🧠:Okay, so I need to prove that ( a^2 + b^2 = 1 ) given the equation ( a sqrt{1 - b^2} + b sqrt{1 - a^2} = 1 ), where both ( a ) and ( b ) are between 0 and 1 inclusive. Hmm, let's start by understanding what the problem is asking. We have this equation involving square roots and variables a and b, and we need to show that the sum of their squares is 1. First, maybe I can recall some trigonometric identities because the square roots remind me of sine and cosine. If I let ( a = sin alpha ) and ( b = sin beta ), then ( sqrt{1 - a^2} = cos alpha ) and ( sqrt{1 - b^2} = cos beta ). Substituting these into the original equation gives:( sin alpha cos beta + sin beta cos alpha = 1 )Hmm, that's the sine addition formula! The left side is ( sin(alpha + beta) ). So, ( sin(alpha + beta) = 1 ). The maximum value of sine is 1, so this implies that ( alpha + beta = frac{pi}{2} + 2pi k ), where k is an integer. But since ( a ) and ( b ) are between 0 and 1, ( alpha ) and ( beta ) would be between 0 and ( frac{pi}{2} ). Therefore, ( alpha + beta = frac{pi}{2} ). If ( alpha + beta = frac{pi}{2} ), then ( beta = frac{pi}{2} - alpha ). So, ( b = sin beta = sin(frac{pi}{2} - alpha) = cos alpha ). But ( a = sin alpha ), so ( cos alpha = sqrt{1 - a^2} ). Therefore, ( b = sqrt{1 - a^2} ). Then, ( a^2 + b^2 = a^2 + (1 - a^2) = 1 ). Wait, that seems straightforward. But did I make any assumptions here? I assumed trigonometric substitutions, but maybe there's another way without trigonometry? Let me try an algebraic approach to verify.Starting with the original equation:( a sqrt{1 - b^2} + b sqrt{1 - a^2} = 1 )Let me square both sides to eliminate the square roots. That might complicate things, but let's see:( [a sqrt{1 - b^2} + b sqrt{1 - a^2}]^2 = 1^2 )Expanding the left side:( a^2 (1 - b^2) + 2ab sqrt{(1 - b^2)(1 - a^2)} + b^2 (1 - a^2) = 1 )Simplify the terms:( a^2 - a^2 b^2 + b^2 - a^2 b^2 + 2ab sqrt{(1 - a^2)(1 - b^2)} = 1 )Combine like terms:( a^2 + b^2 - 2a^2 b^2 + 2ab sqrt{(1 - a^2)(1 - b^2)} = 1 )Hmm, okay, so if we denote ( a^2 + b^2 = S ) and ( ab = P ), maybe that could help. Let's try substituting ( S = a^2 + b^2 ). Then, ( 1 - a^2 = 1 - a^2 ) and ( 1 - b^2 = 1 - b^2 ), so the product ( (1 - a^2)(1 - b^2) = (1 - a^2 - b^2 + a^2 b^2) = 1 - S + a^2 b^2 ).But substituting back into the equation:( S - 2a^2 b^2 + 2ab sqrt{1 - S + a^2 b^2} = 1 )This looks complicated. Maybe there's a better way. Let me think. If I can assume that ( a^2 + b^2 = 1 ), then ( sqrt{1 - b^2} = a ) and ( sqrt{1 - a^2} = b ), so substituting into the original equation gives ( a cdot a + b cdot b = a^2 + b^2 = 1 ), which matches the right-hand side. So that shows that if ( a^2 + b^2 = 1 ), then the original equation holds. But the problem is to show the converse: if the original equation holds, then ( a^2 + b^2 = 1 ).So squaring might still be the way, but we need to handle the square roots. Let's denote ( x = a ), ( y = b ). Then the equation is:( x sqrt{1 - y^2} + y sqrt{1 - x^2} = 1 )Square both sides:( x^2 (1 - y^2) + y^2 (1 - x^2) + 2xy sqrt{(1 - x^2)(1 - y^2)} = 1 )Simplify:( x^2 - x^2 y^2 + y^2 - x^2 y^2 + 2xy sqrt{(1 - x^2)(1 - y^2)} = 1 )Which becomes:( x^2 + y^2 - 2x^2 y^2 + 2xy sqrt{(1 - x^2)(1 - y^2)} = 1 )Let me rearrange this:( (x^2 + y^2) + 2xy sqrt{(1 - x^2)(1 - y^2)} = 1 + 2x^2 y^2 )Hmm, this still seems messy. Maybe if I let ( x^2 + y^2 = 1 ), then substitute into the equation to check if both sides are equal. But I need to prove that ( x^2 + y^2 = 1 ), so I can't assume that yet. Alternatively, maybe let’s set ( a = sin theta ) and ( b = cos phi ), but that might complicate. Wait, in the trigonometric substitution earlier, I assumed ( a = sin alpha ), ( b = sin beta ), leading to ( sin(alpha + beta) = 1 ), which gives ( alpha + beta = pi/2 ), hence ( beta = pi/2 - alpha ), so ( b = cos alpha ), and since ( a = sin alpha ), then ( a^2 + b^2 = sin^2 alpha + cos^2 alpha = 1 ). That works. But is this substitution valid? Since ( 0 leq a, b leq 1 ), then ( alpha ) and ( beta ) can be chosen in the first quadrant, so yes. But maybe someone might object that not all a and b in [0,1] can be expressed as sine or cosine of some angle? Wait, actually, since ( sin theta ) and ( cos theta ) both range between 0 and 1 when ( theta ) is between 0 and ( pi/2 ), which covers all possible a and b in [0,1]. So that substitution is valid. Therefore, this seems like a valid approach. But to make sure, let's check with an example. Suppose a = 0, then the equation becomes 0 * sqrt(1 - b^2) + b * sqrt(1 - 0) = b = 1. So b = 1, then a^2 + b^2 = 0 + 1 = 1. Similarly, if b = 0, then a must be 1. If a = 1, then sqrt(1 - b^2) = sqrt(1 - b^2), and the equation becomes sqrt(1 - b^2) + 0 = 1, so sqrt(1 - b^2) = 1, so 1 - b^2 = 1, so b = 0. So that checks out. What about a case where a and b are both 1/√2. Then a^2 + b^2 = 1/2 + 1/2 = 1. Plugging into the original equation: (1/√2) * sqrt(1 - (1/2)) + (1/√2) * sqrt(1 - (1/2)) = (1/√2)(sqrt(1/2)) + (1/√2)(sqrt(1/2)) = (1/√2)(1/√2) + (1/√2)(1/√2) = 1/2 + 1/2 = 1. So that works. Alternatively, take a = 3/5, then to satisfy a^2 + b^2 = 1, b should be 4/5. Let's check: (3/5) * sqrt(1 - (16/25)) + (4/5) * sqrt(1 - 9/25) = (3/5) * (3/5) + (4/5) * (4/5) = 9/25 + 16/25 = 25/25 = 1. So that works. But how do we know there are no other solutions where a^2 + b^2 ≠ 1? Suppose someone picks a and b such that a^2 + b^2 ≠ 1, but the equation holds. Let's test with a = 1/2 and b = 1/2. Then the left side is (1/2) * sqrt(1 - 1/4) + (1/2) * sqrt(1 - 1/4) = (1/2)(sqrt(3)/2) + (1/2)(sqrt(3)/2) = sqrt(3)/4 + sqrt(3)/4 = sqrt(3)/2 ≈ 0.866, which is less than 1. So the equation does not hold. Another test: a = 0.6, b = 0.8. Then a^2 + b^2 = 0.36 + 0.64 = 1. The original equation: 0.6 * sqrt(1 - 0.64) + 0.8 * sqrt(1 - 0.36) = 0.6 * sqrt(0.36) + 0.8 * sqrt(0.64) = 0.6*0.6 + 0.8*0.8 = 0.36 + 0.64 = 1. So that works. If I pick a = 0.8, b = 0.6, same result. What if I take a = sqrt(1/2), b = sqrt(1/2). Then a^2 + b^2 = 1, and the original equation gives 1 as shown before. But is there a case where a^2 + b^2 ≠ 1 but the equation holds? Let me try to suppose there is. Let’s assume a^2 + b^2 = k ≠ 1 and see if it's possible. Suppose a^2 + b^2 = k. Then we can write the original equation as:a sqrt(1 - b^2) + b sqrt(1 - a^2) = 1But sqrt(1 - b^2) = sqrt( (1 - k) + a^2 ), and sqrt(1 - a^2) = sqrt( (1 - k) + b^2 ). Hmm, not sure if that helps. Maybe express in terms of k. Let’s denote s = a^2 + b^2. Then we have:Original equation squared gives:s - 2a^2 b^2 + 2ab sqrt{(1 - a^2)(1 - b^2)} = 1But if s = 1, then 1 - 2a^2 b^2 + 2ab sqrt{(1 - a^2)(1 - b^2)} = 1, which implies that -2a^2 b^2 + 2ab sqrt{(1 - a^2)(1 - b^2)} = 0. Dividing both sides by 2ab (assuming ab ≠ 0):- ab + sqrt{(1 - a^2)(1 - b^2)} = 0Which implies sqrt{(1 - a^2)(1 - b^2)} = abSquaring both sides:(1 - a^2)(1 - b^2) = a^2 b^2Expanding left side:1 - a^2 - b^2 + a^2 b^2 = a^2 b^2Simplify:1 - a^2 - b^2 = 0 → a^2 + b^2 = 1So this shows that if s = 1, then the equation holds, but also, if the original equation holds, then squaring leads us to a^2 + b^2 = 1. Wait, let me verify that step again.Starting from the squared equation:s - 2a^2 b^2 + 2ab sqrt{(1 - a^2)(1 - b^2)} = 1If we assume the original equation holds, then we derived this equation. Then if s = a^2 + b^2 = 1, then substituting s = 1 gives:1 - 2a^2 b^2 + 2ab sqrt{(1 - a^2)(1 - b^2)} = 1Subtracting 1:-2a^2 b^2 + 2ab sqrt{(1 - a^2)(1 - b^2)} = 0Which simplifies to:sqrt{(1 - a^2)(1 - b^2)} = abThen squaring:(1 - a^2)(1 - b^2) = a^2 b^2Expanding:1 - a^2 - b^2 + a^2 b^2 = a^2 b^2So:1 - a^2 - b^2 = 0 → a^2 + b^2 = 1Therefore, this shows that if the squared equation holds (which it does because we squared the original equation), then a^2 + b^2 = 1. However, squaring can sometimes introduce extraneous solutions. But in this case, since we started with the original equation and squared it, the reverse implication requires that any solution to the squared equation must satisfy the original equation. Wait, but actually, squaring is a one-way implication. If the original equation holds, then the squared equation holds, but not necessarily vice versa. Therefore, when we square, we might get solutions that don't satisfy the original equation. However, in our case, after squaring and simplifying, we arrive at a^2 + b^2 = 1. Therefore, any solution to the original equation must satisfy a^2 + b^2 = 1. But does every solution of a^2 + b^2 = 1 satisfy the original equation? Yes, as we saw earlier. For example, substituting a^2 + b^2 = 1 into the original equation gives:a sqrt(1 - b^2) + b sqrt(1 - a^2) = a * a + b * b = a^2 + b^2 = 1Therefore, the equation a sqrt(1 - b^2) + b sqrt(1 - a^2) = 1 is equivalent to a^2 + b^2 = 1. Hence, the original equation holds if and only if a^2 + b^2 = 1, given the constraints on a and b.Therefore, the proof is established through both trigonometric substitution and algebraic manipulation with squaring. The key was recognizing that squaring the original equation leads directly to the necessity of a^2 + b^2 = 1 after simplification, and conversely, if a^2 + b^2 = 1, then the original equation holds. Another way to think about it is to consider that the left-hand side of the original equation resembles a dot product of two vectors. If we let vector u = (a, b) and vector v = (sqrt(1 - b^2), sqrt(1 - a^2)), then the equation u • v = 1. The Cauchy-Schwarz inequality states that u • v ≤ ||u|| ||v||. Here, ||u|| = sqrt(a^2 + b^2), and ||v|| = sqrt( (1 - b^2) + (1 - a^2) ) = sqrt(2 - a^2 - b^2). So, u • v ≤ sqrt(a^2 + b^2) * sqrt(2 - a^2 - b^2). But we know that u • v = 1. Therefore:sqrt(a^2 + b^2) * sqrt(2 - a^2 - b^2) ≥ 1Let’s square both sides:(a^2 + b^2)(2 - a^2 - b^2) ≥ 1Let S = a^2 + b^2. Then:S(2 - S) ≥ 1→ 2S - S² ≥ 1→ -S² + 2S - 1 ≥ 0Multiply both sides by -1 (inequality reverses):S² - 2S + 1 ≤ 0→ (S - 1)^2 ≤ 0Since a square is always non-negative, the only solution is S - 1 = 0 → S = 1. Therefore, a^2 + b^2 = 1. This uses the Cauchy-Schwarz inequality to demonstrate that equality holds only when the vectors are parallel, but in this case, the equality in the Cauchy-Schwarz inequality combined with the specific structure of the vectors leads directly to S = 1. This method also confirms that a^2 + b^2 must equal 1 for the original equation to hold, given the constraints on a and b. Therefore, through multiple approaches—trigonometric substitution, algebraic manipulation with squaring, and application of the Cauchy-Schwarz inequality—we arrive at the conclusion that ( a^2 + b^2 = 1 ).

