answer：First, I need to determine the total amount of money Maria has already saved and received from her mother.Maria has saved 120, and her mother has offered her 250. Adding these together gives:120 + 250 = 370Next, Maria needs to earn an additional 230 to afford the bike. To find the total cost of the bike, I add the amount she already has to the amount she still needs to earn:370 + 230 = 600Therefore, the retail price of the bike at the bike shop is 600.

answer：Okay, so I have this system of equations to solve. It looks pretty complicated with four variables: x, y, u, and v. There are four equations, which is good because it means I might be able to find a unique solution or at least some specific solutions. Let me write them down again to make sure I have everything:1. ( x^2 + y^2 + u^2 + v^2 = 4 )2. ( xu + yv + xv + yu = 0 )3. ( xyu + yuv + uvx + vxy = -2 )4. ( xyuv = -1 )Alright, starting with the first equation, it's the sum of squares equal to 4. That reminds me of the equation of a sphere in four dimensions, but since we're dealing with real solutions, it's just a constraint on the variables.The second equation is a bit more complex. It has cross terms: xu, yv, xv, yu. Maybe I can factor this somehow. Let me see: xu + yv + xv + yu. Hmm, grouping terms: (xu + xv) + (yv + yu) = x(u + v) + y(u + v) = (x + y)(u + v). Oh, that's nice! So the second equation simplifies to (x + y)(u + v) = 0. That means either x + y = 0 or u + v = 0. So, I have two cases to consider:**Case 1:** x + y = 0**Case 2:** u + v = 0I think I'll start with Case 2 because it might lead to simpler substitutions.**Case 2: u + v = 0**If u + v = 0, then v = -u. Let's substitute v = -u into all the equations.1. ( x^2 + y^2 + u^2 + (-u)^2 = 4 ) simplifies to ( x^2 + y^2 + 2u^2 = 4 )2. The second equation is already satisfied because u + v = 0.3. Substitute v = -u into the third equation: ( xyu + y(-u)u + u(-u)x + (-u)xy ). Let's compute each term:- ( xyu ) remains as is.- ( y(-u)u = -y u^2 )- ( u(-u)x = -u^2 x )- ( (-u)xy = -u xy )So, putting it all together: ( xyu - y u^2 - u^2 x - u xy ). Let's combine like terms:- ( xyu - u xy = 0 ) because they cancel each other out.- ( - y u^2 - u^2 x = -u^2 (x + y) )So the third equation becomes: ( -u^2 (x + y) = -2 ). Let me write that as:( u^2 (x + y) = 2 )4. Substitute v = -u into the fourth equation: ( xy(-u)u = -1 ) which simplifies to ( -xy u^2 = -1 ) or ( xy u^2 = 1 )So now, from equation 4, I have ( xy u^2 = 1 ). Let me denote this as equation 4'.From equation 3', I have ( u^2 (x + y) = 2 ). Let me denote this as equation 3'.So, now I have:- Equation 1: ( x^2 + y^2 + 2u^2 = 4 )- Equation 3': ( u^2 (x + y) = 2 )- Equation 4': ( xy u^2 = 1 )Hmm, so I have three equations with variables x, y, u. Maybe I can express x + y and xy in terms of u^2.Let me denote s = x + y and p = xy. Then, equation 3' becomes ( u^2 s = 2 ) and equation 4' becomes ( p u^2 = 1 ).So, from equation 3': ( s = 2 / u^2 )From equation 4': ( p = 1 / u^2 )Now, equation 1 can be rewritten using s and p. Since ( x^2 + y^2 = (x + y)^2 - 2xy = s^2 - 2p ). So,( x^2 + y^2 + 2u^2 = (s^2 - 2p) + 2u^2 = 4 )Substituting s and p:( ( (2 / u^2)^2 - 2*(1 / u^2) ) + 2u^2 = 4 )Let me compute each term:First term: ( (4 / u^4) - (2 / u^2) )So, the entire equation becomes:( (4 / u^4 - 2 / u^2) + 2u^2 = 4 )Let me combine the terms:( 4 / u^4 - 2 / u^2 + 2u^2 = 4 )To make this easier, let me multiply every term by u^4 to eliminate denominators:( 4 - 2u^2 + 2u^6 = 4u^4 )Bring all terms to one side:( 2u^6 - 4u^4 - 2u^2 + 4 = 0 )Hmm, that's a sixth-degree equation. Maybe I can factor it or make a substitution.Let me let z = u^2. Then, the equation becomes:( 2z^3 - 4z^2 - 2z + 4 = 0 )Let me factor this cubic equation. Let's try possible rational roots using Rational Root Theorem. Possible roots are ±1, ±2, ±4, considering factors of 4 over factors of 2.Testing z=1: 2 - 4 - 2 + 4 = 0. So z=1 is a root.So, I can factor out (z - 1):Using polynomial division or synthetic division:Divide 2z^3 - 4z^2 - 2z + 4 by (z - 1):Coefficients: 2 | -4 | -2 | 4Bring down 2.Multiply by 1: 2.Add to next coefficient: -4 + 2 = -2.Multiply by 1: -2.Add to next coefficient: -2 + (-2) = -4.Multiply by 1: -4.Add to last coefficient: 4 + (-4) = 0. Perfect.So, the cubic factors as (z - 1)(2z^2 - 2z - 4) = 0.Now, set each factor equal to zero:1. z - 1 = 0 => z = 12. 2z^2 - 2z - 4 = 0 => Divide by 2: z^2 - z - 2 = 0 => z = [1 ± sqrt(1 + 8)] / 2 = [1 ± 3]/2 => z = 2 or z = -1But z = u^2, which must be non-negative, so z = 2 or z = 1.So, possible z values: 1 and 2.Now, let's consider each case.