sin A + sin CBut sin(π - x) = sin x, so:2 sin(A + C) = sin A + sin CHmm, maybe using trigonometric identities here. Let’s compute sin(A + C):sin(A + C) = sin A cos C + cos A sin CThus, 2[sin A cos C + cos A sin C] = sin A + sin CSo,2 sin A cos C + 2 cos A sin C = sin A + sin CBring all terms to the left:2 sin A cos C + 2 cos A sin C - sin A - sin C = 0Factor sin A and sin C:sin A (2 cos C -1) + sin C (2 cos A -1) = 0Hmm, not sure if this helps. Alternatively, perhaps we can divide both sides by sin A sin C or something. Let me see.Alternatively, rearrange the equation:2 sin A cos C - sin A + 2 cos A sin C - sin C = 0Factor terms:sin A (2 cos C -1) + sin C (2 cos A -1) =0This seems a bit complicated. Maybe we can use the formula for sin A + sin C. Alternatively, we can use sum to product.Wait, sin A + sin C = 2 sin[(A + C)/2] cos[(A - C)/2]Similarly, sin(A + C) = sin(π - B) = sin B. Wait, earlier we had 2 sin(A + C) = sin A + sin C, which would imply 2 sin B = sin A + sin C. But from Law of Sines, since a + c = 2b, then:From Law of Sines, a = 2R sin A, c = 2R sin C, so a + c = 2R (sin A + sin C) = 2b. Therefore, 2R (sin A + sin C) = 2b => R (sin A + sin C) = b. But also, from Law of Sines, b = 2R sin B. Therefore, 2R sin B = R (sin A + sin C) => 2 sin B = sin A + sin C. So that's consistent with what we had before.But perhaps this isn't directly helpful. Let me think. Maybe combine this with some other identity.Alternatively, since we need to find 5 cos A -4 cos A cos C +5 cos C. Let's denote this expression as E. So E = 5 cos A +5 cos C -4 cos A cos C. Let's factor out 5: E =5 (cos A + cos C) -4 cos A cos C.Perhaps we can express cos A + cos C and cos A cos C in terms of other trigonometric identities. Let's recall that in a triangle, A + B + C = π, so A + C = π - B. So, cos(A + C) = cos(π - B) = -cos B.Also, we can use the identity:cos A + cos C = 2 cos[(A + C)/2] cos[(A - C)/2]But since A + C = π - B, then:cos A + cos C = 2 cos[(π - B)/2] cos[(A - C)/2] = 2 sin(B/2) cos[(A - C)/2]Similarly, cos A cos C can be expressed using product-to-sum formulas:cos A cos C = [cos(A + C) + cos(A - C)] / 2 = [ -cos B + cos(A - C) ] / 2But I don't know if this helps. Let me see. So,E =5 (cos A + cos C) -4 cos A cos CExpressed in terms of B and (A - C):=5*2 sin(B/2) cos[(A - C)/2] -4*( [ -cos B + cos(A - C) ] /2 )=10 sin(B/2) cos[(A - C)/2] +2 cos B -2 cos(A - C)Hmm, this seems more complicated. Maybe this approach is not the best.Alternatively, let's consider expressing E in terms of sin B, since we have some relation from 2 sin B = sin A + sin C.But I need another relation. Maybe if we can relate angles A and C.Alternatively, let's think of variables. Let me let x = A and y = C. Then, B = π -x -y.From the earlier relation, 2 sin B = sin x + sin y.So,2 sin(π -x -y) = sin x + sin yWhich is,2 sin(x + y) = sin x + sin yBut sin(x + y) = sin x cos y + cos x sin yThus,2 sin x cos y + 2 cos x sin y = sin x + sin yRearranged:2 sin x cos y - sin x + 2 cos x sin y - sin y =0sin x (2 cos y -1) + sin y (2 cos x -1)=0Hmm, similar to what I had before. Not sure.Alternatively, divide both sides by sin y sin x:(2 cos y -1)/sin y + (2 cos x -1)/sin x =0But not sure.Alternatively, let me use substitution. Let’s set variables in terms of t.Let’s suppose that angle A = angle C + t. Then, angle C = angle A - t.But since A + B + C = π, then (C + t) + B + C = π => B = π - 2C - tBut maybe not helpful.Alternatively, assume that angles A and C have some relation. But this seems vague.Alternatively, go back to the Law of Cosines expressions.We have expressions for cos A and cos C in terms of sides, but since sides are related by 2b = a + c, maybe we can use that to express a in terms of c and b, substitute into the expressions for cos A and cos C, and then compute E.Wait, but earlier when we parameterized the sides as a = b - d and c = b + d, we found that E simplifies to 4 regardless of the value of d. That suggests that the expression E is always equal to 4 for any triangle where 2b = a + c. Therefore, the answer is 4. Let me check with a concrete example to confirm.Suppose we take a specific triangle where 2b = a + c. Let's choose b = 5. Then, a + c =10. Let's pick a =3 and c=7 (since 3 +7=10). Let's check if this forms a valid triangle. The sides are 3,5,7. Check triangle inequalities:3 +5 >7: 8>7, yes.5 +7>3:12>3, yes.3 +7>5:10>5, yes. So valid triangle.Now, compute angles A and C using Law of Cosines.Compute cos A:cos A = (b² + c² - a²)/(2bc) = (25 +49 -9)/(2*5*7) = (65)/(70) = 13/14 ≈0.9286Compute cos C:cos C = (a² + b² - c²)/(2ab) = (9 +25 -49)/(2*3*5) = (-15)/30 = -0.5Now compute E =5 cos A -4 cos A cos C +5 cos C:5*(13/14) -4*(13/14)*(-0.5) +5*(-0.5)Compute each term:5*(13/14) = 65/14 ≈4.6429-4*(13/14)*(-0.5) = -4*( -6.5/14 ) = 26/14 ≈1.85715*(-0.5) = -2.5Add them together:65/14 +26/14 -35/14 = (65 +26 -35)/14 = 56/14 =4. Exactly 4.So, this example confirms that E=4. Let's try another example.Take b=4, so a +c=8. Let me choose a=2 and c=6. Check triangle inequalities:2+4>6? 6>6? No, 6=6, which is degenerate. Not valid. So pick a=3, c=5. Then sides 3,4,5. Which is a right-angled triangle.Check angles:cos A = (4² +5² -3²)/(2*4*5) = (16 +25 -9)/40 =32/40=0.8cos C = (3² +4² -5²)/(2*3*4)= (9+16 -25)/24=0/24=0Compute E=5*0.8 -4*0.8*0 +5*0 =4 -0 +0=4. So again, E=4.Another example. Take b=6, so a +c=12. Let’s choose a=5, c=7. Check triangle inequalities: 5+6>7, 6+7>5, 5+7>6. All valid.Compute cos A:cos A=(6² +7² -5²)/(2*6*7)= (36 +49 -25)/84=60/84=5/7≈0.7143cos C=(5² +6² -7²)/(2*5*6)= (25 +36 -49)/60=12/60=0.2E=5*(5/7) -4*(5/7)*(0.2) +5*(0.2)=25/7 -4*(1/7) +1=25/7 -4/7 +7/7=(25 -4 +7)/7=28/7=4Again, E=4. So this confirms that regardless of the triangle, as long as 2b=a+c, the value is 4. Therefore, the answer is 4.Final Answerboxed{4}</think>Given a triangle ( triangle ABC ) with sides ( a ), ( b ), and ( c ) opposite angles ( A ), ( B ), and ( C ) respectively, where ( 2b = a + c ). We need to find the value of ( 5 cos A - 4 cos A cos C + 5 cos C ).First, we parameterize the sides ( a ) and ( c ) in terms of ( b ) and a variable ( d ):- Let ( a = b - d )- Let ( c = b + d )Using the Law of Cosines for angles ( A ) and ( C ):- For angle ( A ): [ cos A = frac{(b + 4d)}{2(b + d)} ]- For angle ( C ): [ cos C = frac{(b - 4d)}{2(b - d)} ]Next, we compute the expression ( 5 cos A - 4 cos A cos C + 5 cos C ):1. Calculate ( 5 cos A ) and ( 5 cos C ): [ 5 cos A = 5 cdot frac{b + 4d}{2(b + d)} = frac{5(b + 4d)}{2(b + d)} ] [ 5 cos C = 5 cdot frac{b - 4d}{2(b - d)} = frac{5(b - 4d)}{2(b - d)} ]2. Calculate ( -4 cos A cos C ): [ cos A cos C = frac{(b + 4d)(b - 4d)}{4(b + d)(b - d)} = frac{b^2 - 16d^2}{4(b^2 - d^2)} ] [ -4 cos A cos C = -4 cdot frac{b^2 - 16d^2}{4(b^2 - d^2)} = -frac{b^2 - 16d^2}{b^2 - d^2} = frac{16d^2 - b^2}{b^2 - d^2} ]3. Combine all terms: [ 5 cos A - 4 cos A cos C + 5 cos C = frac{5(b + 4d)}{2(b + d)} + frac{5(b - 4d)}{2(b - d)} + frac{16d^2 - b^2}{b^2 - d^2} ]Simplifying the combined terms:- The first two terms combine to: [ frac{5(b^2 - 4d^2)}{b^2 - d^2} ]- Adding the third term: [ frac{5(b^2 - 4d^2) + 16d^2 - b^2}{b^2 - d^2} = frac{4b^2 - 4d^2}{b^2 - d^2} = 4 ]Thus, the value of ( 5 cos A - 4 cos A cos C + 5 cos C ) is (boxed{4}).]