✅:To prove that ( a^2 + b^2 = 1 ) given the equation ( a sqrt{1 - b^2} + b sqrt{1 - a^2} = 1 ) with ( 0 leq a, b leq 1 ), we proceed as follows: Key Steps and Proof:1. Trigonometric Substitution: - Let ( a = sin alpha ) and ( b = sin beta ), where ( alpha, beta in [0, frac{pi}{2}] ). - Substitute into the equation: [ sin alpha cos beta + sin beta cos alpha = 1 ] - Recognize the sine addition formula: ( sin(alpha + beta) = 1 ). - Since ( sin(theta) = 1 ) implies ( theta = frac{pi}{2} ), we have ( alpha + beta = frac{pi}{2} ). - Thus, ( beta = frac{pi}{2} - alpha ), so ( b = sin beta = cos alpha ). - Therefore, ( a^2 + b^2 = sin^2 alpha + cos^2 alpha = 1 ).2. Algebraic Squaring: - Square both sides of the original equation: [ a^2(1 - b^2) + b^2(1 - a^2) + 2ab sqrt{(1 - a^2)(1 - b^2)} = 1 ] - Simplify to: [ a^2 + b^2 - 2a^2b^2 + 2ab sqrt{(1 - a^2)(1 - b^2)} = 1 ] - Let ( S = a^2 + b^2 ). Substitute ( S ) and rearrange: [ S - 2a^2b^2 + 2ab sqrt{(1 - S + a^2b^2)} = 1 ] - Isolate the square root term and square again to eliminate it. This leads to: [ (S - 1)^2 = 0 implies S = 1 ] - Hence, ( a^2 + b^2 = 1 ).3. Cauchy-Schwarz Inequality: - Interpret the left-hand side as a dot product: [ mathbf{u} cdot mathbf{v} = a sqrt{1 - b^2} + b sqrt{1 - a^2} = 1 ] - Apply the Cauchy-Schwarz inequality: [ 1 leq sqrt{a^2 + b^2} cdot sqrt{(1 - b^2) + (1 - a^2)} = sqrt{S(2 - S)} ] - Square both sides and solve: [ S(2 - S) geq 1 implies (S - 1)^2 leq 0 implies S = 1 ] Conclusion:All methods confirm that the given equation ( a sqrt{1 - b^2} + b sqrt{1 - a^2} = 1 ) implies ( a^2 + b^2 = 1 ).Final Answer: boxed{a^2 + b^2 = 1}

🔑:Given the equation: [ a sqrt{1-b^{2}} + b sqrt{1-a^{2}} = 1 ]and the constraints:[ 0 leq a leq 1 quad text{and} quad 0 leq b leq 1, ]Our goal is to prove: [ a^2 + b^2 = 1. ]1. Construct a Circle and a Quadrilateral: Let’s construct a circle with a diameter (AC = 1). 2. Create Chords: Construct chords (AB = b) and (AD = a) within the circle. Connect points (B) and (D) to a point (C) on the boundary of the circle such that (angle ACB = 90^circ).3. Label the Lengths: From the construction, (DC = sqrt{1 - a^2}) and (BC = sqrt{1 - b^2}).4. Apply Ptolemy’s Theorem: Applying Ptolemy's theorem to cyclic quadrilateral (ABCD), we have: [ AD cdot BC + AB cdot DC = AC cdot BD ] Substituting the known lengths: [ a cdot sqrt{1 - b^2} + b cdot sqrt{1 - a^2} = 1 cdot BD ] Given in the problem, we have: [ a sqrt{1 - b^2} + b sqrt{1 - a^2} = 1 ] Thus: [ BD = 1 ]5. Relationship from the Circle: Since (BD = 1), (BD) is also a diameter of the circle. Therefore, the following relation holds for points on the circle: [ BD = sqrt{a^2 + b^2} ] Since we found (BD = 1): [ 1 = sqrt{a^2 + b^2} ]6. Square Both Sides: Squaring both sides of the equation: [ 1^2 = (sqrt{a^2 + b^2})^2 ] This simplifies to: [ 1 = a^2 + b^2 ] Conclusion:[boxed{a^2 + b^2 = 1}]

🧠:Okay, so I need to figure out how many ways there are to color a circle divided into n sectors using m colors, such that no two adjacent sectors have the same color. Hmm, this reminds me of graph coloring problems, where each sector is a node connected to its neighbors, forming a cycle graph. Let me think step by step.First, if the sectors were in a straight line (like a path instead of a cycle), the number of colorings would be easier. For a line of n sectors, the first sector can be colored in m ways, and each subsequent sector can be colored in (m-1) ways since it just needs to be different from the previous one. So that would be m*(m-1)^(n-1). But since it's a circle, the first and last sectors are also adjacent, so we have to account for that additional constraint. That complicates things.I remember that for circular arrangements, the formula is different from linear ones. Maybe this is related to the concept of chromatic polynomials? Let me recall. The chromatic polynomial counts the number of ways to color a graph with a given number of colors. For a cycle graph C_n, the chromatic polynomial should be (m-1)^n + (-1)^n*(m-1). Let me check if that makes sense.Wait, when n=3 (a triangle), the number of colorings should be m*(m-1)*(m-2). Let's plug n=3 into the formula: (m-1)^3 + (-1)^3*(m-1) = (m^3 - 3m^2 + 3m -1) - (m -1) = m^3 - 3m^2 + 2m. Hmm, m*(m-1)*(m-2) is also m^3 - 3m^2 + 2m. Okay, that matches. For n=4, a square, the number of colorings should be m*(m-1)^3 - m*(m-1). Let me see: Using the formula (m-1)^4 + (-1)^4*(m-1) = (m^4 -4m^3 +6m^2 -4m +1) + (m -1) = m^4 -4m^3 +6m^2 -3m. Alternatively, if I compute it another way: first color the first sector m ways, then the second m-1, the third m-1 (since it's not adjacent to the first), the fourth has to be different from the third and the first. Wait, depending on whether the first and third are the same or different. Maybe this approach is getting complicated. Let me think again.Alternatively, maybe use recurrence relations. For a line of n sectors, the number is m*(m-1)^(n-1). For a circle, it's equal to the number for a line minus the number of colorings where the first and last sectors are the same color. So, if we can find the number of linear colorings where the first and last are the same, then subtract that from the total linear colorings to get the circular colorings.Let me denote the number of circular colorings as C(n, m). Then, C(n, m) = number of linear colorings - number of linear colorings with first and last sectors the same. Let's denote the number of linear colorings as L(n, m) = m*(m-1)^(n-1). Then, let S(n, m) be the number of linear colorings where the first and last sectors are the same. Then, C(n, m) = L(n, m) - S(n, m).So how do we find S(n, m)? Let's think recursively. Suppose we fix the first and last sector to color c1. Then the second sector can be colored in (m-1) ways, the third in (m-1) ways, etc., up to the (n-1)th sector. However, the (n-1)th sector must also be different from the last sector, which is color c1. So, if we fix the first and last to c1, the problem reduces to coloring the middle n-2 sectors such that the (n-1)th sector is not c1. But wait, this is similar to a linear coloring problem where the first sector is color c1, the next n-2 sectors are colored with the constraint that adjacent are different, and the last sector (n-1th) is not c1.Alternatively, maybe there's a better way. Let me consider that for S(n, m), the first and last sectors are the same. Then, we can think of this as a circular arrangement where the first and last are glued together, forming a circle. Wait, but that might not directly help. Alternatively, think of the linear arrangement with the first and last fixed to the same color. Let's denote S(n, m). For the first sector, there are m choices. The last sector must be the same as the first. The sectors in between (positions 2 to n-1) must form a valid coloring where the second sector is different from the first, the third different from the second, ..., and the (n-1)th different from the (n-2)th. Additionally, the (n-1)th sector must be different from the last sector (which is the same as the first). Therefore, the (n-1)th sector must be different from both the (n-2)th sector and the first sector.Wait, so S(n, m) = m * [number of colorings for positions 2 to n-1 where each is different from the previous, and the (n-1)th is different from the first]. Let's denote positions 2 to n-1 as a line of length n-2, with the first position (sector 2) adjacent to sector 1 (color c1), and the last position (sector n-1) must be different from c1. Hmm. So, the problem reduces to coloring a line of length n-2 sectors, where the first can be colored in (m-1) ways (different from c1), each subsequent in (m-1) ways, but the last one must not be c1. Wait, so perhaps the number of such colorings is (m-1)^(n-2) - the number of colorings where the last sector is c1. Let me see.Suppose we have a line of length k (sectors 2 to n-1 is k = n-2). The first sector (sector 2) can be colored in (m-1) colors (different from c1). Each subsequent sector (3 to n-1) can be colored in (m-1) ways (different from previous). So total colorings for the middle sectors: (m-1)^(n-2). However, we need the last sector (n-1) to be different from c1. So from the total (m-1)^(n-2), subtract the number of colorings where the last sector is c1.Let me denote T(k) as the number of colorings of a line of length k where the first sector is colored in (m-1) ways (different from c1), each subsequent different from the previous, and the last sector is c1. So S(n, m) = m * [ (m-1)^(n-2) - T(n-2) ].To find T(k), the number of colorings of a line of length k where the first can be any color except c1, each subsequent different from the previous, and the last is c1. Let's think recursively. For k=1, we have a single sector (sector 2) which must be different from c1, but it also needs to be c1. So T(1)=0. For k=2, we have two sectors: the first must be different from c1, the second must be c1 and different from the first. So the first can be any of (m-1) colors, the second must be c1. So T(2) = (m-1). For k=3, the first sector is different from c1, the second different from the first, the third is c1 and different from the second. So the number is (m-1)*(m-2). Similarly, for k=4, it's (m-1)*(m-2)^2. Wait, this seems like T(k) = (m-1)*(m-2)^{k-1} when k >=2. Wait, but let's check for k=2: T(2) = (m-1)*(m-2)^{1} = (m-1)*(m-2). But earlier I thought T(2) = m-1. That contradicts. Hmm, maybe my initial thought is wrong.Wait, when k=2, the first sector (sector 2) is colored in (m-1) ways (not c1), then sector 3 (which is sector n-1 in the original problem) must be c1 and different from sector 2. Since sector 2 is not c1, sector 3 can be c1 as long as it's different from sector 2. But since sector 2 is not c1, then sector 3 can be c1. So the number of colorings is indeed (m-1). So T(2) = m-1. For k=3: first sector (sector 2) is colored in (m-1) ways, second sector (sector 3) must be different from sector 2, so (m-2) choices. Third sector (sector 4) must be c1 and different from sector 3. If sector 3 is not c1, then sector 4 can be c1. So the number is (m-1)*(m-2)*(1) = (m-1)*(m-2). Similarly, for k=4: (m-1)*(m-2)*(m-2)*1 = (m-1)*(m-2)^2. So the pattern is T(k) = (m-1)*(m-2)^{k-2} for k >=2.Therefore, T(k) = (m-1)*(m-2)^{k-2} when k >=2, and T(1)=0. Therefore, going back, for S(n, m):If n >=3, then S(n, m) = m * [ (m-1)^{n-2} - T(n-2) ]But n-2 is k. So if k = n - 2 >=1, so n >=3.Case 1: If n=2. Then the circle has two sectors adjacent to each other. The number of colorings is m*(m-1). But according to the formula C(n, m) = L(n, m) - S(n, m). If n=2, L(2, m) = m*(m-1). S(2, m) is the number of linear colorings where first and last are the same. Since it's a line of two sectors, the number of colorings where both are the same is m (each color). So S(2, m) = m. Therefore, C(2, m) = m*(m-1) - m = m(m-2). But wait, when n=2, the circle requires that both sectors are different. So the number should be m*(m-1). Wait, this contradicts. Wait, no, if n=2, in a circle, the two sectors are adjacent to each other, so the number of colorings is m*(m-1), same as the linear case because the first and last are adjacent in both cases. Wait, but according to the formula C(n, m) = L(n, m) - S(n, m). For n=2, L(2, m)=m*(m-1), S(2, m)=m (colorings where first and last are the same). Therefore, C(2, m)=m(m-1) - m = m(m-2). But that's incorrect because in reality, for n=2, the number should be m*(m-1). Wait, there's a mistake here. So my approach might be wrong.Alternatively, maybe the formula C(n, m) = (m-1)^n + (-1)^n*(m-1). Let's check for n=2. (m-1)^2 + (1)*(m-1) = m^2 - 2m +1 + m -1 = m^2 - m. Which is m(m-1). Correct. But according to the previous approach, C(n, m) = L(n, m) - S(n, m). For n=2, L(n, m)=m*(m-1), S(n, m)=m. Then C(n, m)=m(m-1)-m=m(m-2), which is wrong. So my approach is flawed. Therefore, perhaps the initial formula is different.Wait, maybe the correct formula is indeed (m-1)^n + (-1)^n*(m-1). Let me check for n=3. (m-1)^3 + (-1)^3*(m-1) = (m^3 - 3m^2 + 3m -1) - (m -1) = m^3 -3m^2 + 2m, which is m(m-1)(m-2). Correct. For n=4: (m-1)^4 + (m-1) = (m^4 -4m^3 +6m^2 -4m +1) + (m -1) = m^4 -4m^3 +6m^2 -3m. Let me compute manually for n=4. The number of colorings should be: first sector m, second m-1, third m-1, fourth must be different from third and first. If first and third are the same, then fourth has m-1 choices. If first and third are different, fourth has m-2 choices. So we need to split into cases.Number of colorings where first and third are the same: first sector m, second sector m-1, third sector must equal first, so 1 choice, fourth sector m-1. Total: m*(m-1)*1*(m-1)=m*(m-1)^2.Number of colorings where first and third are different: first sector m, second sector m-1, third sector m-2 (different from second and first), fourth sector m-2 (different from third and first). Total: m*(m-1)*(m-2)*(m-2).Total colorings: m*(m-1)^2 + m*(m-1)*(m-2)^2 = m*(m-1)*[ (m-1) + (m-2)^2 ].Simplify inside the brackets: (m-1) + (m^2 -4m +4) = m^2 -3m +3. Therefore, total colorings: m*(m-1)*(m^2 -3m +3). Let's compare to the formula (m-1)^4 + (m-1):(m-1)^4 + (m-1) = (m^4 -4m^3 +6m^2 -4m +1) + (m -1) = m^4 -4m^3 +6m^2 -3m.But m*(m-1)*(m^2 -3m +3) = m*(m^3 -3m^2 +3m - m^2 +3m -3) = m*(m^3 -4m^2 +6m -3) = m^4 -4m^3 +6m^2 -3m. Yes, same result. So the formula holds.Therefore, the general formula seems to be (m-1)^n + (-1)^n*(m-1). Let me check for n=2: (m-1)^2 + (m-1) = m^2 -2m +1 +m -1 = m^2 -m = m(m-1). Correct. For n=3: (m-1)^3 - (m-1) = m^3 -3m^2 +3m -1 -m +1 = m^3 -3m^2 +2m. Correct. So the formula works. Therefore, the number of colorings is (m-1)^n + (-1)^n*(m-1).Alternatively, this can be written as (m-1)[(m-1)^{n-1} + (-1)^n]. Hmm, but the original formula is better.Alternatively, using inclusion-exclusion. The total number of colorings for a line is m*(m-1)^{n-1}. For the circle, we need to subtract the colorings where the first and last are the same. Let's denote A as the set of colorings where the first and last sectors are the same. Then |A| = m*(m-1)^{n-2}. Wait, why? Because fix the first and last to the same color (m choices), then color the remaining n-2 sectors with each adjacent different. The second sector has (m-1) choices, third (m-1), etc., up to the (n-1)th sector. But wait, the (n-1)th sector is adjacent to the nth sector, which is the same as the first. So the (n-1)th sector just needs to be different from the (n-2)th and different from the first. Wait, but in this case, the first and last are fixed, so the (n-1)th sector must be different from the (n-2)th and the first (which is the same as the last). So the number of colorings is m * [ (m-2)*(m-1)^{n-3} ]? Hmm, not sure. This approach might be more complicated.Alternatively, using recurrence relations. Let me recall that the chromatic polynomial for a cycle graph C_n is (m-1)^n + (-1)^n*(m-1). That seems to hold as we checked for n=2,3,4. Therefore, the answer should be (m-1)^n + (-1)^n*(m-1). Therefore, the number of colorings is (m-1)^n + (-1)^n*(m-1).Alternatively, sometimes this formula is written as (m-1)^n + (-1)^n*(m-1). For example, when m=3, n=3, we get 2^3 + (-1)^3*2 =8 -2=6, which is 3*2*1=6. Correct. For m=3, n=4, we have 2^4 + 2=16+2=18. Let's check manually: first sector 3 choices, second 2, third 2, fourth must be different from third and first. If first and third are the same, fourth has 2 choices. If first and third are different, fourth has 1 choice. Number of colorings where first and third are the same: 3*2*1*2=12. Number where first and third are different: 3*2*1*1=6. Total 18, which matches the formula. Correct.Therefore, the formula seems valid. Therefore, the answer is (m-1)^n + (-1)^n*(m-1). This can also be written as (m-1)[(m-1)^{n-1} + (-1)^n]. But perhaps it's best left in the original form.Alternatively, sometimes this is presented using the chromatic polynomial formula for a cycle graph, which is indeed (m-1)^n + (-1)^n*(m-1). Therefore, the number of colorings is (m-1)^n + (-1)^n*(m-1).So, final answer: boxed{(m-1)^n + (-1)^n (m-1)}Final AnswerThe number of different ways to color the circle is boxed{(m-1)^n + (-1)^n (m-1)}.

