🧠:[<think>Okay, so I have these two problems to solve. Let's start with part a). It says that for any real numbers a and b with a + b ≥ 0, the inequality 4(a³ + b³) ≥ (a + b)³ holds. Hmm, I need to show this. Let me think about how to approach it. Maybe expanding both sides and comparing terms? Or perhaps using some known inequalities like AM-GM or Cauchy-Schwarz? Let's try expanding both sides first.First, let's recall that a³ + b³ can be factored as (a + b)(a² - ab + b²). So 4(a³ + b³) would be 4(a + b)(a² - ab + b²). On the other hand, (a + b)³ is a³ + 3a²b + 3ab² + b³. So maybe if I can compare these two expressions?Let me write out both sides:Left side: 4(a³ + b³) = 4a³ + 4b³Right side: (a + b)³ = a³ + 3a²b + 3ab² + b³So subtract the right side from the left side:4a³ + 4b³ - (a³ + 3a²b + 3ab² + b³) = 3a³ + 3b³ - 3a²b - 3ab²Factor out 3:3(a³ + b³ - a²b - ab²) = 3[a²(a - b) + b²(b - a)] = 3(a - b)(a² - b²) = 3(a - b)^2(a + b)Wait, let me check that step again. Let me factor a³ - a²b and b³ - ab².a³ - a²b = a²(a - b)b³ - ab² = b²(b - a) = -b²(a - b)So, combining these, we have a²(a - b) - b²(a - b) = (a - b)(a² - b²) = (a - b)^2(a + b)Therefore, the difference between the left and right sides is 3(a - b)^2(a + b). Since a + b is non-negative (given in the problem statement), and (a - b)^2 is always non-negative, the entire expression 3(a - b)^2(a + b) is non-negative. Therefore, 4(a³ + b³) - (a + b)³ ≥ 0, which implies that 4(a³ + b³) ≥ (a + b)³. So that works out. Therefore, part a) is proved. Okay, that wasn't too bad once I expanded and factored properly.Now part b) says that positive real numbers x, y satisfy the relation x + y = 3∛(x + 1) + 3∛(y + 3). We need to determine the maximum value of the sum x + y. Hmm, interesting. Let's parse the problem again. x and y are positive reals, and they satisfy x + y = 3 times the cube root of (x + 1) plus 3 times the cube root of (y + 3). Find the maximum of x + y.So we have to maximize S = x + y, given that S = 3∛(x + 1) + 3∛(y + 3). So S is expressed in terms of x and y, but we need to express it in a way that can be maximized. Maybe introduce substitutions? Let me think.Let me denote u = ∛(x + 1) and v = ∛(y + 3). Then x = u³ - 1 and y = v³ - 3. Then the sum S = x + y = u³ - 1 + v³ - 3 = u³ + v³ - 4. Also, the original equation is S = 3u + 3v. Therefore, we have:u³ + v³ - 4 = 3u + 3vSo u³ + v³ = 3u + 3v + 4We need to maximize S = 3u + 3v + 4. Wait, no, S is equal to 3u + 3v. Wait, let's re-express that.Wait, S is both equal to x + y and to 3u + 3v. From the problem statement: x + y = 3u + 3v. But also, S = x + y = u³ + v³ - 4. Therefore, we have two expressions for S:S = 3u + 3vandS = u³ + v³ - 4Therefore, setting them equal:3u + 3v = u³ + v³ - 4So u³ + v³ - 3u - 3v - 4 = 0Hmm, so the equation is u³ + v³ - 3u - 3v - 4 = 0. We need to find the maximum value of S = 3u + 3v. So essentially, we need to maximize S = 3(u + v) given that u³ + v³ - 3u - 3v - 4 = 0. So perhaps we can express u³ + v³ in terms of S.Alternatively, maybe use part a) here? Let me check. In part a), we had 4(a³ + b³) ≥ (a + b)³. If we set a and b such that a + b is non-negative, which they are since u and v are cube roots, hence real numbers (since x and y are positive, x + 1 and y + 3 are positive, so their cube roots are positive real numbers). Therefore, a = u and b = v would satisfy a + b ≥ 0. Then part a) would give 4(u³ + v³) ≥ (u + v)^3.But from our equation, u³ + v³ = 3u + 3v + 4. Let's denote t = u + v. Then u³ + v³ can be written as t³ - 3uvt. So:t³ - 3uvt = 3(u + v) + 4 = 3t + 4Hence:t³ - 3uvt - 3t - 4 = 0But also, from part a), 4(u³ + v³) ≥ t³But u³ + v³ = 3t + 4, so substituting:4(3t + 4) ≥ t³Which simplifies to:12t + 16 ≥ t³Or:t³ - 12t - 16 ≤ 0So we need to find t such that t³ - 12t - 16 ≤ 0. Let's solve t³ - 12t - 16 = 0. Maybe factor this equation? Let's try possible integer roots. The possible rational roots are factors of 16 over 1, so ±1, ±2, ±4, ±8, ±16.Testing t = 4: 64 - 48 - 16 = 0. Yes, t = 4 is a root. Then we can factor t³ -12t -16 = (t - 4)(t² + 4t + 4). Let's check:(t - 4)(t² + 4t + 4) = t³ +4t² +4t -4t² -16t -16 = t³ -12t -16. Correct.Therefore, the equation factors as (t - 4)(t² + 4t + 4) = 0. The quadratic part t² + 4t +4 = (t + 2)², which has a double root at t = -2. However, since u and v are positive real numbers (as they are cube roots of positive numbers), t = u + v must be positive. Therefore, the only real positive root is t = 4. So the inequality t³ -12t -16 ≤ 0 can be written as (t - 4)(t + 2)^2 ≤ 0. Since (t + 2)^2 is always non-negative, the inequality holds when t - 4 ≤ 0, i.e., t ≤ 4. So the maximum possible value of t is 4. Therefore, t = u + v ≤ 4. Hence, the maximum value of S = 3t = 12. But we need