🔑:begin{array}{l}text { 18. } 2 b=a+c Leftrightarrow 2 sin B=sin A+sin C Leftrightarrow 2 cos frac{A+C}{2}=cos frac{A-C}{2} Leftrightarrow tan frac{A}{2} tan frac{C}{2}=frac{1}{2} Leftrightarrow sqrt{frac{1-cos A}{1+cos A}} cdot sqrt{frac{1-cos C}{1+cos C}}=frac{1}{3} Leftrightarrow 5 cos A-4 cos A cos C+5 cos C=4end{array}

🧠:[<think>Okay, let's try to figure out how many different broadcasting sequences are possible for the television station's advertisements. The problem says there are 5 advertisements in a row: 3 different commercial ads and 2 different National Games promotional ads. The requirements are that the last advertisement must be a National Games one, and the two National Games ads can't be played consecutively. Hmm, okay.First, let me restate the problem to make sure I understand. We have 3 distinct commercials (let's call them C1, C2, C3) and 2 distinct National Games ads (let's say N1, N2). They need to be arranged in a sequence of 5 ads. The last position must be either N1 or N2, and also, the two National Games ads can't be next to each other anywhere in the sequence. So, N1 and N2 can't be adjacent, and the last ad has to be National Games.So, how do I approach this? Maybe break it down into steps. Since the last position must be a National Games ad, maybe I should fix the last position first and then arrange the remaining four positions with the remaining ads, making sure that the two National Games ads aren't next to each other.Let me try that. The last position has to be either N1 or N2. There are 2 choices for the last position. Then, the first four positions need to contain the other National Games ad and the three commercial ads. But wait, hold on: if we've already placed one National Games ad in the last position, then the remaining National Games ad has to be somewhere in the first four positions. But we need to ensure that the two National Games ads aren't adjacent. Since the last position is a National Games ad, the fourth position (position 4) cannot be the other National Games ad. So, the remaining National Games ad has to be placed in positions 1, 2, 3, or 4, but not in position 4 if position 5 is already National Games. Wait, no: if position 5 is National Games, then position 4 cannot be National Games because that would make them consecutive. Therefore, in the first four positions, the remaining National Games ad can be in positions 1, 2, or 3, but not in position 4. Hmm, is that correct?Wait, no. Let's think again. Suppose we have the last position fixed as National Games (either N1 or N2). Then, the other National Games ad (if it's N2 or N1, depending on what's in the last position) must be placed in one of the first four positions, but not adjacent to the last position. Since the last position is 5, position 4 can't be National Games. Therefore, the remaining National Games ad must be placed in positions 1, 2, 3, or 4, but position 4 is adjacent to position 5. So, position 4 cannot be National Games. Therefore, the remaining National Games ad must be in positions 1, 2, or 3. Wait, but the first four positions can be positions 1, 2, 3, 4. But since position 4 can't be National Games, the remaining National Games ad has to be in positions 1, 2, or 3. Then, the other three positions (excluding the position where the second National Games ad is placed) would be filled with the three commercials.But also, the commercials are distinct, so order matters for all ads.This is getting a bit confusing. Maybe another approach is better.Since the last position must be a National Games ad, let's first handle that. There are 2 choices for the last position (N1 or N2). Then, we need to arrange the remaining 4 positions with the remaining 1 National Games ad and the 3 commercial ads, making sure that the National Games ad isn't adjacent to the last position (i.e., it can't be in position 4). Also, since we can't have the two National Games ads consecutive anywhere else, but since we already fixed the last position as National Games, the only possible place where the two National Games ads could be consecutive is if the remaining National Games ad is in position 4. But since we have to avoid that, the remaining National Games ad can't be in position 4.Therefore, the remaining National Games ad must be in positions 1, 2, or 3. Then, the other three positions are filled with commercials. But since the commercials are distinct, we need to consider permutations.Alternatively, maybe think of it as:Total number of sequences where the last ad is National Games and the two National Games ads are not consecutive.First, fix the last position as a National Games ad: 2 choices (N1 or N2). Then, the remaining National Games ad (the other one) needs to be placed in one of the first four positions, but not in position 4 (to avoid being adjacent to the last position). So, positions 1, 2, or 3. Therefore, there are 3 choices for the position of the remaining National Games ad. Then, the remaining three positions (since we've already placed two National Games ads in positions x and 5, where x is 1, 2, or 3) are filled with the three distinct commercials. The commercials can be arranged in 3! = 6 ways. So, for each choice of the last ad (2 choices), each choice of position for the remaining National Games ad (3 choices), and each permutation of the commercials (6 ways), the total number of sequences would be 2 * 3 * 6 = 36.Wait, but is that correct? Let me check.Alternatively, perhaps a better way is to use permutations with restrictions. Let's consider arranging the 5 ads with the constraints:1. Last position must be N1 or N2.2. N1 and N2 cannot be consecutive.So, first, handle the last position. There are 2 choices (N1 or N2). Let's say we choose N1 for the last position. Then, N2 must be placed in one of the first four positions, but not adjacent to the last position. Since the last position is 5, position 4 cannot be N2. So, N2 can be in positions 1, 2, or 3. Then, the remaining three positions (positions 4, and the two positions not occupied by N2) are filled with the three commercials. Since the commercials are distinct, the number of ways is 3! = 6. So, for N1 in the last position: positions for N2: 3 choices, commercials: 6 permutations, so 3*6=18. Similarly, if the last position is N2, then N1 must be in positions 1, 2, or 3. Again, 3 choices for N1 and 6 permutations for commercials, so another 18. Total 18+18=36.Therefore, the answer is 36.But let me verify this with another method to be sure.Alternative approach: Total number of sequences where last ad is National Games and two National Games ads are not consecutive.First, total number of sequences where last ad is National Games, without considering the non-consecutive constraint: choose the last ad (2 choices), then arrange the remaining 4 ads (1 National Games and 3 commercials). The number of ways is 2 * (4! / (1! 3!)) * 3! * 1! ? Wait, maybe not.Wait, if we fix the last position as National Games (2 choices), then we have to arrange the remaining 4 positions with 1 National Games ad and 3 commercial ads. The number of permutations is 4! / (1! 3!) * permutations of commercials. Wait, but commercials are distinct, so perhaps it's (number of ways to choose position for the remaining National Games ad) multiplied by permutations of commercials.So, in the first four positions, we have to place 1 National Games ad and 3 commercials. The number of ways to choose the position for the National Games ad is 4. Then, assign the National Games ad (1 choice, since the other one is already used in the last position? Wait, no. Wait, if the last position is N1, then the remaining National Games ad is N2, or vice versa. Wait, but are the National Games ads distinguishable? Yes, they are different. So, if the last position is N1, then the other National Games ad is N2, and we have to place N2 in one of the first four positions. Similarly, if the last position is N2, the other is N1.Therefore, the number of sequences without considering the consecutive constraint is:For each choice of last ad (2), number of ways to place the other National Games ad in the first four positions (4 positions), and then arrange the three commercials in the remaining three positions (3! ways). So total is 2 * 4 * 6 = 48.But then, from these 48, we need to subtract the number of sequences where the two National Games ads are consecutive. How many such sequences are there?If the last position is a National Games ad (N1 or N2), and the other National Games ad is in position 4 (the one right before the last), then they would be consecutive. So, how many sequences have the two National Games ads in positions 4 and 5?If we fix the last position as N1, then the other National Games ad is N2. If N2 is in position 4, then positions 4 and 5 are N2 and N1, which are consecutive. Similarly, if the last position is N2, then N1 in position 4 would be consecutive. So, for each choice of last ad (2), there's 1 position (position 4) where placing the other National Games ad would make them consecutive. So, the number of invalid sequences is 2 * 1 * 3! = 2 * 1 * 6 = 12.Therefore, total valid sequences are 48 - 12 = 36. Which matches the previous result. So, the answer should be 36.Alternatively, another way: think of arranging the commercials first and then placing the National Games ads in the available slots.But since the last position has to be National Games, maybe this complicates things. Let's try.First, arrange the three commercials. There are 3! = 6 ways to arrange them. Then, we need to place the two National Games ads in the remaining two positions, with the condition that the last position is one of them and they are not consecutive.Wait, but if we arrange the commercials first, we can think of the possible slots where the National Games ads can be placed. However, since one of the National Games ads has to be at the end, let's fix that first.Alternatively, if we have three commercials arranged, which creates 4 gaps (before, between, and after the commercials) where National Games ads can be placed. But since one National Games ad must be at the end, perhaps we need a different approach.Wait, let me think. If we arrange the three commercials, which can be done in 3! ways. Then, we have four possible slots to place the National Games ads: _ C _ C _ C _, where the underscores represent possible slots. But since one National Games ad must be at the end (position 5), which is the last slot after the third commercial. So, we have to place one National Games ad there. Then, the other National Games ad has to be placed in one of the remaining slots, but not adjacent to the last slot.However, the slots adjacent to the last slot would be the slot before the last commercial. Wait, this might not be straightforward.Alternatively, considering that after arranging the three commercials, the structure is:positions: 1, 2, 3, 4, 5commercials are in three positions, leaving two positions for National Games, one of which must be position 5. The other National Games must be in positions 1, 2, 3, or 4, but not adjacent to position 5. Since position 5 is fixed as National Games, position 4 cannot be National Games.So, if we first fix position 5 as National Games, then the other National Games can be in positions 1, 2, or 3. The commercials are in positions 4 and the remaining three positions. Wait, no.Wait, actually, if position 5 is National Games, and the other National Games is in positions 1, 2, or 3, then the commercials are in the remaining four positions except the two National Games. Wait, but there are three commercials. So, maybe this approach is getting tangled.Wait, perhaps my first two methods were better. The first method calculated 2 (choices for last position) * 3 (positions for the other National Games ad) * 3! (arrangements of commercials) = 2*3*6=36. The second method calculated total without restriction (48) minus invalid ones (12) = 36. Both methods arrived at 36. So, that's probably the answer.But to be thorough, let's try another approach. Suppose we list all possibilities step by step.First, fix the last position as N1 or N2: 2 options.Case 1: Last position is N1.Then, we have to place N2 somewhere in positions 1-4, but not adjacent to position 5 (which is N1). Therefore, N2 can't be in position 4. So, N2 can be in positions 1, 2, or 3. So, 3 choices.Once N2 is placed, the remaining three positions are filled with C1, C2, C3. Since they are distinct, the number of arrangements is 3! = 6.Therefore, for Case 1: 3 * 6 = 18.Case 2: Last position is N2.Similarly, N1 must be placed in positions 1, 2, or 3. Again, 3 choices. Then, the commercials are arranged in the remaining three positions: 3! = 6.Therefore, Case 2: 3 * 6 = 18.Total: 18 + 18 = 36.Same result. So, this seems consistent.Therefore, the answer is 36.Final AnswerThe number of different broadcasting sequences possible is boxed{36}.</think>The problem requires determining the number of different broadcasting sequences for 5 advertisements (3 distinct commercials and 2 distinct National Games promotions) with the constraints that the last advertisement must be a National Games promotion and the two National Games promotions cannot be consecutive.1. Fix the last position as a National Games promotion: There are 2 choices (N1 or N2).2. Place the remaining National Games promotion: The remaining National Games promotion (N2 or N1) must be placed in one of the first three positions (positions 1, 2, or 3) to avoid being adjacent to the last position. This gives 3 choices.3. Arrange the commercials: The remaining three positions are filled with the three distinct commercials, which can be arranged in (3!) (6) ways.For each choice of the last position (2 choices), the remaining National Games promotion can be placed in 3 positions, and the commercials can be arranged in 6 ways. Therefore, the total number of sequences is calculated as follows:[2 text{ (choices for last position)} times 3 text{ (positions for the other National Games promotion)} times 6 text{ (arrangements of commercials)} = 2 times 3 times 6 = 36]Thus, the number of different broadcasting sequences possible is (boxed{36}).,<think>Okay, so I have this problem where a television station is broadcasting 5 advertisements in a row. There are 3 different commercial ads and 2 different National Games promotional ads. The requirements are that the last advertisement must be a National Games one, and the two National Games ads can't be played consecutively. I need to figure out how many different broadcasting sequences are possible under these constraints. Hmm, let's break this down step by step.First, let me understand the problem. We have 5 slots for advertisements. The last slot (position 5) must be one of the National Games promotional ads. There are 2 different National Games ads, so the last position has 2 choices. Also, the two National Games ads cannot be played consecutively. So, the other National Games ad (the one not in the last position) can't be right before the last one. Additionally, since there are two National Games ads, they need to be placed in two different positions among the first four, but not adjacent to each other or to the last one.Wait, hold on. Wait, the National Games ads cannot be played consecutively. So, if the last ad is a National Games ad, the one before it (position 4) can't be a National Games ad. So position 4 must be a commercial ad. That might be a key point here.So let me list the constraints again:1. Position 5 must be a National Games ad (2 choices).2. The two National Games ads cannot be consecutive. Since one is in position 5, the other National Games ad must be placed in positions 1, 2, or 3. Because if it's in position 4, then positions 4 and 5 would both be National Games ads, which is not allowed.Therefore, the other National Games ad (the first one) must be in positions 1, 2, or 3, and the remaining three positions (excluding position 5 and the position of the first National Games ad) must be filled with the three different commercial ads.Wait, but the commercials are 3 different ones. So we have 3 distinct commercials and 2 distinct National Games ads.So total, we have 5 ads: 3 commercials (let's call them C1, C2, C3) and 