**Subcase 2.1: z = 1 (i.e., u^2 = 1 => u = 1 or u = -1)**From equation 4': ( p = 1 / u^2 = 1 / 1 = 1 )From equation 3': ( s = 2 / u^2 = 2 / 1 = 2 )So, s = x + y = 2 and p = xy = 1.So, x and y are roots of the quadratic equation t^2 - st + p = 0 => t^2 - 2t + 1 = 0.This factors as (t - 1)^2 = 0, so x = y = 1.So, in this subcase, x = y = 1, u^2 = 1, so u = 1 or u = -1.But since v = -u, if u = 1, then v = -1; if u = -1, then v = 1.So, we have two solutions here:1. (x, y, u, v) = (1, 1, 1, -1)2. (x, y, u, v) = (1, 1, -1, 1)Wait, but let me check if these satisfy all equations.First, let's check the first solution: (1, 1, 1, -1)1. (1^2 + 1^2 + 1^2 + (-1)^2 = 1 + 1 + 1 + 1 = 4) ✔️2. (1*1 + 1*(-1) + 1*1 + 1*(-1) = 1 -1 + 1 -1 = 0) ✔️3. (1*1*1 + 1*(-1)*1 + 1*(-1)*1 + (-1)*1*1 = 1 -1 -1 -1 = -2) ✔️4. (1*1*1*(-1) = -1) ✔️Good. Now the second solution: (1, 1, -1, 1)1. Same as above: 1 + 1 + 1 + 1 = 4 ✔️2. (1*(-1) + 1*1 + 1*(-1) + 1*1 = -1 + 1 -1 + 1 = 0) ✔️3. (1*1*(-1) + 1*1*(-1) + (-1)*1*1 + 1*1*1 = -1 -1 -1 +1 = -2) ✔️4. (1*1*(-1)*1 = -1) ✔️Perfect, both solutions work.**Subcase 2.2: z = 2 (i.e., u^2 = 2 => u = sqrt(2) or u = -sqrt(2))**From equation 4': ( p = 1 / u^2 = 1 / 2 )From equation 3': ( s = 2 / u^2 = 2 / 2 = 1 )So, s = x + y = 1 and p = xy = 1/2.So, x and y are roots of the quadratic equation t^2 - t + 1/2 = 0.Let me compute the discriminant: D = 1 - 4*(1/2) = 1 - 2 = -1.Since the discriminant is negative, there are no real solutions for x and y in this case. So, we discard z = 2.Therefore, in Case 2, we have two real solutions: (1, 1, 1, -1) and (1, 1, -1, 1).Now, let's go back to Case 1: x + y = 0.**Case 1: x + y = 0**If x + y = 0, then y = -x. Let's substitute y = -x into all equations.1. ( x^2 + (-x)^2 + u^2 + v^2 = 4 ) simplifies to ( 2x^2 + u^2 + v^2 = 4 )2. The second equation is already satisfied because x + y = 0.3. Substitute y = -x into the third equation: ( x*(-x)*u + (-x)*u*v + u*v*x + v*x*(-x) ). Let's compute each term:- ( x*(-x)*u = -x^2 u )- ( (-x)*u*v = -x u v )- ( u*v*x = x u v )- ( v*x*(-x) = -x^2 v )So, putting it all together: ( -x^2 u - x u v + x u v - x^2 v ). Let's combine like terms:- ( -x^2 u - x^2 v = -x^2 (u + v) )- ( -x u v + x u v = 0 )So, the third equation simplifies to ( -x^2 (u + v) = -2 ). Let me write that as:( x^2 (u + v) = 2 )4. Substitute y = -x into the fourth equation: ( x*(-x)*u*v = -1 ) which simplifies to ( -x^2 u v = -1 ) or ( x^2 u v = 1 )So now, from equation 4, I have ( x^2 u v = 1 ). Let me denote this as equation 4''.From equation 3'', I have ( x^2 (u + v) = 2 ). Let me denote this as equation 3''.So, now I have:- Equation 1: ( 2x^2 + u^2 + v^2 = 4 )- Equation 3'': ( x^2 (u + v) = 2 )- Equation 4'': ( x^2 u v = 1 )Let me denote s = u + v and p = u v. Then, equation 3'' becomes ( x^2 s = 2 ) and equation 4'' becomes ( x^2 p = 1 ).So, from equation 3'': ( s = 2 / x^2 )From equation 4'': ( p = 1 / x^2 )Now, equation 1 can be rewritten using s and p. Since ( u^2 + v^2 = (u + v)^2 - 2uv = s^2 - 2p ). So,( 2x^2 + s^2 - 2p = 4 )Substituting s and p:( 2x^2 + ( (2 / x^2)^2 ) - 2*(1 / x^2 ) = 4 )Let me compute each term:First term: 2x^2Second term: ( 4 / x^4 )Third term: ( -2 / x^2 )So, the entire equation becomes:( 2x^2 + 4 / x^4 - 2 / x^2 = 4 )Let me combine the terms:( 2x^2 - 2 / x^2 + 4 / x^4 = 4 )To make this easier, let me multiply every term by x^4 to eliminate denominators:( 2x^6 - 2x^2 + 4 = 4x^4 )Bring all terms to one side:( 2x^6 - 4x^4 - 2x^2 + 4 = 0 )Hmm, that's a sixth-degree equation. Maybe I can factor it or make a substitution.Let me let z = x^2. Then, the equation becomes:( 2z^3 - 4z^2 - 2z + 4 = 0 )Wait, this is the same cubic equation as before! So, we can factor it similarly.From before, we know that z = 1 is a root. Let's factor it:Divide 2z^3 - 4z^2 - 2z + 4 by (z - 1):Using synthetic division:Coefficients: 2 | -4 | -2 | 4Bring down 2.Multiply by 1: 2.Add to next coefficient: -4 + 2 = -2.Multiply by 1: -2.Add to next coefficient: -2 + (-2) = -4.Multiply by 1: -4.Add to last coefficient: 4 + (-4) = 0.So, the cubic factors as (z - 1)(2z^2 - 2z - 4) = 0.Set each factor equal to zero:1. z - 1 = 0 => z = 12. 2z^2 - 2z - 4 = 0 => Divide by 2: z^2 - z - 2 = 0 => z = [1 ± sqrt(1 + 8)] / 2 = [1 ± 3]/2 => z = 2 or z = -1Again, z = x^2 must be non-negative, so z = 1 or z = 2.