✅:To determine the number of ways to color a circle divided into ( n ) sectors using ( m ) colors such that adjacent sectors have different colors, we need to consider the constraints of circular arrangements. First, for a linear arrangement (path) of ( n ) sectors, the number of valid colorings is ( m times (m-1)^{n-1} ). However, for a circular arrangement, the first and last sectors are adjacent, adding an additional constraint.The chromatic polynomial for a cycle graph ( C_n ) provides the solution. This polynomial is given by:[(m-1)^n + (-1)^n (m-1)] Key Steps:1. Linear Arrangement: The number of colorings for a line of ( n ) sectors is ( m times (m-1)^{n-1} ).2. Circular Constraint: Adjust for the circular nature by considering the chromatic polynomial for a cycle graph.3. Chromatic Polynomial: For a cycle graph ( C_n ), the chromatic polynomial is derived using inclusion-exclusion principles and recurrence relations, resulting in the formula: [ (m-1)^n + (-1)^n (m-1) ] Verification:- Case ( n = 2 ): [ (m-1)^2 + (m-1) = m(m-1) ] Correct, as two adjacent sectors must be different colors.- Case ( n = 3 ): [ (m-1)^3 - (m-1) = m(m-1)(m-2) ] Correct, as each sector must be different from its two neighbors.- Case ( n = 4 ): [ (m-1)^4 + (m-1) = m^4 - 4m^3 + 6m^2 - 3m ] Verified by manual calculation considering different cases for adjacent sectors.Thus, the number of different ways to color the circle is:[boxed{(m-1)^n + (-1)^n (m-1)}]

🔑:We are given a problem of coloring the sectors of a circle, with the stipulation that adjacent sectors must have different colors. Let's denote the number of ways to color the circle with ( n ) sectors using ( m ) colors as ( a_n(m) ).1. Base Case: When ( n = 2 ), there are ( m ) ways to color the first sector. For the second sector, we have ( m-1 ) choices to ensure that it is a different color from the first sector. Thus,[ a_2(m) = m(m-1) ]2. Recurrence Relation: For ( n geq 2 ), consider the following: - For the first sector ( S_1 ), there are ( m ) possible colors. - For ( S_2 ), there are ( m-1 ) choices (since it must be different from ( S_1 )). - This process continues similarly for sectors ( S_3, S_4, ldots, S_{n-1} ). - For ( S_n ), there are again ( m-1 ) choices to ensure ( S_n ) is different from ( S_{n-1} ), but it is not necessarily different from ( S_1 ). Therefore, this gives us ( m(m-1)^{n-1} ) ways to color the sectors with no consideration for whether ( S_1 ) and ( S_n ) are different.3. Classification: Among ( m(m-1)^{n-1} ) ways, we can classify them into: - Cases where ( S_n neq S_1 ): This is precisely ( a_n(m) ). - Cases where ( S_n = S_1 ). If ( S_n = S_1 ), we can merge ( S_1 ) and ( S_n ) and consider them as a single sector. This reduces the problem to coloring ( n-1 ) sectors. Consequently, this count is given by ( a_{n-1}(m) ). Therefore, we have the equation:[ a_n(m) + a_{n-1}(m) = m(m-1)^{n} ]4. Solving the Recurrence: Define ( b_n(m) = frac{a_n(m)}{(m-1)^n} ). Thus, the recurrence relation becomes:[ b_n(m) + frac{1}{m-1}b_{n-1}(m) = frac{m}{m-1} ][ b_n(m) - 1 = -frac{1}{m-1}(b_{n-1}(m) - 1) ] We can solve this by recognizing it as a geometric progression. Finding the terms recursively:[ b_2(m) = frac{a_2(m)}{(m-1)^2} = frac{m(m-1)}{(m-1)^2} = frac{m}{m-1} ][ b_n(m) - 1 = left(b_2(m) - 1right) left(-frac{1}{m-1}right)^{n-2} ][ b_2(m) - 1 = frac{m}{m-1} - 1 = frac{m - (m-1)}{m-1} = frac{1}{m-1} ][ b_n(m) - 1 = left(frac{1}{m-1}right) left(-frac{1}{m-1}right)^{n-2} = (-1)^{n} frac{1}{(m-1)^{n-1}} ][ b_n(m) = 1 + (-1)^{n} frac{1}{(m-1)^{n-1}} ] Substituting back for ( a_n(m) ):[ a_n(m) = (m-1)^n b_n(m) = (m-1)^n left(1 + (-1)^{n} frac{1}{(m-1)^{n-1}}right) ][ a_n(m) = (m-1)^n + (-1)^n (m-1) ]Conclusion:[ boxed{(m-1)^n + (-1)^n (m-1)} ]

🧠:Okay, let's see. I have these two equations: y + 4 = (x - 2)² and x + 4 = (y - 2)². And I need to find x² + y² given that x ≠ y. Hmm. Alright, so first, maybe I should try to solve these equations simultaneously. Let me write them down again:1. y + 4 = (x - 2)²2. x + 4 = (y - 2)²Since both equations are quadratic, this might result in a system that's a bit complex, but maybe there's a way to substitute one into the other. Let me try to express y from the first equation and plug it into the second. Or maybe express x from the second equation and plug into the first. Let's see.Starting with the first equation: y = (x - 2)² - 4. Let's expand that. (x - 2)² is x² - 4x + 4, so y = x² - 4x + 4 - 4 = x² - 4x. So y = x² - 4x. Okay, so that's y in terms of x. Now, substitute this expression for y into the second equation.Second equation is x + 4 = (y - 2)². Substituting y gives:x + 4 = ( (x² - 4x) - 2 )²Simplify inside the square: (x² - 4x - 2)². So:x + 4 = (x² - 4x - 2)²Now, that's a quartic equation, which might be a bit complicated, but let's try expanding the right-hand side.First, let me denote A = x² - 4x - 2. Then A² is (x² - 4x - 2)(x² - 4x - 2). Let's compute that:Multiply term by term:First term: x² * x² = x⁴x² * (-4x) = -4x³x² * (-2) = -2x²Next term: -4x * x² = -4x³-4x * (-4x) = 16x²-4x * (-2) = 8xNext term: -2 * x² = -2x²-2 * (-4x) = 8x-2 * (-2) = 4So summing all these terms:x⁴ -4x³ -2x² -4x³ +16x² +8x -2x² +8x +4Combine like terms:x⁴-4x³ -4x³ = -8x³-2x² +16x² -2x² = 12x²8x +8x = 16x+4So the expansion is x⁴ -8x³ +12x² +16x +4.Therefore, the equation becomes:x + 4 = x⁴ -8x³ +12x² +16x +4Bring all terms to the left-hand side:0 = x⁴ -8x³ +12x² +16x +4 - x -4Simplify:x⁴ -8x³ +12x² +15x = 0Factor out an x:x(x³ -8x² +12x +15) = 0So, either x = 0 or x³ -8x² +12x +15 = 0.But x=0 would mean, from the first equation, y = 0² -4*0 = 0. Then check if x ≠ y. If x=0, y=0, so x=y=0. But the problem states x ≠ y, so we can discard x=0. Therefore, we need to solve the cubic equation x³ -8x² +12x +15 = 0.Hmm, solving cubic equations can be tricky. Maybe we can try rational roots. The possible rational roots are factors of 15 divided by factors of 1, so ±1, ±3, ±5, ±15.Let's test x=1: 1 -8 +12 +15 = 1 -8= -7 +12=5 +15=20 ≠0x= -1: -1 -8(-1)^2= -1 -8= -9 +12(-1)= -9 -12= -21 +15= -6 ≠0x=3: 27 -8*9 +12*3 +15 =27 -72= -45 +36= -9 +15=6 ≠0x=5: 125 -8*25 +12*5 +15=125 -200= -75 +60= -15 +15=0. Oh! x=5 is a root.Great, so x=5 is a root. Then we can factor the cubic as (x -5)(quadratic). Let's perform polynomial division or use synthetic division.Using synthetic division for x=5:Coefficients: 1 | -8 | 12 | 15Bring down the 1.Multiply 1 by 5: 5. Add to -8: -3Multiply -3 by5: -15. Add to 12: -3Multiply -3 by5: -15. Add to15:0. Perfect.So the cubic factors as (x -5)(x² -3x -3). Therefore, the equation is x(x -5)(x² -3x -3)=0.So solutions are x=0, x=5, and roots of x² -3x -3=0. But as before, x=0 gives y=0 which is invalid. x=5, let's check. Then from y =x² -4x, y=25 -20=5. So x=5, y=5. But again, x=y=5, which is invalid because x≠y. So those solutions are invalid.Therefore, we need to look at the quadratic equation x² -3x -3=0. Let's solve that:x = [3 ± sqrt(9 +12)] / 2 = [3 ± sqrt(21)] /2.So x = [3 + sqrt(21)]/2 or x = [3 - sqrt(21)]/2.Now, we need to find the corresponding y for each x and ensure that x ≠ y.First, take x = [3 + sqrt(21)]/2. Then y = x² -4x. Let's compute that.First, compute x²:x² = ([3 + sqrt(21)]/2)^2 = (9 +6sqrt(21) +21)/4 = (30 +6sqrt(21))/4 = (15 +3sqrt(21))/2.Then compute 4x:4x =4*(3 + sqrt(21))/2 = 2*(3 + sqrt(21)) =6 +2sqrt(21).Therefore, y = x² -4x = (15 +3sqrt(21))/2 -6 -2sqrt(21).Convert 6 to 12/2 and 2sqrt(21) to (4sqrt(21))/2:y = (15 +3sqrt(21))/2 -12/2 -4sqrt(21)/2 = [15 -12 +3sqrt(21) -4sqrt(21)]/2 = (3 - sqrt(21))/2.So y = (3 - sqrt(21))/2.Now, check if x ≠ y. x = [3 + sqrt(21)]/2 and y = [3 - sqrt(21)]/2. Since sqrt(21) is about 4.583, so x ≈ (3 +4.583)/2 ≈7.583/2≈3.7915, and y≈(3 -4.583)/2≈(-1.583)/2≈-0.7915. So definitely x ≠ y.Similarly, for x = [3 - sqrt(21)]/2, compute y:x = [3 - sqrt(21)]/2.x² = ([3 - sqrt(21)]/2)^2 = (9 -6sqrt(21) +21)/4 = (30 -6sqrt(21))/4 = (15 -3sqrt(21))/2.4x =4*(3 - sqrt(21))/2 =2*(3 - sqrt(21))=6 -2sqrt(21).Thus, y =x² -4x = (15 -3sqrt(21))/2 -6 +2sqrt(21).Again, convert 6 to 12/2 and 2sqrt(21) to (4sqrt(21))/2:y = (15 -3sqrt(21))/2 -12/2 +4sqrt(21)/2 = [15 -12 + (-3sqrt(21) +4sqrt(21))]/2 = (3 + sqrt(21))/2.So y = (3 + sqrt(21))/2. Again, since x is [3 - sqrt(21)]/2 and y is [3 + sqrt(21)]/2, x ≈-0.7915 and y≈3.7915, so x ≠ y.Therefore, the two valid solutions are (x, y) = ([3 + sqrt(21)]/2, [3 - sqrt(21)]/2) and (x, y) = ([3 - sqrt(21)]/2, [3 + sqrt(21)]/2). Now, the problem asks for x² + y². Let's compute that for one of the solutions, since due to symmetry, both will give the same result.Take x = [3 + sqrt(21)]/2 and y = [3 - sqrt(21)]/2.Compute x² + y²:First, x² = ([3 + sqrt(21)]/2)^2 = (9 +6sqrt(21) +21)/4 = (30 +6sqrt(21))/4 = (15 +3sqrt(21))/2.Similarly, y² = ([3 - sqrt(21)]/2)^2 = (9 -6sqrt(21) +21)/4 = (30 -6sqrt(21))/4 = (15 -3sqrt(21))/2.Adding them together:x² + y² = (15 +3sqrt(21))/2 + (15 -3sqrt(21))/2 = (15 +15)/2 = 30/2 =15.Wait, 15? But 15 is one of the options (B). However, let me check again. Wait, let's compute again.Wait, x² is (3 + sqrt(21))² /4 = (9 +6sqrt(21)+21)/4 =30 +6sqrt(21)/4 = (15 +3sqrt(21))/2.Similarly, y² is (3 - sqrt(21))² /4 = (9 -6sqrt(21)+21)/4 =30 -6sqrt(21)/4 = (15 -3sqrt(21))/2.Adding x² + y²: (15 +3sqrt(21))/2 + (15 -3sqrt(21))/2 = (15 +15)/2 + (3sqrt(21)-3sqrt(21))/2 = 30/2 +0 =15.So x² + y² =15. But 15 is option (B). But wait, I thought the answer might be 25. Let me check my steps again.Wait, maybe there's a mistake. Let me think. Alternatively, perhaps there is another approach.Alternatively, instead of solving for x and y directly, maybe we can use the equations to find x² + y² more cleverly.Let me consider the two equations:1. y +4 = (x -2)²2. x +4 = (y -2)²Let me expand both equations:First equation: y +4 = x² -4x +4 → y = x² -4x.Second equation: x +4 = y² -4y +4 → x = y² -4y.So we have:y = x² -4xx = y² -4yNow, substitute y from the first equation into the second:x = (x² -4x)² -4(x² -4x)Let me compute (x² -4x)^2:(x² -4x)^2 = x⁴ -8x³ +16x²Then subtract 4(x² -4x):x⁴ -8x³ +16x² -4x² +16xSimplify:x⁴ -8x³ +12x² +16xSo the equation is:x = x⁴ -8x³ +12x² +16xBring x to the left-hand side:0 = x⁴ -8x³ +12x² +15xWhich is the same equation as before: x⁴ -8x³ +12x² +15x =0 → x(x³ -8x² +12x +15)=0. So same as before. So that's consistent. So proceeding as before, leading to x² + y²=15. Hmm.But looking back at the answer choices, (B) is 15. However, I need to verify once again because sometimes in quadratic systems, other solutions might exist.Wait, but when we solved for x and y, we found two solutions where x and y are swapped, and in both cases, x² + y²=15. But let's check with actual numbers.Take x ≈ (3 + 4.583)/2 ≈ 3.7915 and y ≈ (3 -4.583)/2 ≈ -0.7915.Compute x² ≈ (3.7915)^2 ≈14.375, y²≈ (-0.7915)^2≈0.626. Sum≈14.375 +0.626≈15.001. Which is approximately 15. Similarly, the other way around. So that seems correct.But wait, the options given are 10,15,20,25,30. So 15 is an option, so the answer should be (B) 15.But hold on, the problem is from a past exam? Maybe I made a miscalculation. Wait, let me try another approach. Let's compute x + y and x*y and use the identity x² + y² = (x + y)^2 -2xy.Let me see. From the equations:From first equation: y = x² -4x.From second equation: x = y² -4y.Let me denote S =x + y and P=xy.Our goal is to find x² + y² = S² -2P.So if we can find S and P, we can compute x² + y².But how?Alternatively, let's subtract the two original equations:First equation: y +4 = (x -2)²Second equation: x +4 = (y -2)²Subtracting: y +4 - (x +4) = (x -2)² - (y -2)²Simplify left side: y -x = (x -2)² - (y -2)²Right side: Use difference of squares: [(x -2) - (y -2)][(x -2) + (y -2)] = (x - y)(x + y -4)So:y -x = (x - y)(x + y -4)Factor left side: -(x - y) = (x - y)(x + y -4)Bring all terms to left:-(x - y) - (x - y)(x + y -4)=0Factor out (x - y):(x - y)[ -1 - (x + y -4) ] =0Thus:(x - y)(-1 -x -y +4)=0 → (x - y)(3 -x -y)=0Therefore, either x - y=0 or 3 -x -y=0.But given that x ≠ y, we discard x - y=0. So 3 -x -y=0 → x + y=3.So x + y=3.Ah! So now we have S =x + y=3.Now, since we have y =x² -4x, substitute into x + y=3:x + (x² -4x)=3 →x² -3x -3=0.Which is the quadratic we had earlier. So x² -3x -3=0, leading to x=(3 ±sqrt(21))/2. So same as before.Therefore, with x + y=3, and we need x² + y².Since x + y=3, then x² + y²= (x + y)^2 -2xy=9 -2xy.Therefore, if we can find xy, then we can compute x² + y².So let's find P=xy.From the quadratic equation x² -3x -3=0. Let's note that for the quadratic equation x² -3x -3=0, the sum of roots is 3 (from x + y=3, since x and y are the roots here? Wait, no, in the quadratic x² -3x -3=0, sum of roots is 3, product is -3.Wait, but in this quadratic, the roots are x values. However, y is determined by x. Wait, maybe not. Let me think.Alternatively, since x + y=3, then y=3 -x. Then substituting into y =x² -4x:3 -x =x² -4x →x² -3x -3=0. So yes, the roots of this quadratic are x, and y=3 -x.Therefore, in this context, the product of the roots of the quadratic equation x² -3x -3=0 is x1*x2= -3. But wait, how does that relate to our variables? Wait, here, each solution for x gives a corresponding y, but since x and y are related by y=3 -x, the product xy would be x(3 -x)=3x -x². But since x²=3x +3 (from the quadratic equation x²=3x +3), then substituting:xy=3x -x²=3x -(3x +3)= -3.Therefore, xy= -3.Therefore, x² + y²=9 -2*(-3)=9 +6=15.Therefore, x² + y²=15. So the answer is (B) 15.So that's a more straightforward way. By subtracting the original equations, we found that x + y=3, then through substitution, found that xy= -3, leading to x² + y²=15. This avoids having to compute the messy square roots and confirmations. So that's probably the smarter approach.Therefore, the correct answer is B) 15.Final Answerboxed{B}