to check if this is achievable. When t = 4, the equation becomes:u³ + v³ = 3*4 + 4 = 16But also, t = u + v = 4. So if u + v = 4 and u³ + v³ = 16, can we find u and v such that these hold?Let me compute u³ + v³. Since u + v = 4, we can write u³ + v³ = (u + v)^3 - 3uv(u + v) = 64 - 12uv. So 64 - 12uv = 16, which implies 12uv = 48, so uv = 4. So we have u + v = 4 and uv = 4. Then the quadratic equation would be x² - 4x + 4 = 0, which factors as (x - 2)^2 = 0. Therefore, u = v = 2. Therefore, u = v = 2. Then x = u³ - 1 = 8 -1 =7, y = v³ -3 =8 -3=5. Therefore, x + y = 12, which is indeed equal to 3∛(x +1) + 3∛(y +3) = 3∛8 + 3∛8 = 3*2 +3*2=6+6=12. So this works. Therefore, the maximum value of x + y is 12. Therefore, the answer is boxed{12}.But wait, let me double-check everything to make sure.First, part a) we proved by expanding and factoring. Then part b), we substituted u and v, derived the equation, connected it to part a)'s inequality, found the maximum t = 4, checked that it's achievable when u = v = 2, leading to x =7, y=5, which satisfy the original equation. So yes, that seems correct.Final Answera) boxed{4left(a^{3}+b^{3}right) geq(a+b)^{3}}b) boxed{12}</think>a) To show that for any real numbers (a, b) with (a + b geq 0), the inequality (4(a^3 + b^3) geq (a + b)^3) holds, we start by expanding both sides:Left side: (4(a^3 + b^3))Right side: ((a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3)Subtracting the right side from the left side, we get:[4(a^3 + b^3) - (a^3 + 3a^2b + 3ab^2 + b^3) = 3a^3 + 3b^3 - 3a^2b - 3ab^2]Factoring out 3, we have:[3(a^3 + b^3 - a^2b - ab^2)]This can be further factored as:[3[(a^3 - a^2b) + (b^3 - ab^2)] = 3[a^2(a - b) - b^2(a - b)] = 3(a - b)(a^2 - b^2) = 3(a - b)^2(a + b)]Since (a + b geq 0) and ((a - b)^2 geq 0), the entire expression is non-negative. Therefore, (4(a^3 + b^3) geq (a + b)^3) holds.b) Given the relation (x + y = 3 sqrt[3]{x + 1} + 3 sqrt[3]{y + 3}) for positive real numbers (x, y), we need to determine the maximum value of (x + y).Introduce substitutions (u = sqrt[3]{x + 1}) and (v = sqrt[3]{y + 3}). Then (x = u^3 - 1) and (y = v^3 - 3). The sum (x + y) becomes:[u^3 + v^3 - 4]The given relation translates to:[x + y = 3u + 3v]Thus, we have:[u^3 + v^3 - 4 = 3u + 3v]Rewriting, we get:[u^3 + v^3 = 3u + 3v + 4]Let (t = u + v). Then:[u^3 + v^3 = t^3 - 3uvt]Substituting into the equation, we get:[t^3 - 3uvt = 3t + 4]Using the inequality from part a), (4(u^3 + v^3) geq (u + v)^3), we substitute (u^3 + v^3 = 3t + 4):[4(3t + 4) geq t^3]This simplifies to:[t^3 - 12t - 16 leq 0]Solving (t^3 - 12t - 16 = 0), we find (t = 4) is a root. Factoring, we get:[(t - 4)(t + 2)^2 = 0]Since (t = u + v) must be positive, the maximum value of (t) is 4. Thus, (S = 3t = 12).Checking if (t = 4) is achievable, we find (u = v = 2), leading to (x = 7) and (y = 5), which satisfy the original equation. Therefore, the maximum value of (x + y) is (boxed{12}).,<think>Alright, let's tackle these two problems step by step. Starting with part 1a. So, we need to show that for any real numbers ( a, b ) with ( a + b geq 0 ), the inequality ( 4(a^3 + b^3) geq (a + b)^3 ) holds. Hmm, okay. Let me recall some inequalities that might be useful here. Maybe expanding the right-hand side and comparing term by term? Let's try expanding both sides.First, expand ( (a + b)^3 ). That's ( a^3 + 3a^2b + 3ab^2 + b^3 ). So the inequality becomes:( 4(a^3 + b^3) geq a^3 + 3a^2b + 3ab^2 + b^3 )Subtracting ( a^3 + b^3 ) from both sides:( 3(a^3 + b^3) geq 3a^2b + 3ab^2 )Divide both sides by 3:( a^3 + b^3 geq a^2b + ab^2 )Hmm, okay, so simplifying, we have to show ( a^3 + b^3 geq a^2b + ab^2 ). Wait, this looks familiar. Isn't this related to the rearrangement inequality or perhaps the AM-GM inequality? Let me see. Alternatively, factor the left-hand side. ( a^3 + b^3 = (a + b)(a^2 - ab + b^2) ). The right-hand side is ( ab(a + b) ). So if we factor out ( (a + b) ), assuming ( a + b neq 0 ), then the inequality becomes ( (a^2 - ab + b^2) geq ab ). Which simplifies to ( a^2 - 2ab + b^2 geq 0 ), which is ( (a - b)^2 geq 0 ). Oh! That's always true. So since squares are non-negative, this holds. But wait, we assumed ( a + b neq 0 ). What if ( a + b = 0 )? Then ( a + b = 0 ), but the original condition is ( a + b geq 0 ), so ( a + b = 0 ) is allowed. Let's check the original inequality in that case. If ( a + b = 0 ), then the right-hand side ( (a + b)^3 = 0 ), and the left-hand side is ( 4(a^3 + b^3) ). But since ( b = -a ), then ( a^3 + b^3 = a^3 + (-a)^3 = a^3 - a^3 = 0 ). So both sides are zero. So equality holds here. Therefore, in all cases where ( a + b geq 0 ), the inequality ( 4(a^3 + b^3) geq (a + b)^3 ) holds. That seems to work. So part 1a is proven by reducing