**Subcase 1.1: z = 1 (i.e., x^2 = 1 => x = 1 or x = -1)**From equation 4'': ( p = 1 / x^2 = 1 / 1 = 1 )From equation 3'': ( s = 2 / x^2 = 2 / 1 = 2 )So, s = u + v = 2 and p = u v = 1.So, u and v are roots of the quadratic equation t^2 - st + p = 0 => t^2 - 2t + 1 = 0.This factors as (t - 1)^2 = 0, so u = v = 1.But wait, from Case 1, y = -x. So, if x = 1, then y = -1; if x = -1, then y = 1.So, we have two solutions here:1. (x, y, u, v) = (1, -1, 1, 1)2. (x, y, u, v) = (-1, 1, 1, 1)Wait, let me check if these satisfy all equations.First, let's check the first solution: (1, -1, 1, 1)1. (1^2 + (-1)^2 + 1^2 + 1^2 = 1 + 1 + 1 + 1 = 4) ✔️2. (1*1 + (-1)*1 + 1*1 + (-1)*1 = 1 -1 + 1 -1 = 0) ✔️3. (1*(-1)*1 + (-1)*1*1 + 1*1*1 + 1*1*(-1) = -1 -1 + 1 -1 = -2) ✔️4. (1*(-1)*1*1 = -1) ✔️Good. Now the second solution: (-1, 1, 1, 1)1. Same as above: 1 + 1 + 1 + 1 = 4 ✔️2. ((-1)*1 + 1*1 + (-1)*1 + 1*1 = -1 + 1 -1 + 1 = 0) ✔️3. ((-1)*1*1 + 1*1*1 + 1*1*(-1) + 1*(-1)*1 = -1 + 1 -1 -1 = -2) ✔️4. ((-1)*1*1*1 = -1) ✔️Perfect, both solutions work.**Subcase 1.2: z = 2 (i.e., x^2 = 2 => x = sqrt(2) or x = -sqrt(2))**From equation 4'': ( p = 1 / x^2 = 1 / 2 )From equation 3'': ( s = 2 / x^2 = 2 / 2 = 1 )So, s = u + v = 1 and p = u v = 1/2.So, u and v are roots of the quadratic equation t^2 - t + 1/2 = 0.Let me compute the discriminant: D = 1 - 4*(1/2) = 1 - 2 = -1.Again, the discriminant is negative, so there are no real solutions for u and v in this case. So, we discard z = 2.Therefore, in Case 1, we have two real solutions: (1, -1, 1, 1) and (-1, 1, 1, 1).Wait a minute, but in Case 2, we had solutions where u and v were 1 and -1, and in Case 1, we have solutions where x and y are 1 and -1. So, altogether, we have four solutions:1. (1, 1, 1, -1)2. (1, 1, -1, 1)3. (1, -1, 1, 1)4. (-1, 1, 1, 1)But wait, let me check if these are all distinct. Actually, solutions 3 and 4 are similar to solutions 1 and 2 but with different variable assignments. However, they are distinct solutions because the variables are assigned differently.But let me verify if all these solutions are valid.Wait, in Case 2, when u = 1, v = -1, and x = y = 1, that's solution 1.When u = -1, v = 1, and x = y = 1, that's solution 2.In Case 1, when x = 1, y = -1, and u = v = 1, that's solution 3.When x = -1, y = 1, and u = v = 1, that's solution 4.But wait, in solution 3 and 4, u and v are both 1, so v = 1, but in Case 2, v = -u. So, in Case 1, u and v are both 1, which is allowed because in Case 1, we have x + y = 0, not u + v = 0.So, all four solutions are valid and distinct.But wait, let me check if there are more solutions. For example, in Case 2, when u = 1, v = -1, and x = y = 1, but what if u = -1, v = 1, and x = y = 1, that's another solution.Similarly, in Case 1, when x = 1, y = -1, and u = v = 1, and when x = -1, y = 1, and u = v = 1, those are two more solutions.So, in total, we have four solutions.But wait, let me check if there are any other possibilities. For example, in Case 2, when u = 1, v = -1, and x = y = 1, but could u be something else? No, because in Case 2, u^2 = 1, so u can only be 1 or -1.Similarly, in Case 1, x^2 = 1, so x can only be 1 or -1.Therefore, all possible real solutions are the four I found.But wait, let me think again. In Case 2, when u = 1, v = -1, and x = y = 1, that's one solution. When u = -1, v = 1, and x = y = 1, that's another solution. Similarly, in Case 1, when x = 1, y = -1, and u = v = 1, that's another solution, and when x = -1, y = 1, and u = v = 1, that's the fourth solution.But wait, in Case 1, when x = 1, y = -1, and u = v = 1, that's (1, -1, 1, 1). Similarly, when x = -1, y = 1, and u = v = 1, that's (-1, 1, 1, 1). So, these are valid.But let me check if there are any other solutions where, for example, u and v are different from 1 and -1. But in Case 2, u^2 can only be 1 or 2, but 2 leads to no real solutions, so only u^2 = 1 is valid. Similarly, in Case 1, x^2 can only be 1 or 2, but 2 leads to no real solutions, so only x^2 = 1 is valid.Therefore, the only real solutions are the four I found.But wait, let me check if these solutions are unique or if there are more. For example, in Case 2, when u = 1, v = -1, and x = y = 1, that's one solution. Similarly, when u = -1, v = 1, and x = y = 1, that's another solution. In Case 1, when x = 1, y = -1, and u = v = 1, that's another solution, and when x = -1, y = 1, and u = v = 1, that's the fourth solution.Wait, but in Case 1, when x = 1, y = -1, and u = v = 1, that's (1, -1, 1, 1). Similarly, when x = -1, y = 1, and u = v = 1, that's (-1, 1, 1, 1). So, these are valid.But wait, in Case 2, when u = 1, v = -1, and x = y = 1, that's (1, 1, 1, -1). Similarly, when u = -1, v = 1, and x = y = 1, that's (1, 1, -1, 1).So, in total, we have four solutions:1. (1, 1, 1, -1)2. (1, 1, -1, 1)3. (1, -1, 1, 1)4. (-1, 1, 1, 1)But wait, let me check if these are all distinct. For example, solution 3 is (1, -1, 1, 1), and solution 4 is (-1, 1, 1, 1). These are distinct because the positions of x and y are swapped, but since x and y are different variables, these are different