✅:Given the equations ( y + 4 = (x - 2)^2 ) and ( x + 4 = (y - 2)^2 ) with ( x neq y ), we need to find the value of ( x^2 + y^2 ).First, we express ( y ) and ( x ) in terms of each other:1. ( y = (x - 2)^2 - 4 )2. ( x = (y - 2)^2 - 4 )Substituting ( y = x^2 - 4x ) into the second equation:[ x = (x^2 - 4x - 2)^2 - 4 ]Expanding and simplifying this equation leads to a quartic equation:[ x(x^3 - 8x^2 + 12x + 15) = 0 ]Since ( x neq 0 ), we solve the cubic equation ( x^3 - 8x^2 + 12x + 15 = 0 ). Using the Rational Root Theorem, we find ( x = 5 ) is a root. Factoring out ( (x - 5) ), we get:[ (x - 5)(x^2 - 3x - 3) = 0 ]Solving ( x^2 - 3x - 3 = 0 ) gives roots:[ x = frac{3 pm sqrt{21}}{2} ]For these roots, we find corresponding ( y ) values:- If ( x = frac{3 + sqrt{21}}{2} ), then ( y = frac{3 - sqrt{21}}{2} )- If ( x = frac{3 - sqrt{21}}{2} ), then ( y = frac{3 + sqrt{21}}{2} )Next, we use the identity ( x^2 + y^2 = (x + y)^2 - 2xy ). By subtracting the original equations, we find ( x + y = 3 ). Using the quadratic equation ( x^2 - 3x - 3 = 0 ), we find the product ( xy = -3 ).Thus:[ x^2 + y^2 = (x + y)^2 - 2xy = 3^2 - 2(-3) = 9 + 6 = 15 ]The value of ( x^2 + y^2 ) is (boxed{B}).

🔑:: Let's solve the problem step by step:1. Given Equations: [ y + 4 = (x - 2)^2 ] [ x + 4 = (y - 2)^2 ]2. Rearranging If Form: [ (x-2)^2 = y + 4 implies x^2 - 4x + 4 = y + 4 implies x^2 - 4x = y ] [ (y-2)^2 = x + 4 implies y^2 - 4y + 4 = x + 4 implies y^2 - 4y = x ]3. Combining the Equations: Add both equations: [ x^2 - 4x + y^2 - 4y = x + y ] Which simplifies to: [ x^2 + y^2 = 5(x + y) ]4. Formulating Another Equation: Subtract the second equation from the first: [ x^2 - y^2 - 4x + 4y = y - x ] Factorize the difference of squares: [ (x - y)(x + y) - 4(x - y) = y - x ] Since ( x neq y ), we can factor out ( (x - y) ): [ (x - y)(x + y - 4) = 3(y - x) ] Thus, rewrite and simplify: [ x + y - 4 = 3 implies x + y = 7 ]5. Using the Results to Find ( x^2 + y^2 ): Substitute ( x + y = 3 ) into: [ x^2 + y^2 = 5(x + y) ] [ x^2 + y^2 = 5 times 3 = 15 ] Conclusion:[boxed{15}]

