🧠:Okay, so I have this problem where I need to show that the difference (x_k^2 - 3y_k^2) is constant for all terms of the two sequences with the same index. The sequences are defined starting from (x_1 = 2) and (y_1 = 1), and then for each (k geq 2), they follow the recursive formulas:[x_k = 2x_{k-1} + 3y_{k-1}][y_k = x_{k-1} + 2y_{k-1}]First, I need to understand what exactly is being asked. The problem wants me to prove that (x_k^2 - 3y_k^2) doesn't change as (k) increases; it's the same for all (k). That means if I compute this expression for (k=1), then for (k=2), (k=3), and so on, they should all equal the same constant value.Let me start by computing the value of (x_1^2 - 3y_1^2) to see what the constant might be. Plugging in the given initial values:(x_1^2 - 3y_1^2 = 2^2 - 3(1)^2 = 4 - 3 = 1).So, if this difference is indeed constant, then for all (k), (x_k^2 - 3y_k^2 = 1). That seems to be the case we need to prove.Now, the challenge is to show that regardless of (k), this expression remains 1. Since the sequences are defined recursively, induction might be a good approach here. Mathematical induction often works well with recursively defined sequences because we can assume the statement holds for (k-1) and then prove it for (k).Let me recall the principle of mathematical induction. To prove a statement for all natural numbers (k), we need to do two steps:1. Base Case: Verify the statement is true for the initial value (usually (k=1)).2. Inductive Step: Assume the statement holds for some (k = n), and then prove it for (k = n+1).In this problem, the base case is already computed as 1, which is (x_1^2 - 3y_1^2 = 1). So that's established.Now, for the inductive step. Let's assume that for some (k = n), the equation (x_n^2 - 3y_n^2 = 1) holds true. Then, we need to show that (x_{n+1}^2 - 3y_{n+1}^2 = 1) as well.Given the recursive definitions:[x_{n+1} = 2x_n + 3y_n][y_{n+1} = x_n + 2y_n]So, substituting these into the expression (x_{n+1}^2 - 3y_{n+1}^2):First, compute (x_{n+1}^2):[(2x_n + 3y_n)^2 = 4x_n^2 + 12x_ny_n + 9y_n^2]Then compute (3y_{n+1}^2):[3(x_n + 2y_n)^2 = 3(x_n^2 + 4x_ny_n + 4y_n^2) = 3x_n^2 + 12x_ny_n + 12y_n^2]Subtracting the second expression from the first:[x_{n+1}^2 - 3y_{n+1}^2 = (4x_n^2 + 12x_ny_n + 9y_n^2) - (3x_n^2 + 12x_ny_n + 12y_n^2)]Simplifying term by term:- (4x_n^2 - 3x_n^2 = x_n^2)- (12x_ny_n - 12x_ny_n = 0)- (9y_n^2 - 12y_n^2 = -3y_n^2)So combining these:[x_{n+1}^2 - 3y_{n+1}^2 = x_n^2 - 3y_n^2]But by the inductive hypothesis, (x_n^2 - 3y_n^2 = 1). Therefore:[x_{n+1}^2 - 3y_{n+1}^2 = 1]Thus, if the statement holds for (k = n), it also holds for (k = n+1). Since the base case is true and the inductive step is proven, by mathematical induction, the expression (x_k^2 - 3y_k^2) is constant and equal to 1 for all (k geq 1).Wait, let me verify this with a specific example to ensure there's no mistake. Let's compute the next term in the sequences and check the difference.For (k=2):[x_2 = 2x_1 + 3y_1 = 2*2 + 3*1 = 4 + 3 = 7][y_2 = x_1 + 2y_1 = 2 + 2*1 = 2 + 2 = 4]Then, (x_2^2 - 3y_2^2 = 7^2 - 3*4^2 = 49 - 3*16 = 49 - 48 = 1). That's correct.Now (k=3):[x_3 = 2x_2 + 3y_2 = 2*7 + 3*4 = 14 + 12 = 26][y_3 = x_2 + 2y_2 = 7 + 2*4 = 7 + 8 = 15]Compute the difference: (26^2 - 3*15^2 = 676 - 3*225 = 676 - 675 = 1). Still 1.One more term, (k=4):[x_4 = 2x_3 + 3y_3 = 2*26 + 3*15 = 52 + 45 = 97][y_4 = x_3 + 2y_3 = 26 + 2*15 = 26 + 30 = 56]Check the difference: (97^2 - 3*56^2 = 9409 - 3*3136 = 9409 - 9408 = 1). Yep, still 1. So the induction seems to hold, and the examples confirm it.Alternatively, maybe there's another way to see this without induction. Let me think. The recursive relations resemble linear transformations. Perhaps we can model this as a matrix equation and analyze the eigenvalues or something, but that might be overcomplicating. However, noticing that the expression (x_k^2 - 3y_k^2) is reminiscent of a quadratic form, which can sometimes be preserved under certain linear transformations.Looking at the recursion:[begin{pmatrix}x_k y_kend{pmatrix}= begin{pmatrix}2 & 3 1 & 2end{pmatrix}begin{pmatrix}x_{k-1} y_{k-1}end{pmatrix}]So each step is a multiplication by the matrix (A = begin{pmatrix}2 & 3 1 & 2end{pmatrix}). The quadratic form (x^2 - 3y^2) can be represented as (mathbf{v}^T Q mathbf{v}) where (Q = begin{pmatrix}1 & 0 0 & -3end{pmatrix}). If the transformation preserves this quadratic form, then:[(Amathbf{v})^T Q (Amathbf{v}) = mathbf{v}^T (A^T Q A) mathbf{v} = mathbf{v}^T Q mathbf{v}]Which implies that (A^T Q A = Q). Let's check if this is true.Compute (A^T Q A):First, (A^T = begin{pmatrix}2 & 1 3 & 2end{pmatrix}).Multiply (A^T Q):[begin{pmatrix}2 & 1 3 & 2end{pmatrix}begin{pmatrix}1 & 0 0 & -3end{pmatrix}=begin{pmatrix}2*1 + 1*0 & 2*0 + 1*(-3) 3*1 + 2*0 & 3*0 + 2*(-3)end{pmatrix}=begin{pmatrix}2 & -3 3 & -6end{pmatrix}]Then multiply by A:[begin{pmatrix}2 & -3 3 & -6end{pmatrix}begin{pmatrix}2 & 3 1 & 2end{pmatrix}=begin{pmatrix}2*2 + (-3)*1 & 2*3 + (-3)*2 3*2 + (-6)*1 & 3*3 + (-6)*2end{pmatrix}]Compute each element:First row, first column: (4 - 3 = 1)First row, second column: (6 - 6 = 0)Second row, first column: (6 - 6 = 0)Second row, second column: (9 - 12 = -3)So the result is:[begin{pmatrix}1 & 0 0 & -3end{pmatrix}= Q]Therefore, (A^T Q A = Q), which confirms that the quadratic form (x^2 - 3y^2) is preserved under the transformation by matrix (A). Therefore, each application of the recursion maintains the value of (x_k^2 - 3y_k^2), so it remains constant for all (k).This gives another perspective, using linear algebra, to confirm why the difference remains constant. The matrix approach shows that the transformation is an isometry with respect to the quadratic form (x^2 - 3y^2), meaning it preserves the value of this form. Hence, regardless of how many times we apply the transformation (i.e., regardless of (k)), the value of (x_k^2 - 3y_k^2) stays the same as the initial value.Therefore, combining both the inductive proof and the matrix quadratic form preservation, we can be confident that the difference (x_k^2 - 3y_k^2) is indeed constant and equal to 1 for all (k).Another way to think about it might be recognizing that the recursion resembles the recurrence relations for solutions to Pell's equation. Pell's equation is (x^2 - Dy^2 = N), where D is a non-square positive integer and N is an integer. In this case, D is 3 and N is 1, so the equation (x^2 - 3y^2 = 1) is indeed Pell's equation. The sequences generated here are essentially generating solutions to Pell's equation, and each subsequent term is a larger solution. The recursive formulas here are similar to the method of generating new solutions from existing ones in Pell's equation theory, often involving linear combinations of previous solutions. This connects the problem to number theory and Pell equations, providing another angle to understand why the difference remains constant.In summary, multiple approaches confirm the result: mathematical induction directly shows the constancy through the recursive step, linear algebra via quadratic forms demonstrates the invariance under the transformation matrix, and connections to Pell's equation theory contextualize the problem within a broader mathematical framework. All these methods converge on the conclusion that (x_k^2 - 3y_k^2 = 1) for all (k), hence the difference is indeed constant.Final AnswerThe difference (x_{k}^{2} - 3 y_{k}^{2}) is constant and equal to (boxed{1}).

✅:To show that the difference (x_k^2 - 3y_k^2) is constant for all terms of the sequences defined by the given recursive rules, we start by verifying the base case and then use mathematical induction.Base Case: For (k = 1):[x_1^2 - 3y_1^2 = 2^2 - 3 cdot 1^2 = 4 - 3 = 1]Inductive Step: Assume for some (k = n), the statement (x_n^2 - 3y_n^2 = 1) holds. We need to show it holds for (k = n + 1).Using the recursive definitions:[x_{n+1} = 2x_n + 3y_n][y_{n+1} = x_n + 2y_n]Substitute these into the expression (x_{n+1}^2 - 3y_{n+1}^2):[x_{n+1}^2 = (2x_n + 3y_n)^2 = 4x_n^2 + 12x_ny_n + 9y_n^2][3y_{n+1}^2 = 3(x_n + 2y_n)^2 = 3(x_n^2 + 4x_ny_n + 4y_n^2) = 3x_n^2 + 12x_ny_n + 12y_n^2]Subtracting these:[x_{n+1}^2 - 3y_{n+1}^2 = (4x_n^2 + 12x_ny_n + 9y_n^2) - (3x_n^2 + 12x_ny_n + 12y_n^2)][= x_n^2 - 3y_n^2]By the inductive hypothesis, (x_n^2 - 3y_n^2 = 1). Therefore:[x_{n+1}^2 - 3y_{n+1}^2 = 1]This completes the inductive step. By mathematical induction, the expression (x_k^2 - 3y_k^2) is constant for all (k geq 1).Additionally, using linear algebra, the transformation matrix (A = begin{pmatrix}2 & 3 1 & 2end{pmatrix}) preserves the quadratic form (x^2 - 3y^2), confirming the invariance.Thus, the difference (x_k^2 - 3y_k^2) is constant and equal to (boxed{1}).