it to the square of ( (a - b) ), which is always non-negative.Moving on to part 1b: The positive real numbers ( x, y ) satisfy ( x + y = 3 sqrt[3]{x + 1} + 3 sqrt[3]{y + 3} ). We need to determine the maximum value of the sum ( x + y ).So the problem is to maximize ( S = x + y ) given the constraint ( S = 3 sqrt[3]{x + 1} + 3 sqrt[3]{y + 3} ). Wait, since ( S = x + y ), so substituting, we have ( S = 3(sqrt[3]{x + 1} + sqrt[3]{y + 3}) ). Hmm. So we need to maximize ( S ), given that ( S = 3(sqrt[3]{x + 1} + sqrt[3]{y + 3}) ), and ( x, y > 0 ).But how can we approach this? Maybe express ( x ) and ( y ) in terms of other variables. Let me denote ( a = sqrt[3]{x + 1} ) and ( b = sqrt[3]{y + 3} ). Then ( x = a^3 - 1 ) and ( y = b^3 - 3 ). Then the sum ( S = x + y = a^3 + b^3 - 4 ). But we also have the equation ( S = 3(a + b) ). Therefore, combining these, we get:( a^3 + b^3 - 4 = 3(a + b) )So ( a^3 + b^3 = 3(a + b) + 4 ). Now, we need to maximize ( S = 3(a + b) ). So if we can express ( a^3 + b^3 ) in terms of ( a + b ), perhaps we can find the maximum value of ( a + b ), which in turn gives the maximum ( S ).Let’s denote ( t = a + b ). Then we can express ( a^3 + b^3 ) as ( (a + b)^3 - 3ab(a + b) ). Therefore:( (a + b)^3 - 3ab(a + b) = 3(a + b) + 4 )Substituting ( t ):( t^3 - 3ab t = 3t + 4 )Rearrange:( t^3 - 3ab t - 3t - 4 = 0 )Or:( t^3 - 3t(ab + 1) - 4 = 0 )Hmm, but we have two variables here: ( t ) and ( ab ). However, we might recall that for fixed ( t = a + b ), the product ( ab ) is maximized when ( a = b ), by AM-GM. But here, we might need to relate ( ab ) in terms of ( t ). Alternatively, maybe we can find a relation to bound ( ab ). Let me think.Alternatively, perhaps we can use the inequality from part 1a here. Since part 1a was about ( 4(a^3 + b^3) geq (a + b)^3 ). Let's see if that can be applied here. Let me check:Given that ( a^3 + b^3 = 3(a + b) + 4 ), we can write:From 1a, ( 4(a^3 + b^3) geq (a + b)^3 )Substituting the expression for ( a^3 + b^3 ):( 4[3t + 4] geq t^3 )So ( 12t + 16 geq t^3 )Thus, ( t^3 - 12t - 16 leq 0 )We need to solve ( t^3 - 12t - 16 leq 0 ). Let's find the roots of the equation ( t^3 - 12t - 16 = 0 ). Maybe try integer roots. Let's test t = 4: 64 - 48 -16 = 0. Yes, t = 4 is a root. Then factor the polynomial. Divide by (t - 4):Using polynomial division or synthetic division. Coefficients: 1 (t^3), 0 (t^2), -12 (t), -16.Divide by (t - 4):4 | 1 0 -12 -16Bring down 1.Multiply 4*1 = 4. Add to next term: 0 +4=4.Multiply 4*4=16. Add to -12: 4.Multiply 4*4=16. Add to -16: 0.So the polynomial factors as (t - 4)(t^2 + 4t + 4). Which is (t - 4)(t + 2)^2.Therefore, ( t^3 - 12t - 16 = (t - 4)(t + 2)^2 ). Therefore, the inequality ( t^3 - 12t - 16 leq 0 ) becomes ( (t - 4)(t + 2)^2 leq 0 ).The term ( (t + 2)^2 ) is always non-negative. So the inequality depends on ( t - 4 leq 0 ), since ( (t + 2)^2 geq 0 ). So equality holds when ( t = 4 ) or ( t = -2 ). But ( t = a + b ), and ( a = sqrt[3]{x + 1} ), ( b = sqrt[3]{y + 3} ). Since ( x, y ) are positive real numbers, ( x + 1 > 1 ), so ( a > sqrt[3]{1} = 1 ), similarly ( y + 3 > 3 ), so ( b > sqrt[3]{3} approx 1.442 ). Therefore, ( t = a + b > 1 + 1.442 = 2.442 ). So ( t ) is positive, so ( t + 2 ) is positive, so the inequality ( (t - 4)(t + 2)^2 leq 0 ) implies ( t - 4 leq 0 ), hence ( t leq 4 ). Therefore, the maximum possible value of ( t = a + b ) is 4. Therefore, the maximum value of ( S = 3t = 3*4 = 12 ).But we need to check if equality is achievable. When does equality hold in the inequality from part 1a? From part 1a, equality holds when ( a = b ), since the inequality ( 4(a^3 + b^3) geq (a + b)^3 ) reduces to equality when ( a = b ). Wait, but actually in part 1a, we saw that the inequality reduces to ( (a - b)^2 geq 0 ), which holds for all real numbers, with equality when ( a = b ). So in our application of part 1a's inequality here, equality holds when ( a = b ).But in this case, when we applied the inequality from part 1a, we substituted ( a^3 + b^3 = 3t + 4 ), so the equality in 1a would hold when ( a = b ). Let's check if that's possible.If ( a = b ), then since ( t = a + b = 2a ), so ( a = t/2 ). Then substituting into the equation ( a^3 + b^3 = 3t + 4 ), which becomes ( 2a^3 = 3(2a) + 4 ). So ( 2a^3 = 6a + 4 ), divide both sides by 2: ( a^3 = 3a + 2 ). Then ( a^3 - 3a - 2 = 0 ). Let's find the roots. Trying a=2: 8 -6 -2=0. So a=2 is a root. Factor: (a - 2)(a^2 + 2a + 1) = 0. Which is (a - 2)(a + 1)^2 = 0. Therefore, a=2 or a=-1. But since a = sqrt[3]{x + 1} > 1, so a=2 is acceptable. So when a=2, then x +1 = a^3 = 8, so x=7. Then since a = b =2, so b=2 implies y +3 = b^3=8, so y=5. Therefore, x=7, y=5. Then x + y =12. So this satisfies the original equation: x + y =12, and 3(sqrt[3]{x +1} + sqrt[3]{y +3})=3(2 + 2)=12. So yes, equality is achievable. Therefore, the maximum value of x + y is 12.Therefore, the answer is 12.Final Answera) boxed{4left(a^{3}+b^{3}right) geq(a+b)^{3}} is proven for all real numbers (a, b) with (a + b geq 0).b) The maximum value of (x + y) is boxed{12}.