2 National Games (NG1, NG2). The problem is about arranging these 5 ads in a sequence of 5 positions with the given constraints.But wait, are the commercials all different and need to be considered as distinct? Yes, the problem says 3 different commercial advertisements and 2 different National Games promotional advertisements. So each ad is unique.Therefore, the total number of possible sequences without any constraints would be 5! = 120. But with the constraints, we need to adjust.But let's not get ahead of ourselves. Let's think about the constraints step by step.First, the last position must be a National Games ad. So position 5 has 2 choices: NG1 or NG2.Then, the other National Games ad must be placed in positions 1, 2, or 3 (since position 4 cannot be National Games as it would be consecutive to position 5). So the first National Games ad can be in positions 1, 2, or 3. Let's denote the positions as 1, 2, 3, 4, 5.So, the two National Games ads are in positions i and 5, where i is 1, 2, or 3. Also, since the National Games ads are different, we need to consider permutations (NG1 in position i and NG2 in 5, or NG2 in position i and NG1 in 5).Therefore, for each valid position i (1, 2, 3), we have 2 possibilities for the National Games ads.Once we've placed the two National Games ads, the remaining three positions must be filled with the three different commercials. Since the commercials are distinct, the number of ways to arrange them is 3! = 6 for each arrangement of National Games ads.So, the total number of sequences would be: number of valid positions for the first National Games ad * number of permutations of National Games ads * number of permutations of commercials.Let's compute that step by step.Number of valid positions for the first National Games ad: 3 (positions 1, 2, 3).For each of these positions, the two National Games ads can be arranged in 2 ways (NG1 in position i and NG2 in 5, or NG2 in position i and NG1 in 5).So that's 3 * 2 = 6 ways to place the National Games ads.Then, for each of these 6 ways, the remaining 3 positions (the ones not occupied by National Games ads) need to be filled with the 3 distinct commercials. Since there are 3! = 6 permutations for the commercials, the total number of sequences would be 6 * 6 = 36.Wait, but let me confirm this. Let me take an example. Suppose the first National Games ad is in position 1. Then the National Games ads are in positions 1 and 5. The remaining positions 2, 3, 4 need to be filled with commercials C1, C2, C3. Each of these can be arranged in 3! = 6 ways. Similarly, if the first National Games ad is in position 2, then positions 2 and 5 have National Games, and the remaining positions 1, 3, 4 have commercials. Similarly for position 3. So each of the 3 positions for the first National Games ad leads to 2 * 6 = 12 sequences? Wait, no. Wait, for each position (1, 2, 3), there are 2 choices for which National Games ad is there, and then 6 arrangements for the commercials. So per position, 2 * 6 = 12. Then 3 positions: 3 * 12 = 36. Yes, that seems correct.But wait, let me check if there's any overlap or overcounting. For example, if National Games ads are placed in position 1 and 5, and another arrangement where they are placed in position 2 and 5, these are separate cases. So no overlap. Each case is unique.But another way to think about it is:First, choose the position for the first National Games ad (positions 1, 2, 3). For each such position, assign one of the two National Games ads to that position and the other to position 5. Then, arrange the three commercials in the remaining three positions. That would indeed be 3 positions * 2 permutations of National Games * 6 permutations of commercials = 36. So that's 36 possible sequences.But wait, is there another constraint? The problem says "the two National Games adverts cannot be played consecutively". So even if they are not in consecutive positions, except if one is in position 4 and 5. Wait, but position 4 is already a commercial because we considered that if the first National Games is in positions 1, 2, or 3, then the remaining positions (including position 4) are commercials. So position 4 is always a commercial in this calculation, right?Wait, hold on. Wait, if the first National Games ad is in position 3, then the positions are 3 and 5. So positions 4 is in between, which is a commercial. So even between 3 and 5, there's a commercial in position 4. So they are separated. Similarly, if the first National Games is in position 2, then positions 2 and 5, with positions 3 and 4 being commercials. So in all cases, the two National Games ads are separated by at least one commercial. So the constraint is satisfied. And if the first National Games is in position 1, then positions 1 and 5, with positions 2, 3, 4 as commercials. So again, no consecutive National Games ads.So this approach seems to satisfy the non-consecutive National Games ads requirement.Therefore, the total number of sequences is 3 * 2 * 6 = 36. So 36 different broadcasting sequences.But wait, let me check this again with another approach to verify.Alternative approach: Total number of sequences where the last ad is National Games, and the two National Games ads are not consecutive.First, fix position 5 as a National Games ad. There are 2 choices for this.Then, we need to place the other National Games ad in positions 1 to 4, but not in position 4 (since that would make positions 4 and 5 both National Games, which are consecutive). So the other National Games ad can be in positions 1, 2, or 3. So 3 choices.So for the National Games ads: 2 choices for position 5, and 3 positions for the other National Games ad, and then 2 ways to assign the two National Games ads (since they are different). Wait, maybe.Wait, if we fix position 5 first: 2 choices (NG1 or NG2). Then, the other National Games ad (either NG2 or NG1) has to go into positions 1, 2, or 3. So that's 3 positions. So for each of the 2 choices in position 5, we have 3 positions for the other National Games ad, and since the ads are different, once position 5 is chosen, the other ad is fixed. For example, if position 5 is NG1, then the other National Games ad must be NG2 in positions 1, 2, or 3. Similarly, if position 5 is NG2, the other is NG1 in positions 1, 2, or 3. So total National Games arrangements: 2 (for position 5) * 3 (positions 1-3) = 6. Then, the remaining 3 positions are commercials, which are 3 distinct commercials, so 3! = 6. Therefore total sequences: 6 * 6 = 36. Same result as before. So that seems to confirm.Another way: Use permutations with restrictions.Total number of sequences where last ad is National Games: First, choose the ad for position 5: 2 options. Then, arrange the remaining 4 ads (3 commercials and 1 National Games) in positions 1-4, with the condition that the National Games ad in positions 1-4 cannot be adjacent to position 5. Wait, but position 4 is adjacent to position 5, so the National Games ad cannot be in position 4. So it's similar to the previous reasoning. So the National Games ad in positions 1-4 must be in positions 1, 2, or 3. So for each choice of National Games in position 5 (2 options), we have 3 positions for the other National Games ad. Then, the commercials are arranged in the remaining 3 spots. So 2 * 3 * 3! = 2 * 3 * 6 = 36. Same answer.Alternatively, let's use inclusion-exclusion. First, calculate all sequences where the last ad is National Games, then subtract those where the two National Games ads are consecutive.Total sequences with last ad National Games: Choose the last ad (2 options). Then arrange the remaining 4 ads (3 commercials and 1 National Games) in the first 4 positions. The number of arrangements is 4! / 1! (since all are distinct). Wait, but the commercials are 3 distinct and the National Games is 1 distinct. So total permutations: 4! = 24. So total sequences: 2 * 24 = 48.But now, from these 48 sequences, we need to subtract those where the two National Games ads are consecutive.How many sequences have the two National Games ads consecutive? If the last ad is National Games, and the previous ad (position 4) is also National Games. But wait, position 5 is fixed as National Games. So the only way the two National Games ads are consecutive is if position 4 is also National Games. But in our problem, the other National Games ad is in positions 1-4, but if it's in position 4, then it's consecutive to position 5.So the number of sequences where National Games ads are consecutive is equal to the number of sequences where position 4 is National Games and position 5 is National Games.So, how many such sequences are there?First, choose the ad for position 5: 2 options. Then, position 4 must be the other National Games ad. So once position 5 is chosen (say NG1), position 4 must be NG2. Then, the remaining 3 positions (1-3) must be filled with the 3 commercials. The number of permutations is 3! = 6. So for each choice of position 5, we have 1 choice for position 4 (the other National Games ad) and 6 permutations for the commercials. Therefore, total bad sequences: 2 * 1 * 6 = 12.Therefore, total valid sequences = total sequences with last ad National Games (48) minus bad sequences where National Games are consecutive (12) = 36. So same result.Therefore, the answer should be 36.But let me just think if there's another angle. For example, treating the problem as arranging the commercials first and then placing the National Games ads in the available slots.But given the constraints, maybe that's more complicated. Let's see.If we first arrange the commercials, then place the National Games ads in the gaps.But since the last position must be a National Games ad, perhaps this approach would need adjustment.Alternatively, since position 5 is fixed as a National Games ad, perhaps we can think of arranging the other ads first.So fix position 5 as NG. Then, we have to place the other NG ad in positions 1-4, not adjacent to position 5 (i.e., not in position 4). So positions 1, 2, 3. Then arrange the commercials in the remaining positions.Alternatively, arranging the commercials first. Let's try that.Suppose we first place the three commercials. The commercials are C1, C2, C3. They need to occupy three positions among 1-4, because position 5 is already taken by NG.Wait, no. Wait, position 5 is NG, so the remaining four positions (1-4) must contain three commercials and one NG. But the NG in positions 1-4 cannot be in position 4 (as it would be adjacent to position 5). So the NG must be in positions 1-3. Therefore, in positions 1-4, we have three commercials and one NG in positions 1-3.So first, arrange the three commercials and one NG in positions 1-4, with the NG in positions 1-3.The number of ways to choose the position for the NG in 1-3: 3 choices.Once the position is chosen, assign the NG ad (since there are two different NGs, but one is already used in position 5). Wait, hold on. Wait, position 5 is fixed as one NG, and the other NG is in positions 1-3. Since the NGs are different, the NG in positions 1-3 must be the other one. So if position 5 is NG1, then positions 1-3 must be NG2, and vice versa.Wait, no. Wait, if position 5 is NG1, then the other NG (NG2) must be in positions 1-3. Similarly, if position 5 is NG2, the other NG (NG1) must be in positions 1-3. Therefore, once we fix position 5 as either NG1 or NG2, the other NG is determined.Therefore, the number of ways is: choose position 5 (2 options), choose position for the other NG in 1-3 (3 options), arrange the three commercials in the remaining three positions (3! = 6). Therefore, total sequences: 2 * 3 * 6 = 36. Same result again.Therefore, all approaches lead to 36. So I think that's the correct answer.But let me just try one more way for thoroughness.Suppose we use the multiplication principle step by step.Step 1: Decide the National Games ad for position 5. There are 2 choices.Step 2: Decide where to place the other National Games ad. It can't be in position 4, so it must be in positions 1, 2, or 3. That's 3 choices.Step 3: Assign the specific National Games ad to that position. Since one is already used in position 5, the other one is fixed. So for each choice in Step 1 and Step 2, there's only 1 choice for the National Games ad in the selected position.Wait, but in Step 1, we already chose which National Games ad is in position 5. Then, the other National Games ad is uniquely determined. So for example, if we choose NG1 for position 5, then the other National Games ad must be NG2, placed in positions 1, 2, or 3. Similarly, if we choose NG2 for position 5, the other must be NG1. Therefore, once Step 1 is done, the ad in the other position is fixed. Therefore, in Step 2, when choosing the position, there is no additional choice for the ad, just placement.Therefore, the number of ways to place the National Games ads is 2 (choices for position 5) * 3 (positions for the other NG ad) = 6.Step 4: Arrange the three commercials in the remaining three positions. Since the commercials are distinct, this is 3! = 6 permutations.Therefore, total sequences: 6 * 6 = 36.Yep, same answer. So I think it's solid.Another angle: Let's think of the two National Games ads as A and B. The last position must be either A or B. Suppose it's A. Then the other National Games ad B must be placed in positions 1, 2, or 3. For each of these, the three commercials can be arranged in 3! ways. Similarly, if the last position is B, then the other National Games ad A must be in positions 1, 2, or 3. So total sequences: 2 * 3 * 3! = 2 * 3 * 6 = 36.All right, so after multiple approaches and verifications, I think 36 is the correct answer. I don't see any oversight in the reasoning. Each step accounts for the constraints: last position must be National Games, the two National Games cannot be consecutive (so the other one can't be in position 4), and all ads are distinct, so permutations are considered accordingly.Final AnswerThe number of different broadcasting sequences is boxed{36}.</think>The television station needs to broadcast 5 advertisements, including 3 different commercial ads and 2 different National Games promotional ads. The constraints are that the last advertisement must be a National Games ad, and the two National Games ads cannot be played consecutively.1. Position 5 (last advertisement): There are 2 choices for the National Games ad (NG1 or NG2).2. Position for the other National Games ad: This must be in positions 1, 2, or 3 to avoid consecutive placement with position 5. There are 3 choices for this position.3. Arranging the National Games ads: For each choice of position 5, the other National Games ad is fixed (since they are distinct). Therefore, there are 2 (choices for position 5) * 3 (positions for the other National Games ad) = 6 ways to place the National Games ads.4. Arranging the commercials: The remaining 3 positions must be filled with the 3 distinct commercials, which can be arranged in 3! = 6 ways.Multiplying these together, we get:[ 2 text{ (choices for position 5)} times 3 text{ (positions for the other National Games ad)} times 3! text{ (arrangements of commercials)} = 2 times 3 times 6 = 36 ]Thus, the number of different broadcasting sequences is (boxed{36}).]

🔑:As the problem is fundamentally about counting the number of valid sequences, we should consider the given conditions one step at a time. According to the conditions:1. The last advertisement must be a National Games promotional ad. This condition can be satisfied in 2 ways, since there are 2 different National Games adverts to choose from.2. Since the 2 National Games adverts cannot be placed consecutively, the remaining National Games advert must be placed in one of the first 3 slots. There are 3 choices for where to place this advert.3. The remaining 3 commercial advertisements can then be arranged in the remaining 3 positions without any additional restrictions, which allows for 3! (factorial) permutations.Combining these steps, the total number of different broadcasting sequences is determined by the product of the number of choices in each step. Hence, the calculation will be as follows:text{Total number of sequences} = 2 times C^1_3 times 3! = 2 times 3 times 6 = boxed{36}.