solutions.Similarly, solutions 1 and 2 are distinct because u and v are swapped.Therefore, all four solutions are valid and distinct.Wait, but let me check if there are any other solutions where, for example, both x + y = 0 and u + v = 0. Is that possible? Because in the second equation, we have (x + y)(u + v) = 0, so it's possible that both x + y = 0 and u + v = 0.So, let's consider **Case 3: x + y = 0 and u + v = 0**In this case, y = -x and v = -u.Let's substitute y = -x and v = -u into the equations.1. ( x^2 + (-x)^2 + u^2 + (-u)^2 = 4 ) simplifies to ( 2x^2 + 2u^2 = 4 ) => ( x^2 + u^2 = 2 )2. The second equation is satisfied because both x + y and u + v are zero.3. Substitute y = -x and v = -u into the third equation: ( x*(-x)*u + (-x)*u*(-u) + u*(-u)*x + (-u)*x*(-x) ). Let's compute each term:- ( x*(-x)*u = -x^2 u )- ( (-x)*u*(-u) = x u^2 )- ( u*(-u)*x = -u^2 x )- ( (-u)*x*(-x) = u x^2 )So, putting it all together: ( -x^2 u + x u^2 - u^2 x + u x^2 ). Let's combine like terms:- ( -x^2 u + u x^2 = 0 )- ( x u^2 - u^2 x = 0 )So, the third equation simplifies to 0 = -2, which is a contradiction. Therefore, there are no solutions in this case.Therefore, Case 3 does not yield any solutions.So, in conclusion, the only real solutions are the four I found earlier:1. (1, 1, 1, -1)2. (1, 1, -1, 1)3. (1, -1, 1, 1)4. (-1, 1, 1, 1)But wait, let me check if these are all distinct. For example, solution 3 is (1, -1, 1, 1), and solution 4 is (-1, 1, 1, 1). These are distinct because the positions of x and y are swapped, but since x and y are different variables, these are different solutions.Similarly, solutions 1 and 2 are distinct because u and v are swapped.Therefore, all four solutions are valid and distinct.But wait, let me check if there are any other solutions where, for example, both x + y = 0 and u + v = 0. But as I saw earlier, that leads to a contradiction, so no solutions there.Therefore, the only real solutions are the four I found.Wait, but let me check if I missed any other possibilities. For example, in Case 2, when u = 1, v = -1, and x = y = 1, that's one solution. Similarly, when u = -1, v = 1, and x = y = 1, that's another solution. In Case 1, when x = 1, y = -1, and u = v = 1, that's another solution, and when x = -1, y = 1, and u = v = 1, that's the fourth solution.So, yes, these are all the solutions.Therefore, the real solutions are:(1, 1, 1, -1), (1, 1, -1, 1), (1, -1, 1, 1), and (-1, 1, 1, 1).But wait, let me check if these are all distinct. For example, (1, 1, 1, -1) and (1, 1, -1, 1) are distinct because v is different. Similarly, (1, -1, 1, 1) and (-1, 1, 1, 1) are distinct because x and y are swapped.Therefore, all four solutions are valid and distinct.So, in conclusion, the real solutions to the system are:(1, 1, 1, -1), (1, 1, -1, 1), (1, -1, 1, 1), and (-1, 1, 1, 1).But wait, let me check if these are all distinct. For example, (1, 1, 1, -1) and (1, 1, -1, 1) are distinct because v is different. Similarly, (1, -1, 1, 1) and (-1, 1, 1, 1) are distinct because x and y are swapped.Therefore, all four solutions are valid and distinct.So, the final answer is these four solutions.

answer：First, I'll denote the incomes of P1 and P2 as I₁ and I₂, respectively, and their expenditures as E₁ and E₂, respectively.From the problem, I know that the ratio of their expenditures is 3:2, which means E₁/E₂ = 3/2.Both P1 and P2 save Rs. 2200 each year. This means that:E₁ = I₁ - 2200E₂ = I₂ - 2200Given that P1's income (I₁) is Rs. 5500, I can calculate E₁:E₁ = 5500 - 2200 = 3300Now, using the ratio of expenditures:E₁/E₂ = 3/23300/E₂ = 3/2Cross-multiplying to solve for E₂:2 * 3300 = 3 * E₂6600 = 3E₂E₂ = 6600 / 3E₂ = 2200With E₂ known, I can find P2's income (I₂):E₂ = I₂ - 22002200 = I₂ - 2200I₂ = 2200 + 2200I₂ = 4400Now, I have both incomes:I₁ = 5500I₂ = 4400To find the ratio of their incomes, I'll divide both by their greatest common divisor, which is 1100:5500 / 1100 : 4400 / 1100 = 5 : 4Therefore, the ratio of their incomes is 5:4.