</think> Summarya) To show that (4(a^3 + b^3) geq (a + b)^3) for any real numbers (a, b) with (a + b geq 0):1. Expand ((a + b)^3) to get (a^3 + 3a^2b + 3ab^2 + b^3).2. Subtract (a^3 + b^3) from both sides of the inequality, leading to (3(a^3 + b^3) geq 3a^2b + 3ab^2).3. Divide by 3 to get (a^3 + b^3 geq a^2b + ab^2).4. Factor the left-hand side as ((a + b)(a^2 - ab + b^2)) and the right-hand side as (ab(a + b)).5. Simplify to ((a - b)^2 geq 0), which is always true. Equality holds when (a = b) or (a + b = 0).Thus, the inequality (4(a^3 + b^3) geq (a + b)^3) holds for all real numbers (a, b) with (a + b geq 0).b) To determine the maximum value of (x + y) given (x + y = 3 sqrt[3]{x + 1} + 3 sqrt[3]{y + 3}):1. Let (a = sqrt[3]{x + 1}) and (b = sqrt[3]{y + 3}), then (x = a^3 - 1) and (y = b^3 - 3).2. Express the sum (S = x + y) as (a^3 + b^3 - 4) and set it equal to (3(a + b)).3. Substitute (t = a + b) to get (t^3 - 3abt = 3t + 4).4. Use the inequality from part (a) to show (4(a^3 + b^3) geq (a + b)^3), leading to (t^3 - 12t - 16 leq 0).5. Solve the cubic equation (t^3 - 12t - 16 = 0) to find (t = 4) as the valid root.6. Check that equality is achievable when (a = b = 2), leading to (x = 7) and (y = 5).Thus, the maximum value of (x + y) is (boxed{12}).]

🔑:Solution. a) The inequality is equivalent to:4 a^{3}+4 b^{3} geq a^{3}+b^{3}+3 a b(a+b) Leftrightarrow a^{3}+b^{3} geq a b(a+b) Leftrightarrow(a+b)(a-b)^{2} geq 0, true.b) If we denote sqrt[3]{x+1}=a and sqrt[3]{y+3}=b Rightarrow x=a^{3}-1 and y=b^{3}-3. The equality from the statement becomes:a^{3}+b^{3}=3(a+b)+4 stackrel{a)}{Rightarrow} 4left(a^{3}+b^{3}right)=12(a+b)+16 geq(a+b)^{3} Rightarrow(a+b)^{3}-12(a+b)-16 leq 0Leftrightarrow(a+b-4)(a+b+2)^{2} leq 0 Leftrightarrow a+b leq 4 Rightarrow x+y=a^{3}+b^{3}-4=3(a+b) leq 12 Rightarrow x+y leq 12. In conclusion, the maximum value of the sum x+y is 12. The maximum is achieved for a=b=2, i.e., x=7, y=5.## Scoring Rubric.| a) Proof of the inequality | 3 mathrm{p} || :--- | :--- || b) Prove x+y leq 12 | 3 mathrm{p} || Finalize that the maximum value of the sum x+y is 12 for x=7, y=5 | 1 mathrm{p} |

🧠:[<think>Okay, so the problem is about a cube where each vertex has a number, all distinct. A number is "happy" if it equals the sum of the three adjacent numbers. We need to find the maximum number of happy numbers possible. Hmm, let's try to visualize this.First, a cube has 8 vertices, each connected to three others. So each vertex is adjacent to three others. If a number at a vertex is equal to the sum of its three neighbors, it's happy. We need to maximize the number of such happy numbers. All numbers are distinct, which might complicate things because if we have too many happy numbers, maybe they would have to repeat?Let me start by thinking about smaller cases. Suppose we have one happy number. Then that number is the sum of its three neighbors. The other numbers can be arbitrary, as long as they are distinct. But if we want multiple happy numbers, their relationships might conflict.Suppose two adjacent vertices are both happy. Then each of them is the sum of their neighbors, which includes each other. Let's denote two adjacent vertices as A and B. If A is happy, then A = sum of its three neighbors, which includes B and two others. Similarly, B = sum of its three neighbors, which includes A and two others. But if A and B are both happy, substituting one into the other might lead to equations that require certain relationships between the other numbers.For example, let's say vertex A is connected to B, C, D. Then A = B + C + D. Vertex B is connected to A, E, F. So B = A + E + F. Substitute A from the first equation into the second: B = (B + C + D) + E + F. Rearranging gives 0 = C + D + E + F. So the sum of these four numbers must be zero. But all numbers are distinct, so possible? Maybe, but then how does this affect other vertices?This seems like a system of equations. If we have multiple happy numbers, each imposes an equation. The cube's structure will determine how these equations interact. Let me try to model this.Each vertex can be represented as a variable. Let's name the vertices as follows: imagine the cube with top and bottom faces. Let’s name the top face vertices as A, B, C, D going around, and the bottom face directly below them as E, F, G, H. So A is connected to B, D, and E; B is connected to A, C, F; etc.So variables: A, B, C, D, E, F, G, H.Each vertex