🔑:To demonstrate that the difference ( x_k^2 - 3 y_k^2 ) is constant for all ( k ), follow these steps.1. Expressing ( x_k ) and ( y_k ) in terms of ( x_{k-1} ) and ( y_{k-1} ): Use the given recurrence relations to find ( x_k ) and ( y_k ) in terms of ( x_{k-1} ) and ( y_{k-1} ): [ x_k = 2 x_{k-1} + 3 y_{k-1} ] [ y_k = x_{k-1} + 2 y_{k-1} ]2. Substitute into ( x_k^2 - 3 y_k^2 ): Calculate ( x_k^2 - 3 y_k^2 ) using the expressions for ( x_k ) and ( y_k ): [ x_k^2 - 3 y_k^2 = (2 x_{k-1} + 3 y_{k-1})^2 - 3(x_{k-1} + 2 y_{k-1})^2 ]3. Expand both terms: [ (2 x_{k-1} + 3 y_{k-1})^2 = 4 x_{k-1}^2 + 12 x_{k-1} y_{k-1} + 9 y_{k-1}^2 ] [ 3 (x_{k-1} + 2 y_{k-1})^2 = 3 (x_{k-1}^2 + 4 x_{k-1} y_{k-1} + 4 y_{k-1}^2) = 3 x_{k-1}^2 + 12 x_{k-1} y_{k-1} + 12 y_{k-1}^2 ]4. Combine and simplify the expressions: [ x_k^2 - 3 y_k^2 = (4 x_{k-1}^2 + 12 x_{k-1} y_{k-1} + 9 y_{k-1}^2) - (3 x_{k-1}^2 + 12 x_{k-1} y_{k-1} + 12 y_{k-1}^2) ] Simplifying this: [ x_k^2 - 3 y_k^2 = (4 x_{k-1}^2 - 3 x_{k-1}^2) + (12 x_{k-1} y_{k-1} - 12 x_{k-1} y_{k-1}) + (9 y_{k-1}^2 - 12 y_{k-1}^2) ] Results in: [ x_k^2 - 3 y_k^2 = x_{k-1}^2 - 3 y_{k-1}^2 ]5. Determine initial condition: Compute the expression for the initial values ( x_1 = 2 ) and ( y_1 = 1 ): [ x_1^2 - 3 y_1^2 = 2^2 - 3 cdot 1^2 = 4 - 3 = 1 ]6. Conclusion: Since it is proven by induction that ( x_k^2 - 3 y_k^2 = x_{k-1}^2 - 3 y_{k-1}^2 ), and the initial value ( x_1^2 - 3 y_1^2 = 1 ), the expression ( x_k^2 - 3 y_k^2 ) remains constant and equal to 1 for every ( k ): [ x_k^2 - 3 y_k^2 = 1 , text{for all}, k. ] [ boxed{1} ]