answer：Okay, so I have this problem with nine squares, and I need to find the area of square I. They've given me the areas of squares A and B, which are 1 cm² and 81 cm² respectively. Hmm, let me try to visualize the figure they're talking about. Since it's nine squares, maybe it's arranged in a 3x3 grid? That seems likely. So, squares A, B, and I are part of this grid, and I need to figure out the area of square I based on the areas of A and B.First, let's recall that the area of a square is side length squared. So, if square A has an area of 1 cm², its side length must be 1 cm because 1 squared is 1. Similarly, square B has an area of 81 cm², so its side length is the square root of 81, which is 9 cm. Okay, so square A is 1 cm on each side, and square B is 9 cm on each side.Now, I need to figure out how these squares relate to each other in the 3x3 grid. Maybe squares A and B are adjacent or diagonal to each other? If it's a grid, then each square is connected to its neighbors. Let me think about how the side lengths might add up or relate in such a grid.If square A is 1 cm and square B is 9 cm, perhaps they are placed next to each other in some way. Maybe square A is in one corner, and square B is in another corner, and the other squares in between have side lengths that are combinations or differences of these. Hmm, this is a bit abstract without seeing the figure, but I'll try to proceed.Let me consider that in a 3x3 grid, each row and column must add up to the same total length if it's a perfect grid. So, if square A is 1 cm and square B is 9 cm, maybe the total length of the row containing A and the total length of the column containing B are related.Wait, maybe it's like a magic square where the sums of the sides in each row and column are equal? That could be a possibility. If that's the case, then the sum of the side lengths in each row and each column would be the same.Let me assign variables to the unknown squares. Let's say square A is in the top-left corner, square B is in the top-right corner, and square I is in the bottom-right corner. Then, the other squares would be labeled B, C, D, E, F, G, H, and I. Wait, but actually, the problem only mentions squares A, B, and I, so maybe the others are just part of the figure but not labeled.Alternatively, perhaps the figure is constructed in such a way that the side lengths of the squares form a sequence or follow a particular pattern. For example, each subsequent square could have a side length that's a multiple or a function of the previous one.Given that square A is 1 cm and square B is 9 cm, maybe the side lengths increase by a certain factor or follow an arithmetic progression. Let's see, from 1 to 9, that's an increase of 8. If it's an arithmetic sequence, the common difference would be 8, but that seems too large for a 3x3 grid. Alternatively, it could be a geometric sequence where each term is multiplied by a factor. From 1 to 9, the factor would be 9, but that also seems too large.Wait, maybe the side lengths are related in a way that each square's side is the sum of the sides of the squares adjacent to it. For example, if square A is 1 cm and square B is 9 cm, then the square adjacent to both might have a side length of 1 + 9 = 10 cm. But I'm not sure if that's the case here.Alternatively, perhaps the figure is constructed such that each square's side length is the difference between the sides of larger squares. For instance, if square B is 9 cm and square A is 1 cm, then the square between them might be 9 - 1 = 8 cm. That could make sense if the squares are arranged in a way where they overlap or are nested.Let me try to break it down step by step. Since square A is 1 cm², its side is 1 cm. Square B is 81 cm², so its side is 9 cm. If these squares are adjacent in the grid, maybe the other squares' sides are determined by adding or subtracting these lengths.Suppose square A is in the top-left corner, and square B is in the top-right corner. Then, the square in the middle of the top row might have a side length equal to the sum of square A and square B, which would be 1 + 9 = 10 cm. Similarly, the square below square A might have a side length related to square A and another square.Alternatively, if square A is 1 cm and square B is 9 cm, the square adjacent to both might have a side length that's the difference, 9 - 1 = 8 cm. That could be the case if the squares are arranged in a way where they share a common side.Wait, maybe the figure is a combination of squares where each square's side is the sum or difference of the sides of the squares it's adjacent to. For example, if square A is 1 cm and square B is 9 cm, then the square between them could be 1 + 9 = 10 cm, and the square below square A could be 1 + something else.But without seeing the figure, it's a bit challenging. Maybe I should consider that the area of square I is determined by some pattern or relationship between squares A and B. Since the options are all perfect squares (196, 256, 289, 324, 361), which correspond to side lengths of 14, 16, 17, 18, and 19 cm respectively, I need to figure out which one fits.Let me think about the possible relationships. If square A is 1 cm and square B is 9 cm, maybe square I is related to both. Perhaps it's the sum of their side lengths squared? That would be (1 + 9)² = 100, but 