🧠:Okay, so I need to figure out the value of B in this 9x9 grid game where two players take turns placing crosses (first player) and noughts (second player). When the grid is full, they count K, the number of rows and columns where crosses outnumber noughts, and H, the number where noughts outnumber crosses. Then B = K - H. The question is asking for the value of B such that the first player can guarantee at least B, and the second player can prevent the first player from getting more than B. Hmm, interesting.Let me start by understanding the game mechanics. Each row and column is a separate contest between crosses and noughts. Since the grid is 9x9, there are 9 rows and 9 columns, making a total of 18 lines (rows + columns) that contribute to K and H. Each line has 9 cells. For a row or column to count towards K, there must be more crosses than noughts in that line. Similarly, for H, more noughts than crosses. If there's an equal number, neither K nor H is incremented. So B is the difference between these two counts. The first player wants to maximize B, the second wants to minimize it.Since the game is played on a 9x9 grid, which is odd-sized, each row and column has an odd number of cells. That means in each line, there can't be a tie; one symbol must outnumber the other. Wait, no. Wait, 9 cells: if each player takes turns, the total number of crosses and noughts in each line will depend on how many each placed there. But since the entire grid is filled, each line has 9 cells. If both players play optimally, how does the count work?Wait, actually, the total number of crosses and noughts in the entire grid will be 81 cells. Since the first player starts, they will have placed one more cross than noughts. Because 81 is odd: 41 crosses and 40 noughts. So overall, crosses have a one-cell advantage. But how does this translate to individual rows and columns?But the problem counts K and H as the number of rows and columns where crosses outnumber noughts and vice versa. So each row and column is evaluated individually. Since each row has 9 cells, to outnumber the other symbol, you need at least 5 of your symbol in that line.Therefore, for each row and column, if there are 5 or more crosses, it's a K; if 5 or more noughts, it's an H. Since the total crosses are 41 and noughts 40, maybe the first player can arrange their crosses to maximize K while minimizing H, and the second player does the opposite.But how does the interaction between rows and columns work? Because a cell is part of both a row and a column. So when a player places a cross or a nought, it affects both a row and a column. This interdependency complicates things.Let me think of a simpler case. Suppose the grid is 1x1. Then the first player places a cross, so K is 2 (the single row and column), H is 0, so B=2. But that's trivial. If the grid is 3x3. Let's see. First player can place crosses, and second noughts. Total crosses: 5, noughts: 4. Each row and column has 3 cells. To have more crosses, need at least 2 in a line. Wait, 3 cells: majority is 2. So in each line, crosses need 2 to win. But since the total crosses are 5, spread over 3 rows and 3 columns. But how to maximize K. Hmm. Maybe in the 3x3 grid, the first player can secure a certain number of rows and columns. But this is just an example; the problem is 9x9.Back to 9x9. Each line (row or column) has 9 cells. To have a majority, need at least 5. So the first player wants as many lines as possible with at least 5 crosses, and the second player wants to prevent that by having as many lines with at least 5 noughts.But since the first player has one more cross overall, maybe there's a balance in how these are distributed. The key is that each cross placed affects both a row and a column. Similarly for noughts. So the players are trying to influence both rows and columns with each move.I need to find the minimal B such that the first player can ensure at least B, and the second can prevent more than B. So B is the equilibrium point where both can enforce it.Let me consider the total number of lines: 18. For each line, either K or H increments, but not both. So K + H <= 18, but actually, since each line is either K or H, K + H = 18. Because in each line, one of them must outnumber the other. Wait, but no: if a line has exactly 5 crosses and 4 noughts, then K increments. If it's 5 noughts and 4 crosses, H increments. But in the case of 5-4, it's a difference of 1. However, since the total number of crosses is 41 and noughts 40, each cross has a slight edge.But how does this total relate to the counts in individual lines?Wait, maybe there's a relationship between the total number of crosses and the sum over all lines of the majority counts. Let's see. Each cross is in one row and one column. Similarly, each nought is in one row and one column. So when we count, for each line, whether crosses or noughts have the majority, we are effectively counting for each line which player has more symbols there.But the total number of crosses is 41 and noughts 40. How does this affect the total number of lines where crosses dominate versus noughts?Suppose we consider that each cross contributes to one row and one column, and similarly for noughts. But the exact distribution depends on how the players place their symbols.But perhaps we can model this as a zero-sum game where the first player is trying to maximize B = K - H, and the second player is trying to minimize it.Alternatively, since each line is a separate contest, but each move affects two lines (a row and a column), the players are balancing their influence on multiple lines at once.Another approach: since the total number of crosses is 41 and noughts 40, the first player has an overall advantage. The question is how much of this advantage can be translated into dominating more lines (rows and columns).But how to distribute 41 crosses in the grid such that as many rows and columns as possible have at least 5 crosses. Similarly, the second player is trying to place 40 noughts to block as many rows and columns as possible.But given that each cross and nought is placed in a cell that's part of both a row and a column, it's a matter of efficiently allocating your symbols to maximize your influence.Let me think about the minimum number of crosses needed to dominate a line. For a line (row or column) to have crosses outnumbering noughts, you need at least 5 crosses. So 5 crosses and 4 noughts. Similarly, for noughts to dominate, 5 noughts and 4 crosses.Each cross placed in a line contributes towards dominating that line, but since each cross is in one row and one column, placing a cross in a cell affects both its row and column.Similarly, noughts affect both a row and column.So the problem is akin to a bipartite graph where players are assigning weights (crosses or noughts) to edges (cells) to influence nodes (rows and columns). The goal is to maximize the number of nodes where your weight is dominant.But I'm not sure if that's helpful. Maybe another angle: the total number of crosses is 41. To dominate a row, you need 5 crosses. Similarly for a column. But each cross can contribute to one row and one column.Therefore, if you want to maximize the number of dominated rows and columns, you need to distribute the crosses efficiently.But since the second player is trying to block by placing noughts, the first player can't just freely place crosses wherever. It's a turn-based game with alternating moves.Wait, the game is that players take turns, first player places crosses, second places noughts, until the grid is filled. So each of the 81 cells is filled, with 41 crosses and 40 noughts.So the first player has the first move advantage, but since they have one more cross, perhaps they can leverage that to control more lines.But how to model the optimal play? Maybe consider that for each line, the difference in the number of crosses and noughts is either +1 (if crosses dominate) or -1 (if noughts dominate). But since each line has 9 cells, the difference can be odd numbers from -9 to +9. However, since the total crosses are 41 and noughts 40, the total difference across all lines is?Wait, each cross is counted in one row and one column, so the total number of crosses in all lines is 41*2 = 82. Similarly, noughts are 40*2 = 80. Therefore, the total number of crosses across all lines is 82, and noughts 80. Each line has 9 cells, so 18 lines, each with 9 cells, total cells 162, which is 81*2, so that checks out.Now, for each line, the difference between crosses and noughts. Let’s denote for each line, D_i = (number of crosses - number of noughts). Since each line has 9 cells, D_i can be odd numbers from -9 to +9. For the line to be a K, D_i >= 1; for H, D_i <= -1. The total sum of all D_i across all 18 lines is 82 - 80 = 2. So the total sum of D_i is 2.Therefore, the sum of all D_i is 2. Let's denote K is the number of lines where D_i >=1, and H is the number where D_i <= -1. Then, since each D_i is at least 1 for K lines, and at most -1 for H lines, and the remaining lines (if any) have D_i =0. But wait, since each line has an odd number of cells, D_i must be odd. Because 9 is odd, the difference between crosses and noughts is odd. So D_i is odd. Therefore, D_i can be 1,3,5,7,9 for K lines, and -1,-3,-5,-7,-9 for H lines. There can't be any lines with D_i=0 because that would require equal crosses and noughts, which is impossible in an odd-length line.Therefore, all 18 lines are either K or H. Therefore, K + H = 18. And sum of D_i = sum over K lines of (positive odd numbers) + sum over H lines of (negative odd numbers) = 2.Let’s denote S = sum of D_i = 2. Let’s let K be the number of lines where D_i >=1 (i.e., K lines with D_i =1,3,5,7,9) and H =18 - K lines where D_i <=-1 (i.e., D_i =-1,-3,...,-9). Then, S = sum_{K lines} D_i + sum_{H lines} D_i = 2.Our goal is to find the maximum possible B = K - H such that S = 2. But since H =18 - K, B = K - (18 - K) = 2K -18. So B = 2K -18. Hence, maximizing B is equivalent to maximizing K.But subject to the constraint that the sum of all D_i is 2. Let's try to model this. Let’s assume that the minimal positive D_i is 1 and the minimal negative D_i is -1. To maximize K, we need as many lines as possible with D_i=1, and the remaining lines with as small negative D_i as possible. Because each K line contributes at least 1, and each H line contributes at least -1 (but could be more negative). Since the total sum must be 2, we can set up the equation:Let’s assume that K lines have D_i=1, and H lines have D_i=-1. Then total sum would be K*1 + H*(-1) = K - H = B. But since H =18 - K, this becomes K - (18 - K) = 2K -18. However, the total sum would also be 2. Therefore, 2K -18 = 2 => 2K =20 => K=10. So B=2*10 -18=2.But wait, this would require that all K lines have D_i=1 and all H lines have D_i=-1. However, the sum of D_i would be 10*1 +8*(-1)=10 -8=2, which matches. But is this possible? That would mean that there are 10 lines where crosses have 5 and noughts 4 (since D_i=1), and 8 lines where noughts have 5 and crosses 4 (D_i=-1). But let's check if this is feasible given the total number of crosses and noughts.Each K line (10 lines) has 5 crosses and 4 noughts. Each H line (8 lines) has 4 crosses and 5 noughts. Let's compute total crosses and noughts.For the rows: Suppose there are R rows with crosses leading and C columns with crosses leading. Similarly, (9 - R) rows and (9 - C) columns with noughts leading. Wait, but the problem counts rows and columns together. So K = R + C and H = (9 - R) + (9 - C) =18 - (R + C). Therefore, K = R + C, H =18 - K. So total crosses in rows: R*5 + (9 - R)*4. Similarly, total crosses in columns: C*5 + (9 - C)*4. But since each cross is in one row and one column, the total crosses would be R*5 + (9 - R)*4 + C*5 + (9 - C)*4. Wait, no. Each cross is counted in both a row and a column. Therefore, total crosses would be (sum over all rows of crosses in rows) = sum over all columns of crosses in columns). So total crosses is equal to both the sum of crosses in rows and the sum of crosses in columns. But since each cross is in exactly one row and one column, the total number of crosses is equal to the sum over rows of crosses in rows, which is equal to the sum over columns of crosses in columns. Therefore, the total crosses can be computed as:Total crosses = sum_{rows} crosses in each row + sum_{columns} crosses in each column. Wait, no. Each cross is in one row and one column. Therefore, the total number of crosses is the same as the sum of crosses in each row (which counts all crosses), and also the same as the sum of crosses in each column. Therefore, total crosses = sum_{rows} crosses in row = sum_{columns} crosses in column.Similarly for noughts. Therefore, if we have R rows with 5 crosses and 9 - R rows with 4 crosses, then total crosses from rows would be 5R +4(9 - R)=5R +36 -4R= R +36. Similarly, columns: if we have C columns with 5 crosses and 9 - C columns with 4 crosses, total crosses from columns would be 5C +4(9 - C)=5C +36 -4C= C +36. But wait, this can't be right, because the total crosses should be equal when summed by rows and by columns. But actually, the total crosses are counted once in rows and once in columns. Wait, no. Each cross is in one row and one column, so the total number of crosses is 41. But the sum over rows of crosses in rows is 41, and the sum over columns of crosses in columns is also 41. Therefore, if we have R rows with 5 crosses and 9 - R rows with 4 crosses, then total crosses from rows would be 5R +4(9 - R)= R +36. Setting this equal to 41: R +36=41 => R=5. Similarly, for columns: C +36=41 => C=5. So R=5 and C=5. Therefore, total K= R + C=10. Then H=18 -10=8. Then B=10 -8=2.But wait, this suggests that if you have 5 rows with 5 crosses and 4 rows with 4 crosses, and similarly 5 columns with 5 crosses and 4 columns with 4 crosses, then total crosses would be 5*5 +4*4 =25 +16=41. Wait, no: rows contribute 5*5 +4*4=25 +16=41 crosses? Wait, no: each row is either 5 or 4 crosses. If there are R=5 rows with 5 crosses, and 4 rows with 4 crosses, total crosses from rows is 5*5 +4*4=25 +16=41. Similarly for columns: 5 columns with 5 crosses, 4 columns with 4 crosses, total crosses from columns is 5*5 +4*4=41. But since each cross is in one row and one column, the total crosses must be 41. So this works out.Therefore, this configuration is possible: 5 rows with 5 crosses each, 4 rows with 4 crosses each; 5 columns with 5 crosses each, 4 columns with 4 crosses each. However, the placement of crosses must be such that each cell is in one row and one column. But is this possible? For example, arranging the grid so that there are 5 rows each with 5 crosses and 4 rows with 4 crosses, and similarly for columns. This seems like a combinatorial design problem.But even if such a configuration exists, does it mean that the first player can enforce it regardless of the second player's moves? Or is this just a theoretical upper bound?Wait, the first player wants to maximize K - H, the second player wants to minimize it. If the first player can arrange 10 K lines and 8 H lines, achieving B=2, and the second player can prevent B from being higher than 2, then B=2 is the answer.But we need to verify if this is indeed achievable. Let's think about the game dynamics.The first player starts by placing crosses, and the second player responds with noughts. Since both players are trying to influence the rows and columns, the optimal strategy would involve controlling key cells that affect multiple lines. For example, placing a cross in a cell that is part of a row and column that the first player wants to win, while the second player would block by placing noughts in cells that are critical for multiple lines.However, since the grid is symmetric, perhaps the first player can mirror the second player's moves in some way to maintain a balance. Wait, but since the grid is odd-sized, there's a central cell which might be important.Alternatively, think of the game as a collection of independent lines, but they are not independent because each move affects two lines. Therefore, it's a matter of pairing up the lines and using the extra cross to gain an advantage.Another approach: since there's an odd number of cells in each line, and the first player has an extra move, the first player can ensure that they have the last move in each line. Wait, but each line has 9 cells, so the first player would place the 1st, 3rd, 5th, 7th, and 9th symbols in each line. Therefore, in each line, the first player can make 5 moves (since 9 is odd), and the second player makes 4 moves. Therefore, in each line, the first player can potentially place 5 crosses, but the second player is trying to disrupt this.Wait, but the players alternate turns globally, not per line. So if the first player focuses on a particular row, they can try to place their cross in that row, but the second player can choose to place their nought in a different row or column. So the first player can't necessarily control all cells in a single line because the second player can interfere.However, perhaps the first player can use a strategy of pairing rows and columns and ensuring that for each pair, they get a majority in one of them. But this is vague.Alternatively, since the total advantage is 1 cross, and each line requires a majority of 1 (5-4), perhaps the first player can convert their global advantage into a local advantage in a certain number of lines. The total difference in all lines is 2 (from earlier, sum D_i=2). If each K line contributes +1 and each H line contributes -1, then with 10 K lines and 8 H lines, the total is 10*1 +8*(-1)=2. This seems to fit. Therefore, the maximum possible B is 2.But can the first player ensure that? And can the second player prevent B from being higher than 2?To verify, let's think about the strategies.First player's strategy: try to create as many lines where they have a 5-4 majority. Since they have an extra cross, they can potentially do this in 10 lines (5 rows and 5 columns), but the math above shows that with 5 rows and 5 columns (total 10 lines), each with 5 crosses, the total crosses would be 5*5 +4*4=41. Wait, no, that's rows and columns separately. But each cross is in both a row and a column. So if 5 rows have 5 crosses each, that's 25 crosses. The remaining 4 rows have 4 crosses each, 16 crosses. Total rows:25+16=41 crosses. Similarly, columns:5 columns with 5 crosses each (25) and 4 columns with 4 crosses each (16), total 41. However, the crosses in the rows and columns must overlap. For example, the 5 crosses in a row must be placed in columns that also have their own crosses. This is possible via a design where the intersections of the 5 high-cross rows and 5 high-cross columns have extra crosses. But does such a design exist?This is similar to a combinatorial problem where you want a 9x9 matrix with 5 rows and 5 columns each having 5 crosses, and the rest having 4 crosses, such that the total number of crosses is 41. Let's check:If we have 5 rows with 5 crosses each: 5*5=25 crosses. The remaining 4 rows have 4 crosses each:4*4=16. Total rows:25+16=41. Similarly, columns:5 columns with 5 crosses, 4 columns with 4 crosses:25+16=41. However, the overlapping cells must be counted in both rows and columns. This is similar to a bipartite graph where rows and columns are nodes and cells are edges. We need to assign 41 crosses such that 5 rows have degree 5, 4 rows have degree 4, 5 columns have degree 5, 4 columns have degree 4. This is possible if the bipartite graph is possible. According to the Gale-Ryser theorem, for a bipartite graph with degree sequences (rows: five 5s and four 4s; columns: five 5s and four 4s), the necessary and sufficient condition is that the sum of row degrees equals sum of column degrees (which they do:41) and for every k, the sum of the largest k row degrees is at most the sum of the largest k column degrees plus the sum of the smallest remaining column degrees. This might hold here, but I'm not sure. It might be complicated, but intuitively, since both row and column degree sequences are similar, it's likely possible. Therefore, such a matrix exists.Therefore, if the first player can arrange their crosses in such a configuration, they can achieve B=2. However, the question is whether they can do this regardless of the second player's actions. Because the second player is actively trying to prevent this.Similarly, the second player would want to arrange their noughts to maximize H and minimize K. But since they have fewer noughts (40), they might not be able to prevent the first player from getting at least B=2.But how?Perhaps by mirroring the first player's moves in some symmetric fashion. However, since the grid is of odd size, there is a central cell. If the second player uses a pairing strategy, pairing each cell with its symmetric counterpart and mirroring the first player's moves, they can limit the first player's advantage. However, because the total number of cells is odd, the first player gets to place the central cell, which might give them an extra advantage.Alternatively, the second player can adopt a strategy where for every cross the first player places, they place a nought in a different row and column, disrupting the first player's attempts to build majorities.But this is getting too vague. Let's think about the total difference. Since the total sum of D_i is 2, the first player can't achieve a higher total. If they try to make more than 10 K lines, say 11, then H=7, so B=22-18=4. But the total sum would be 11*1 +7*(-1)=11-7=4, which contradicts the total sum of 2. Therefore, it's impossible. Hence, the maximum possible B is 2, given the total sum constraint.Therefore, regardless of strategies, B cannot exceed 2 because the total sum of differences is fixed at 2. Therefore, the first player cannot guarantee more than 2, and the second player can ensure that B does not exceed 2. Conversely, the first player can ensure at least 2 by arranging their crosses to achieve the total sum of 2, which corresponds to B=2. Therefore, B=2 is the answer.But wait, let me double-check. The total sum S=2, which is the sum over all D_i. Since each D_i is odd, and there are 18 terms (all odd), the sum of 18 odd numbers is even (since 18 is even). 2 is even, so that's okay. If we have K lines with D_i=1 and H lines with D_i=-1, then the sum is K - H = B =2. But K + H =18, so B=2K -18. Solving 2K -18=2 gives K=10, H=8. This works. If any line has a higher difference, say D_i=3, then another line must compensate with a more negative difference. For example, if one line has D_i=3, then to keep the total sum at 2, another line must have D_i=-1. But then K remains the same (since 3 is still a K line), but the total sum would still require the same number of K and H lines. Wait, no: if you have one line with D_i=3 instead of 1, and another line with D_i=-3 instead of -1, the total sum remains the same: (3 -3) instead of (1 -1). Therefore, varying the differences doesn't change the total sum. Therefore, the total B is still determined by the number of K and H lines, not their individual differences. Therefore, as long as the sum of D_i is 2, B=2K -18=2. Therefore, regardless of how the differences are distributed among the lines, B=2.Wait, but if some lines have higher differences, but the total sum remains 2, does that affect B? No, because B is the difference between the number of K and H lines, not the magnitude of the differences. Therefore, even if some lines have larger differences, B is still K - H =2K -18. And since the total sum S=2= sum D_i = sum over K lines D_i + sum over H lines D_i. But the sum of D_i for K lines must be at least K (if each has D_i=1), and the sum of D_i for H lines must be at most -H (if each has D_i=-1). Therefore, the maximum possible sum S is at least K - H = B, but we know S=2. Therefore, B cannot exceed 2. Conversely, if K - H =2, then S=2. Therefore, B=2 is the only possible value that satisfies both conditions: the first player can ensure at least 2, and the second player can prevent it from being higher than 2.Hence, the value of B is 2.Final Answerboxed{2}

✅:The game takes place on a 9 times 9 grid where two players take turns placing crosses (first player) and noughts (second player). The goal is to determine the value of B such that the first player can ensure a win of at least B, and the second player can ensure the first player's score is no more than B.1. Game Mechanics: - Each row and column is a contest between crosses and noughts. - A row or column contributes to K if crosses outnumber noughts, and to H if noughts outnumber crosses. - The total number of cells is 81, so the first player places 41 crosses and the second player places 40 noughts.2. Key Observations: - Each row and column has 9 cells, requiring at least 5 crosses or noughts to outnumber. - The total number of lines (rows and columns) is 18. - The difference B = K - H must be calculated.3. Sum of Differences: - Each cross contributes to one row and one column, leading to a total of 82 crosses and 80 noughts across all lines. - The total difference between crosses and noughts across all lines is 82 - 80 = 2.4. Line Contributions: - Each line (row or column) contributes either +1 (crosses dominate) or -1 (noughts dominate). - The sum of these contributions must equal 2.5. Maximizing B: - Let K be the number of lines where crosses dominate and H be the number where noughts dominate. - K + H = 18 and K - H = B. - Solving K - H = 2 (total sum) and K + H = 18 gives K = 10 and H = 8, resulting in B = 2.6. Conclusion: - The first player can ensure B geq 2 by optimally placing crosses. - The second player can prevent B from exceeding 2 by strategically placing noughts.Thus, the value of B is (boxed{2}).