🧠:[<think>Okay, let's see. I need to arrange the numbers 1 through 6 in a row such that:1. The first number isn't 1 (a₁ ≠ 1)2. The third number isn't 3 (a₃ ≠ 3)3. The fifth number isn't 5 (a₅ ≠ 5)4. The first number is less than the third number, which is less than the fifth number (a₁ < a₃ < a₅)And I need to find how many such arrangements are possible. Hmm. Let me break this down step by step.First, normally, without any restrictions, there are 6! = 720 permutations of the numbers 1 to 6. But here, we have several constraints. Let's tackle them one by one.The first constraint: a₁ ≠ 1. So, the first position can be any number from 2 to 6. That gives 5 choices instead of 6 for the first position.Second constraint: a₃ ≠ 3. So, the third position can't be 3. But we also have the third number must be greater than the first number. So, this might interact with the third constraint.Third constraint: a₅ ≠ 5. The fifth position can't be 5. Also, the fifth number must be greater than the third number. So, similar to the previous, this is part of the ordering a₁ < a₃ < a₅.Fourth constraint: a₁ < a₃ < a₅. So, the first, third, and fifth positions must form an increasing sequence. But each of these positions also has their own restrictions (a₁ ≠ 1, a₃ ≠ 3, a₅ ≠ 5).So, maybe the way to approach this is to first choose the numbers for positions 1, 3, 5 such that they satisfy a₁ < a₃ < a₅, and a₁ ≠ 1, a₃ ≠ 3, a₅ ≠ 5. Then, arrange the remaining numbers in the remaining positions (2, 4, 6) without any restrictions.Therefore, the total number of arrangements would be equal to the number of valid triples (a₁, a₃, a₅) multiplied by the number of permutations of the remaining 3 numbers in the other positions.So, first, let's find the number of valid triples (a₁, a₃, a₅) where a₁ < a₃ < a₅, a₁ ≠ 1, a₃ ≠ 3, a₅ ≠ 5.To do this, let's consider all possible triples from the numbers 1 to 6 where a₁ < a₃ < a₅, and then subtract those triples that violate the conditions a₁ ≠ 1, a₃ ≠ 3, or a₅ ≠ 5.But maybe inclusion-exclusion principle would work here. Alternatively, maybe it's simpler to count directly the number of valid triples by considering possible choices.First, let's note that a₁, a₃, a₅ are three distinct numbers since they are in increasing order. So, the total number of triples without any restrictions (other than increasing) is C(6, 3) = 20. But we need to subtract those triples where a₁ = 1, a₃ = 3, or a₅ = 5.But actually, the constraints are:- a₁ ≠ 1- a₃ ≠ 3- a₅ ≠ 5So, we need the number of increasing triples (a₁, a₃, a₅) where none of the three positions have those forbidden numbers. Let's compute this.First, compute total number of increasing triples: C(6,3) = 20.Now, subtract the triples where a₁ = 1. How many such triples are there?If a₁ = 1, then a₃ and a₅ must be from the remaining numbers 2,3,4,5,6 with a₃ < a₅. So, the number of such triples is C(5,2) = 10. But wait, in addition, even if a₁ =1, we need to check if other constraints are satisfied? Wait no. The constraints are a₁ ≠1, a₃ ≠3, a₅ ≠5. So when we subtract, we subtract all triples where a₁=1 regardless of other conditions. Similarly for the others.But actually, we need to compute the number of triples that satisfy a₁ ≠1, a₃≠3, a₅≠5. So, using inclusion-exclusion:Let A be the set of triples with a₁=1, B with a₃=3, C with a₅=5.We need to compute |total| - |A ∪ B ∪ C|.By inclusion-exclusion:|A ∪ B ∪ C| = |A| + |B| + |C| - |A∩B| - |A∩C| - |B∩C| + |A∩B∩C|So, let's compute each term.First, |A|: number of triples where a₁=1. As above, that's C(5,2)=10.|B|: number of triples where a₃=3. So, a₁ < 3 < a₅. So a₁ can be 1 or 2, and a₅ can be 4,5,6. So, the number of triples where a₃=3 is the number of possible a₁ and a₅ such that a₁ <3 <a₅. So a₁ can be 1 or 2 (2 choices), a₅ can be 4,5,6 (3 choices). So total is 2*3=6.Wait, but in an increasing triple (a₁, a₃, a₅), if a₃=3, then a₁ can be 1 or 2, and a₅ can be 4,5,6. So, yes, 2*3=6.Similarly, |C|: number of triples where a₅=5. Then a₁ < a₃ <5. So a₁ and a₃ can be any numbers from 1-4 with a₁ <a₃ <5. The number of such triples is C(4,2)=6.Wait, because if a₅=5, then a₁ and a₃ are selected from 1,2,3,4 with a₁ <a₃. So that's 4 choose 2, which is 6. So |C|=6.Now, |A∩B|: triples where a₁=1 and a₃=3. Then a₅ must be greater than 3, so 4,5,6. So 3 triples.Similarly, |A∩C|: triples where a₁=1 and a₅=5. Then a₃ must be between 1 and 5, so a₃ can be 2,3,4. So the number is 3 triples: (1,2,5), (1,3,5), (1,4,5). Wait, but a₃ cannot be 3 because of the original problem's constraints? Wait no, in this case, we are computing |A∩C|, which is triples where a₁=1 and a₅=5. The a₃ can be 2,3,4. However, in the original problem, a₃ ≠3 is a separate constraint, but here in inclusion-exclusion, we are counting the forbidden triples (those that have a₁=1, a₃=3, a₅=5). Wait, no. Wait, inclusion-exclusion here is to count the number of triples that have at least one of the forbidden positions. So, when we compute |A∩C|, it's triples where a₁=1 AND a₅=5. These triples might still have a₃=3 or not. Wait, but in the set A∩C, a₃ can be any number as long as a₁=1 and a₅=5. So, for example, in this case, the triple is (1, x, 5) where x is between 1 and 5, and 1 < x <5. So x can be 2,3,4. Therefore, |A∩C|=3.Similarly, |B∩C|: triples where a₃=3 and a₅=5. Then a₁ must be less than 3, so a₁ can be 1 or 2. So there are 2 triples: (1,3,5) and (2,3,5). So |B∩C|=2.Lastly, |A∩B∩C|: triples where a₁=1, a₃=3, a₅=5. There's only one such triple: (1,3,5). So |A∩B∩C|=1.Therefore, applying inclusion-exclusion:|A ∪ B ∪ C| = |A| + |B| + |C| - |A∩B| - |A∩C| - |B∩C| + |A∩B∩C| = 10 +6 +6 -3 -3 -2 +1 = 10+6+6=22; 22 -3-3-2=22-8=14; 14 +1=15.So |A ∪ B ∪ C|=15. Therefore, the number of valid triples is total triples (20) minus forbidden triples (15) = 5? Wait, that can't be, because 20-15=5. But that seems very low. Let me check the calculations again.Wait, maybe I made an error in the values.Let's recompute:|A| = 10 (triples with a₁=1)|B| = 6 (triples with a₃=3)|C| =6 (triples with a₅=5)|A∩B| =3 (triples with a₁=1, a₃=3)|A∩C| =3 (triples with a₁=1, a₅=5)|B∩C|=2 (triples with a₃=3, a₅=5)|A∩B∩C|=1 (triple (1,3,5))Therefore,|A ∪ B ∪ C| = 10 +6 +6 -3 -3 -2 +1 = (10+6+6) =22; ( -3-3-2)= -8; 22 -8=14; 14 +1=15. So yes, 15 forbidden triples.Total triples:20. So valid triples:20 -15=5. So only 5 valid triples? Let's list them to confirm.The valid triples must have a₁ ≠1, a₃ ≠3, a₅≠5, and a₁ <a₃ <a₅.Let me list all possible triples and check.The numbers are 1,2,3,4,5,6. So possible increasing triples:1. (1,2,3): invalid (a₁=1)2. (1,2,4): invalid (a₁=1)3. (1,2,5): invalid (a₁=1)4. (1,2,6): invalid (a₁=1)5. (1,3,4): invalid (a₁=1, a₃=3)6. (1,3,5): invalid (a₁=1, a₃=3, a₅=5)7. (1,3,6): invalid (a₁=1, a₃=3)8. (1,4,5): invalid (a₁=1)9. (1,4,6): invalid (a₁=1)10. (1,5,6): invalid (a₁=1)11. (2,3,4): invalid (a₃=3)12. (2,3,5): invalid (a₃=3, a₅=5)13. (2,3,6): invalid (a₃=3)14. (2,4,5): invalid (a₅=5)15. (2,4,6): valid? Let's see: a₁=2≠1, a₃=4≠3, a₅=6≠5. Yes, this is valid.16. (2,5,6): a₅=5 invalid17. (3,4,5): a₅=5 invalid18. (3,4,6): a₃=3 invalid19. (3,5,6): a₃=3 and a₅=5 invalid20. (4,5,6): a₅=5 invalidWait, so only (2,4,6) is valid? But according to inclusion-exclusion, we had 5 valid triples. That contradicts the enumeration here. So, where is the mistake?Wait, hold on. When I listed all 20 triples, only (2,4,6) is valid? But that can't be. Let me check again.Wait, for example:Triple (2,3,4): a₃=3 invalid.Triple (2,3,5): a₃=3 and a₅=5 invalid.Triple (2,3,6): a₃=3 invalid.Triple (2,4,5): a₅=5 invalid.Triple (2,4,6): valid.Triple (2,5,6): a₅=5 invalid.Triple (3,4,5): a₃=3 and a₅=5 invalid.Triple (3,4,6): a₃=3 invalid.Triple (3,5,6): a₃=3 and a₅=5 invalid.Triple (4,5,6): a₅=5 invalid.Wait, so only one valid triple? But according to inclusion-exclusion, we had 5. That seems like a problem.Wait, so clearly there is a mistake here. Let me check my inclusion-exclusion again.Wait, maybe the mistake is in the definition. When we compute the forbidden triples via inclusion-exclusion, we need to take into account that even if a triple has a₁=1, it's forbidden regardless of other constraints. Similarly for a₃=3 and a₅=5. But when we subtract these forbidden triples, perhaps the inclusion-exclusion counts all triples that have a₁=1 OR a₃=3 OR a₅=5. But when I listed all the triples, only (2,4,6) is left. So why does inclusion-exclusion say 20 -15=5? There must be a mistake in the inclusion-exclusion calculation.Wait, let's recount |A|, |B|, |C|.|A|: number of triples with a₁=1. That's C(5,2)=10. Correct.|B|: number of triples with a₃=3. As computed before, 2*3=6. Correct.|C|: number of triples with a₅=5. C(4,2)=6. Correct.Then the intersections:|A∩B|: triples with a₁=1 and a₃=3. Then a₅ must be >3, so 4,5,6. So 3 triples. Correct.|A∩C|: triples with a₁=1 and a₅=5. Then a₃ must be 2,3,4. So 3 triples. Correct.|B∩C|: triples with a₃=3 and a₅=5. Then a₁ must be 1 or 2. So 2 triples. Correct.|A∩B∩C|: the triple (1,3,5). Correct.Thus, inclusion-exclusion formula gives 10+6+6 -3 -3 -2 +1=15 forbidden triples. Total triples 20, so 20-15=5. But manual counting shows only 1 valid triple. So there's a contradiction here.Therefore, my mistake must be in the inclusion-exclusion reasoning.Wait, maybe when considering the triples where a₃=3, the numbers a₁ and a₅ are not just any numbers but must be part of an increasing triple.But earlier, when calculating |B| as 6, that was correct. Similarly for |C| as 6.Wait, but perhaps when we list all the triples, even if a₅=5, for example, but a₃=3, which is forbidden, but in the inclusion-exclusion we subtract all triples where a₅=5, regardless of a₃. Wait, but when we subtract |C|, which is triples where a₅=5, but in those triples, a₃ might be 3 or not. Similarly, overlapping with other forbidden positions.But according to the manual count, the only valid triple is (2,4,6). Let's check why inclusion-exclusion gives 5.Wait, perhaps the problem is that the forbidden conditions are not just the positions but the numbers in those positions. So, for example, in the original problem, even if a₃ is 3, but if the position is not 3, but in our case, the position is fixed. Wait, no. The problem defines a_i as the i-th number. So:- The first number cannot be 1.- The third number cannot be 3.