is connected to three others. Let me note the connections:A: B, D, EB: A, C, FC: B, D, GD: A, C, HE: A, F, HF: B, E, GG: C, F, HH: D, E, GNow, if we want to set up equations where some variables are the sum of their neighbors. Let's suppose that some variables are happy, meaning they equal the sum of their neighbors.Suppose we try to set as many variables as possible to be happy. Let's see how many equations we can set up.But each happy variable introduces an equation. Since all variables are distinct, the equations must be consistent with distinct values.But how many equations can we have without over-constraining the system? A cube's graph has cycles, so we need to be careful about dependencies.Let me try a simple case. Suppose we set all variables to be happy. Then each variable is equal to the sum of its three neighbors. Let's see if that's possible.If all variables are happy, then:A = B + D + EB = A + C + FC = B + D + GD = A + C + HE = A + F + HF = B + E + GG = C + F + HH = D + E + GThis is a system of 8 equations. Let's try to solve it.First, from equation A: A = B + D + EFrom equation B: B = A + C + FSubstitute A from equation A into equation B: B = (B + D + E) + C + F => 0 = D + E + C + FSimilarly, equation C: C = B + D + GEquation D: D = A + C + HEquation E: E = A + F + HEquation F: F = B + E + GEquation G: G = C + F + HEquation H: H = D + E + GThis seems quite interconnected. Let me see if there's a non-trivial solution where all variables are zero? But the problem states that all numbers are distinct, so zero isn't allowed for all. Also, even if non-zero, having all variables satisfy these equations might require certain relationships.Alternatively, maybe all variables are zero except some? But again, numbers must be distinct. So this seems impossible. Therefore, it's not possible to have all 8 numbers happy.Similarly, maybe 7? Probably not, since the equations would be too interdependent. Let's think step by step.Perhaps start with fewer equations and see if we can find a system with as many happy numbers as possible.Suppose we try to set four numbers as happy. Maybe two on the top face and two on the bottom. Let's pick A, C, F, H.Wait, but each happy number imposes equations that relate others. Let me see.If A is happy: A = B + D + EIf C is happy: C = B + D + GIf F is happy: F = B + E + GIf H is happy: H = D + E + GBut now, let's see if these equations can be satisfied with distinct numbers.From A = B + D + EC = B + D + GF = B + E + GH = D + E + GLet’s try to express variables in terms of B, D, E, G.From A = B + D + EC = B + D + GF = B + E + GH = D + E + GSo H is equal to F + D - B? Because H = D + E + G, and F = B + E + G. So H = D + (F - B - G) + G = D + F - B. Therefore, H = F + D - B.But we also have other variables: Let's check if we can solve for these.Also, variables not yet mentioned: variables B, D, E, G, and then A, C, F, H as above. The other variables are not constrained yet. Wait, but all variables must be distinct. Let's suppose B, D, E, G are parameters, then A, C, F, H are determined by them. Then the remaining variables (which ones? Wait, the cube has 8 variables: A, B, C, D, E, F, G, H. So if we set B, D, E, G as parameters, then A, C, F, H are determined. Then we still have variables not mentioned: variables? Wait, no, all variables are included. A is determined by B, D, E. C is determined by B, D, G. F is determined by B, E, G. H is determined by D, E, G. So all variables are expressed in terms of B, D, E, G. So the system reduces to 4 parameters. But we need all eight variables to be distinct. Let's see if we can choose B, D, E, G such that all variables are distinct.But we need to also make sure that other variables (A, C, F, H) are distinct from each other and from B, D, E, G.This might be possible, but maybe we can find such numbers. Alternatively, perhaps there's a contradiction in these equations.Wait, from above, H = F + D - B. But H is also determined as D + E + G. And F = B + E + G. So substituting F into H's equation: H = (B + E + G) + D - B = D + E + G, which matches the original equation. So no contradiction there.But we can have consistency. So with four happy numbers (A, C, F, H), the system is consistent and allows the other variables to be expressed in terms of B, D, E, G. Now, can we choose B, D, E, G such that all variables are distinct?For example, let's assign some numbers to B, D, E, G. Let's pick small distinct integers.Let me try B=1, D=2, E=3, G=4.Then:A = B + D + E = 1 + 2 + 3 = 6C = B + D + G = 1 + 2 + 4 = 7F = B + E + G = 1 + 3 + 4 = 8H = D + E + G = 2 + 3 + 4 = 9So variables:A=6, B=1, C=7, D=2, E=3, F=8, G=4, H=9All numbers are distinct. So in this case, we have four happy numbers: A, C, F, H. Each