100 isn't one of the options. Alternatively, maybe it's the product of their areas, which would be 1 * 81 = 81, but that's already square B.Alternatively, maybe it's the sum of their areas, 1 + 81 = 82, but that's not an option either. Hmm, maybe it's something else.Wait, perhaps the figure is constructed such that each square's side length is the sum of the sides of the squares to its left and above it. So, if square A is 1 cm and square B is 9 cm, then the square to the right of A would be 1 + something, and the square below A would be 1 + something else.Alternatively, maybe it's a Pythagorean theorem situation, where square I is the hypotenuse of a right triangle formed by squares A and B. But that would be sqrt(1² + 9²) = sqrt(82), which isn't an integer, so that doesn't fit.Wait, maybe the figure is a larger square made up of smaller squares, and square I is the largest square in the figure. If square B is 9 cm, and square A is 1 cm, maybe square I is built by combining multiple squares, including A and B.Alternatively, perhaps the side lengths of the squares form a sequence where each term is the sum of the previous two. Like the Fibonacci sequence. Starting with 1 and 9, the next term would be 10, then 19, then 29, etc. But I don't know if that's the case here.Wait, let's think about the possible side lengths of square I. The options are 14, 16, 17, 18, 19 cm. If square I is the largest square, maybe it's the sum of square B and another square. If square B is 9 cm, and another square is, say, 9 cm as well, then 9 + 9 = 18 cm, which is one of the options. So, maybe square I is 18 cm, making its area 324 cm².Alternatively, if square I is the sum of square A and square B, that would be 1 + 9 = 10 cm, but 10² is 100, which isn't an option. So that doesn't fit.Wait, maybe square I is the result of some other relationship. If square A is 1 cm and square B is 9 cm, maybe square I is the square of the sum of their side lengths, which would be (1 + 9)² = 100, but again, that's not an option.Alternatively, maybe square I is the square of the difference of their side lengths, which would be (9 - 1)² = 64, but that's not an option either.Hmm, maybe I need to think about the figure differently. If it's a 3x3 grid, then each row and column must add up to the same total. So, if square A is 1 cm and square B is 9 cm, maybe the total length of the top row is 1 + something + 9, and the total length of the first column is 1 + something + something else.Wait, perhaps the figure is such that the side lengths of the squares form an arithmetic progression. If square A is 1 cm and square B is 9 cm, and there are three squares in a row, then the middle square would have a side length of (1 + 9)/2 = 5 cm. But I don't know if that's the case.Alternatively, maybe it's a geometric progression. From 1 to 9, the common ratio would be 9, but that seems too large for a 3x3 grid.Wait, maybe the figure is constructed such that each square's side length is the sum of the sides of the squares to its left and above it. So, if square A is 1 cm and square B is 9 cm, then the square to the right of A would be 1 + something, and the square below A would be 1 + something else.But without seeing the figure, it's hard to be precise. Maybe I should look for a pattern in the options. The options are 196, 256, 289, 324, 361, which are squares of 14, 16, 17, 18, and 19. These are all relatively large numbers, so square I must be a large square.Given that square B is 9 cm, which is already quite large, maybe square I is even larger, perhaps 18 cm, making its area 324 cm². That seems plausible.Alternatively, maybe square I is the sum of square B and another square. If square B is 9 cm, and another square is 9 cm, then 9 + 9 = 18 cm, so area 324 cm².Wait, another approach: if the figure is a 3x3 grid, then the total area would be the sum of all nine squares. But since we only know two of them, that might not help directly. However, if square I is the largest square, it might be the sum of the areas of squares A and B plus others, but that's too vague.Alternatively, maybe the side length of square I is the sum of the side lengths of squares A and B, which would be 1 + 9 = 10 cm, but 10² is 100, not an option. So that doesn't fit.Wait, perhaps the side length of square I is the difference of the side lengths of squares B and A, which would be 9 - 1 = 8 cm, but 8² is 64, not an option.Alternatively, maybe it's the product of the side lengths, 1 * 9 = 9 cm, but that's square B itself.Hmm, I'm stuck. Maybe I need to think about the figure differently. If it's a 3x3 grid, and square A is 1 cm, square B is 9 cm, then perhaps the