🔑:1. First Player's Strategy: The first player should place their mark (cross) in the center of the board. Let's denote the center of the 9x9 board as ( (5, 5) ). Following this, every time the second player makes a move, the first player should respond by placing a mark in the cell that is symmetric relative to the center of the board. For instance, if the second player places a mark at ( (i, j) ), then the first player should place a mark at ( (10-i, 10-j) ). This ensures that no matter how the second player plays, the board will maintain symmetry concerning the center. 2. Symmetry Maintenance: By maintaining the described symmetry, every move by the second player will be countered in a way that the positions ( (i, j) ) and ( (10-i, 10-j) ) will hold a cross and a circle, respectively. 3. Final Board Configuration Analysis: In the final position, in every pair of cells symmetric to each other with respect to the center, there will be one cross and one circle. This symmetry implies that there will be more crosses in the central row and the central column as these rows and columns are counted twice in the symmetry strategy. Thus, the central row and column will favor crosses over circles.4. Calculation of mathbf{B}: At the end of the game, let's analyze the rows and columns: - Central row ( (5, j) ) for ( j = 1, 2, ldots, 9 ) will contain more crosses than circles. - Central column ( (i, 5) ) for ( i = 1, 2, ldots, 9 ) will also contain more crosses than circles. - For every non-central row and column, there exists a symmetric counterpart that balances the number of crosses and circles. Therefore, we conclude that there will be exactly two advantageous lines (central row and column) for the first player.5. Second Player's Strategy to Ensure mathbf{B} Does Not Exceed 2: The second player can assure that the first player’s advantage does not exceed 2 by following this strategy: - If the second player has the opportunity to place a mark in the cell symmetric to the one currently occupied by the first player, they should do so. - If the central cell has already been occupied, the second player should place their marks such that each move counteracts the symmetry maintained by the first player. This continues until the final stages, ensuring that the first player cannot achieve an advantage in any extra lines beyond the central row and column.6. Conclusion: Following these strategies, the first player can always ensure a winning advantage of 2 points, and the second player can always restrict the first player’s advantage to no more than 2 points.[boxed{B = 2}]

🧠:Alright, let me try to work through this problem step by step. It's about alloys and the force they exert on the bottom of a container when submerged in water. Hmm, okay, so first, I need to understand what's being asked here. We have two different alloys, A and B, each made of two metals. The third alloy is a combination of the original alloys, and we need to find the force it exerts on the bottom of the container when placed in water.Let me start by recalling some physics concepts. When an object is placed in water, it displaces some volume of water, which results in a buoyant force. The force exerted on the bottom of the container would then be the actual weight of the object minus the buoyant force. Wait, but actually, if the object is submerged and resting on the bottom, the force it exerts on the bottom might just be its actual weight, because the buoyant force would reduce the apparent weight, but if it's sitting on the bottom, the normal force (which is what we're measuring as the force on the bottom) would be the actual weight minus the buoyant force. Or is it different? Hmm, maybe I need to clarify this.Wait, the problem states that when the alloys are placed in a container of water, they exert a certain force on the bottom. So, if the alloys are submerged, the buoyant force would act upwards, reducing the normal force from the container's bottom. Therefore, the force exerted on the bottom (which is the normal force) would be equal to the weight of the alloy minus the buoyant force. Alternatively, if the alloy is floating, the buoyant force would equal the weight, so the normal force would be zero. But since both alloys are exerting some force, they must be submerged, not floating. Therefore, the force on the bottom is the actual weight minus the buoyant force. Wait, but that might not be the case. If the container is filled with water and the alloy is placed in, causing displacement, but maybe the container is not filled to the brim. Alternatively, maybe the problem is considering the total force on the bottom, which would include the weight of the water plus the submerged alloy. Wait, but the problem says "exerts a force of 30 N on the bottom," which might just be the normal force from the alloy's weight adjusted by buoyancy. Hmm, this is a bit confusing. Let me think again.Alternatively, maybe the force exerted on the bottom is simply the weight of the alloy minus the buoyant force. Because the buoyant force is the upward force, so the net force on the bottom would be the weight minus that buoyant force. So, F = Weight - Buoyant Force. But the problem states the force is 30 N for Alloy A and 10 N for Alloy B. So, using this formula, we can write equations for each alloy. Let's note that.First, the weight of Alloy A is mass * gravity. Its mass is 6 kg, so weight is 6 kg * 9.8 m/s² ≈ 58.8 N. But the force on the bottom is 30 N. If that's the case, then according to F = Weight - Buoyant Force, the buoyant force would be 58.8 N - 30 N = 28.8 N. Similarly, for Alloy B, mass is 3 kg, weight is 3 * 9.8 ≈ 29.4 N. Force on the bottom is 10 N, so buoyant force is 29.4 - 10 = 19.4 N.But buoyant force is equal to the weight of the displaced water, which is density_water * volume_displaced * gravity. So, Buoyant Force = ρ_water * V * g. Therefore, we can find the volume displaced by each alloy. Let's compute that.For Alloy A: 28.8 N = 1000 kg/m³ * V_A * 9.8 m/s². Solving for V_A: V_A = 28.8 / (1000 * 9.8) ≈ 28.8 / 9800 ≈ 0.0029388 m³.For Alloy B: 19.4 N = 1000 * V_B * 9.8. So V_B = 19.4 / 9800 ≈ 0.0019796 m³.Now, the volume of each alloy is equal to the volume of displaced water, right? Because when submerged, the volume displaced is equal to the volume of the object. So, the volume of Alloy A is 0.0029388 m³, and Alloy B is 0.0019796 m³.But each alloy is a combination of two metals. Let me note that Alloy A has the first metal twice as abundant as the second. So, if the total mass is 6 kg, then let's denote the mass of the second metal as m, so the first metal is 2m. Then 2m + m = 6 kg => 3m = 6 kg => m = 2 kg. Therefore, Alloy A is 4 kg of metal 1 and 2 kg of metal 2.Similarly, Alloy B has the first metal five times less abundant than the second. So, the mass of the first metal is (1/5) that of the second. Let me denote the mass of the first metal as m', so the second metal is 5m'. Total mass is m' + 5m' = 6m' = 3 kg. Therefore, m' = 0.5 kg. So, Alloy B is 0.5 kg of metal 1 and 2.5 kg of metal 2.Now, the key here is probably to find the densities of the two metals. Because the volume of each alloy is the sum of the volumes of the constituent metals. Since density is mass/volume, volume is mass/density.Let me denote the density of metal 1 as ρ₁ and metal 2 as ρ₂.For Alloy A: Volume V_A = (4 kg)/ρ₁ + (2 kg)/ρ₂ ≈ 0.0029388 m³.For Alloy B: Volume V_B = (0.5 kg)/ρ₁ + (2.5 kg)/ρ₂ ≈ 0.0019796 m³.So, we have two equations:1) 4/ρ₁ + 2/ρ₂ = 0.00293882) 0.5/ρ₁ + 2.5/ρ₂ = 0.0019796We need to solve these two equations for ρ₁ and ρ₂. Then, once we have ρ₁ and ρ₂, we can find the density of the third alloy (which is a combination of Alloy A and Alloy B), compute its volume, find the buoyant force, and then compute the force on the bottom.Wait, but actually, the third alloy is obtained by combining the original alloys. So, we need to combine Alloy A and Alloy B. The problem doesn't specify in what ratio, but maybe it's combining both alloys together? So total mass would be 6 kg + 3 kg = 9 kg. The composition would be the sum of the masses of each metal from both alloys. From Alloy A: 4 kg metal 1 and 2 kg metal 2. From Alloy B: 0.5 kg metal 1 and 2.5 kg metal 2. So total in the third alloy: 4 + 0.5 = 4.5 kg metal 1, and 2 + 2.5 = 4.5 kg metal 2. So the third alloy has equal masses of metal 1 and metal 2, each 4.5 kg.Therefore, the third alloy has a total mass of 9 kg, composed of 4.5 kg each of metal 1 and metal 2. Then, we need to compute the volume of this third alloy, which would be (4.5 kg)/ρ₁ + (4.5 kg)/ρ₂. Then, the buoyant force would be ρ_water * V * g, and the force on the bottom would be the weight of the alloy (9 kg * 9.8 m/s²) minus the buoyant force.But to do all that, we first need to find ρ₁ and ρ₂. So, let's focus on solving the two equations.First, let me convert the volumes into exact values:For Alloy A: Volume V_A = 28.8 N / (1000 kg/m³ * 9.8 m/s²) = 28.8 / 9800 m³Similarly, V_A = 28.8 / 9800 = 0.0029387755 m³ ≈ 0.0029388 m³Similarly, V_B = 19.4 / 9800 ≈ 0.0019795918 m³ ≈ 0.0019796 m³So, the equations are:Equation 1: 4/ρ₁ + 2/ρ₂ = 28.8 / 9800Equation 2: 0.5/ρ₁ + 2.5/ρ₂ = 19.4 / 9800Let me write them as:1) 4/ρ₁ + 2/ρ₂ = 28.8 / 98002) 0.5/ρ₁ + 2.5/ρ₂ = 19.4 / 9800Let me denote x = 1/ρ₁ and y = 1/ρ₂. Then the equations become:1) 4x + 2y = 28.8 / 98002) 0.5x + 2.5y = 19.4 / 9800Now, let's compute the right-hand sides numerically:28.8 / 9800 ≈ 0.002938775519.4 / 9800 ≈ 0.0019795918So, equations:1) 4x + 2y = 0.00293877552) 0.5x + 2.5y = 0.0019795918Let me solve these two equations for x and y.Multiply the second equation by 8 to make coefficients more manageable:Equation 1: 4x + 2y = 0.0029387755Equation 2 scaled: 4x + 20y = 0.0158367344Subtract Equation 1 from the scaled Equation 2:(4x + 20y) - (4x + 2y) = 0.0158367344 - 0.002938775518y = 0.0128979589Therefore, y = 0.0128979589 / 18 ≈ 0.0007165533Then, substitute back into Equation 1 to find x.From Equation 1: 4x + 2y = 0.00293877554x = 0.0029387755 - 2*0.0007165533 ≈ 0.0029387755 - 0.0014331066 ≈ 0.0015056689Thus, x = 0.0015056689 / 4 ≈ 0.0003764172So, x ≈ 0.0003764172 (which is 1/ρ₁) and y ≈ 0.0007165533 (which is 1/ρ₂)Therefore, ρ₁ = 1 / x ≈ 1 / 0.0003764172 ≈ 2657.3 kg/m³Similarly, ρ₂ = 1 / y ≈ 1 / 0.0007165533 ≈ 1395.3 kg/m³Hmm, let me check these densities. Metal 1 is about 2657 kg/m³ and metal 2 is about 1395 kg/m³. These seem plausible. For example, aluminum is around 2700 kg/m³, so maybe metal 1 is aluminum. Metal 2 is less dense, maybe magnesium or something else. Anyway, the exact metals don't matter, just their densities.Now, moving on to the third alloy. As previously determined, the third alloy has 4.5 kg of each metal. So, the volume of the third alloy is:V_third = (4.5 kg)/ρ₁ + (4.5 kg)/ρ₂ = 4.5*(1/ρ₁ + 1/ρ₂) = 4.5*(x + y) = 4.5*(0.0003764172 + 0.0007165533) = 4.5*(0.0010929705) ≈ 4.5*0.0010929705 ≈ 0.0049183673 m³Then, the buoyant force on the third alloy is:Buoyant Force = ρ_water * V_third * g = 1000 kg/m³ * 0.0049183673 m³ * 9.8 m/s² ≈ 1000 * 0.0049183673 * 9.8 ≈ 0.0049183673 * 9800 ≈ 48.2 NThe weight of the third alloy is 9 kg * 9.8 m/s² = 88.2 NTherefore, the force exerted on the bottom is Weight - Buoyant Force = 88.2 N - 48.2 N ≈ 40 NWait, but 88.2 - 48.2 is exactly 40 N. Hmm, that's a round number. Let me check the calculations again to be sure.First, let's verify the values of x and y:From the equations:Equation 1: 4x + 2y = 28.8 / 9800 ≈ 0.0029387755Equation 2: 0.5x + 2.5y = 19.4 / 9800 ≈ 0.0019795918After solving, we found x ≈ 0.0003764172 and y ≈ 0.0007165533So, 4x + 2y ≈ 4*0.0003764172 + 2*0.0007165533 ≈ 0.0015056688 + 0.0014331066 ≈ 0.0029387754, which matches the right-hand side. Similarly, 0.5x + 2.5y ≈ 0.5*0.0003764172 + 2.5*0.0007165533 ≈ 0.0001882086 + 0.0017913833 ≈ 0.0019795919, which also matches. So the solution for x and y is correct.Therefore, the densities are approximately ρ₁ ≈ 2657 kg/m³ and ρ₂ ≈ 1395 kg/m³. Then, the volume of the third alloy:4.5 kg / 2657 kg/m³ ≈ 0.001694 m³4.5 kg / 1395 kg/m³ ≈ 0.003225 m³Total volume ≈ 0.001694 + 0.003225 ≈ 0.004919 m³, which matches the previous calculation. Then, buoyant force:0.004919 m³ * 1000 kg/m³ * 9.8 m/s² ≈ 0.004919 * 9800 ≈ 48.2 NWeight is 9 kg * 9.8 ≈ 88.2 NThus, force on the bottom: 88.2 - 48.2 = 40 N. That seems correct.But let me think again—if the third alloy is made by combining Alloy A and Alloy B, then the total mass is 6 + 3 = 9 kg. However, the problem says "obtained by combining the original alloys"—so perhaps it's just the sum of Alloy A and Alloy B? Then, the third alloy's composition would indeed be 4.5 kg each of metal 1 and metal 2. So, my calculation seems right.Alternatively, maybe the problem is considering the combination as mixing the alloys, not the metals. Wait, but Alloy A and Alloy B are themselves alloys of the two metals. So, combining Alloy A and Alloy B would result in a new alloy where the total mass of metal 1 is 4 + 0.5 = 4.5 kg, and metal 2 is 2 + 2.5 = 4.5 kg. So yes, equal parts. Hence, the third alloy is 50% each metal by mass.Therefore, the calculation seems correct. The force exerted on the bottom would be 40 N.But wait, let me just check the logic again. The force on the bottom is equal to the weight minus the buoyant force. Because the object is submerged, displacing water, so the buoyant force reduces the normal force. Therefore, F = mg - ρ_water * V * g. Which is exactly what we did. So, with V being the volume of the alloy, which we calculated based on the densities of the constituent metals.Therefore, all steps seem correct, and the answer should be 40 N.Final AnswerThe third alloy will exert a force of boxed{40} newtons on the bottom.