of these equals the sum of their neighbors. Let's check for example A:A is connected to B, D, E. 1 + 2 + 3 = 6, which is A. Correct.C is connected to B, D, G. 1 + 2 + 4 = 7, which is C. Correct.F is connected to B, E, G. 1 + 3 + 4 = 8, which is F. Correct.H is connected to D, E, G. 2 + 3 + 4 = 9, which is H. Correct.And the other numbers: B=1, D=2, E=3, G=4. All distinct, and no overlaps with A, C, F, H (6,7,8,9). So this works. So we have four happy numbers. Can we get more?Wait, maybe try to set more variables as happy. Let's see. Suppose in the above example, is there another number that happens to be the sum of its neighbors? Let's check variable E. E is connected to A, F, H. A=6, F=8, H=9. Sum is 6+8+9=23. But E is 3, not 23. So E is not happy. Similarly, check B: connected to A, C, F. A=6, C=7, F=8. 6+7+8=21. B=1, not happy. D: connected to A, C, H. A=6, C=7, H=9. Sum=22, D=2. Not happy. G: connected to C, F, H. 7+8+9=24, G=4. Not happy. So only A, C, F, H are happy. So 4.But maybe there's a different configuration where more are happy. Let's try.Suppose we set five variables as happy. Let's pick A, C, F, H, and maybe another one. Let's say G. If G is happy, then G = C + F + H. Wait, but in the previous example, G=4, and C=7, F=8, H=9. 7+8+9=24, not 4. So if we want G to be happy, we need G = C + F + H. But in our previous setup, that would mean changing G. Let's see.But if we set G as happy, then G = C + F + H. But previously, C = B + D + G. So substituting, we have G = (B + D + G) + F + H. Which simplifies to 0 = B + D + F + H. So B + D + F + H = 0. But in our previous assignment, B=1, D=2, F=8, H=9. Sum is 1+2+8+9=20 ≠0. So that introduces a new equation. Therefore, making G happy would require B + D + F + H =0. But in our previous example, that's not the case. So maybe we need to adjust parameters.Alternatively, let's try to set up five happy numbers: A, C, F, H, G.So equations:A = B + D + EC = B + D + GF = B + E + GH = D + E + GG = C + F + HNow, substituting G into the other equations:From C = B + D + G => C = B + D + (C + F + H) => 0 = B + D + F + H.From equation for H: H = D + E + G = D + E + C + F + H => 0 = D + E + C + F.But from the equation above, we have B + D + F + H =0, and D + E + C + F =0.Additionally, A = B + D + EF = B + E + G = B + E + C + F + H => 0 = B + E + C + H.This is getting complicated. Let me see if I can express variables in terms of others.From B + D + F + H =0 (from C's equation)From D + E + C + F =0 (from H's equation)From B + E + C + H =0 (from F's equation)From A = B + D + EG = C + F + HWe also have equations from the original happy numbers. Let me see if we can find a solution.Let’s denote equations:1. B + D + F + H = 02. D + E + C + F = 03. B + E + C + H = 04. A = B + D + E5. G = C + F + HWe can try to solve this system.From equations 1 and 2:Equation1: B + D + F + H =0Equation2: D + E + C + F =0Subtract equation2 from equation1: (B + D + F + H) - (D + E + C + F) = B - E - C + H =0 => B + H = E + C.Similarly, from equations 1 and 3:Equation1: B + D + F + H =0Equation3: B + E + C + H =0Subtract equation3 from equation1: (B + D + F + H) - (B + E + C + H) = D + F - E - C =0 => D + F = E + C.From earlier, B + H = E + C, so D + F = B + H.So that's an interesting relation.Also, from equation2: D + E + C + F =0, and since D + F = E + C, substitute into equation2: (E + C) + E + C = 2E + 2C =0 => E + C =0. So E = -C.Similarly, from D + F = E + C =0, so D + F=0 => F = -D.From equation1: B + D + F + H = B + D + (-D) + H = B + H =0 => H = -B.From equation3: B + E + C + H = B + (-C) + C + (-B) =0, which checks out.From equation4: A = B + D + E = B + D + (-C). But since E = -C.But from E = -C and equation D + F =0, and F = -D.From equation5: G = C + F + H = C + (-D) + (-B) = C - D - B.Now, variables:A = B + D - CE = -CF = -DH = -BG = C - D - BVariables remaining: B, C, D are free variables? Wait, but we have relations.But all variables must be distinct. Let's assign values to B, C, D such that all variables are distinct.Let’s pick B, C, D as distinct numbers, and then compute A, E, F, G, H.Let me try B=1, C=2, D=3.Then:E = -C = -2F = -D = -3H = -B = -1A = B + D - C =1 +3 -2=2But A=2, which is same as C=2. Not allowed, since all numbers must be distinct.So need to choose B, C, D such that A = B + D - C is distinct from all others.Let me try B=1, C=3, D=2.Then:E = -3F = -2H = -1A =1 +2 -3=0G =3 -2 -1=0So G=0, same as A=0. Not allowed.Hmm. Let's try B=2, C=1, D=3.Then:E = -1F = -3H = -2A =2 +3 -1=4G =1 -3 -2= -4Variables:A=4, B=2, C=1, D=3, E=-1, F=-3, G=-4, H=-2Check if all distinct: 4,2,1,3,-1,-3,-4,-2. Yes, all distinct.So let's verify the equations:A = B + D + E => 4 =2 +3 +(-1)=4. Correct.C = B + D + G =>1=2 +3 +(-4)=1. Correct.F = B + E + G =>-3=2 +(-1)+(-4)=-3. Correct.H = D + E + G =>-2=3 +(-1)+(-4)=-2. Correct.G = C + F + H =>-4=1 +(-3)+(-2)=-4. Correct.So all five equations are satisfied. Also, all numbers are distinct: 4,2,1,3,-1,-3,-4,-2. Yes, all are distinct. So in this case, we have five happy numbers: A, C, F, H, G.That's five. Is that the maximum?Wait, let's check the other variables: B, D, E.B is connected to A, C, F. Their sum is 4 +1 +(-3)=2. B is 2. So B is happy as well? Wait, B=2. A=4, C=1, F=-3. 