🧠:Okay, let's try to figure out this matchstick game problem. So we have 10 million matches, and two players, A and B, who take turns removing matches. The rule is that on each turn, a player can remove p^n matches, where p is a prime number and n is a non-negative integer (so n=0,1,2,...). The player who takes the last match wins. The question is: does player A (who goes first) have a winning strategy, or does player B?First, I need to understand the allowed moves. The players can remove p^0, p^1, p^2, etc., where p is prime. Since p^0 is 1 for any prime p, that means each player can always remove 1 match. Then, p^1 is the prime itself, so primes like 2, 3, 5, 7, etc. Then p^2 is squares of primes: 4, 9, 25, 49, etc. Similarly, higher powers like 8 (2^3), 16 (2^4), but wait, primes raised to any exponent. So allowed moves are numbers of the form prime^n, where n is a non-negative integer. So 1, 2, 3, 4, 5, 7, 9, 11, 13, 16, 17, etc. So any number that's a prime power.Wait, but p is a prime, and n is a non-negative integer, so the allowed moves are 1, 2, 3, 4, 5, 7, 8, 9, 11, 13, 16, 17, 19, 23, 25, 27, etc. So numbers that are powers of primes, including 1 (since p^0=1 for any prime p). So 1 is allowed. Therefore, each player can remove 1 match, which is crucial because that means the game can't end in a situation where a player can't move (since even if there's 1 match left, they can take it).But the goal is to take the last match. So, similar to Nim games, but with different move options. So this is a variant of the take-away game. In standard Nim, players can remove any number from 1 up to some limit, but here, moves are restricted to prime powers.The key in such games is to determine the losing positions (or P-positions) and the winning positions (N-positions). A P-position is one where the previous player can force a win, assuming optimal play, and an N-position is where the next player can force a win.Since the starting number of matches is 10,000,000, we need to determine if that's a P-position or an N-position. If it's a N-position, then player A can win; if it's a P-position, then player B can win.To figure this out, we need to analyze the game's Grundy numbers or use some other method to characterize the positions. But with 10 million matches, calculating Grundy numbers directly is impractical. So perhaps there's a pattern or mathematical property that can help.Alternatively, maybe we can model this as a variant of the subtraction game where the allowed moves are the set of prime powers. Then, the Sprague-Grundy theorem tells us that each position (number of matches) has a Grundy number, which is the mex (minimum excludant) of the Grundy numbers of the positions reachable from it.So, let's try to compute some Grundy numbers for small values and see if a pattern emerges.Let's start with n=0 (no matches left). That's a terminal position, so G(0)=0.n=1: You can remove 1 match (since 1 is p^0 for any prime), leading to n=0. So the Grundy number is mex{G(0)} = mex{0} = 1.n=2: You can remove 1 or 2 (since 2 is prime). Removing 1 leads to n=1 (G=1), removing 2 leads to n=0 (G=0). So mex{0,1} = 2.Wait, but wait, mex is the minimum non-negative integer not in the set. So for n=2, the reachable positions are n=1 (G=1) and n=0 (G=0). The mex of {0,1} is 2. Therefore, G(2)=2.n=3: You can remove 1, 3. So from n=3, removing 1 leads to n=2 (G=2), removing 3 leads to n=0 (G=0). So mex{0,2} = 1. Therefore, G(3)=1.n=4: You can remove 1, 2, 4 (since 4 is 2^2). Removing 1 leads to n=3 (G=1), removing 2 leads to n=2 (G=2), removing 4 leads to n=0 (G=0). The mex of {0,1,2} is 3. So G(4)=3.n=5: You can remove 1,5. So removing 1 leads to n=4 (G=3), removing 5 leads to n=0. So mex{0,3} =1. G(5)=1.n=6: Allowed moves: 1, 2, 3, 4, 5, etc., but primes and their powers. Wait, primes are 2,3,5,7,... and their powers. So 1,2,3,4,5,7,8,9, etc. So for n=6, allowed moves are all prime powers less than or equal to 6. So 1,2,3,4,5. So from 6, you can remove 1,2,3,4,5. Thus, leading to n=5,4,3,2,1. The Grundy numbers for these are G(5)=1, G(4)=3, G(3)=1, G(2)=2, G(1)=1. So mex{1,3,1,2,1} which is mex{1,2,3} = 0. Therefore, G(6)=0. So n=6 is a losing position.Wait, this is interesting. So if the starting position is 6, the first player cannot force a win. Let's check:If starting at 6, possible moves: remove 1 to 5. But if the player removes 1, leaving 5 (G=1), then the next player can take 5 and win. If the first player removes 2, leaving 4 (G=3), next player can take 4 and win. Similarly, removing 3 leaves 3 (G=1), next player takes 3 and wins. Removing 4 leaves 2 (G=2), next player takes 2 and wins. Removing 5 leaves 1 (G=1), next player takes 1 and wins. So yes, n=6 is a losing position.So G(6)=0.n=7: Allowed moves: 1, 7 (since 7 is prime). Removing 1 leads to n=6 (G=0), removing 7 leads to n=0. So mex{0, 0} =1. G(7)=1.n=8: Allowed moves: 1, 2, 4, 8 (since 8 is 2^3). So removing 1 leads to n=7 (G=1), removing 2 leads to n=6 (G=0), removing 4 leads to n=4 (G=3), removing 8 leads to n=0. So mex{0,1,3} =2. Therefore, G(8)=2.n=9: Allowed moves: 1, 3, 9 (3^2). Removing 1 leads to n=8 (G=2), removing 3 leads to n=6 (G=0), removing 9 leads to n=0. So mex{0,2} =1. G(9)=1.n=10: Allowed moves: prime powers up to 10. So 1,2,3,4,5,7,9. So removing these leads to n=9,8,7,6,5,3,1. Their G-values: G(9)=1, G(8)=2, G(7)=1, G(6)=0, G(5)=1, G(3)=1, G(1)=1. So mex{0,1,2} =3. Therefore, G(10)=3.n=11: Allowed moves 1, 11. Removing 1 leads to n=10 (G=3), removing 11 leads to n=0. So mex{0,3} =1. G(11)=1.n=12: Allowed moves: 1,2,3,4,5,7,9,11. Removing these leads to n=11,10,9,8,7,5,3,1. Their G-values: 1,3,1,2,1,1,1,1. So mex{1,2,3} =0. Therefore, G(12)=0.Hmm, so n=12 is another losing position. So pattern seems that every 6, 12, 18... maybe multiples of 6 are losing positions? Let's check n=18.But before that, let's see:n=6: G=0n=12: G=0Let's check n=18. Let's compute G(13) to G(18):But this might take too long. Let's see if the losing positions are multiples of 6. If so, then the Grundy numbers repeat every 6, which is possible. Let's check n=6,12,18.But let's check n=13: Allowed moves 1,13. So remove 1 to 12 (G=0) or 13 to 0. mex{0,0}=1. So G(13)=1.n=14: Allowed moves 1,2,7, etc. Let's see. All prime powers <=14: 1,2,3,4,5,7,9,11,13. So removing these leads to n=13,12,11,10,9,7,5,3,1. G-values: 1,0,1,3,1,1,1,1,1. mex{0,1,3}=2. So G(14)=2.n=15: Allowed moves: 1,2,3,4,5,7,9,11,13. Removing these leads to n=14,13,12,11,10,8,6,4,2. G-values:2,1,0,1,3,2,0,3,2. So mex{0,1,2,3}=4. Wait, but mex is the minimum non-negative integer not present. The set is {0,1,2,3}, so mex=4. Therefore, G(15)=4. Hmm, this complicates things. So n=15 has a Grundy number 4.But maybe my initial thought about multiples of 6 is wrong. So n=6,12 are losing positions (G=0), but n=18 might not be. Let's check n=18.But computing up to n=18 would take a while. Alternatively, let's check n=6,12,18. If the pattern is that multiples of 6 are losing positions (G=0), then 10,000,000 divided by 6 is 1,666,666.666..., so 10,000,000 is not a multiple of 6. Therefore, if the losing positions are multiples of 6, then 10,000,000 is not a losing position, so player A can win. But wait, but 10,000,000 modulo 6 is 10,000,000 /6 = 1,666,666*6 = 9,999,996, remainder 4. So 10,000,000 ≡4 mod6. If the losing positions are multiples of 6, then 10,000,000 is 4 mod6, so it's a winning position. Therefore, player A can remove 4 matches to leave 9,999,996, which is a multiple of 6, putting B in a losing position.But wait, but in our earlier analysis, n=4 has G=3, so maybe this approach is oversimplified. Wait, the problem is that if the losing positions are multiples of 6, then the strategy is to bring the opponent to a multiple of 6. But in our calculation, n=6 is a losing position (G=0), n=12 is also G=0. Let's check n=18.To compute G(18), we need to look at all possible moves from 18. The allowed moves are all prime powers up to 18. So prime powers <=18 are: 1,2,3,4,5,7,8,9,11,13,16,17. So removing these amounts:If you remove 1, you get n=17 (G=1)Remove 2: n=16 (need to compute G(16))Remove 3: n=15 (G=4)Remove 4: n=14 (G=2)Remove 5: n=13 (G=1)Remove 7: n=11 (G=1)Remove 8: n=10 (G=3)Remove 9: n=9 (G=1)Remove 11: n=7 (G=1)Remove 13: n=5 (G=1)Remove 16: n=2 (G=2)Remove 17: n=1 (G=1)So the Grundy numbers of the reachable positions are: 1 (from 17), G(16) (unknown yet), 4 (from15), 2 (from14),1 (from13),1 (from11),3 (from10),1 (from9),1 (from7),1 (from5),2 (from2),1 (from1).Wait, but we need to compute G(16) first.G(16): Allowed moves are prime powers <=16:1,2,3,4,5,7,8,9,11,13,16. Removing these gives:n=15 (G=4), n=14 (G=2), n=13 (G=1), n=12 (G=0), n=11 (G=1), n=9 (G=1), n=8 (G=2), n=7 (G=1), n=5 (G=1), n=3 (G=1), n=0. So the reachable G-values are 4,2,1,0,1,1,2,1,1,1,0. So mex{0,1,2,4}=3. Therefore, G(16)=3.So back to G(18), the reachable positions include G(16)=3. So the set of G-values from n=18 is:1 (17), 3 (16),4 (15),2 (14),1 (13),1 (11),3 (10),1 (9),1 (7),1 (5),2 (2),1 (1). So the set is {1,2,3,4}. Therefore, mex is 0. Hence, G(18)=0.Ah, so n=18 is also a losing position (G=0). So it seems that multiples of 6 are losing positions. Let's verify n=24, but this might take too long. But assuming the pattern continues, then every multiple of 6 is a losing position. Therefore, if the starting number is not a multiple of 6, the first player can remove some number of matches to leave a multiple of 6, forcing the second player into a losing position.But wait, when we saw n=4, which is not a multiple of 6, it had G=3, which is a winning position. However, if the losing positions are multiples of 6, then the strategy is to reduce the pile to the nearest multiple of 6 below the current number. For example, if the pile is 10,000,000, which is 4 mod6, the first player can remove 4 matches to leave 9,999,996, which is a multiple of 6. Then, whatever number the second player removes (which is a prime power), the first player can remove 6 minus that number, but wait, but 6 is not a prime power. Hmm, actually, the problem is that 6 is not a prime power. So the standard approach of pairing moves to reach the next multiple of 6 might not work here because the moves are restricted to prime powers.Wait, let's think again. Suppose the losing positions are multiples of 6. If player A can always remove k matches such that the remaining matches are a multiple of 6, then B is in a losing position. But for that to work, k must be a prime power, and 10,000,000 - k must be divisible by 6. So 10,000,000 mod6 is 4, so k must be congruent to 4 mod6. But k is a prime power. So we need to find a prime power k such that k ≡4 mod6.Let's see: primes are 2,3,5,7,11, etc.Prime powers:2^0=1:1 mod6=12^1=2:2 mod6=22^2=4:4 mod6=42^3=8:8 mod6=22^4=16:16 mod6=43^0=1:13^1=3:33^2=9:35^0=1:15^1=5:57^0=1:17^1=7:111^1=11:513^1=13:1Etc.So the possible prime powers mod6:Looking for k ≡4 mod6. From the above, 2^2=4, which is 4 mod6. Similarly, 2^4=16≡4, 2^6=64≡4, etc. So 2^{2n} where n≥1: 4,16,64,... which are ≡4 mod6. Also, 5^1=5≡5 mod6, 5^2=25≡1, 5^3=125≡5, etc. 7^1=7≡1, 7^2=49≡1, etc. 11≡5, 13≡1, etc.So the prime powers congruent to 4 mod6 are only the powers of 2 where the exponent is even and at least 2. Because 2^2=4, 2^4=16, etc. So 4,16,64,256,...Therefore, in order to reduce 10,000,000 to a multiple of 6, player A needs to remove 4,16,64,... matches. The question is, does 10,000,000 - 6m = k, where k is such a prime power. Specifically, 10,000,000 mod6=4, so we need to subtract 4 mod6, which as above, k=4,16,64,... etc.Is 4 a valid move? Yes, since 4=2^2. So player A can remove 4 matches, leaving 9,999,996, which is divisible by6. Then, whatever prime power player B removes (say, x), player A can remove (6 - x) mod6, but since (6 - x) needs to be a prime power. Wait, this might not hold because x is a prime power, and 6 - x needs to also be a prime power. Let's check possible x:If B removes 1 (which is always allowed), then A needs to remove 5. But 5 is a prime, so that's allowed. If B removes 2, A removes 4. If B removes 3, A removes 3. If B removes 4, A removes 2. If B removes 5, A removes 1. But 6 - x must be a prime power. Let's check:If B removes 1 (prime power), then A removes 5 (prime power). Yes.If B removes 2, A removes 4 (prime power). Yes.If B removes 3, A removes 3 (prime power). Yes.If B removes 4, A removes 2 (prime power). Yes.If B removes 5, A removes 1 (prime power). Yes.Wait a minute! Because 6 - x is also a prime power if x is a prime power removed by B. Let's verify:For x=1: 6-1=5, which is prime.x=2:6-2=4=2², which is a prime power.x=3:6-3=3, which is prime.x=4:6-4=2, which is prime.x=5:6-5=1, which is a prime power (any p^0=1).x=7: But B cannot remove 7 because that would take the pile below zero. Wait, after the first move, the pile is 9,999,996, which is a multiple of6. Let's say the pile is 6m. When B removes x (prime power), then A removes y=6 -x. But wait, 6m -x -y =6m -x - (6 -x)=6m -6=6(m-1). So A can always respond by removing 6 -x, which is a prime power if x is a prime power. Therefore, as long as for every prime power x ≤6m, 6 -x is also a prime power, then this strategy works.But let's check all possible x:Possible prime powers x that B can remove from a pile of 6m:x can be 1,2,3,4,5, etc., as long as x ≤6m. So if B removes x, then A removes y=6 -x.But we need to ensure that y is a prime power. Let's check for x from 1 to 5 (since 6 -x must be positive, so x <6):x=1: y=5 (prime) – valid.x=2: y=4 (prime power) – yes.x=3: y=3 (prime) – yes.x=4: y=2 (prime) – yes.x=5: y=1 (prime power) – yes.x=7: but x=7 can only be removed if 7 ≤6m. But since the pile is 6m, if m≥2, then 6m≥12, so B can remove 7. Then y=6 -7= -1, which is invalid. Wait, this is a problem.Wait, the earlier logic assumes that B can only remove x such that x ≤6m, but y=6 -x must also be a valid move (i.e., a prime power). However, if B removes x >6, then 6 -x would be negative, which is impossible. Therefore, the strategy only works if B is forced to remove x ≤6. But if the pile is 6m, and B removes x >6, then A cannot respond with y=6 -x because x >6 would make y negative.Therefore, the initial assumption that multiples of6 are losing positions is flawed because B can remove a prime power greater than6, leaving A unable to mirror the move.For example, consider pile size 12 (a multiple of6). If B removes 7 (a prime), leaving 5. Then A has to remove from 5. But 5 is a prime power, so A can remove all 5 and win. Wait, but in that case, B's move was 7 from 12, leaving 5. Then A takes 5 and wins. So in this case, B's move was bad. But maybe B would not make such a move.Wait, but if the pile is a multiple of6, and B removes a prime power greater than6, then the remaining pile is less than6m -6, but A can still win by taking the remaining matches if possible. However, the key is whether B can make a move that A cannot counter.Wait, let's take pile=12. If B removes 7, leaving 5. Then A can take 5 and win. So B would not choose to remove 7. Similarly, if B removes 11, leaving 1, A takes 1 and wins. Therefore, B's best move is to remove a prime power such that the remaining pile is a losing position. But if 6m is a losing position, then B should not be able to make a move to another losing position. Therefore, our earlier assumption that multiples of6 are losing positions might still hold, but we need to verify.Wait, let's take pile=6. If player A takes 4, leaving 2. Then player B can take 2 and win. But wait, no. If pile=6, it's B's turn? Wait, no: pile=6 is a losing position, so if it's B's turn, B cannot force a win. Wait, if the pile is6 and it's B's turn, B has to remove a prime power. Let's say B removes1, leaving5. Then A removes5 and wins. If B removes2, leaves4. A removes4 and wins. If B removes3, leaves3. A removes3 and wins. If B removes4, leaves2. A removes2 and wins. If B removes5, leaves1. A removes1 and wins. So yes, if the pile is6 and it's B's turn, B loses.Similarly, for pile=12. If it's B's turn, whatever B removes, A can respond by taking the complement to6. Wait, but in pile=12, if B removes x, A removes y=6 -x. But 12 -x -y=12 -x - (6 -x)=6. So then the pile is6, which is a losing position for B. Wait, this is different. Wait, let's clarify:If the pile is12, and B removes x (prime power), A removes y=6 -x (prime power), resulting in pile=12 -x -y=6. Then it's B's turn again with pile=6, which is a losing position. Therefore, A can force B into the losing position.But this requires that y=6 -x is a valid move (prime power) whenever B removes x from pile=12. For example:If B removes7 from12, leaving5. Then A can take5 (prime power) and win. But according to the earlier strategy, A should take y=6 -x=6 -7=-1, which is invalid. Therefore, the mirroring strategy only works if x ≤6. If B takes x >6, then A cannot mirror, but might still have a winning move.But in the case of pile=12, if B removes7 (leaving5), A can take5 and win. If B removes11 (leaving1), A takes1 and wins. If B removes a prime power like 9 (leaving3), A takes3 and wins. Similarly, B removing8 (2^3) from12 leaves4, A takes4 and wins. So even if B takes a prime power >6, A can still take the remaining matches and win.Therefore, in pile=12 (multiple of6), if it's B's turn, any move B makes allows A to win immediately. Hence, multiples of6 are indeed losing positions.Therefore, the strategy is: if the starting number is a multiple of6, the second player can win; otherwise, the first player can win by reducing the pile to the nearest multiple of6.Given that 10,000,000 divided by6 is 1,666,666*6=9,999,996, with a remainder of4. Therefore, 10,000,000 is not a multiple of6, so player A can remove4 matches (since4 is a prime power, 2²), leaving9,999,996, which is a multiple of6. Then, whatever B removes, A can remove the complement to6, maintaining the pile as a multiple of6 after each pair of moves. Eventually, the pile will be reduced to6, then B is forced to leave A with a winning move.But wait, we need to confirm that player A can always respond to B's moves by maintaining the multiple of6. Let's formalize this:If the pile is6m, and B removesx (a prime power), then A removes6 -x (which must also be a prime power). As checked earlier:Forx=1, y=5 (prime)x=2, y=4 (prime power)x=3, y=3 (prime)x=4, y=2 (prime)x=5, y=1 (prime power)These are all valid moves.But ifx >6, then6 -x would be negative, but in such a case, when B removes a prime powerx >6 from6m, the remaining pile is6m -x. Sincex >6,6m -x <6(m -1). However, player A can then remove the entire remaining pile if it's a prime power. For example, ifm=2 (pile=12), and B removes7, leaving5. A removes5 and wins. If B removes11, leaving1. A removes1 and wins. If B removes9, leaving3. A removes3 and wins. Similarly, if B removes8 (2^3), leaving4. A removes4 and wins. Therefore, even if B removes a large prime power, A can finish the game. Hence, multiples of6 are indeed losing positions because any move from6m allows the opponent to win either by reducing to6(m -1) or by taking the rest.Therefore, the conclusion is that the losing positions are multiples of6. Since10,000,000 is not a multiple of6, player A can remove4 matches (2²) to make it9,999,996, which is a multiple of6, and then mirror B's moves to maintain the multiple, ensuring victory.But let's verify this strategy with the example we computed earlier:n=6 is a losing position. n=12 is a losing position. So if the starting pile is10,000,000 ≡4 mod6, player A removes4, making it9,999,996. Then, whatever B removes (x), A removes6 -x. For example:If B removes1, A removes5.If B removes2, A removes4.If B removes3, A removes3.If B removes4, A removes2.If B removes5, A removes1.If B removes a higher prime power, like7, then A removes5 (since6 -7 is invalid, but in reality, the pile is9,999,996 -7=9,999,989. But wait, this disrupts the multiple of6. Wait, no. If B removes7 from9,999,996, the new pile is9,999,996 -7=9,999,989. But9,999,989 mod6= (9,999,996 -7) mod6=0 -7 mod6= (-7) mod6=5. Then A needs to remove5 to make it9,999,984, which is divisible by6. But5 is a prime power, so yes.Wait, in this case, the mirroring strategy is not exactly6 -x, but rather removing a number that brings the pile back to a multiple of6. The amount to remove is(current pile mod6). Because if the current pile is6m +r, and B removesx, then the new pile is6m +r -x. A needs to remove(r -x) mod6 to make it6(m -1) + (r -x - ((r -x) mod6)). Wait, this might be complicated.Alternatively, since the initial move by A reduces the pile to a multiple of6, any move by B will make the pile6m -x. Then, A can remove(6 -x mod6) to make it6(m -1). For example:If B removesx from6m, the new pile is6m -x. A removesy=(6m -x) -6(m -1)=6m -x -6m +6=6 -x. Therefore, y=6 -x. As long as y is a prime power, A can do this.But ifx >6, then6 -x is negative, but in reality, y should be such that6m -x -y=6(m -1). Therefore,y=6m -x -6(m -1)=6 -x. However, ifx >6, then6 -x is negative, which is impossible. Therefore, this strategy only works ifx ≤6. Ifx >6, then A cannot use this strategy and must find another prime powery to remove such that6m -x -y=6(m -1). Which would requirey=6 -x +6k for somek, but this is unclear.Wait, this indicates that the initial assumption that multiples of6 are losing positions might not hold when players can remove prime powers greater than6. However, in our earlier small cases (n=6,12,18), even if B removes a large prime power, A can still win by taking the remaining matches. For example, in n=12, if B removes7, leaving5, A takes5 and wins. So the key is that if the pile is a multiple of6, any move by B either allows A to immediately win (by taking the rest) or to reduce back to a multiple of6.But how can we generalize this? Let's think recursively. If a pile is6m, any move by B (removingx) will leave6m -x. A can respond by:- Ifx <6m: If6m -x is a prime power, A can take it all and win.- Otherwise, A removesy=6 - (x mod6), provided thaty is a prime power, to bring the pile back to6(m -1) + (x mod6) -y. Wait, this might not be straightforward.Alternatively, perhaps the multiples of6 are losing positions because no matter what B does, A can always win by either taking the remainder or adjusting to the next multiple of6. However, this needs to be verified.Let's test with n=18. If B removes7 from18, leaving11. A can remove11 (prime) and win. If B removes13, leaving5. A removes5 and wins. If B removes5, leaving13. A removes13 and wins. If B removes1, leaving17. A removes5 (since17 -5=12, which is a multiple of6). Wait, but5 is a prime power. So17 -5=12. Then B is faced with12, which is a losing position. Similarly, if B removes2 from18, leaving16. A removes4 (prime power) to reach12. If B removes3, leaving15. A removes3 (prime power) to reach12. If B removes4, leaving14. A removes2 to reach12. If B removes5, leaving13. A removes13 to win. So in all cases, A can either win immediately or reduce to12, a multiple of6.Therefore, the strategy holds: if the pile is a multiple of6, any move by B allows A to either win immediately or reduce to the next lower multiple of6. Hence, multiples of6 are indeed losing positions.Therefore, the conclusion is that the first player (A) can win by reducing the pile to the nearest multiple of6 below10,000,000, which is9,999,996, by removing4 matches (since10,000,000 -4=9,999,996). Then, whatever B removes, A can mirror the move to maintain the pile as a multiple of6, eventually forcing B to lose.But we need to confirm that4 is a prime power. Yes,4=2², so it's allowed. Therefore, player A's first move is to remove4 matches.