✅:To determine the force exerted by the third alloy on the bottom of a container when submerged in water, we need to analyze the given information about Alloy A and Alloy B, and then combine them to form the third alloy.1. Force Analysis for Alloy A and Alloy B: - Alloy A: Mass = 6 kg, Force = 30 N - Weight = (6 times 9.8 = 58.8 , text{N}) - Buoyant Force = (58.8 - 30 = 28.8 , text{N}) - Volume (V_A = frac{28.8}{1000 times 9.8} approx 0.0029388 , text{m}^3) - Alloy B: Mass = 3 kg, Force = 10 N - Weight = (3 times 9.8 = 29.4 , text{N}) - Buoyant Force = (29.4 - 10 = 19.4 , text{N}) - Volume (V_B = frac{19.4}{1000 times 9.8} approx 0.0019796 , text{m}^3)2. Composition of Alloys: - Alloy A: First metal (4 kg), Second metal (2 kg) - Alloy B: First metal (0.5 kg), Second metal (2.5 kg)3. Density Calculations: - Let (rho_1) and (rho_2) be the densities of the first and second metals. - For Alloy A: (frac{4}{rho_1} + frac{2}{rho_2} = 0.0029388) - For Alloy B: (frac{0.5}{rho_1} + frac{2.5}{rho_2} = 0.0019796) - Solving these equations, we find (rho_1 approx 2657.3 , text{kg/m}^3) and (rho_2 approx 1395.3 , text{kg/m}^3).4. Third Alloy Composition: - Combining Alloy A and Alloy B: Total mass = 9 kg (4.5 kg each of the first and second metals) - Volume of the third alloy: (V_{text{third}} = frac{4.5}{rho_1} + frac{4.5}{rho_2} approx 0.004918 , text{m}^3)5. Force Calculation for Third Alloy: - Weight = (9 times 9.8 = 88.2 , text{N}) - Buoyant Force = (1000 times 0.004918 times 9.8 approx 48.2 , text{N}) - Force on the bottom = (88.2 - 48.2 = 40 , text{N})Thus, the force exerted by the third alloy on the bottom of the container is (boxed{40}) newtons.

🔑:1. Determine the masses of each metal in each alloy: For alloy A: - Total mass = 6 kg - Let m_1 be the mass of the first metal, and m_2 be the mass of the second metal. - Given: m_1 = 2m_2 - Therefore, m_1 + m_2 = 6 - Substitute m_1 = 2m_2 into the equation: 2m_2 + m_2 = 6 which simplifies to 3m_2 = 6. - Thus, m_2 = 2 kg and m_1 = 4 kg. For alloy B: - Total mass = 3 kg - Let m_3 be the mass of the first metal, and m_4 be the mass of the second metal. - Given: m_3 = frac{1}{5} m_4 - Therefore, m_3 + m_4 = 3 - Substitute m_3 = frac{1}{5}m_4 into the equation: frac{1}{5}m_4 + m_4 = 3 which simplifies to frac{6}{5}m_4 = 3. - Multiplying both sides by 5/6 gives m_4 = 2.5 kg and m_3 = 0.5 kg.2. Combine the masses of the metals from alloys A and B to form a new alloy: Combined masses of metals: - First metal: m_1 + m_3 = 4 text{ kg} + 0.5 text{ kg} = 4.5 text{ kg} - Second metal: m_2 + m_4 = 2 text{ kg} + 2.5 text{ kg} = 4.5 text{ kg}3. Determine the total mass of the new alloy: - Total mass of the new alloy: 4.5 text{ kg} + 4.5 text{ kg} = 9 text{ kg}4. Determine the pressure force exerted by the new alloy in water: - The pressure force exerted by an object in water due to gravity is equal to the weight of the object. - Weight is given by W = m cdot g where g = 9.8 m/s². - Since in both original alloys the gravitational force was described as pressure force on the bottom (10 N for 3 kg and 30 N for 6 kg), we use the sum of these pressure forces. - Pressure force from new alloy due to its weight: 30 text{ N} + 10 text{ N} = 40 text{ N} Conclusion:The pressure force created by the third alloy when placed in a vessel with water is ( boxed{40 text{ N}} ).

🧠:Okay, let's try to tackle these questions one by one. I need to be careful and methodical here. Let's start with part a) about the license plate 729.First, for each day, there are specific rules. The car can be driven on days where it meets the criteria. Let's check each day's rule against plate 729.Monday: Odd-numbered plates. License plates are three-digit numbers from 000 to 999. An odd-numbered plate would end with an odd digit. The plate 729 ends with 9, which is odd. So, Monday is allowed. Check mark for Monday.Tuesday: Sum of digits >= 11. Let's compute the sum: 7 + 2 + 9 = 18. 18 is definitely greater than 11. So, Tuesday is okay. Another check.Wednesday: Multiples of 3. To check if 729 is a multiple of 3, we can sum the digits. Wait, 7+2+9=18, and since 18 is divisible by 3, 729 is a multiple of 3. So, Wednesday is good.Thursday: Sum of digits <=14. Wait, the sum is 18, which is greater than 14. So, Thursday is not allowed. So, Thursday is out.Friday: At least two identical digits. Looking at 7, 2, 9. All digits are different. So, no two identical digits. Therefore, Friday is not allowed.Saturday: Plate strictly less than 500. 729 is greater than 500, so Saturday is a no-go.Sunday: All digits <=5. Digits are 7, 2, 9. 7 and 9 are both greater than 5. So, Sunday is not allowed.So compiling the days: Monday, Tuesday, Wednesday. The rest are excluded. So part a) answer is Monday, Tuesday, Wednesday.Wait, let me double-check each day:Monday: Ends with 9 (odd) – yes. Tuesday: Sum 18 >=11 – yes. Wednesday: 729 ÷3 = 243, no remainder – yes. Thursday: 18 >14 – no. Friday: All digits unique – no. Saturday: 729 >500 – no. Sunday: 7 and 9 exceed 5 – no.Yes, that's correct. So part a) is Monday, Tuesday, Wednesday.Moving to part b): Maria wants a plate that can be driven every day except Sunday. So needs to satisfy all days from Monday to Saturday, but not Sunday. Let's figure out the constraints.First, each day's requirement:Monday: Odd-numbered (ends with 1,3,5,7,9).Tuesday: Sum of digits >=11.Wednesday: Multiple of 3.Thursday: Sum of digits <=14. Wait, since Tuesday requires sum >=11 and Thursday requires sum <=14. So the sum must be between 11 and 14 inclusive.Friday: At least two identical digits.Saturday: Plate <500 (i.e., first digit 0,1,2,3,4).But exclude Sunday: At least one digit >5. Because Sunday requires all digits <=5; so to not be allowed on Sunday, at least one digit must be >5.So the plate must satisfy:1. Ends with odd digit (Monday). 2. Sum of digits between 11 and 14 inclusive (Tuesday and Thursday). 3. Multiple of 3 (Wednesday). 4. At least two identical digits (Friday). 5. First digit 0-4 (Saturday). 6. At least one digit >5 (to exclude Sunday). So, let's try to find such a plate. Let's consider possible numbers starting with digits 0-4, ending with odd digits, have sum 11-14, multiple of 3, with at least two identical digits, and at least one digit >5.This seems a bit involved. Maybe we can break it down step by step.First, let's think about the constraints:- First digit: 0,1,2,3,4. - Last digit: 1,3,5,7,9. - Sum of digits: 11,12,13,14. - Sum must be divisible by 3 (since the number itself must be divisible by 3). So sum is 12 or 15, but wait, the sum has to be between 11-14, so possible sums are 11,12,13,14. Among these, the sum must be divisible by 3. So possible sums: 12. Because 12 is the only sum in 11-14 divisible by 3. Wait, 12 and 15, but 15 is over. So only sum=12 satisfies both sum >=11, sum <=14, and divisible by 3. So sum must be 12.Wait, but the sum is required to be >=11 (Tuesday) and <=14 (Thursday). So possible sums are 11,12,13,14. But it also needs to be a multiple of 3 (Wednesday). Therefore, sum must be 12. Because 12 is the only number in 11-14 divisible by 3. 11: 11 mod3=2; 12 mod3=0; 13 mod3=1;14 mod3=2. So sum must be 12. So that's a key point. Therefore, sum=12.So the plate number must have digits adding up to 12, first digit 0-4, last digit odd, at least two digits the same, and at least one digit >5.So let's think of such numbers.First digit is 0-4 (let's denote as a), middle digit (b), last digit (c) is 1,3,5,7,9. a + b + c =12. Also, at least one digit >5 (since can't be all <=5, otherwise would be allowed on Sunday). Also, must have at least two identical digits.Possible approaches: Let's fix a (0-4), c (1,3,5,7,9), then compute b=12 -a -c. Then check if b is between 0-9, and check if at least two digits are same, and at least one digit >5.This might take time. Let's proceed step by step.Start with first digit a=0:Then c can be 1,3,5,7,9. Then b=12 -0 -c =12 -c. So possible values:For c=1: b=11. But b must be a single digit (0-9). 11 invalid. c=3: b=9. Then digits are 0,9,3. Check duplicates: 0,9,3 all different. At least one digit >5: 9 is >5. So this plate is 093. Does 093 satisfy all?First digit 0 (so plate 093 is 093, which is less than 500 (Saturday). Ends with 3 (odd, Monday). Sum 0+9+3=12 (Tuesday and Thursday). 093 is divisible by 3 (since sum is 12). At least two identical digits: 0,9,3 are all different. So doesn't satisfy Friday. So invalid.Next c=5: b=12 -5=7. Digits: 0,7,5. All different. At least one digit >5: 7. So plate 075. But again, all digits different. So Friday's condition not met. Invalid.c=7: b=12 -7=5. Digits 0,5,7. All different. Again, same issue. Plate 057. Not meeting Friday.c=9: b=12 -9=3. Digits 0,3,9. All different. Plate 039. Same problem. So no a=0 possible.Next a=1:c can be 1,3,5,7,9. Then b=12 -1 -c=11 -c.Check possible c:c=1: b=10. Invalid. c=3: b=8. Digits 1,8,3. All different. At least one digit >5: 8. Plate 183. But all digits different. c=5: b=6. Digits 1,6,5. All different. Plate 165. c=7: b=4. Digits 1,4,7. All different. c=9: b=2. Digits 1,2,9. All different. All these have unique digits. So a=1 gives no solution.a=2:b=12 -2 -c=10 -c.c=1: b=9. Digits 2,9,1. All different. c=3: b=7. 2,7,3. Different. c=5: b=5. Digits 2,5,5. Here, two 5s. So duplicate. So plate 255. Check: sum=2+5+5=12. First digit 2 (less than 500). Ends with 5 (odd). At least two digits same (two 5s). At least one digit >5: 5 is not >5. Wait, all digits are 2,5,5. 5 is equal to 5, but not greater. So no digit >5. Therefore, this would be allowed on Sunday, which is not desired. So plate 255 would be allowed on Sunday, which Maria wants to exclude. So invalid.c=7: b=3. Digits 2,3,7. All different. c=9: b=1. Digits 2,1,9. All different. So only c=5 gives a duplicate, but that plate has no digit >5, so it's excluded.a=3:b=12 -3 -c=9 -c.Possible c:c=1: b=8. Digits 3,8,1. All different. c=3: b=6. Digits 3,6,3. Duplicate 3s. Plate 363. Check: sum=3+6+3=12. First digit 3 (ok). Ends with 3 (odd). At least two digits same (two 3s). Now, check if any digit >5: 6 is >5. So plate 363. This seems to satisfy all conditions except Sunday. Because 6 is >5, so not allowed on Sunday. Let's check:Sunday: all digits <=5? 3,6,3. 6 >5, so not allowed. So 363 is allowed Monday-Saturday, except Sunday. Wait, but let's confirm all other days:Monday: ends with 3 – yes. Tuesday: sum 12 >=11 – yes. Wednesday: 363 ÷3=121 – yes. Thursday: sum 12 <=14 – yes. Friday: two 3s – yes. Saturday: first digit 3 <5 – yes. Sunday: has a 6 – not allowed. So this plate 363 is a candidate.Is there another plate? Let's check other c for a=3.c=5: b=4. Digits 3,4,5. All different. c=7: b=2. Digits 3,2,7. All different. c=9: b=0. Digits 3,0,9. All different. Only c=3 gives 363. So 363 is a possible plate.a=4:b=12 -4 -c=8 -c.Possible c:c=1: b=7. Digits 4,7,1. All different. c=3: b=5. Digits 4,5,3. All different. c=5: b=3. Digits 4,3,5. All different. c=7: b=1. Digits 4,1,7. All different. c=9: b= -1. Invalid. Wait, c=9 gives b=8 -9= -1. Not possible. So no.So only possible plate found is 363. Let's check if there are others.Wait, maybe another a=3, c=9: no. Wait, a=3, c=9 would be b=9 -9=0. So digits 3,0,9. All different. No duplicates.Alternatively, let's see if there's another plate with a different a. For a=4, perhaps:Wait, a=4, c=5: sum=4 + b +5=12 → b=3. So digits 4,3,5. All different. a=4, c=7: b=1. Digits 4,1,7. All different. No duplicates here.Another approach: Maybe there are plates with two identical digits, sum 12, first digit 0-4, last digit odd, and at least one digit >5.Wait, let's think of another example. For example, 399. But first digit 3. Sum:3+9+9=21. Not 12. Not valid.Or 255: sum 2+5+5=12, but digits are 2,5,5. First digit 2. Ends with 5. But as before, all digits <=5 except 5 itself. Wait, 5 is equal to 5, not greater. So no digit >5, so would be allowed on Sunday. Not desired.Another idea: 336. First digit 3. Ends with 6, which is even. Not allowed on Monday.Wait, maybe 354. First digit 3. Ends with 4, which is even. No.Wait, maybe 471. First digit 4. Ends with 1. Sum 4+7+1=12. Digits: 4,7,1. All different. But has a 7, which is >5. So allowed on all days except Sunday. But does it have duplicates? No, all unique. So Friday not satisfied.Hmm. What about 399? Sum is 21, which is over.Wait, 363 is the only one I found. Let's check if there's another. Let's try a=2 again.a=2. c=5: b=5. Plate 255. But as discussed, no digit >5. So invalid.What about a=4, c=7: plate 417. Sum 4+1+7=12. Ends with 7 (odd). First digit 4. Digits 4,1,7. All different. But has 7>5. So this would satisfy:Monday: ends with 7 – yes. Tuesday: sum 12 – yes. Wednesday: divisible by 3 (sum 12) – yes. Thursday: sum 12 <=14 – yes. Friday: needs two identical digits – no. So fails Friday.So no. Another attempt: a=3, c=3. Then b=6. Plate 363. As before.Is there another plate with two identical digits, sum 12, first digit 0-4, last digit odd, and at least one digit >5?For example, let's consider a plate like 444. First digit 4. Ends with 4 (even). Not allowed on Monday.Or 333: sum 9. Not 12.How about 345: sum 12. Ends with 5. First digit 3. Digits 3,4,5. All different. And 5 is not >5. So no.Wait, 372: sum 3+7+2=12. Ends with 2 (even). Not allowed on Monday.Hmm. Maybe 561: but first digit 5 is invalid for Saturday.Wait, maybe 390: ends with 0 (even). No.Alternatively, 381: sum 3+8+1=12. Ends with 1. First digit 3. Digits 3,8,1. All different. So no duplicate. So Friday not satisfied.Alternatively, 453: sum 4+5+3=12. Ends with 3. First digit 4. Digits 4,5,3. All different. And 5 is not >5. Wait, 5 is equal to 5, so plate 453 would have all digits <=5 except none. Wait, 5 is allowed on Sunday? Wait, Sunday requires all digits <=5. If a plate has all digits <=5, it can't be driven on Sunday. Wait, no. Wait, Sunday's rule is "only cars with plates where all three digits are less than or equal to 5". So if a plate has all digits <=5, it can be driven on Sunday. Maria wants a plate that can be driven every day except Sunday, so her plate must NOT be allowed on Sunday, meaning at least one digit >5.So plate 453: digits 4,5,3. All digits <=5 except none. All digits <=5. Therefore, it would be allowed on Sunday, which is not desired. So invalid.So back to 363. That seems to be the only possible plate. Let's confirm:363: digits 3,6,3. Sum 12. Ends with 3 (odd). First digit 3. Duplicates: two 3s. At least one digit >5: 6. So satisfies all conditions. Therefore, Maria should get plate 363.Wait, but maybe there's another plate. Let's check a=4, c=9. But sum would need to be 12. Wait, a=4, c=9: sum=4 + b +9 =13 +b. To get sum=12, 13 +b=12 → b=-1. Invalid.Alternatively, a=4, c=5: sum=4 +5 +b=9 +b=12 → b=3. Plate 435. Digits 4,3,5. All different. And 5 is not >5. So allowed on Sunday. Not desired.Hmm. So maybe 363 is the only one. Let's check another case where two digits are the same and one digit >5.For example, plate 399: sum=3+9+9=21. Not 12. Invalid.Plate 345: sum=12, but all digits unique. Wait, 3+4+5=12. But digits are unique. So no.Wait, maybe plate 336: sum=3+3+6=12. Ends with 6 (even). Not allowed on Monday.Plate 318: sum=3+1+8=12. Ends with 8 (even). No.Plate 372: sum=12, ends with 2 (even). No.Hmm. It's challenging. Maybe 363 is the only one. Let's confirm again:363: - Monday: ends with 3 (odd) – yes. - Tuesday: sum 12 – yes. - Wednesday: divisible by 3 – yes. - Thursday: sum 12 <=14 – yes. - Friday: two 3s – yes. - Saturday: first digit 3 <5 – yes. - Sunday: has 6 >5 – excluded – yes. Yes, this works. So answer to part b) is 363.Moving to part c): Mayor Pietro needs a plate that allows him to drive every day. So the plate must satisfy all the day's requirements, including Sunday.So constraints:Every day's rules:- Monday: Ends with odd digit. - Tuesday: Sum >=11. - Wednesday: Multiple of 3. - Thursday: Sum <=14. - Friday: At least two identical digits. - Saturday: Plate <500 (first digit 0-4). - Sunday: All digits <=5. So, the plate must satisfy all these:1. Ends with odd digit. 2. Sum of digits >=11 and <=14 (from Tuesday and Thursday). 3. Divisible by 3. 4. At least two identical digits. 5. First digit 0-4. 6. All digits <=5. 7. So the digits can only be 0-5, but sum must be between 11-14. Wait, since all digits <=5 (Sunday's rule), the maximum possible sum is 5+5+5=15. But Thursday requires sum <=14, so sum must be 11,12,13,14. Also, Tuesday requires sum >=11. So sum between 11-14. Also, the number must be a multiple of 3 (sum divisible by 3). Therefore, sum must be 12 or 15, but 15 exceeds Thursday's limit. So sum must be 12.So sum=12, all digits <=5, first digit 0-4, ends with odd digit, at least two identical digits.Let's find such a plate.Digits: all <=5. So digits can be 0,1,2,3,4,5.Sum=12. Possible combinations.Possible digits a,b,c where a + b + c=12, each <=5, a (first digit) 0-4, c (last digit) odd (1,3,5). Also, at least two digits the same.Let's try to find such numbers.First, a can be 0,1,2,3,4.c must be 1,3,5.Let's consider possible combinations.Start with a=4 (the highest possible first digit):c=5: Then b=12 -4 -5=3. So digits 4,3,5. All digits <=5. Check duplicates: all unique. Not acceptable (needs at least two same). c=3: b=12 -4 -3=5. Digits 4,5,3. All unique. c=1: b=12 -4 -1=7. But b=7 >5. Invalid.Next a=3:c=5: b=12 -3 -5=4. Digits 3,4,5. All unique. c=3: b=12 -3 -3=6. b=6 >5. Invalid. c=1: b=12 -3 -1=8. Invalid.a=2:c=5: b=12 -2 -5=5. Digits 2,5,5. Two 5s. So duplicates. Plate 255. Check: - Ends with 5 (odd). - Sum=12. - Divisible by 3 (sum 12). - At least two 5s. - First digit 2 <5. - All digits <=5. So 255 satisfies all conditions. Let's confirm each day:Monday: ends with 5 – yes. Tuesday: sum 12 >=11 – yes. Wednesday: divisible by 3 – yes. Thursday: sum 12 <=14 – yes. Friday: two 5s – yes. Saturday: first digit 2 – yes. Sunday: all digits <=5 – yes. Perfect. So 255 is a valid plate. Let's check if there are others.a=2, c=5: 255. c=3: b=12 -2 -3=7. Invalid. c=1: b=12 -2 -1=9. Invalid.a=1:c=5: b=12 -1 -5=6. Invalid. c=3: b=12 -1 -3=8. Invalid. c=1: b=12 -1 -1=10. Invalid.a=0:c=5: b=12 -0 -5=7. Invalid. c=3: b=12 -0 -3=9. Invalid. c=1: b=12 -0 -1=11. Invalid.So the only possible plate is 255.Wait, but let's check another possibility. For example, a=4, but digits must be <=5.Wait, what about plate 453? No, digits must all be <=5. 453 has 4,5,3 – all <=5. But sum is 12. Ends with 3 (odd). But first digit 4. Wait, 453: - Monday: ends with 3 – yes. - Tuesday: sum 12 – yes. - Wednesday: divisible by 3 – yes. - Thursday: sum 12 – yes. - Friday: all digits unique – no. So fails Friday. So invalid.Another example: 345. Ends with 5. Sum 12. First digit 3. But digits are unique. Fails Friday.What about 333: sum 9. Not 12. Invalid.444: sum 12. Wait, 4+4+4=12. Plate 444. - Ends with 4 (even) – fails Monday. So invalid.Another idea: 552. Sum 5+5+2=12. Ends with 2 (even). Invalid.525: ends with 5. Sum 12. First digit 5. But first digit must be 0-4. So invalid.So 255 seems to be the only one. Let me confirm again:255: - Monday: 5 is odd – yes. - Tuesday: 12 >=11 – yes. - Wednesday: 255 ÷3=85 – yes. - Thursday: 12 <=14 – yes. - Friday: two 5s – yes. - Saturday: first digit 2 – yes. - Sunday: all digits <=5 – yes. Yes, satisfies all. So answer is 255.Finally, part d): Why can all inhabitants of Autolândia drive at least once a week?We need to show that every possible plate from 000 to 999 is allowed on at least one day. That is, for any plate, it meets at least one day's criteria.Possible approach: For any plate, check if it meets at least one of the seven days' rules.We need to argue that every three-digit number (000-999) meets at least one of the following:1. Ends with odd digit (Monday). 2. Sum of digits >=11 (Tuesday). 3. Multiple of 3 (Wednesday). 4. Sum of digits <=14 (Thursday). 5. At least two identical digits (Friday). 6. Plate <500 (Saturday). 7. All digits <=5 (Sunday). We need to show that every number satisfies at least one of these.Let's consider a plate that doesn't satisfy any of the rules. Suppose such a plate exists; then:- It ends with even digit (not Monday). - Sum of digits <11 (not Tuesday). - Not a multiple of 3 (not Wednesday). - Sum of digits >14 (not Thursday). - All digits distinct (not Friday). - Plate >=500 (not Saturday). - At least one digit >5 (not Sunday). We need to show that such a plate cannot exist. If we can prove that no such plate exists, then all inhabitants can drive at least once.Assume a plate with:- Ends with even digit. - Sum <11. - Not divisible by 3. - Sum >14. Wait, sum <11 and sum >14? Contradiction. Wait, if sum <11, then sum cannot be >14. So this is impossible. Wait, Thursday's rule is sum <=14, so not Thursday would be sum >14. But Tuesday's rule is sum >=11. So if a plate doesn't meet Tuesday (sum <11) and doesn't meet Thursday (sum >14), it's impossible. Because sum cannot be both <11 and >14. Therefore, any plate must satisfy either Tuesday or Thursday (or both). Wait, no. The negation is: a plate that doesn't meet Tuesday and doesn't meet Thursday must have sum <11 AND sum >14, which is impossible. Therefore, every plate must satisfy at least one of Tuesday or Thursday. Therefore, every plate can be driven on either Tuesday or Thursday, hence at least once a week.Wait, that's a key point. Because Tuesday is sum >=11, Thursday is sum <=14. The intersection is 11 <= sum <=14. But if a plate has sum <11, then it must meet Thursday's rule (sum <=14) only if sum <=14, which it is. Wait, no. Wait, Thursday's rule is sum <=14. So if a plate has sum <11, then it automatically satisfies Thursday's rule (sum <=14). Therefore, every plate with sum <11 can drive on Thursday. Similarly, every plate with sum >14 can drive on Tuesday (since sum >14 implies sum >=15 >=11). But sum can't be both <11 and >14. Wait, no: sum can be in three regions:- sum <11: can drive on Thursday. - 11 <= sum <=14: can drive on both Tuesday and Thursday. - sum >14: can drive on Tuesday. Thus, every plate satisfies at least one of Tuesday or Thursday. Therefore, every plate can be driven on at least Tuesday or Thursday. Hence, all inhabitants can drive at least once a week.Therefore, the answer is that every plate must satisfy either Tuesday's or Thursday's criteria (or both), so all can drive at least on those days.But let me check if there's any plate that might not satisfy either. For example, a plate with sum=10. Then it's not Tuesday (sum <11), but Thursday requires sum <=14, which is true. So can drive on Thursday. Similarly, a plate with sum=15: can drive on Tuesday. So no plate exists that doesn't satisfy at least one of Tuesday or Thursday. Hence, part d)'s answer is that every plate meets either Tuesday's or Thursday's criteria, ensuring all can drive at least once.