- The fifth number cannot be 5.So, in the triples (a₁, a₃, a₅), the first element (a₁) cannot be 1, the second element (a₃) cannot be 3, and the third element (a₅) cannot be 5. So, when we count the forbidden triples, it's triples where the first element is 1, OR the second element is 3, OR the third element is 5. So, for example, a triple like (2,3,6) is forbidden because the second element is 3. Similarly, (2,4,5) is forbidden because the third element is 5.So, when we list all triples:From the list above:Valid triples must have:- First element ≠1,- Second element ≠3,- Third element ≠5,and elements in increasing order.Looking at all 20 triples:1. (1,2,3): invalid (a₁=1, a₅=3? Wait no, the triple is (1,2,3), so a₁=1, a₃=2, a₅=3. Wait, in this case, a₁=1 invalid, a₃=2 is okay, a₅=3≠5. So invalid because a₁=1.Wait, no, the original problem says a₁ is the first number, a₃ is the third number, a₅ is the fifth number, but in the permutation, the positions are 1,3,5. So, when we are considering the triple (a₁, a₃, a₅), these are the numbers in positions 1, 3, 5 of the permutation. So, in the triple (1,2,3), that would mean a₁=1 (invalid), a₃=2 (valid, since a₃≠3), a₅=3 (valid, since a₅≠5). But since a₁=1, this triple is invalid.Similarly, triple (2,3,4) would have a₃=3, which is invalid.Triple (2,4,5) has a₅=5, which is invalid.Triple (2,4,6): all valid.Triple (3,4,6): a₁=3, which is allowed (since a₁≠1), a₃=4≠3, a₅=6≠5. So, this is also valid. Wait, in my previous manual count, I thought (3,4,6) was invalid because a₃=3. Wait, no, in this triple, the third number is 4, which is not 3. Wait, no, in the triple (3,4,6), a₁=3, a₃=4, a₅=6. So a₁≠1, a₃≠3, a₅≠5. So this is valid. But in my earlier list, when I listed all triples, I considered (3,4,6) as invalid because a₃=3? Wait, no. Wait, the original list was:19. (3,5,6): a₃=3 and a₅=5 invalidWait, but (3,4,6) is number 18. Wait, in my earlier list:18. (3,4,6): a₃=3 invalidWait, no. Wait, in the triple (3,4,6), the positions 1,3,5 are 3,4,6. So, the third position is 4, which is not 3. So, a₃=4≠3, so it's okay. So why did I mark it as invalid? That was my mistake.Similarly, let's go back through the list:11. (2,3,4): a₃=3 invalid12. (2,3,5): a₃=3 invalid, a₅=5 invalid13. (2,3,6): a₃=3 invalid14. (2,4,5): a₅=5 invalid15. (2,4,6): valid16. (2,5,6): a₅=5 invalid17. (3,4,5): a₅=5 invalid18. (3,4,6): valid (a₁=3≠1, a₃=4≠3, a₅=6≠5)19. (3,5,6): a₅=5 invalid20. (4,5,6): a₅=5 invalidSo actually, there are two valid triples: (2,4,6) and (3,4,6). Wait, but that's still not 5. Hmm. Wait, let's check more carefully.Wait, the triples are ordered as (a₁, a₃, a₅). So, in the original list:Triple 15: (2,4,6) – valid.Triple 18: (3,4,6) – valid.Triple 19: (3,5,6) – invalid (a₅=5)Triple 20: (4,5,6) – invalid (a₅=5)Wait, but where are the other triples?Wait, let's check triples where a₁=2, a₃=5, but a₅=6: (2,5,6). But a₅=5 is invalid, but here a₅=6, so actually, (2,5,6) would be a₅=6≠5. Wait, no, the triple (2,5,6): a₁=2, a₃=5, a₅=6. So a₃=5, which is allowed because the constraint is a₃≠3. So a₃ can be anything except 3. So (2,5,6): a₁=2≠1, a₃=5≠3, a₅=6≠5. Therefore, this is valid. But in my previous count, I considered (2,5,6) as invalid. Wait, no. Wait, the triple (2,5,6) has a₅=6, which is not 5, so it's okay. So that's another valid triple.Wait, so let's check:Triple (2,5,6): a₁=2≠1, a₃=5≠3, a₅=6≠5. So valid. So that's another valid triple.Similarly, let's check others:Triple (3,5,6): a₃=5≠3, a₅=6≠5, a₁=3≠1. So this is valid? Wait, (3,5,6): a₁=3≠1, a₃=5≠3, a₅=6≠5. So that's valid.But in the original list, I thought (3,5,6) was invalid because a₅=5, but a₅ is 6. Wait, that was my mistake. So (3,5,6) is valid.Similarly, triple (4,5,6) is invalid because a₅=5, but in (3,5,6), a₅=6. So (3,5,6) is valid.So let's check all triples again:1. (1,2,3): invalid (a₁=1)2. (1,2,4): invalid (a₁=1)3. (1,2,5): invalid (a₁=1)4. (1,2,6): invalid (a₁=1)5. (1,3,4): invalid (a₁=1, a₃=3)6. (1,3,5): invalid (a₁=1, a₃=3, a₅=5)7. (1,3,6): invalid (a₁=1, a₃=3)8. (1,4,5): invalid (a₁=1)9. (1,4,6): invalid (a₁=1)10. (1,5,6): invalid (a₁=1)11. (2,3,4): invalid (a₃=3)12. (2,3,5): invalid (a₃=3, a₅=5)13. (2,3,6): invalid (a₃=3)14. (2,4,5): invalid (a₅=5)15. (2,4,6): valid16. (2,5,6): valid (a₁=2, a₃=5, a₅=6)17. (3,4,5): invalid (a₅=5)18. (3,4,6): valid19. (3,5,6): valid (a₁=3, a₃=5, a₅=6)20. (4,5,6): invalid (a₅=5)So actually, the valid triples are numbers 15, 16, 18, 19. That's four. But inclusion-exclusion gave 5. Wait, so where is the fifth one?Wait, let's check again. Did I miss any triple?Triple (3,4,6): valid.Triple (2,4,6): valid.Triple (2,5,6): valid.Triple (3,5,6): valid.Is there another one? Let's see:Wait, what about (4,5,6)? Invalid. (3,4,5)? Invalid. (2,4,5)? Invalid. (3,6, something)? Wait, the triples are strictly increasing.Wait, maybe (4,5,6) is invalid. How about (2,6, something)? Wait, no. Wait, the triples are 3 elements: a₁, a₃, a₅. So in the list above, numbers 15,16,18,19 are valid. So that's four. But inclusion-exclusion said 5. Hmm. So discrepancy.Wait, maybe my inclusion-exclusion was wrong. Let's recalculate:Total triples:20Forbidden triples:|A ∪ B ∪ C|=15. Therefore, 20-15=5. But manual count gives 4. So, where's the mistake?Wait, maybe when I considered |A|, |B|, |C|, but in reality, some triples are being subtracted multiple times, but maybe the counts are incorrect.Wait, let's try to list the forbidden triples according to the inclusion-exclusion.Forbidden triples are those with a₁=1 OR a₃=3 OR a₅=5.Total forbidden triples:15. Let's list them:From the 20 triples, the forbidden ones are:1. (1,2,3): A2. (1,2,4): A3. (1,2,5): A and C4. (1,2,6): A5. (1,3,4): A and B6. (1,3,5): A and B and C7. (1,3,6): A and B8. (1,4,5): A and C9. (1,4,6): A10. (1,5,6): A11. (2,3,4): B12. (2,3,5): B and C13. (2,3,6): B14. (2,4,5): C15. (3,4,5): C16. (3,5,6): ?Wait, hold on, the total forbidden triples according to inclusion-exclusion are 15. But in reality, from the list above, triples 1-10 are A-related, 11-13 are B-related, 14-15 are C-related. But how many is that? Let's count:Triple numbers 1-10: 10 triples (all A)Triple numbers 11-13: 3 triples (B)Triple numbers 14-15: 2 triples (C)Total forbidden triples:10+3+2=15. But the actual forbidden triples as per the list are:Triples 1-10 (all with a₁=1), triples 11-13 (all with a₃=3), triples 14-15 (all with a₅=5). Wait, but triples 16-20 are:16. (2,5,6): valid17. (3,4,5): forbidden (C)18. (3,4,6): valid19. (3,5,6): valid20. (4,5,6): forbidden (C)Wait, but according to inclusion-exclusion, triples 14-15 are C, and triple 17 and 20 are also C. So, triples with a₅=5 are:3. (1,2,5)6. (1,3,5)8. (1,4,5)12. (2,3,5)14. (2,4,5)17. (3,4,5)20. (4,5,6) – a₅=5, so this is also forbidden.Wait, but in the initial listing, triple 20 is (4,5,6), which has a₅=5? Wait, no. Wait, the fifth element is 6, but the a₅=5 would be if the fifth element is 5. Wait, triple (4,5,6): a₁=4, a₃=5, a₅=6. So a₅=6≠5. So that's allowed. Wait, but in the original problem, the fifth position cannot be 5. So in the triple (4,5,6), a₅=6≠5, so it's allowed. So my mistake was in considering triple 20 as forbidden. So triple 20 is actually valid? Wait, no:Wait, triple (4,5,6): the numbers in positions 1,3,5 are 4,5,6. So, the fifth position is 6, which is not 5. So a₅=6≠5, so it's allowed. But in the original problem, the fifth position is a₅, which cannot be 5. So (4,5,6) is valid because a₅=6. But in the inclusion-exclusion, |C| is the number of triples where a₅=5. So (4,5,6) is not in |C|. Therefore, triples with a₅=5 are:Triple (1,2,5), (1,3,5), (1,4,5), (2,3,5), (2,4,5), (3,4,5). That's 6 triples. Which is |C|=6. So that's correct.So triple 17: (3,4,5): a₅=5, forbidden. So triple 17 is forbidden.Triple 20: (4,5,6): a₅=6≠5, so allowed. So triple 20 is valid. But in the manual count above, I considered it as invalid. So my mistake. So (4,5,6) is a valid triple:a₁=4≠1, a₃=5≠3, a₅=6≠5. So yes, it's valid. So actually, there are five valid triples:15. (2,4,6)16. (2,5,6)18. (3,4,6)19. (3,5,6)20. (4,5,6)Wait, but according to the numbering earlier, triple 20 is (4,5,6). Let's check:a₁=4≠1, a₃=5≠3, a₅=6≠5. So yes, valid. So that's the fifth valid triple. So inclusion-exclusion was correct.Earlier, when I listed the triples, I missed that (4,5,6) is valid. So that's the fifth one. So the valid triples are:(2,4,6), (2,5,6), (3,4,6), (3,5,6), (4,5,6). Five triples. So the inclusion-exclusion was correct.Therefore, the number of valid triples is 5.Therefore, for each of these 5 triples, the remaining numbers (those not in the triple) can be arranged freely in the remaining positions (positions 2,4,6). There are 3! =6 permutations for each triple. So total arrangements would be 5*6=30. Therefore, the answer is 30.But wait, hold on. Wait, in the problem statement, it says "Arrange the numbers 1,2,3,4,5,6 in a row...". So, we have 6 positions. The constraints are on positions 1,3,5 (a₁, a₃, a₅). So, once we fix the numbers in positions 1,3,5 as one of the valid triples, the remaining numbers can be placed in positions 2,4,6 in any order.Yes, so for each valid triple (a₁, a₃, a₅), the remaining three numbers can be arranged in 3! =6 ways. Therefore, total number of arrangements is 5*6=30.But wait, let me double-check with an example. Take the triple (2,4,6). The remaining numbers are 1,3,5. These can go in positions 2,4,6. So, for example: [2,1,4,3,6,5] or [2,3,4,5,6,1], etc. Each of these is a valid permutation because a₁=2≠1, a₃=4≠3, a₅=6≠5, and a₁ <a₃ <a₅.Similarly, for the triple (4,5,6), the remaining numbers are 1,2,3. These can be arranged in positions 2,4,6 in any order. So, for example, [4,1,5,2,6,3], which satisfies all conditions.Therefore, 5 triples, each with 6 permutations: 5*6=30.But let me cross-verify with another approach.Alternative approach:First, select three numbers for positions 1,3,5 such that they are in increasing order and satisfy a₁ ≠1, a₃≠3, a₅≠5.As we found, there are 5 such triples.Alternatively, think of choosing three numbers from the set {1,2,3,4,5,6} excluding the forbidden numbers in each position. Wait, but it's not straightforward because the forbidden numbers are per position, not per number.Wait, but since the triples must be increasing, and the forbidden numbers are 1 for position 1, 3 for position 3, 5 for position 5.So, perhaps we can model this as choosing three numbers, none of which are in the forbidden positions. Wait, no. For example, number 1 can't be in position 1, but can be in other positions. Similarly, number 3 can't be in position 3, but can be elsewhere. Number 5 can't be in position 5, but can be elsewhere.But in our case, since we are choosing a triple (a₁, a₃, a₅) which is increasing, with a₁ ≠1, a₃≠3, a₅≠5.So, maybe another way is:We need to choose three numbers for positions 1,3,5 such that:- The first number is not 1,- The third number is not 3,- The fifth number is not 5,- And they are in increasing order.Alternatively, the numbers selected for positions 1,3,5 must form an increasing sequence with the constraints on each position.But this seems similar to what we did before.Alternatively, we can think of generating all possible triples where the numbers are not the forbidden ones in their respective positions and are increasing.For example:- For position 1: available numbers {2,3,4,5,6}- For position 3: available numbers {1,2,4,5,6}- For position 5: available numbers {1,2,3,4,6}But since the triple must be increasing, we need to choose a₁ < a₃ < a₅ with a₁ from {2,3,4,5,6}, a₃ from {1,2,4,5,6}, a₅ from {1,2,3,4,6}, but a₁ < a₃ < a₅.But this is getting complicated. Maybe it's better to list the 5 valid triples and verify.We found:1. (2,4,6)2. (2,5,6)3. (3,4,6)4. (3,5,6)5. (4,5,6)Each of these triples satisfies:- a₁ ≠1: For example, in (4,5,6), a₁=4≠1.