4+1+(-3)=2. So B=2. So B is happy too. Wait, but in our setup, we only set A, C, F, H, G as happy. But according to the equations, B is also happy. Wait, but we didn't include B in our initial five. Wait, let me check.In the equations, we set A, C, F, H, G as happy. But when we solved the system, it turned out that B is also equal to the sum of its neighbors. So in reality, we have six happy numbers? Let's check.Variables:A=4 (happy)B=2 (sum of A=4, C=1, F=-3: 4+1-3=2. So B is happy.C=1 (happy)D=3: connected to A=4, C=1, H=-2. Sum:4+1-2=3. So D is 3. So D is happy.E=-1: connected to A=4, F=-3, H=-2. Sum:4 -3 -2=-1. So E is happy.F=-3 (happy)G=-4 (happy)H=-2 (happy)Wait, so actually all eight variables are happy! But that contradicts our earlier thought that all variables can't be happy. But in this case, they all satisfy the condition. But all numbers are distinct: 4,2,1,3,-1,-3,-4,-2. All distinct. So, but wait, this seems contradictory. How come?But if all eight numbers are happy and distinct, that would mean the answer is 8. But the problem states "not necessarily positive" and "all eight numbers are distinct". But in my previous example, assigning those values gives all numbers as happy. But let's verify again.Wait, let's check each variable:A=4: neighbors B=2, D=3, E=-1. 2 +3 +(-1)=4. Correct.B=2: neighbors A=4, C=1, F=-3. 4 +1 +(-3)=2. Correct.C=1: neighbors B=2, D=3, G=-4. 2 +3 +(-4)=1. Correct.D=3: neighbors A=4, C=1, H=-2. 4 +1 +(-2)=3. Correct.E=-1: neighbors A=4, F=-3, H=-2. 4 +(-3) +(-2)=-1. Correct.F=-3: neighbors B=2, E=-1, G=-4. 2 +(-1) +(-4)=-3. Correct.G=-4: neighbors C=1, F=-3, H=-2. 1 +(-3) +(-2)=-4. Correct.H=-2: neighbors D=3, E=-1, G=-4. 3 +(-1) +(-4)=-2. Correct.Wow, so all eight numbers are happy and distinct. But this contradicts my initial thought that it's impossible. So maybe the answer is 8? But wait, the problem says "not necessarily positive" and "all eight numbers are distinct". So in this case, all numbers are distinct, some positive, some negative, and all are happy. So why is the question asking for the maximum number of happy numbers? If 8 is possible, then the answer should be 8. But maybe there's a mistake here.Wait, but in the example I constructed, all variables are happy. Let me check again. Assign variables as follows:A=4, B=2, C=1, D=3, E=-1, F=-3, G=-4, H=-2.Each variable is the sum of its neighbors, and all are distinct. So according to this, all eight can be happy. Therefore, is the answer 8? But the problem seems to suggest that it's not possible, hence asking for the maximum. Maybe I made a mistake in the assignment.Wait, but in the system of equations, if all variables are happy, then they satisfy the equations. But in my specific example, they do. So unless there is a restriction I'm missing. The problem says "numbers, not necessarily positive. All eight numbers are distinct." In my example, numbers are 4,2,1,3,-1,-3,-4,-2. All distinct. No repetition. So maybe the answer is 8? But I need to verify.Wait, let me check the cube structure again. Each vertex is connected to three others. In the example, each number is indeed the sum of the three connected numbers. So it works. Therefore, the maximum number is 8. But the problem says "If a number is equal to the sum of the three numbers living in the adjacent vertices, then it is happy." So all numbers can be happy. Therefore, the answer is 8.But this seems counterintuitive. How can all numbers be both distinct and each the sum of their neighbors? Because in a cycle, if each node is the sum of neighbors, you can get repeating numbers or zero. But in this 3D structure, maybe it's possible with positive and negative numbers. The example shows it is possible.Wait, but in the equations we set up, we ended up with a system where variables are expressed in terms of others, and by choosing B, C, D appropriately, we could get distinct numbers. So maybe 8 is possible. Then why does the problem ask for the maximum? Maybe I made a miscalculation.Wait, let me check the example again. A=4, B=2, C=1, D=3, E=-1, F=-3, G=-4, H=-2.Each is the sum of its neighbors. So they are all happy. Therefore, 8 is possible. Thus, the answer should be 8. But why does the problem exist then? Maybe there's a catch.Wait, let