✅:a) The car with plate 729 can be driven on Monday, Tuesday, and Wednesday. - Monday: Ends with 9 (odd). - Tuesday: Sum = 7+2+9 = 18 (≥11). - Wednesday: 729 is divisible by 3. - Excluded on other days due to sum >14 (Thursday), unique digits (Friday), plate ≥500 (Saturday), and digits >5 (Sunday).b) Maria should choose plate 363. - Monday: Ends with 3 (odd). - Tuesday/Thursday: Sum = 3+6+3 = 12 (11–14). - Wednesday: Divisible by 3. - Friday: Two 3s. - Saturday: First digit 3 (<500). - Sunday: Digit 6 (>5) excludes it. c) Mayor Pietro should choose plate 255. - Monday: Ends with 5 (odd). - Tuesday/Thursday: Sum = 2+5+5 = 12 (11–14). - Wednesday: Divisible by 3. - Friday: Two 5s. - Saturday: First digit 2 (<500). - Sunday: All digits ≤5. d) All inhabitants can drive at least once a week because every plate satisfies either Tuesday’s (sum ≥11) or Thursday’s (sum ≤14) rule. - Plates with sum <11 drive on Thursday. - Plates with sum >14 drive on Tuesday. - All other plates (sum 11–14) drive on both days. - Thus, no plate is excluded from all days. Final Answers a) Monday, Tuesday, Wednesday b) 363 c) 255 d) All plates satisfy either Tuesday’s or Thursday’s criteria, ensuring everyone can drive at least once.

🔑:Part (a):To determine the days the car with the license plate number 729 can circulate, we verify if it meets the criteria for each day of the week:1. Monday: Only cars with an odd-numbered plate can circulate. Since 729 is odd (ends in 9), the car can circulate on Monday. 2. Tuesday: Only cars whose digits sum to 11 or greater can circulate. The sum of the digits of 729 is: [ 7 + 2 + 9 = 18 geq 11 ] Thus, the car can circulate on Tuesday.3. Wednesday: Only cars whose number is a multiple of 3 can circulate. We check if 729 is divisible by 3: [ 7 + 2 + 9 = 18 quad text{and} quad 18 div 3 = 6 ] So, 729 is divisible by 3, and the car can circulate on Wednesday. 4. Thursday: Only cars whose digits sum to 14 or less can circulate. We already calculated that: [ 7 + 2 + 9 = 18 notleq 14 ] Thus, the car cannot circulate on Thursday. 5. Friday: Only cars with at least two identical digits can circulate. The digits of 729 are all different (7, 2, 9), so the car cannot circulate on Friday. 6. Saturday: Only cars whose numbers are strictly less than 500 can circulate. Since: [ 729 > 500 ] The car cannot circulate on Saturday. 7. Sunday: Only cars whose digits are all less than or equal to 5 can circulate. The second digit of 729 is 9, which is greater than 5, so the car cannot circulate on Sunday.Conclusion:[text{The car with license plate 729 can circulate on Monday, Tuesday, and Wednesday.}]boxed{Monday, Tuesday, Wednesday}Part (b):Maria wants a car that can circulate every day except Sunday. We need to find an appropriate plate number by examining the conditions for each day:1. Monday: The plate must be an odd number. 2. Tuesday and Thursday: The sum of the digits ( A + B + C ) must be between 11 and 14 inclusive. [ 11 leq A + B + C leq 14 ] 3. Wednesday: The plate must be a multiple of 3. 4. Friday: The plate must contain at least two identical digits. 5. Saturday: The plate number must be less than 500. Let ( A, B, C ) represent the digits of the plate. Due to the Saturday criteria, ( A in {0, 1, 2, 3, 4} ).We'll analyze potential cases:Case ( A = 3 ):If ( A = 3 ), then:[B + C = A + B + C - 3 in {8, 9, 10, 11}]Since the sum must also be a range ( 11 leq A + B + C leq 14 ), the possibilities are narrowed.Possible combinations:- If ( B = 3 ): [ 3 + B + C - 3 = B + C = 8, 9, 10, 11 ] Since ( 363 ) would work: [ 3 + 6 + 3 = 12 quad (text{in given range}) quad text{and}quad text{, multiple of 3} ]Other ( A = 3 ) combinations:- ( 373: 3 + 7 + 3 = 13), possible. Case analysis:After checking, it turns 363 satisfies all:- (3+6+3=12)Conclusion:[boxed{363}]is the valid plate number.Part (c):For Pietro, we need a plate that circulates every day. Looking for a similar computation:Circulate:1. Monday: Odd2. Tuesday & Thursday: The sum of digits ( {11, 12, 13, 14})3. Wednesday: Multiple 34. Friday: Two identical5. Saturday and Sunday cars (0-5).Analyze ( A, B, C ):Possible (A=2):If ( A = 2 ):1. Not multiple of 3 less2. Sum within (9-12 ll 11-14)!Possible ( A =3 )[255 (Correct, sum = 15, must check)Thus, 123, 445 doesnt fit, so.boxed{255}}```