other squares have side lengths that are combinations of these. For example, the square adjacent to A and B might be 1 + 9 = 10 cm, and then square I could be built from that.Alternatively, maybe it's a fractal-like structure where each square is built from smaller squares, and square I is the largest one encompassing all others.Wait, another idea: if square A is 1 cm and square B is 9 cm, maybe the figure is such that square I is the square whose side is the sum of the sides of squares A and B, but in a different arrangement. For example, if square A is placed next to square B, the combined length would be 1 + 9 = 10 cm, but that's not an option.Alternatively, maybe square I is the square whose side is the difference of the sides of squares B and A, which is 9 - 1 = 8 cm, but again, 8² is 64, not an option.Wait, maybe the figure is constructed such that each square's side length is the sum of the previous two squares' side lengths, similar to the Fibonacci sequence. Starting with 1 and 9, the next would be 10, then 19, then 29, etc. But 29² is 841, which is way too big.Alternatively, maybe it's a geometric progression where each term is multiplied by a factor. From 1 to 9, the factor is 9, so the next term would be 81, but that's already square B.Wait, maybe the figure is such that square I is the square whose side length is the sum of the side lengths of squares A and B multiplied by some factor. For example, (1 + 9) * 2 = 20 cm, but 20² is 400, not an option.Alternatively, maybe it's (9 - 1) * something. (9 - 1) = 8, 8 * 2 = 16, which is an option. 16² is 256, which is option (b). Hmm, that could be.But why would it be multiplied by 2? Maybe because it's two steps away or something. I'm not sure.Wait, another approach: if square A is 1 cm and square B is 9 cm, maybe the figure is such that square I is the square whose side length is the sum of the side lengths of squares A and B plus another square. For example, 1 + 9 + something. But without knowing the something, it's hard to say.Alternatively, maybe square I is related to square B in a way that its side length is triple that of square B. 9 * 3 = 27, but 27² is 729, not an option.Wait, maybe it's the sum of the areas of squares A and B, which is 1 + 81 = 82, but 82 isn't a perfect square.Alternatively, maybe it's the product of the areas, 1 * 81 = 81, which is square B.Hmm, I'm going in circles here. Let me try to think of it differently. Since the options are all perfect squares, and the side lengths are 14, 16, 17, 18, 19, maybe square I is built by combining multiple squares, including A and B, in a way that the side length is the sum of several side lengths.For example, if square I is built by adding square B (9 cm) and another square of 9 cm, that would make 18 cm, which is an option. So, 18² = 324 cm², which is option (d).Alternatively, if square I is built by adding square B (9 cm) and square A (1 cm) and another square of 8 cm, that would make 9 + 1 + 8 = 18 cm, again leading to 324 cm².Wait, that seems plausible. If square I is the combination of square B, square A, and another square, the total side length would be 18 cm, making the area 324 cm².Alternatively, maybe it's built by arranging squares A and B in a way that their sides add up to the side of square I. For example, if square A is placed next to square B, the combined length would be 1 + 9 = 10 cm, but that's not an option. However, if they are arranged in a different configuration, maybe the side length of square I is larger.Wait, another idea: if the figure is a right triangle where squares A and B are on the legs, then square I would be on the hypotenuse. But the area of square I would then be the sum of the areas of squares A and B, which is 1 + 81 = 82, but that's not a perfect square.Alternatively, maybe it's a different kind of figure where square I encompasses both squares A and B, making its side length the sum of their sides, but as I thought earlier, that would be 10 cm, not an option.Wait, maybe the figure is a 3x3 grid where each row and column adds up to the same total. If square A is 1 cm and square B is 9 cm, then the total length of the top row would be 1 + something + 9, and the total length of the first column would be 1 + something + something else. If the grid is uniform, then all rows and columns would add up to the same total.Let me assume that the total length of each row and column is the same. Let's call this total length T. Then, the top row would be 1 + x + 9 = T, and the first column would be 1 + y + z = T, where x, y, z are the side lengths of the other squares.But without more information, it's hard to solve for T. Maybe I need to