- a₃ ≠3: For example, in (2,5,6), a₃=5≠3.- a₅ ≠5: For example, in (3,5,6), a₅=6≠5.- And they are in increasing order.Yes, so these five triples are correct.Therefore, each of these five triples can be arranged in the first, third, fifth positions, and the remaining three numbers can be arranged in the other three positions in 3! ways. So total permutations:5 * 3! * 1 =5*6=30.Therefore, the answer should be 30.But wait, let me check if there's any restriction on the other positions. The problem says nothing about positions 2,4,6. They can be any numbers, right? So, as long as positions 1,3,5 satisfy the constraints, the rest can be anything. So yes, the remaining three numbers can be arranged freely.Therefore, 5 triples *6 permutations each=30. So the answer is 30.But just to make sure, let's check one of the triples and see how many arrangements it generates.Take the triple (2,4,6). The remaining numbers are 1,3,5. These can be placed in positions 2,4,6 in any order:Positions:1:22:1,3,53:44:1,3,55:66:1,3,5So, for example, permutation [2,1,4,3,6,5] is valid. Similarly, [2,3,4,5,6,1], [2,5,4,1,6,3], etc. Each of these is a unique permutation, and there are 6 of them. Similarly, for each of the five triples, we have 6 permutations. So total 5*6=30.Therefore, the number of different arrangements is 30. So the answer is 30, which in boxed form is boxed{30}.Final Answerboxed{30}</think>To solve the problem of arranging the numbers 1, 2, 3, 4, 5, 6 in a row such that (a_1 neq 1), (a_3 neq 3), (a_5 neq 5), and (a_1 < a_3 < a_5), we need to follow these steps:1. Identify valid triples ((a_1, a_3, a_5)): - The triples must be increasing ((a_1 < a_3 < a_5)). - (a_1 neq 1), (a_3 neq 3), (a_5 neq 5).2. Count valid triples using inclusion-exclusion principle: - Total number of increasing triples from 6 numbers: (binom{6}{3} = 20). - Subtract triples violating the constraints: - Triples with (a_1 = 1): (binom{5}{2} = 10). - Triples with (a_3 = 3): (2 times 3 = 6). - Triples with (a_5 = 5): (binom{4}{2} = 6). - Intersections of these sets: - (a_1 = 1) and (a_3 = 3): 3 triples. - (a_1 = 1) and (a_5 = 5): 3 triples. - (a_3 = 3) and (a_5 = 5): 2 triples. - (a_1 = 1), (a_3 = 3), (a_5 = 5): 1 triple. - Using inclusion-exclusion, the number of invalid triples is (10 + 6 + 6 - 3 - 3 - 2 + 1 = 15). - Valid triples: (20 - 15 = 5).3. Verify valid triples: - The valid triples are ((2, 4, 6)), ((2, 5, 6)), ((3, 4, 6)), ((3, 5, 6)), and ((4, 5, 6)).4. Calculate permutations: - Each valid triple leaves 3 remaining numbers, which can be arranged in (3!) ways. - Total permutations: (5 times 3! = 5 times 6 = 30).Thus, the number of different arrangements is (boxed{30}).,<think>Alright, let's see. So the problem is to arrange the numbers 1 through 6 in a row, right? And there are some conditions. Let me break them down. First, the constraints:1. a₁ ≠ 1. So the first number can't be 1.2. a₃ ≠ 3. The third number can't be 3.3. a₅ ≠ 5. The fifth number can't be 5.4. Also, we need a₁ < a₃ < a₅. So the first, third, and fifth positions have to be in increasing order, but they themselves can't be 1, 3, 5 respectively.Our goal is to find how many different arrangements satisfy all these conditions.Hmm. Let's think about how to approach this. Since we're dealing with permutations with certain restrictions and ordering constraints, maybe inclusion-exclusion could work here? Or perhaps breaking it down step by step.Wait, another thought: maybe first select the numbers for positions 1, 3, 5 such that they satisfy a₁ < a₃ < a₅ and none of them are 1, 3, 5 respectively. Then arrange the remaining numbers in the even positions. That might be a way to go.Let me structure the problem:Positions 1, 3, 5 (odd positions) need to have numbers not equal to 1, 3, 5 respectively. Also, the numbers in these positions must be strictly increasing: a₁ < a₃ < a₅.Once we choose these three numbers, the remaining numbers (which include 1, 3, 5 if they weren't chosen) go into positions 2, 4, 6, which can be arranged freely.So first, the key is to count how many valid triples (a₁, a₃, a₅) there are such that:- a₁ ≠ 1- a₃ ≠ 3- a₅ ≠ 5- a₁ < a₃ < a₅And then multiply by the number of permutations of the remaining three numbers in the even positions, which is 3! = 6.Therefore, total arrangements = number of valid triples * 6.So the main task is to compute the number of valid triples.Let me tackle that. First, note that the numbers 1, 2, 3, 4, 5, 6 are all distinct, so the triples (a₁, a₃, a₅) are strictly increasing, meaning they are combinations of 3 distinct numbers in increasing order. But each position has a forbidden number: a₁ ≠ 1, a₃ ≠ 3, a₅ ≠ 5.So we need to count all 3-element subsets of {1,2,3,4,5,6} that do not have 1 in the first position, 3 in the second position, or 5 in the third position when arranged in increasing order.Wait, but since the subset is arranged in increasing order, the first element is the smallest, the second is the middle, and the third is the largest. So in the context of the triple (a₁, a₃, a₅), since a₁ < a₃ < a₅, the first element of the triple is the smallest, then the middle, then the largest. So in this case, the original positions (1,3,5) correspond to the order of the elements.Therefore, when we pick a combination of three numbers, say {x, y, z} where x < y < z, then automatically a₁ = x, a₃ = y, a₅ = z. So the constraints become:- x ≠ 1 (since a₁ ≠ 1)- y ≠ 3 (since a₃ ≠ 3)- z ≠ 5 (since a₅ ≠ 5)So essentially, we need to count all 3-element subsets {x, y, z} with x < y < z, such that x ≠ 1, y ≠ 3, z ≠ 5.Therefore, the problem reduces to counting the number of such subsets. Then multiply by 6 (for the permutations of the remaining numbers in the even positions).So let's calculate the number of valid triples.First, total number of 3-element subsets from 6 elements is C(6,3) = 20.But we need to subtract those subsets that violate the constraints:1. Subsets where x = 1 (i.e., contain 1 as the smallest element)2. Subsets where y = 3 (i.e., contain 3 as the middle element)3. Subsets where z = 5 (i.e., contain 5 as the largest element)But we need to be careful with overlaps. For example, a subset could have both x=1 and y=3, so subtracting them separately would double-count. So inclusion-exclusion is necessary here.Let's define:A: subsets with x = 1 (first element is 1)B: subsets with y = 3 (middle element is 3)C: subsets with z = 5 (last element is 5)We need |A ∪ B ∪ C|, and subtract this from the total number of subsets to get the valid ones.By inclusion-exclusion:|A ∪ B ∪ C| = |A| + |B| + |C| - |A∩B| - |A∩C| - |B∩C| + |A∩B∩C|Then the number of valid subsets = C(6,3) - |A ∪ B ∪ C|So let's compute each term.First, |A|: number of subsets with x=1. Since x is the smallest element, which is 1. So the other two elements must be from {2,3,4,5,6}, and they must be larger than 1. So the number is C(5,2) = 10.Wait, but in the subset {1, a, b}, arranged in order, so the other two elements can be any two from the remaining 5, so yes, C(5,2)=10.|B|: subsets with y=3. The middle element is 3. So the smallest element must be from {1,2}, and the largest element must be from {4,5,6}. So number of such subsets is C(2,1)*C(3,1)=2*3=6.Wait, explanation: To form a subset where the middle element is 3, we need one element less than 3 and one element greater than 3. The elements less than 3 are {1,2}, so choose 1 from them, and elements greater than 3 are {4,5,6}, choose 1 from them. So 2*3=6. That's correct.|C|: subsets with z=5. The largest element is 5. So the other two elements must be from {1,2,3,4}, and they must be less than 5. So number of such subsets is C(4,2)=6.Because choosing any two elements from {1,2,3,4}, which are 4 elements, so C(4,2)=6.Now, |A∩B|: subsets where x=1 and y=3. So the three elements are {1,3,z}, where z >3. So z can be 4,5,6. Therefore, there are 3 subsets: {1,3,4}, {1,3,5}, {1,3,6}. So |A∩B|=3.Similarly, |A∩C|: subsets where x=1 and z=5. So the three elements are {1, y, 5} where 1 < y <5. So y can be 2,3,4. So subsets: {1,2,5}, {1,3,5}, {1,4,5}. So |A∩C|=3.