✅:Solution:Player A has a winning strategy. The key lies in reducing the pile to a multiple of 6 on each turn. Here's the step-by-step reasoning:1. Prime Power Moves: Players can remove any prime power (numbers like 1, 2, 3, 4, 5, 7, 9, etc.). Notably, 1 is allowed (as (p^0) for any prime (p)).2. Losing Positions: Positions (pile sizes) that are multiples of 6 are losing positions. If a player faces a multiple of 6 and both play optimally, they will lose. This is because any move they make can be countered by the opponent to return to the next lower multiple of 6.3. Initial Position: Starting with 10,000,000 matches, which is not a multiple of 6 (since (10,000,000 mod 6 = 4)), Player A can force a win.4. First Move: Player A removes 4 matches (a prime power, as (4 = 2^2)), leaving (9,999,996), which is a multiple of 6.5. Maintaining the Strategy: After each move by Player B, Player A removes a number of matches that complements B's move to maintain the pile as a multiple of 6. For example: - If B removes (x), A removes (6 - x) (if (x leq 5)). - If B removes a larger prime power (e.g., 7), A removes the remaining matches (e.g., 5 if B removes 7 from 12, leaving 5).Conclusion: Player A can always force the game into a losing position for B by reducing the pile to multiples of 6. Hence, Player A has a winning strategy, starting by removing 4 matches.boxed{A}