make some assumptions. If the grid is such that each row and column adds up to the same total, and square I is in the bottom-right corner, then its side length would be part of the last row and last column.Alternatively, maybe the figure is constructed such that each square's side length is the sum of the sides of the squares to its left and above it. So, if square A is 1 cm and square B is 9 cm, then the square to the right of A would be 1 + something, and the square below A would be 1 + something else.But again, without seeing the figure, it's challenging to be precise. Maybe I should look for a pattern in the options. The options are 196, 256, 289, 324, 361, which are squares of 14, 16, 17, 18, 19. These are all relatively large numbers, so square I must be a large square.Given that square B is 9 cm, which is already quite large, maybe square I is even larger, perhaps 18 cm, making its area 324 cm². That seems plausible.Alternatively, maybe square I is the result of some other relationship. If square A is 1 cm and square B is 9 cm, maybe square I is the square of the sum of their side lengths, which would be (1 + 9)² = 100, but again, that's not an option.Wait, another idea: if the figure is a 3x3 grid where each square's side length is the sum of the sides of the squares to its left and above it, then starting with square A as 1 cm and square B as 9 cm, the square to the right of A would be 1 + something, and the square below A would be 1 + something else. Continuing this pattern, square I would be the sum of all previous squares, leading to a larger side length.But without knowing the exact configuration, it's hard to calculate. However, given the options, 324 cm² seems like a reasonable answer, as it's a perfect square and fits the pattern of being larger than square B.So, after considering various possibilities and the given options, I think the area of square I is 324 cm².

answer：First, I recognize that the binomial coefficient (dbinom{12}{3}) represents the number of ways to choose 3 elements out of 12 without considering the order.To calculate this, I use the formula for binomial coefficients:[dbinom{n}{k} = frac{n!}{k!(n - k)!}]Plugging in the values:[dbinom{12}{3} = frac{12!}{3!(12 - 3)!} = frac{12!}{3! times 9!}]Next, I simplify the factorials by canceling out the common terms in the numerator and the denominator:[frac{12 times 11 times 10 times 9!}{3! times 9!} = frac{12 times 11 times 10}{3!} = frac{12 times 11 times 10}{6}]Finally, I perform the division to find the result:[frac{1320}{6} = 220]Therefore, the value of (dbinom{12}{3}) is 220.

answer：Okay, so I need to find an integer n, where n is between 0 and 9, such that n is congruent to -4321 modulo 10. Hmm, let me think about what that means. Modulo 10 means we're looking at the remainder when -4321 is divided by 10. So, if I can find the remainder when -4321 is divided by 10, that should give me the value of n. But wait, negative numbers can be a bit tricky when dealing with remainders. I remember that when you divide a negative number by a positive number, the remainder should still be positive. So, how do I handle that? Let me recall the division algorithm. It states that for any integers a and b (with b > 0), there exist unique integers q and r such that a = bq + r, where 0 ≤ r < b. In this case, a is -4321 and b is 10. So, I need to find q and r such that -4321 = 10q + r, and r is between 0 and 9.Hmm, okay. Let's try to find q and r. If I divide -4321 by 10, what do I get? Well, 10 times 432 is 4320, which is just 1 less than 4321. So, 10 times (-432) would be -4320. So, if I take -4321 and add 4320, that gives me -1. But I need a positive remainder. So, maybe I can adjust q to make r positive. If I take q as -433 instead of -432, then 10 times (-433) is -4330. Now, if I subtract -4330 from -4321, that gives me 9. So, -4321 = 10*(-433) + 9. That fits the division algorithm because 0 ≤ 9 < 10. Therefore, the remainder r is 9, which means that -4321 is congruent to 9 modulo 10. So, n should be 9. Wait, let me double-check that. If I take -4321 and add 4320, I get -1. But since I want a positive remainder, I can add 10 to -1, which gives me 9. That makes sense because -1 and 9 are congruent modulo 10. Another way to think about it is to look at the last digit of the number. For positive numbers, the last digit gives the remainder modulo 10. But for negative numbers, it's a bit different. The last digit of 4321 is 1, so the last digit of -4321 would be -1, but since we want a positive remainder, we add 10 to get 9. So, both methods lead me to the same conclusion. The integer n that satisfies the condition is 9.