|B∩C|: subsets where y=3 and z=5. So the middle element is 3 and the largest is 5. So the elements are {x,3,5} where x <3. So x can be 1 or 2. Therefore, subsets: {1,3,5}, {2,3,5}. So |B∩C|=2.Lastly, |A∩B∩C|: subsets where x=1, y=3, z=5. That's the subset {1,3,5}. Only one such subset. So |A∩B∩C|=1.Putting it all together:|A ∪ B ∪ C| = |A| + |B| + |C| - |A∩B| - |A∩C| - |B∩C| + |A∩B∩C|= 10 + 6 + 6 - 3 - 3 - 2 + 1= (10 + 6 + 6) - (3 + 3 + 2) + 1= 22 - 8 + 1= 15Therefore, the number of invalid subsets is 15. Total subsets are 20, so valid subsets are 20 -15=5.Wait, 20 -15=5? That seems low. Let me check the calculations again.Wait, |A| is 10, |B| 6, |C|6, so 10+6+6=22.Then |A∩B|=3, |A∩C|=3, |B∩C|=2, so sum of these is 3+3+2=8.Then |A∩B∩C|=1.Therefore, 22 -8 +1=15. So |A ∪ B ∪ C|=15.Total subsets=20, so valid subsets=20 -15=5.Hmm, so only 5 valid triples. Let me verify this because 5 seems small. Let's list them.The valid triples {x,y,z} must satisfy x≠1, y≠3, z≠5, and x < y < z.So x can be 2,3,4,5 but not 1. Wait, x is the first element, so x must be at least 2. But since the triple is increasing, x < y < z.But z can't be 5. So z can be 6 or 4, but z has to be the largest. Wait, if z can't be 5, then z must be 6 or 5 is excluded. Wait, z is the largest element, so if z is 5, then it's invalid. So z must be 6. Wait, no. Wait, z can be 4, 5, or 6. Wait, but z ≠5, so z can be 4 or 6? But z has to be the largest element. So if the triple includes numbers up to 6, z can be 4 only if all elements are <=4, but then z would be 4. However, z cannot be 5. So z must be either 4 or 6.Wait, this seems confusing. Let me think. If we have a triple where the largest element is 6, then z=6, which is allowed (since z≠5). If the largest element is 5, then z=5, which is invalid. If the largest element is 4, then z=4, which is allowed. So z can be 4 or 6.Similarly, x must not be 1, so x can be 2,3,4. But x has to be less than y, which is less than z.Wait, perhaps listing all possible triples is a good idea, even though it's tedious.Total possible triples (without any restrictions) are 20. Let's list the invalid ones, which are 15. Then the valid ones must be 5.Wait, but let's see:Invalid triples are those that have x=1, or y=3, or z=5.So first, all triples with x=1: these are the triples starting with 1. There are C(5,2)=10 such triples.Then triples with y=3. These are the triples where 3 is the middle element. There are 2*3=6 such triples: {1,3,4}, {1,3,5}, {1,3,6}, {2,3,4}, {2,3,5}, {2,3,6}.But note that some of these are already counted in the first set (the ones starting with 1). Specifically, {1,3,4}, {1,3,5}, {1,3,6} are in both A and B.Similarly, triples with z=5: these are triples ending with 5. There are C(4,2)=6 such triples: {1,2,5}, {1,3,5}, {1,4,5}, {2,3,5}, {2,4,5}, {3,4,5}.Again, some overlap with A and B.So when we subtract |A| + |B| + |C| =10 +6 +6=22, but then we have overlaps. The overlaps are:A∩B: triples starting with 1 and having 3 as the middle element: {1,3,4}, {1,3,5}, {1,3,6} → 3 triples.A∩C: triples starting with 1 and ending with 5: {1,2,5}, {1,3,5}, {1,4,5} → 3 triples.B∩C: triples with middle element 3 and ending with 5: {1,3,5}, {2,3,5} → 2 triples.A∩B∩C: {1,3,5} → 1 triple.Therefore, inclusion-exclusion gives:Total invalid = 10 +6 +6 -3 -3 -2 +1 =15.So invalid triples are 15. Thus, valid triples are 20 -15=5.So there are 5 valid triples. Let's list them to confirm.The valid triples must not have x=1, y=3, z=5, and must be increasing.So x can be 2,3,4,5; but since x < y < z, and y ≠3, z ≠5.Let me list all possible triples not containing 1 in x, 3 in y, or 5 in z.Possible triples:Start with x=2:- Then y can be 3,4,5,6, but y≠3. So y=4,5,6.But z must be > y and z≠5. So:If x=2, y=4:- z can be 5 or 6. But z≠5, so z=6. So triple {2,4,6}If x=2, y=5:- z must be 6 (since z>5 and z≠5). So triple {2,5,6}If x=2, y=6:- z must be >6, which doesn't exist. So no such triple.Wait, x=2, y=6 is invalid because z would have to be greater than 6, which isn't possible. So only {2,4,6} and {2,5,6} when x=2.Next, x=3:- Then y must be >3 and y≠3 (which is already satisfied). But y also can't be 3, but since x=3, y must be greater than 3. So y can be 4,5,6.But z must be >y and z≠5.So:x=3, y=4:- z can be 5 or 6. z≠5, so z=6. Triple {3,4,6}x=3, y=5:- z must be 6. So triple {3,5,6}x=3, y=6:- z must be >6, invalid.So triples {3,4,6}, {3,5,6}Next, x=4:- Then y must be >4. y can be 5,6.But z must be >y and z≠5. So:x=4, y=5:- z must be 6. So {4,5,6}x=4, y=6:- z must be >6, invalid.So only {4,5,6}x=5:- Wait, x has to be less than y and less than z. If x=5, then y and z must be >5, which is 6. But then y=6, z must be >6, which doesn't exist. So x=5 is invalid.x=6: Similarly impossible.So total valid triples:From x=2: {2,4,6}, {2,5,6}From x=3: {3,4,6}, {3,5,6}From x=4: {4,5,6}Total of 5 triples. Yep, that's 5. So that checks out.Therefore, the number of valid triples is 5.Thus, the total number of arrangements is 5 * 3! =5 *6=30.Wait, 5 triples, each can be combined with 3! permutations of the remaining numbers in positions 2,4,6. So 5*6=30. So is the answer 30?But wait, let me check again.Wait, for each valid triple {x,y,z}, the remaining numbers are the three numbers not in the triple. These can be arranged freely in positions 2,4,6. Since there are 3 numbers left, they can be permuted in 3! ways. So yes, 6 ways.Therefore, total arrangements are 5*6=30.But let's make sure that there are no additional constraints. The problem says "arrange the numbers 1,2,3,4,5,6 in a row" with the given conditions. The conditions are only on a₁, a₃, a₅. The other positions (2,4,6) can be anything, right? So as long as the numbers in positions 1,3,5 satisfy the constraints, the rest can be any permutation. Therefore, 3! for the remaining numbers.Therefore, the answer should be 30. Wait, but wait, let me cross-verify with another approach.Alternative approach:Alternatively, instead of counting valid triples, maybe model it step by step.But given that positions 1,3,5 need to be increasing, with a₁≠1, a₃≠3, a₅≠5.Alternatively, we can think of it as choosing three positions, with the constraints, and then permuting the rest.But maybe inclusion-exclusion on the permutations. However, since the positions 1,3,5 have to be in increasing order, it's similar to selecting a combination (since order is enforced) with restrictions.But I think the first approach is solid. The key was breaking it down into subsets and applying inclusion-exclusion. Since the answer via that method is 5 triples, leading to 30 total arrangements.But wait, let's see. Let's take one of the valid triples and check.Take the triple {2,4,6}. So a₁=2, a₃=4, a₅=6. Then the remaining numbers are {1,3,5}. These can be arranged in positions 2,4,6 as any permutation. So 3! =6. Similarly for the other triples. So each triple contributes 6 arrangements.So 5 triples *6=30. That seems correct.Alternatively, let's compute it as follows:First, choose a₁, which cannot be 1. Then choose a₃ >a₁, and a₃ ≠3. Then choose a₅ >a₃, and a₅≠5.But this step-by-step selection might be more complex, but let's attempt it.First, choose a₁: possible values are 2,3,4,5 (since a₁ ≠1). Wait, but a₁ has to be less than a₃ which is less than a₅, so a₁ can't be too large. For example, if a₁=5, then a₃ and a₅ would have to be larger than 5, but only 6 is left, which is impossible. So a₁ can be 2,3,4.Wait, hold on. Wait, a₁ can be 2,3,4,5 but considering that a₃ has to be greater than a₁ and not equal to 3, and a₅ has to be greater than a₃ and not equal to 5.So let's proceed step by step.Case 1: a₁=2Then a₃ must be greater than 2, and a₃≠3. So possible a₃:4,5,6.Then for each a₃, a₅ must be greater than a₃ and a₅≠5.Case 1.1: a₃=4Then a₅ can be 5 or 6. But a₅≠5, so a₅=6.Case 1.2: a₃=5a₅ must be greater than 5, which is 6. So a₅=6.Case 1.3: a₃=6Then a₅ must be greater than 6, which doesn't exist. So invalid.Thus, for a₁=2, possible (a₃,a₅) pairs: (4,6), (5,6). So two possibilities.Case 2: a₁=3Then a₃ must be greater than 3, and a₃≠3 (which is already satisfied). So a₃ can be 4,5,6.For each a₃:Case 2.1: a₃=4a₅ must be >4 and a₅≠5. So a₅=6.Case 2.2: a₃=5a₅ must be >5, which is 6.Case 2.3: a₃=6a₅ must be >6, invalid.Thus, for a₁=3, possible (a₃,a₅) pairs: (4,6), (5,6). So two possibilities.Case 3: a₁=4Then a₃ must be >4, so a₃=5,6.But a₃≠3 is already satisfied.Case 3.1: a₃=5Then a₅ must be >5 and ≠5. So a₅=6.Case 3.2: a₃=6a₅ must be >6, invalid.Thus, for a₁=4, only one possibility: (5,6).Case 4: a₁=5Then a₃ must be >5, which is 6. Then a₅ must be >6, invalid. So no possibilities.Thus, total number of triples:Case1:2, Case2:2, Case3:1, Case4:0. Total: 2+2+1=5. Which matches the previous count.Therefore, 5 triples. Then multiplied by 3! gives 5*6=30.Therefore, the answer should be 30.But let me check once more with actual enumeration.List all valid triples:From earlier:{2,4,6}, {2,5,6}, {3,4,6}, {3,5,6}, {4,5,6}That's 5 triples. For each, the remaining numbers are:For {2,4,6}: remaining numbers 1,3,5. Permute in positions 2,4,6: 6 ways.Similarly, {2,5,6}: remaining numbers 1,3,4.{3,4,6}: remaining numbers 1,2,5.{3,5,6}: remaining numbers 1,2,4.{4,5,6}: remaining numbers 1,2,3.Each of these gives 6 permutations. So 5*6=30 total arrangements.Therefore, the answer is 30. So I think that's correct.Final AnswerThe number of different arrangements is boxed{30}.</think>To solve the problem of arranging the numbers 1, 2, 3, 4, 5, 6 in a row with the given constraints, we need to consider the following steps:1. Identify the constraints: - (a_1 neq 1) - (a_3 neq 3) - (a_5 neq 5) - (a_1 < a_3 < a_5)2. Count valid triples for positions 1, 3, 5: - We need to count 3-element subsets ({x, y, z}) such that (x < y < z) and (x neq 1), (y neq 3), (z neq 5).3. Use inclusion-exclusion principle: - Calculate the total number of 3-element subsets: ( binom{6}{3} = 20 ). - Subtract subsets violating the constraints using inclusion-exclusion: - ( |A| ) (subsets with (x = 1)): ( binom{5}{2} = 10 ) - ( |B| ) (subsets with (y = 3)): ( 2 times 3 = 6 ) - ( |C| ) (subsets with (z = 5)): ( binom{4}{2} = 6 ) - Intersections: ( |A cap B| = 3 ), ( |A cap C| = 3 ), ( |B cap C| = 2 ), ( |A cap B cap C| = 1 ) - Total invalid subsets: ( 10 + 6 + 6 - 3 - 3 - 2 + 1 = 15 ) - Valid subsets: ( 20 - 15 = 5 )4. List valid triples: - Valid triples are ({2, 4, 6}), ({2, 5, 6}), ({3, 4, 6}), ({3, 5, 6}), ({4, 5, 6})5. Calculate permutations of remaining numbers: - Each valid triple can be combined with permutations of the remaining 3 numbers in positions 2, 4, 6: (3! = 6)6. Total number of arrangements: - Multiply the number of valid triples by permutations: (5 times 6 = 30)Thus, the number of different arrangements is (boxed{30}).]

🔑:This problem is a step-by-step counting problem.There are two steps:1. First, arrange a_1, a_3, a_5. - When a_1=2, there are 2 possibilities; - When a_1=3, there are 2 possibilities; - When a_1=4, there is 1 possibility; - In total, there are 5 possibilities.2. Next, arrange a_2, a_4, a_6. There are A_3^3=6 possibilities.Therefore, the total number of different arrangements is 5 times 6 = 30.Hence, the answer is boxed{30}.Analysis: This problem is a step-by-step counting problem. First, arrange a_1, a_3, a_5 for the cases when a_1=2, a_1=3, and a_1=4, calculating the number of outcomes for these cases. In the second step, arrange a_2, a_4, a_6, and calculate the number of outcomes. The result is obtained by applying the principle of step-by-step counting.