🔑:1. Understanding the Game: We have a pile of 10 million fire sticks. Two players, A and B, take turns picking fire sticks from the pile. Player A goes first. At each turn, a player can take ( p^n ) fire sticks, where ( p ) is a prime number and ( n = 0, 1, 2, ldots ). The player who takes the last stick wins.2. Allowed Moves: Players can only take ( 1, 2, 3, 4, 5 ) fire sticks per turn since taking 6 or any multiple of 6 is not allowed. This is due to the fact that taking ( p^n ) sticks for (p=6) or multiples of (6) is not allowed. 3. First Move Strategy: Player A should take 4 fire sticks on the first move. This will leave (10,000,000 - 4 = 9,999,996) fire sticks.4. Maintaining the Multiple of 6: Once the number of remaining sticks is a multiple of 6 (in this case, (9,999,996) sticks), Player B cannot make any move that will make the remaining number of sticks a multiple of 6 after their turn. This is because they can only remove 1 to 5 sticks, none of which will transform a multiple of 6 into another multiple of 6: - If ( n equiv 0 pmod{6} ), then ( n-1, n-2, n-3, n-4, n-5 notequiv 0 pmod{6} ).5. Ensuring Subsequent Moves Keep Multiple of 6: After Player B takes their turn, the number of sticks left will never be a multiple of 6. Player A will then adjust the number of sticks taken to bring the total back to a multiple of 6. For example, if Player B takes ( k ) sticks, then: - ( n-k = 6m - k ) sticks are left, where ( k ) is between 1 and 5.6. Repeat the Process: Player A continues taking the requisite number of sticks to keep the number of remaining sticks a multiple of 6 after accounting for Player B's removal. They can always do this by taking ( 6 - k ) sticks if Player B has taken ( k ) sticks. 7. Inevitable Victory: This strategy controls the game such that Player A finally takes the last stick. When there are exactly 6 fire sticks left before Player A's turn, Player B can only take 1 to 5 sticks, leaving a non-multiple of 6 for Player A. Thus, Player A can always adjust to bring the number back to a multiple of 6 until only 1 stick remains, which Player A takes to win the game.Conclusion: Player A has a winning strategy if they start by taking 4 fire sticks and then always adjust their subsequent moves to leave a multiple of 6 remaining for Player B after each of Player B's turns. [boxed{text{Player A has a guaranteed winning strategy.}}]

🧠:Okay, let me try to work through this problem step by step. So, we have a jar with 97 marbles, and they can be red, green, or blue. Neil draws two marbles without replacement, and the probability that both are the same color is 5/12. Then, Jerry draws two marbles with replacement, and we need to find the probability that both are the same color. First, I need to figure out how many marbles there are of each color. Let me denote the number of red marbles as R, green as G, and blue as B. So, R + G + B = 97. Neil's probability of drawing two marbles of the same color without replacement is given by the formula for combinations. For each color, the number of ways to choose two marbles is C(n, 2) = n(n-1)/2, where n is the number of marbles of that color. Then, the total number of ways to draw two marbles of the same color is [R(R-1) + G(G-1) + B(B-1)] / 2. The total number of ways to draw any two marbles is C(97, 2) = 97*96/2. So, the probability is [R(R-1) + G(G-1) + B(B-1)] / (97*96) = 5/12. Let me write that equation down:[R(R-1) + G(G-1) + B(B-1)] / (97*96) = 5/12.Multiplying both sides by 97*96 gives:R(R-1) + G(G-1) + B(B-1) = (5/12)*97*96.Calculating the right side:First, 97*96 = 9312.Then, 5/12 of that is (5*9312)/12. Let's compute that:9312 divided by 12 is 776. So, 5*776 = 3880.Therefore, R(R-1) + G(G-1) + B(B-1) = 3880.Now, another equation we have is R + G + B = 97.We need to find R, G, B such that these two equations hold. Then, once we have R, G, B, we can compute the probability for Jerry, which is with replacement. The probability for Jerry would be (R² + G² + B²) / (97²), since when drawing with replacement, the number of ways to get two of the same color is R² + G² + B², and the total is 97².So, we need to find R² + G² + B². Let's see how we can connect this with the given equations.We know that R + G + B = 97. Let's square both sides:(R + G + B)² = 97² = 9409.Expanding the left side:R² + G² + B² + 2(RG + RB + GB) = 9409.So, if we can find RG + RB + GB, then we can compute R² + G² + B².But from the first equation, we have R(R-1) + G(G-1) + B(B-1) = 3880.Let's expand that:R² - R + G² - G + B² - B = 3880.Which can be rewritten as:(R² + G² + B²) - (R + G + B) = 3880.We know that R + G + B = 97, so substituting that in:(R² + G² + B²) - 97 = 3880.Therefore, R² + G² + B² = 3880 + 97 = 3977.Wait, so R² + G² + B² = 3977. Then, plugging this back into the squared sum equation:3977 + 2(RG + RB + GB) = 9409.Therefore, 2(RG + RB + GB) = 9409 - 3977 = 5432.So, RG + RB + GB = 5432 / 2 = 2716.But do we need that? Well, actually, we already found R² + G² + B² = 3977, which is needed for Jerry's probability. Let me check that again.Wait, hold on. If R² + G² + B² is 3977, then the probability for Jerry is 3977 / (97²). Let me compute 97 squared. 97*97 is 9409. So, the probability is 3977 / 9409. Let me see if that reduces. Let me check.First, check if 3977 and 9409 have a common factor. Let's compute GCD(3977, 9409). Let's divide 9409 by 3977:9409 ÷ 3977 = 2 with a remainder of 9409 - 2*3977 = 9409 - 7954 = 1455.Now, GCD(3977, 1455). 3977 ÷ 1455 = 2, remainder 3977 - 2*1455 = 3977 - 2910 = 1067.GCD(1455, 1067). 1455 ÷ 1067 = 1, remainder 388.GCD(1067, 388). 1067 ÷ 388 = 2, remainder 1067 - 776 = 291.GCD(388, 291). 388 ÷ 291 = 1, remainder 97.GCD(291, 97). 291 ÷ 97 = 3, remainder 0. So, GCD is 97.Wait, so GCD(3977, 9409) is 97. Let me check:3977 ÷ 97. 97*40 = 3880. 3977 - 3880 = 97. So, 97*41 = 3977. So, 3977 = 97*41.Similarly, 9409 ÷ 97. 97*97 = 9409. So, 9409 = 97*97.Therefore, 3977/9409 = (97*41)/(97*97) = 41/97.So, the probability is 41/97.Wait, so without even knowing the individual R, G, B, just by using the equations, we can compute R² + G² + B², and thus the probability for Jerry. That's neat. So, we didn't need to find R, G, B specifically. The key was realizing that the given probability for Neil allows us to compute R² + G² + B² through the equations.Let me verify my steps again to make sure I didn't make a mistake.Given:1. R + G + B = 97.2. [R(R-1) + G(G-1) + B(B-1)] = 3880.From equation 2, expanding gives R² - R + G² - G + B² - B = 3880.Which is equal to (R² + G² + B²) - (R + G + B) = 3880.Since R + G + B = 97, substituting gives (R² + G² + B²) - 97 = 3880, so R² + G² + B² = 3977.Then, since Jerry's probability is (R² + G² + B²)/97² = 3977/9409 = 41/97 after simplification.Therefore, the answer should be 41/97.But wait, let me check if the calculation of R(R-1) + G(G-1) + B(B-1) = 3880 is correct.Given that the probability was 5/12. So, [R(R-1) + G(G-1) + B(B-1)] / (97*96/2) = 5/12.Multiply both sides by 97*96/2:R(R-1) + G(G-1) + B(B-1) = (5/12)*(97*96/2).Wait, hold on! Wait, the total number of ways to draw two marbles is C(97, 2) = 97*96/2. So, the probability is [C(R,2) + C(G,2) + C(B,2)] / C(97,2) = 5/12.Therefore, [ (R(R-1)/2 + G(G-1)/2 + B(B-1)/2 ) ] / (97*96/2) = 5/12.Multiplying numerator and denominator by 2 gives [R(R-1) + G(G-1) + B(B-1)] / (97*96) = 5/12.Therefore, R(R-1) + G(G-1) + B(B-1) = (5/12)*97*96.Compute 97*96 first: 97*96 = (100 - 3)*96 = 9600 - 288 = 9312.Then, 5/12 * 9312 = (5 * 9312)/12.Divide 9312 by 12: 12*776 = 9312. So, 5*776 = 3880. Correct. So, R(R-1) + G(G-1) + B(B-1) = 3880.Then, proceeding as before, expand this to R² - R + G² - G + B² - B = 3880.Which is R² + G² + B² - (R + G + B) = 3880.Since R + G + B = 97, substitute that in:R² + G² + B² - 97 = 3880 => R² + G² + B² = 3880 + 97 = 3977. Correct.Then, the probability with replacement is (R² + G² + B²)/97² = 3977/9409. Simplify.Since 3977 = 97*41 and 9409 = 97*97, so 3977/9409 = 41/97. That's correct.Therefore, the probability Jerry gets two marbles of the same color is 41/97.Wait, but let me think again. The answer comes out so neatly. Is there a way this could be a coincidence? Let me check with an example.Suppose all marbles are the same color. Then, R=97, G=B=0. Then, Neil's probability would be C(97,2)/C(97,2) = 1, which is not 5/12. So, that's not the case.Suppose two colors. Let's say R= a, G = b, B = 0. Then, the same color probability would be [a(a-1) + b(b-1)] / (97*96) = 5/12. But in this case, since B=0, R + G = 97. But solving this would require integer solutions. But in the given problem, since the answer comes out so neatly without considering the individual numbers, maybe the answer is indeed 41/97.Alternatively, maybe there's an error in the logic. Let me see. The key step is that from the given probability without replacement, we can compute R² + G² + B². Then, using that, compute the probability with replacement. Since with replacement, the probability is (R² + G² + B²)/97². That seems correct because when you draw with replacement, the number of ways to get two of the same color is R*R + G*G + B*B, since for each color, you can pick the first and second marble as that color. The total possible outcomes are 97*97.Therefore, the logic seems sound. Therefore, the answer is 41/97.Final AnswerThe probability that Jerry draws two marbles of the same color is boxed{dfrac{41}{97}}.

✅:Given a jar with 97 marbles that are either red, green, or blue, we know that the probability of drawing two marbles of the same color without replacement is ( frac{5}{12} ). We need to find the probability of drawing two marbles of the same color with replacement.1. Equations Setup: - Let ( R ), ( G ), and ( B ) be the number of red, green, and blue marbles respectively. - ( R + G + B = 97 ). - The probability of drawing two marbles of the same color without replacement is given by: [ frac{R(R-1) + G(G-1) + B(B-1)}{97 cdot 96} = frac{5}{12} ]2. Calculating the Numerator: - Multiplying both sides by ( 97 cdot 96 ): [ R(R-1) + G(G-1) + B(B-1) = frac{5}{12} cdot 97 cdot 96 = 3880 ]3. Expanding and Simplifying: - Expanding the left-hand side: [ R^2 - R + G^2 - G + B^2 - B = 3880 ] - Using ( R + G + B = 97 ): [ R^2 + G^2 + B^2 - 97 = 3880 implies R^2 + G^2 + B^2 = 3977 ]4. Probability with Replacement: - The probability of drawing two marbles of the same color with replacement is: [ frac{R^2 + G^2 + B^2}{97^2} = frac{3977}{9409} ] - Simplifying the fraction: [ frac{3977}{9409} = frac{41}{97} ]Thus, the probability that Jerry draws two marbles of the same color is (boxed{dfrac{41}{97}}).

🔑:Let us solve the problem step-by-step and provide the detailed steps needed.# Step 1: Understand the given informationThere are a total of 97 marbles in a jar, which can be either red, green, or blue. Neil draws two marbles without replacement and notes that the probability that they are of the same color is (frac{5}{12}). We need to find the probability that Jerry, who draws two marbles with replacement, will draw two marbles of the same color.# Step 2: Define the variablesLet (a, b,) and (c) be the number of red, green, and blue marbles, respectively. We have:[ a + b + c = 97 ]# Step 3: Calculate the probability for Neil's drawNeil's situation involves drawing two marbles without replacement. Thus, the probability of drawing two marbles of the same color is computed as:[ frac{a(a-1) + b(b-1) + c(c-1)}{97 cdot 96} = frac{5}{12} ]Let's rewrite the numerator:[ a(a-1) + b(b-1) + c(c-1) = a^2 - a + b^2 - b + c^2 - c ]Thus, the probability equation is:[ frac{a^2 - a + b^2 - b + c^2 - c}{97 cdot 96} = frac{5}{12} ]# Step 4: Equate the fractions and solve for the sum of squaresTo solve this, cross-multiply to clear the fraction:[ 12(a^2 - a + b^2 - b + c^2 - c) = 5 cdot 97 cdot 96 ]Simplifying the right-hand side, we have:[ 5 cdot 97 cdot 96 = 5 cdot (97 cdot 96) = 46560 ]Thus,[ 12(a^2 - a + b^2 - b + c^2 - c) = 46560 ]Simplify:[ a^2 + b^2 + c^2 - (a + b + c) = 3880 ]Since (a + b + c = 97):[ a^2 + b^2 + c^2 - 97 = 3880 ]Therefore:[ a^2 + b^2 + c^2 = 3977 ]# Step 5: Calculate the probability for Jerry's drawJerry draws two marbles with replacement. The probability that the two marbles are of the same color is given by:[ frac{a^2 + b^2 + c^2}{(a + b + c)^2} = frac{a^2 + b^2 + c^2}{97^2} ]Using ( a^2 + b^2 + c^2 = 3977 ):[ frac{3977}{97^2} = frac{3977}{9409} = frac{41}{97} ]# ConclusionAfter computing all the necessary steps, we conclude that the probability that Jerry draws two marbles of the same color, with replacement, is:[ boxed{frac{41}{97}} ]

🧠:Okay, so I need to find the cosine of the angle between vectors AB and AC. Let me recall how to do that. I think the formula involves the dot product of the two vectors divided by the product of their magnitudes. Yeah, that's right. The cosine of the angle θ between two vectors u and v is given by:cosθ = (u · v) / (||u|| ||v||)So first, I need to find the vectors AB and AC. To get vector AB, I subtract the coordinates of point A from point B, right? Similarly for vector AC, subtract A from C. Let me write down the coordinates again to make sure I have them right.Points given:A(3, 3, -1)B(5, 1, -2)C(4, 1, -3)So vector AB is B - A. Let's compute each component:For AB:x-component: 5 - 3 = 2y-component: 1 - 3 = -2z-component: -2 - (-1) = -2 + 1 = -1So AB = (2, -2, -1)Now for vector AC:x-component: 4 - 3 = 1y-component: 1 - 3 = -2z-component: -3 - (-1) = -3 + 1 = -2So AC = (1, -2, -2)Alright, now I need the dot product of AB and AC. The dot product is computed by multiplying corresponding components and then adding them up.Let's compute AB · AC:(2)(1) + (-2)(-2) + (-1)(-2) = 2 + 4 + 2 = 8Hmm, that seems right. Let me check each term again:First component: 2*1 = 2Second component: -2*-2 = 4Third component: -1*-2 = 2Total: 2 + 4 + 2 = 8. Yep.Next, I need the magnitudes of AB and AC.Starting with ||AB||:||AB|| = sqrt(2^2 + (-2)^2 + (-1)^2) = sqrt(4 + 4 + 1) = sqrt(9) = 3Wait, let me verify each term:2 squared is 4, (-2) squared is 4, (-1) squared is 1. So 4 + 4 + 1 is indeed 9. Square root of 9 is 3. Okay, that's straightforward.Now ||AC||:||AC|| = sqrt(1^2 + (-2)^2 + (-2)^2) = sqrt(1 + 4 + 4) = sqrt(9) = 3Same process here. 1 squared is 1, (-2) squared is 4, another (-2) squared is 4. Sum is 1+4+4=9. Square root is 3. So both vectors have a magnitude of 3. Interesting.So plugging back into the cosine formula:cosθ = (8) / (3 * 3) = 8/9So the cosine of the angle between AB and AC is 8/9. Let me just make sure I didn't make any mistakes in the calculations. Let me recheck each step.First, vectors AB and AC:AB: (5-3, 1-3, -2 - (-1)) = (2, -2, -1). Correct.AC: (4-3, 1-3, -3 - (-1)) = (1, -2, -2). Correct.Dot product: 2*1 + (-2)*(-2) + (-1)*(-2) = 2 + 4 + 2 = 8. Correct.Magnitudes: Both are sqrt(4 + 4 + 1) = sqrt(9) = 3 and sqrt(1 + 4 + 4) = sqrt(9) = 3. Correct.So 8 divided by (3*3) is 8/9. Seems right. Therefore, the answer should be 8/9.Final AnswerThe cosine of the angle between vectors overrightarrow{AB} and overrightarrow{AC} is boxed{dfrac{8}{9}}.

✅:To find the cosine of the angle between vectors overrightarrow{AB} and overrightarrow{AC}, we use the formula:[costheta = frac{overrightarrow{AB} cdot overrightarrow{AC}}{||overrightarrow{AB}|| , ||overrightarrow{AC}||}]First, we determine the vectors overrightarrow{AB} and overrightarrow{AC} by subtracting the coordinates of point A from points B and C respectively:[overrightarrow{AB} = B - A = (5 - 3, 1 - 3, -2 - (-1)) = (2, -2, -1)][overrightarrow{AC} = C - A = (4 - 3, 1 - 3, -3 - (-1)) = (1, -2, -2)]Next, we calculate the dot product of overrightarrow{AB} and overrightarrow{AC}:[overrightarrow{AB} cdot overrightarrow{AC} = (2)(1) + (-2)(-2) + (-1)(-2) = 2 + 4 + 2 = 8]We then find the magnitudes of overrightarrow{AB} and overrightarrow{AC}:[||overrightarrow{AB}|| = sqrt{2^2 + (-2)^2 + (-1)^2} = sqrt{4 + 4 + 1} = sqrt{9} = 3][||overrightarrow{AC}|| = sqrt{1^2 + (-2)^2 + (-2)^2} = sqrt{1 + 4 + 4} = sqrt{9} = 3]Finally, substituting these values into the formula for costheta:[costheta = frac{8}{3 times 3} = frac{8}{9}]The cosine of the angle between vectors overrightarrow{AB} and overrightarrow{AC} is boxed{dfrac{8}{9}}.

🔑:1. Calculate the vectors (overrightarrow{AB}) and (overrightarrow{AC}): [ overrightarrow{AB} = B - A = (5-3, 1-3, -2-(-1)) = (2, -2, -1) ] [ overrightarrow{AC} = C - A = (4-3, 1-3, -3-(-1)) = (1, -2, -2) ]2. Find the dot product ((overrightarrow{AB}, overrightarrow{AC})): [ (overrightarrow{AB}, overrightarrow{AC}) = 2 cdot 1 + (-2) cdot (-2) + (-1) cdot (-2) ] [ = 2 + 4 + 2 = 8 ]3. Calculate the magnitudes (|overrightarrow{AB}|) and (|overrightarrow{AC}|): [ |overrightarrow{AB}| = sqrt{2^2 + (-2)^2 + (-1)^2} = sqrt{4 + 4 + 1} = sqrt{9} = 3 ] [ |overrightarrow{AC}| = sqrt{1^2 + (-2)^2 + (-2)^2} = sqrt{1 + 4 + 4} = sqrt{9} = 3 ]4. Determine the cosine of the angle (varphi) between the vectors: [ cos (overrightarrow{AB}, overrightarrow{AC}) = frac{(overrightarrow{AB}, overrightarrow{AC})}{|overrightarrow{AB}| cdot |overrightarrow{AC}|} ] [ = frac{8}{3 cdot 3} = frac{8}{9} ]5. Conclusion: [ cos(overrightarrow{AB}, overrightarrow{AC}) = frac{8}{9} ] [ boxed{frac{8}{9}} ]