🧠:[<think>Alright, let me try to work through this problem step by step. The expression given is 81^(1/log₅3) + 27^(log₉36) + 3^(4/log₇9). Hmm, that looks a bit intimidating at first, but maybe I can break it down term by term. Let me start with the first term: 81^(1/log₅3). Okay, so 81 is 3^4, right? Let me confirm that: 3^4 = 81. Yep, that's correct. So maybe rewriting 81 as 3^4 could simplify things. Then the first term becomes (3^4)^(1/log₅3). Using exponent rules, that would be 3^(4/log₅3). Hmm. Now, the exponent here is 4 divided by log base 5 of 3. I remember there's a change of base formula for logarithms. The log base a of b can be written as log base c of b divided by log base c of a. So maybe I can use that to simplify the denominator here. Let me think. If I want to express log base 5 of 3 in terms of natural logarithms or logarithms with a different base, maybe base 3 or base 10? Let me try base 3. So log base 5 of 3 is equal to log base 3 of 3 divided by log base 3 of 5. Wait, that's 1 / log base 3 of 5. Because log base a of b is 1 / log base b of a. So log base 5 of 3 is 1 / log base 3 of 5. Therefore, 1 / log base 5 of 3 is just log base 3 of 5. So substituting back into the exponent, 4 / log₅3 = 4 * log base 3 of 5. So the first term is 3^(4 * log₃5). Now, 3 raised to a logarithm with base 3... There's a property where a^(log_a b) = b. So if I have 3^(log₃5), that's 5. But here the exponent is 4 times that. So 3^(4 * log₃5) is equal to (3^(log₃5))^4, which is 5^4. 5^4 is 625. So the first term simplifies to 625. Nice, that wasn't so bad. Moving on to the second term: 27^(log₉36). Let's see. 27 is 3^3, and 9 is 3^2. So maybe rewriting both in terms of base 3 would help. Let's rewrite 27 as 3^3 and 9 as 3^2. So the logarithm part is log base 9 of 36. Let me write that as log_{3^2}36. There's a logarithm identity that log_{a^n} b = (1/n) log_a b. So applying that here, log_{3^2}36 = (1/2) log₃36. Therefore, the exponent becomes (1/2) log₃36. So the second term is (3^3)^[(1/2) log₃36]. Let's simplify the exponent first. 3^3 raised to some power is 3^(3 * exponent). So that's 3^(3 * (1/2) log₃36) = 3^( (3/2) log₃36 ). Again, using the property that a^(log_a b) = b. But here we have a coefficient in front of the logarithm. Remember that c * log_a b = log_a (b^c). So 3^( (3/2) log₃36 ) = 3^( log₃36^(3/2) ) = 36^(3/2). Because 3 raised to the log base 3 of something is just that something. So 36^(3/2). Let's compute that. 36^(1/2) is 6, so 6^3 is 216. Therefore, 36^(3/2) is 6^3 = 216. So the second term is 216. Alright, that's straightforward. Now onto the third term: 3^(4 / log₇9). Let me see. Again, log₇9. Let's express this logarithm in terms of base 3 or base 7? Maybe base 3 since 9 is 3^2. Let me try that. So log₇9 = log₇(3^2) = 2 log₇3. Therefore, 4 / log₇9 = 4 / (2 log₇3) = 2 / log₇3. So the exponent is 2 / log₇3. Similar to the first term, maybe we can manipulate this. Using the reciprocal property, 1 / log₇3 is log base 3 of 7. Because log₇3 = 1 / log₃7. Therefore, 2 / log₇3 = 2 * log₃7. So the exponent becomes 2 log₃7. Therefore, the third term is 3^(2 log₃7). Again, using the same property as before. 3^(log₃7) = 7, so 3^(2 log₃7) = (3^(log₃7))^2 = 7^2 = 49. So the third term is 49. Putting it all together: first term is 625, second term is 216, third term is 49. Adding them up: 625 + 216. Let me compute that first. 625 + 200 is 825, then +16 is 841. Then 841 + 49. 841 + 40 is 881, then +9 is 890. So the total is 890. Wait a second, let me double-check those additions to make sure. 625 + 216: 600 + 200 is 800, 25 + 16 is 41, so 800 + 41 is 841. Then 841 + 49: 840 + 40 is 880, and 1 + 9 is 10, so 880 + 10 is 890. Yep, that seems right. So all three terms simplified to 625, 216, and 49, which add up to 890. Let me just check each term again to ensure I didn't make a mistake. First term: 81^(1/log₅3) becomes 3^4^(1/log₅3) = 3^(4/log₅3). Then since log₅3 = 1/log₃5, so 4/log₅3 = 4 log₃5. Therefore, 3^(4 log₃5) = (3^log₃5)^4 = 5^4 = 625. Correct. Second term: 27^(log₉36) = 3^3^(log_{3^2}36) = 3^(3 * (1/2) log₃36) = 3^( (3/2) log₃36 ) = 36^(3/2) = (6^2)^(3/2) = 6^(2*(3/2)) = 6^3 = 216. Correct. Third term: 3^(4 / log₇9). log₇9 = log₇3^2 = 2 log₇3. So 4 / log₇9 = 2 / log₇3 = 2 log₃7. Then 3^(2 log₃7) = (3^log₃7)^2 = 7^2 = 49. Correct. All steps check out, so adding them together: 625 + 216 + 49 = 890. Therefore, the value of the expression is 890. Final Answerboxed{890}</think>To solve the given expression (81^{frac{1}{log _{5} 3}} + 27^{log _{9} 36} + 3^{frac{4}{log _{7} 9}}), we break it down term by term:1. First term: (81^{frac{1}{log _{5} 3}}) - Rewrite 81 as (3^4): ((3^4)^{frac{1}{log _{5} 3}}) - Simplify the exponent: (3^{frac{4}{log _{5} 3}}) - Use the change of base formula: (log_{5} 3 = frac{1}{log_{3} 5}), so (frac{4}{log_{5} 3} = 4 log_{3} 5) - Apply the property (a^{log_a b} = b): (3^{4 log_{3} 5} = (3^{log_{3} 5})^4 = 5^4 = 625)2. Second term: (27^{log_{9} 36}) - Rewrite 27 as (3^3) and 9 as (3^2): ((3^3)^{log_{3^2} 36}) - Simplify the exponent: (log_{3^2} 36 = frac{1}{2} log_{3} 36), so (3^{3 cdot frac{1}{2} log_{3} 36} = 3^{frac{3}{2} log_{3} 36}) - Apply the property (a^{log_a b} = b): (3^{frac{3}{2} log_{3} 36} = 36^{frac{3}{2}} = (6^2)^{frac{3}{2}} = 6^3 = 216)3. Third term: (3^{frac{4}{log_{7} 9}}) - Rewrite 9 as (3^2): (log_{7} 9 = log_{7} 3^2 = 2 log_{7} 3) - Simplify the exponent: (frac{4}{2 log_{7} 3} = frac{2}{log_{7} 3}) - Use the change of base formula: (frac{2}{log_{7} 3} = 2 log_{3} 7) - Apply the property (a^{log_a b} = b): (3^{2 log_{3} 7} = (3^{log_{3} 7})^2 = 7^2 = 49)Adding all the terms together:[625 + 216 + 49 = 890]Thus, the final answer is (boxed{890}).,<think>Okay, let's see. I need to solve this expression: 81^(1/log₅3) + 27^(log₉36) + 3^(4/log₇9). Hmm, that looks a bit complicated with all the exponents and logarithms. Let me break it down term by term. First, I remember that logarithms can be converted using different bases, maybe that will help. Also, properties of exponents might be useful here. Let me recall some logarithmic identities. There's the change of base formula: log_b a = log_c a / log_c b. Maybe that can be applied here. Also, remember that a^(log_b c) = c^(log_b a). Not sure if that will help, but maybe.Starting with the first term: 81^(1/log₅3). Let's simplify this. 81 is 3^4, right? So 81 = 3^4. Then, 81^(1/log₅3) = (3^4)^(1/log₅3) = 3^(4/log₅3). Hmm, okay. Now, log₅3 is log base 5 of 3. Maybe I can use the change of base formula here. Let's write log₅3 as ln3 / ln5. So then 1/log₅3 is ln5 / ln3. Therefore, 3^(4/log₅3) = 3^(4*(ln5/ln3)) = 3^(4 log₃5). Wait, because ln5/ln3 is log₃5. Yes, that's right. So 3^(4 log₃5). Now, 3^(log₃5) is equal to 5, because a^(log_a b) = b. So, 3^(log₃5) = 5. Therefore, 3^(4 log₃5) = (3^(log₃5))^4 = 5^4 = 625. Oh, that's nice! So the first term simplifies to 625. Okay, that wasn't too bad. Moving on to the second term: 27^(log₉36). Let's see. 27 is 3^3, and 9 is 3^2. So maybe express everything in base 3. Let me rewrite log₉36 as log_(3^2)36. Using the logarithm power rule, log_(a^b)c = (1/b) log_a c. Therefore, log₉36 = (1/2) log₃36. So, log₉36 = (1/2) log₃36. Then, 27^(log₉36) = (3^3)^[(1/2) log₃36] = 3^(3*(1/2) log₃36) = 3^( (3/2) log₃36 ). Again, similar to the first term. 3^(log₃36) would be 36. So here, the exponent is (3/2) log₃36. So that can be written as 3^(log₃36^(3/2)) = 36^(3/2). Wait, is that right? Let me check. Alternatively, using the property a^(k log_a b) = b^k. So if you have 3^( (3/2) log₃36 ), that's equal to 36^(3/2). Yes, that works. So 36^(3/2) is the square root of 36 cubed. The square root of 36 is 6, so 6^3 = 216. So 36^(3/2) is 216. Therefore, the second term is 216. Alright, two terms down. Now the third term: 3^(4/log₇9). Let's tackle this. Again, log₇9 is log base 7 of 9. Let's use the change of base formula again. log₇9 = ln9 / ln7. So 1/log₇9 = ln7 / ln9. Then 4/log₇9 = 4*(ln7 / ln9) = 4*(log₉7). Wait, log₉7 is ln7 / ln9. So 4/log₇9 = 4 log₉7. Therefore, 3^(4 log₉7). Hmm, similar to before.But here, the base of the exponent is 3, and the logarithm is base 9. Let me see if I can relate them. Since 9 is 3^2, log₉7 is log_(3^2)7 = (1/2) log₃7. Therefore, 4 log₉7 = 4*(1/2) log₃7 = 2 log₃7. So 3^(2 log₃7). Then, using the property a^(k log_a b) = b^k, this is 7^2 = 49. So the third term is 49. Putting it all together: 625 + 216 + 49. Let's compute that. 625 + 216 is 841, and 841 + 49 is 890. So the total is 890. Wait a minute, let me double-check each term to make sure I didn't make a mistake. First term: 81^(1/log₅3) becomes 3^(4/log₅3). Then converted log₅3 to ln3/ln5, so reciprocal is ln5/ln3 which is log₃5. So 4 log₃5. Then 3^(4 log₃5) is 5^4 = 625. That seems correct.Second term: 27^(log₉36). Express 27 as 3^3, log₉36 as (1/2) log₃36. So exponent becomes 3*(1/2) log₃36 = (3/2) log₃36. Then 3 raised to that power is 36^(3/2) = 6^3 = 216. Correct.Third term: 3^(4/log₇9). log₇9 is ln9 / ln7, reciprocal is ln7 / ln9 = log₉7. Multiply by 4 gives 4 log₉7. Then express log₉7 as (1/2) log₃7, so 4*(1/2) log₃7 = 2 log₃7. Then 3^(2 log₃7) = 7^2 = 49. Correct.Adding them up: 625 + 216. Let me do 625 + 200 = 825, then +16 is 841. Then 841 +49. 841 +40=881, +9=890. Yep, 890. So I think that's the answer. Let me just make sure there's no alternative way someone might approach this problem and get a different result, but all the steps seem solid. Changing the bases appropriately and using logarithm properties. Yeah, I think that's right.Final Answerboxed{890}</think>To solve the expression (81^{frac{1}{log _{5} 3}} + 27^{log _{9} 36} + 3^{frac{4}{log _{7} 9}}), we break it down term by term:1. First term: (81^{frac{1}{log _{5} 3}}) - Rewrite 81 as (3^4): (81^{frac{1}{log _{5} 3}} = (3^4)^{frac{1}{log _{5} 3}}). - Use the change of base formula: (log _{5} 3 = frac{ln 3}{ln 5}), so (frac{1}{log _{5} 3} = frac{ln 5}{ln 3} = log _{3} 5). - Simplify: (3^{4 log _{3} 5} = (3^{log _{3} 5})^4 = 5^4 = 625).2. Second term: (27^{log _{9} 36}) - Rewrite 27 as (3^3) and 9 as (3^2): (27^{log _{9} 36} = (3^3)^{log _{3^2} 36}). - Use the logarithm power rule: (log _{3^2} 36 = frac{1}{2} log _{3} 36). - Simplify: (3^{3 cdot frac{1}{2} log _{3} 36} = 3^{frac{3}{2} log _{3} 36} = (3^{log _{3} 36})^{frac{3}{2}} = 36^{frac{3}{2}} = 6^3 = 216).3. Third term: (3^{frac{4}{log _{7} 9}}) - Rewrite 9 as (3^2): (log _{7} 9 = log _{7} 3^2 = 2 log _{7} 3). - Use the change of base formula: (frac{1}{log _{7} 9} = frac{1}{2 log _{7} 3} = frac{1}{2} log _{3} 7). - Simplify: (3^{4 cdot frac{1}{2} log _{3} 7} = 3^{2 log _{3} 7} = (3^{log _{3} 7})^2 = 7^2 = 49).Adding all terms together:[625 + 216 + 49 = 890](boxed{890})]

🔑:Solution.81^{frac{1}{log _{5} 3}}+27^{log _{9} 36}+3^{frac{4}{log _{7} 9}}=3^{4 log _{3} 5}+3^{frac{3}{2^{2} log _{3} 36}}+3^{frac{4}{2} log _{3} 7}=5^{4}+36^{frac{3}{2}}+49= =625+216+49=890.Answer: 890.