🧠:Okay, let's see. I need to figure out the percentage of students who received certificates in the Fermat competition. There are 30 boys and 20 girls participating. The problem says that 10% of the boys and 20% of the girls got certificates. Then, I have to find the overall percentage of students who received certificates. The options are given from (A) 14 to (E) 50. Let me break this down step by step.First, let me confirm the total number of students. There are 30 boys and 20 girls. So, 30 + 20 = 50 students in total. That seems straightforward. Now, certificates are awarded to 10% of the boys and 20% of the girls. I need to calculate how many boys and how many girls that is, then add them together to find the total number of certificate recipients. Then, I can find what percentage that total is of the entire student body (which is 50 students).Starting with the boys: 10% of 30 boys. To calculate 10%, I can convert the percentage to a decimal by dividing by 100. So 10% is 0.10. Then multiply 0.10 by 30. Let me compute that: 0.10 × 30 = 3. So, 3 boys received certificates.Next, the girls: 20% of 20 girls. Again, converting 20% to a decimal is 0.20. Multiply that by 20: 0.20 × 20 = 4. So, 4 girls received certificates.Now, total certificate recipients are boys plus girls: 3 + 4 = 7 students.Wait, but the total number of students is 50. So, 7 out of 50 students received certificates. To find the percentage, I can set up the fraction 7/50 and convert that to a percentage. To do that, I divide 7 by 50 and multiply by 100. Let me compute that:7 ÷ 50 = 0.14. Then, 0.14 × 100 = 14%. So that would be 14%, which is option (A). Hmm, but let me double-check because sometimes these problems can be tricky.Wait, let me confirm the calculations again. 10% of 30 boys is 3, 20% of 20 girls is 4. Total certificates: 3+4=7. Total participants: 50. So 7/50 is indeed 0.14, so 14%. That's option (A). But wait, the options given are (A) 14, (B) 15, (C)16, etc. So 14 is an option. So maybe that's correct.But let me think again. Maybe I made a mistake in the percentage calculation. Let's verify step by step.First, boys: 30 students. 10% of 30 is 3. Girls: 20 students. 20% of 20 is 4. Total certificates: 3+4=7. Total students: 50. So 7/50. To convert to percentage: (7/50)*100 = (7*100)/50 = (700)/50 = 14. Yes, that's 14%. So the answer should be (A) 14. Hmm, but why is option (C) 16 here? Is there a possibility that I messed up the percentages?Wait, another way to approach this: Maybe they want the percentage of boys and girls combined in a different way? Let me see. But no, the total number of students is 50. So the certificates are given to 7 out of 50. That's 14%. Hmm.Alternatively, perhaps the question is phrased differently. Wait, the problem says "Of the students who participated, the percentage that received certificates was..." So that's the same as total certificates over total participants, which is 7/50=14%. So that's correct. So maybe the answer is (A) 14. But let me check if I misread the number of boys and girls. The problem says 30 boys and 20 girls. Yes. 30+20=50. 10% of 30 is 3, 20% of 20 is 4. 3+4=7. 7/50=14%. So yeah, (A) is correct.But wait, let me think if there's another interpretation. Maybe they are asking for the percentage of certificate recipients relative to the total number of participants, but considering each gender's contribution? Wait, no, the total participants are 50. So the total certificates are 7, so 7/50 is 14%. That's straightforward.Alternatively, maybe the question is a weighted average? Let's see. The percentage of boys is 30/50 of the total participants, and girls are 20/50. The certificate rates are 10% and 20% respectively. So maybe calculating the weighted average certificate rate. Let's try that approach.So, boys are 30 out of 50, which is 60% of the participants. Girls are 40%. The certificate rates are 10% for boys and 20% for girls. So the overall certificate percentage would be (0.6 * 10%) + (0.4 * 20%) = 6% + 8% = 14%. That gives the same result. So that's another way to compute it, arriving at 14%. So again, (A) 14.Therefore, I think the correct answer is (A) 14. The other options might be there to trick you if you, for example, added 10% and 20% to get 30% (option D) or something else. But no, the correct calculation is 14%. So I'm confident that (A) is the right answer.

✅:To determine the percentage of students who received certificates, we start by calculating the number of boys and girls who were awarded:- Boys: 10% of 30 = (0.10 times 30 = 3).- Girls: 20% of 20 = (0.20 times 20 = 4).Total certificates = (3 + 4 = 7).There are 50 students in total (30 boys + 20 girls). The percentage of students who received certificates is:[frac{7}{50} times 100 = 14%.]Answer: (A) 14.

🔑:1. Calculate the number of boys who received certificates: - Total number of boys = 30 - Percentage of boys who received certificates = 10% [ text{Number of boys who received certificates} = left(frac{10}{100}right) times 30 = 3 ]2. Calculate the number of girls who received certificates: - Total number of girls = 20 - Percentage of girls who received certificates = 20% [ text{Number of girls who received certificates} = left(frac{20}{100}right) times 20 = 4 ]3. Calculate the total number of students who received certificates: [ text{Total number of students who received certificates} = 3 (text{boys}) + 4 (text{girls}) = 7 ]4. Calculate the total number of students who participated: [ text{Total number of students who participated} = 30 (text{boys}) + 20 (text{girls}) = 50 ]5. Calculate the percentage of students who received certificates: [ text{Percentage of students who received certificates} = left(frac{7}{50}right) times 100% ] Perform the division and multiplication: [ left(frac{7}{50}right) times 100 = 14% ] Conclusion:The percentage of students who received certificates is ( boxed{14} ).

🧠:Okay, so I have this problem here: Find the value of ( x ) where ( x ) and ( y ) are real numbers with ( x > y ), and they satisfy the equations ( x^{2} y^{2} + x^{2} + y^{2} + 2xy = 40 ) and ( xy + x + y = 8 ). Hmm, let's try to figure this out step by step.First, let's look at the two equations. The first one is a bit complicated: ( x^{2} y^{2} + x^{2} + y^{2} + 2xy = 40 ). The second one is ( xy + x + y = 8 ). Maybe I can find a substitution or a way to relate these two equations. Let me start with the second equation because it looks simpler. If I let ( S = x + y ) and ( P = xy ), that might be helpful. These are the standard substitutions for symmetric equations. So, substituting into the second equation: ( P + S = 8 ). That gives me a relationship between ( S ) and ( P ). Now, maybe I can express the first equation in terms of ( S ) and ( P ) as well. Let's try expanding or rearranging the first equation. The first term is ( x^{2} y^{2} ), which is ( (xy)^2 = P^2 ). Then we have ( x^{2} + y^{2} ). I know that ( x^{2} + y^{2} = (x + y)^2 - 2xy = S^2 - 2P ). The last term is ( 2xy = 2P ). Putting all these together into the first equation: ( P^2 + (S^2 - 2P) + 2P = 40 ). Let's simplify that. The ( -2P ) and ( +2P ) cancel out, so we get ( P^2 + S^2 = 40 ). So now we have two equations in terms of ( S ) and ( P ):1. ( P + S = 8 )2. ( P^2 + S^2 = 40 )This seems more manageable. Let's try to solve for ( S ) and ( P ). From the first equation, we can express ( P = 8 - S ). Substitute this into the second equation: ( (8 - S)^2 + S^2 = 40 ).Expanding ( (8 - S)^2 ): ( 64 - 16S + S^2 ). So substituting back, we have:( 64 - 16S + S^2 + S^2 = 40 )Combine like terms:( 2S^2 - 16S + 64 = 40 )Subtract 40 from both sides:( 2S^2 - 16S + 24 = 0 )Divide the entire equation by 2 to simplify:( S^2 - 8S + 12 = 0 )Now, solving this quadratic equation for ( S ):The quadratic factors as ( (S - 6)(S - 2) = 0 ), so the solutions are ( S = 6 ) or ( S = 2 ).Now, let's find the corresponding ( P ) for each ( S ):- If ( S = 6 ), then ( P = 8 - 6 = 2 )- If ( S = 2 ), then ( P = 8 - 2 = 6 )So we have two cases:1. ( S = 6 ) and ( P = 2 )2. ( S = 2 ) and ( P = 6 )Now, we need to find ( x ) and ( y ) such that ( x + y = S ) and ( xy = P ), with ( x > y ). Let's tackle each case.Case 1: ( S = 6 ), ( P = 2 )Here, ( x + y = 6 ) and ( xy = 2 ). The quadratic equation for this would be ( t^2 - 6t + 2 = 0 ). Let's find the roots using the quadratic formula:( t = [6 ± sqrt(36 - 8)] / 2 = [6 ± sqrt(28)] / 2 = [6 ± 2*sqrt(7)] / 2 = 3 ± sqrt(7) )So the solutions are ( x = 3 + sqrt(7) ), ( y = 3 - sqrt(7) ). Since ( sqrt(7) ≈ 2.6458 ), ( x ≈ 5.6458 ) and ( y ≈ 0.3542 ), so ( x > y ). This seems valid.Case 2: ( S = 2 ), ( P = 6 )Here, ( x + y = 2 ) and ( xy = 6 ). The quadratic equation is ( t^2 - 2t + 6 = 0 ). Let's compute the discriminant:Discriminant ( D = 4 - 24 = -20 ). Since the discriminant is negative, there are no real solutions in this case. Therefore, this case is invalid because the problem states that ( x ) and ( y ) are real numbers.Therefore, the only valid solution is from Case 1: ( x = 3 + sqrt(7) ), ( y = 3 - sqrt(7) ).But wait, let me verify this solution by plugging back into the original equations to make sure.Verification:First equation: ( xy + x + y = 8 ). From Case 1:( xy = 2 ), ( x + y = 6 ). So ( 2 + 6 = 8 ). Correct.Second equation: ( x^{2} y^{2} + x^{2} + y^{2} + 2xy = 40 ).Compute each term:- ( x^{2} y^{2} = (xy)^2 = 2^2 = 4 )- ( x^{2} + y^{2} = (x + y)^2 - 2xy = 6^2 - 2*2 = 36 - 4 = 32 )- ( 2xy = 4 )Adding them up: 4 + 32 + 4 = 40. Correct.So the solution checks out. Therefore, the value of ( x ) is ( 3 + sqrt{7} ).Wait, just to be thorough, let me check if there's another way this could be approached. Maybe directly manipulating the equations without substitution?Let me see. The second equation is ( xy + x + y = 8 ). If I add 1 to both sides, it becomes ( xy + x + y + 1 = 9 ), which factors into ( (x + 1)(y + 1) = 9 ). Interesting, but not sure if that helps with the first equation. Let's see.But maybe if I set ( a = x + 1 ), ( b = y + 1 ), then ( ab = 9 ). Then, perhaps express the first equation in terms of ( a ) and ( b ). Let's try that.Original first equation: ( x^2 y^2 + x^2 + y^2 + 2xy = 40 ).Express in terms of ( a ) and ( b ):First, ( x = a - 1 ), ( y = b - 1 ). Then:( (a - 1)^2 (b - 1)^2 + (a - 1)^2 + (b - 1)^2 + 2(a - 1)(b - 1) = 40 )Hmm, this looks messy, but maybe expanding it could lead to something. Let's try.First term: ( (a - 1)^2 (b - 1)^2 )Second term: ( (a - 1)^2 )Third term: ( (b - 1)^2 )Fourth term: ( 2(a - 1)(b - 1) )Let me compute each term:First term: ( [(a - 1)(b - 1)]^2 = (ab - a - b + 1)^2 ). Since ( ab = 9 ), substitute that in: ( (9 - a - b + 1)^2 = (10 - a - b)^2 ).Second term: ( (a - 1)^2 = a^2 - 2a + 1 )Third term: ( (b - 1)^2 = b^2 - 2b + 1 )Fourth term: ( 2(a - 1)(b - 1) = 2(ab - a - b + 1) = 2(9 - a - b + 1) = 2(10 - a - b) )Putting all these together into the equation:( (10 - a - b)^2 + (a^2 - 2a + 1) + (b^2 - 2b + 1) + 2(10 - a - b) = 40 )Let's expand ( (10 - a - b)^2 ):( 100 - 20a - 20b + a^2 + 2ab + b^2 )So substituting back:( [100 - 20a - 20b + a^2 + 2ab + b^2] + [a^2 - 2a + 1] + [b^2 - 2b + 1] + [20 - 2a - 2b] = 40 )Now combine all terms:First, the expanded square:100 - 20a - 20b + a^2 + 2ab + b^2Then adding the second term:+ a^2 - 2a + 1Third term:+ b^2 - 2b + 1Fourth term:+ 20 - 2a - 2bNow, let's collect like terms:- a² terms: 1a² + 1a² = 2a²- b² terms: 1b² + 1b² = 2b²- ab term: 2ab- a terms: -20a -2a -2a = -24a- b terms: -20b -2b -2b = -24b- constants: 100 +1 +1 +20 = 122So altogether, the equation becomes:2a² + 2b² + 2ab -24a -24b + 122 = 40Subtract 40 from both sides:2a² + 2b² + 2ab -24a -24b + 82 = 0Divide the entire equation by 2:a² + b² + ab -12a -12b + 41 = 0Hmm, this seems complicated. But since we know that ( ab = 9 ), maybe substitute that in.Let me see:Given ( ab = 9 ), so ( ab = 9 ). Let's also note that ( a + b = (x + 1) + (y + 1) = x + y + 2 = S + 2 ). From earlier, in the valid case, S = 6, so a + b = 8. In the other case, S = 2, so a + b = 4, but that case was invalid. Let's check if this helps.So in the valid case, a + b = 8, ab = 9. So, perhaps we can use these.But maybe we can express the equation a² + b² + ab -12a -12b +41 =0 in terms of a + b and ab.Note that a² + b² = (a + b)^2 - 2ab. So:(a + b)^2 - 2ab + ab -12(a + b) +41 = 0Simplify:(a + b)^2 - ab -12(a + b) +41 = 0We know that in the valid case, a + b =8 and ab=9. Substitute:8² -9 -12*8 +41 = 64 -9 -96 +41 = (64 -96) + (-9 +41) = (-32) + 32 = 0. So it works.Similarly, for the other case where a + b =4 and ab=9:4² -9 -12*4 +41 =16 -9 -48 +41= (16 -48) + (-9 +41)= (-32) +32=0. So even though the variables are complex here, the equation still holds. But since in that case, x and y are not real, the substitution with a and b still satisfies the equation because the algebra works regardless of the variables being real or complex. But this approach didn't really help us find the solution more easily. The substitution with S and P was more straightforward. So, going back, our initial solution seems correct.Another way to check: Let me compute x and y as 3 ± sqrt(7) and verify all terms.Compute x = 3 + sqrt(7), y = 3 - sqrt(7).Compute xy: (3 + sqrt(7))(3 - sqrt(7)) = 9 - (sqrt(7))² = 9 -7 =2. Correct.Compute x + y: 3 + sqrt(7) + 3 - sqrt(7) =6. Correct.Compute x² y²: (xy)^2 = 2²=4.Compute x² + y²: (x + y)^2 - 2xy =36 -4=32.Compute 2xy=4.So total:4 +32 +4=40. Correct.Therefore, the answer is definitely x=3 + sqrt(7).Just to ensure there are no other solutions, even complex ones, but the problem states x and y are real numbers, so the other case is invalid. So yes, the only real solution is x=3 + sqrt(7), and since x > y, that's the correct value.Final AnswerThe value of ( x ) is (boxed{3 + sqrt{7}}).

✅:Given the equations ( x^2 y^2 + x^2 + y^2 + 2xy = 40 ) and ( xy + x + y = 8 ) with ( x > y ), we start by substituting ( S = x + y ) and ( P = xy ). From the second equation, we have:[ P + S = 8 ]For the first equation, we substitute ( x^2 y^2 = P^2 ), ( x^2 + y^2 = S^2 - 2P ), and ( 2xy = 2P ):[ P^2 + (S^2 - 2P) + 2P = 40 ]Simplifying, we get:[ P^2 + S^2 = 40 ]We now have the system:1. ( P + S = 8 )2. ( P^2 + S^2 = 40 )Substituting ( P = 8 - S ) into the second equation:[ (8 - S)^2 + S^2 = 40 ]Expanding and simplifying:[ 64 - 16S + S^2 + S^2 = 40 ][ 2S^2 - 16S + 24 = 0 ][ S^2 - 8S + 12 = 0 ]Factoring the quadratic equation:[ (S - 6)(S - 2) = 0 ]Thus, ( S = 6 ) or ( S = 2 ).For ( S = 6 ), ( P = 8 - 6 = 2 ). For ( S = 2 ), ( P = 8 - 2 = 6 ).Case 1: ( S = 6 ), ( P = 2 )- Solving ( x + y = 6 ) and ( xy = 2 ): - The quadratic equation ( t^2 - 6t + 2 = 0 ) has roots ( t = 3 pm sqrt{7} ). - Since ( x > y ), ( x = 3 + sqrt{7} ) and ( y = 3 - sqrt{7} ).Case 2: ( S = 2 ), ( P = 6 )- Solving ( x + y = 2 ) and ( xy = 6 ): - The discriminant is negative, indicating no real solutions.Verification of Case 1:- ( xy + x + y = 2 + 6 = 8 )- ( x^2 y^2 + x^2 + y^2 + 2xy = 4 + 32 + 4 = 40 )Thus, the value of ( x ) is (boxed{3 + sqrt{7}}).

🔑:1. We start with the given equations: [ x^2 y^2 + x^2 + y^2 + 2xy = 40 ] and [ xy + x + y = 8. ]2. First, square the second equation: [ (xy + x + y)^2 = 8^2. ] Hence: [ (xy + x + y)^2 = 64. ]3. Expand the left-hand side of (xy + x + y)^2: [ (xy + x + y)^2 = (xy)^2 + 2xy(x + y) + (x + y)^2. ] So: [ (xy)^2 + 2xy(x + y) + (x + y)^2 = 64. ]4. Given the other equation, rewrite x^2 y^2 + x^2 + y^2 + 2xy = 40 as: [ (xy)^2 + x^2 + y^2 + 2xy = 40. ]5. Use the identity (x + y)^2 = x^2 + y^2 + 2xy to replace x^2 + y^2: [ x^2 + y^2 = (x + y)^2 - 2xy. ] Therefore: [ (xy)^2 + (x + y)^2 - 2xy + 2xy = 40, ] which simplifies to: [ (xy)^2 + (x + y)^2 = 40. ]6. Now we have two new simplified equations: [ (xy)^2 + (x + y)^2 = 40, ] and from the squared sum equation: [ (xy)^2 + 2xy(x + y) + (x + y)^2 = 64. ] 7. Subtract the first new equation from the squared sum equation: [ [(xy)^2 + 2xy(x + y) + (x + y)^2] - [(xy)^2 + (x + y)^2] = 64 - 40. ] This reduces to: [ 2xy(x + y) = 24. ]8. Therefore: [ xy(x + y) = 12. ]9. Let S = x + y and P = xy. Then: [ xy cdot (x + y) = 12 quad text{and} quad (x + y)^2 + (xy)^2 = 40. ] 10. Substitute S = x + y and P = xy: [ P cdot S = 12, ] [ S^2 + P^2 = 40. ]11. Substitute P = frac{12}{S} into S^2 + P^2 = 40: [ S^2 + left(frac{12}{S}right)^2 = 40. ]12. Simplify and solve for S: [ S^2 + frac{144}{S^2} = 40. ]13. Multiply by S^2 to clear the fraction: [ S^4 + 144 = 40S^2. ]14. Rearrange to form a quadratic equation in terms of S^2: [ S^4 - 40S^2 + 144 = 0. ]15. Let u = S^2. Then the equation becomes: [ u^2 - 40u + 144 = 0. ]16. Solve this quadratic equation using the quadratic formula u = frac{-b pm sqrt{b^2 - 4ac}}{2a}: [ u = frac{40 pm sqrt{40^2 - 4 cdot 1 cdot 144}}{2 cdot 1} = frac{40 pm sqrt{1600 - 576}}{2} = frac{40 pm sqrt{1024}}{2} = frac{40 pm 32}{2}. ]17. Hence: [ u = frac{40 + 32}{2} = 36 quad text{or} quad u = frac{40 - 32}{2} = 4. ]18. Therefore, S^2 = 36 or S^2 = 4, so: [ S = 6 quad text{or} quad S = 2. ]19. If S = 6, then P = frac{12}{S} = 2. If S = 2, then P = frac{12}{S} = 6.20. Given x > y, we consider x + y = 6 and xy = 2. The quadratic t^2 - 6t + 2 = 0 has: [ t = frac{6 pm sqrt{36 - 8}}{2} = frac{6 pm sqrt{28}}{2} = frac{6 pm 2sqrt{7}}{2} = 3 pm sqrt{7}. ]21. Therefore, since x > y, we take the larger root: [ x = 3 + sqrt{7}. ] Conclusion:[boxed{3 + sqrt{7}}]