me check the problem statement again: "All eight numbers are distinct." In my example, they are. "If a number is equal to the sum of the three numbers living in the adjacent vertices, then it is happy." All numbers satisfy this. Therefore, maximum number is 8. So the answer is 8. But maybe the original problem has a different configuration? Or perhaps I misread the problem.Wait, the problem says: "a number is equal to the sum of the three numbers living in the adjacent vertices, then it is happy." So if a number satisfies that condition, it's happy. So in my example, all numbers satisfy that condition, so all are happy. So the answer is 8. But I need to confirm.Wait, but in my system of equations, when I tried to set all variables as happy, the equations led to a consistent solution with distinct numbers. So unless there is a mistake in my calculations, the answer should be 8.But wait, in the literature, I recall something about the cube's graph being bipartite. The cube is a bipartite graph, meaning its vertices can be divided into two sets with no edges within each set. For a cube, it's the even and odd layers, or in our naming, top face (A,B,C,D) and bottom face (E,F,G,H). In a bipartite graph, if all vertices in one set are sources and the other set are sinks, but here we have each vertex being a sum of its neighbors. In bipartite graphs, such a system might require eigenvalues or something. Wait, but maybe the solution I found is a valid one. Let me check.In the example, top face: A=4, B=2, C=1, D=3; bottom face: E=-1, F=-3, G=-4, H=-2. So top face numbers are positive, bottom face negative. Each top face node is connected to three bottom face nodes, and vice versa. Each top face node is the sum of its three bottom face neighbors. For example, A is connected to E, B, D. Wait, no, in the cube structure, each top face node is connected to two adjacent top face nodes and one bottom face node. Wait, in the cube, each vertex is connected to three others: for example, A is connected to B, D, and E. So A is on the top face connected to two top face neighbors (B, D) and one bottom face (E). Similarly, E is connected to A, F, H.In my example, A=4, B=2, D=3, E=-1. So A=2+3+(-1)=4. Correct. Similarly, B is connected to A=4, C=1, F=-3. 4+1+(-3)=2. Correct. But B is on the top face, connected to A (top), C (top), and F (bottom). So in this case, the top face nodes are sums that include both top and bottom face nodes. Similarly, bottom face nodes are sums of their neighbors, which include both bottom and top.But in the example, all nodes are happy, so both top and bottom. So maybe in bipartite graphs, such solutions exist where all nodes can be happy? But usually, in bipartite graphs, if you set one partition to be the sum of the other, but here it's mutual. Each partition's nodes are sums of the other partition's nodes plus their own partition. Wait, in the cube, each node is connected to three nodes: for top face, two on top and one on bottom; for bottom face, two on bottom and one on top. Therefore, when a top face node is the sum of its neighbors, it includes two top and one bottom. Similarly, a bottom face node includes two bottom and one top. Therefore, in the equations, there is a mixture.But in our solution, we have variables in both partitions expressed in terms of others. But through the equations, we found a consistent solution where all variables are happy. Therefore, the answer should be 8.But the problem is presented as a competition problem, likely expecting an answer less than 8, maybe 4 or 5. So where is the mistake?Wait, in my example, are all variables distinct? Let's list them again: A=4, B=2, C=1, D=3, E=-1, F=-3, G=-4, H=-2. Yes, all are distinct. So why is this possible? Maybe the key is that in 3D space, the cube's structure allows such a solution. So if the answer is 8, then the problem's answer is 8. But maybe there's a condition I missed.Wait, the problem says "not necessarily positive". So negative numbers are allowed. In my solution, some numbers are negative. So that's allowed. All numbers are distinct. So that's satisfied. Therefore, according to this example, 8 is possible. Therefore, the answer is 8.But I need to confirm again. Let's take another example with different numbers to see if this is a general solution.Let’s choose B=3, C=2, D=1.Then:E = -C = -2F = -D = -1H = -B = -3A = B + D - C =3 +1 -2=2G = C - D - B=2 -1 -3=-2Wait, but G=-2, which is same as E=-2. Not allowed. So need to choose B, C, D such that G is distinct.Let me try B=4, C=1, D=2.E = -1F = -2H = -4A =4 +2 -1=5G =1 -2 -4= -5Variables:A=5, B=4, C=1, D=2, E=-1, F=-2, G=-5, H=-4All distinct. Check if they satisfy the conditions:A=5: B=4, D=2, E=-1. 4+2-1=5. Correct.B=4: A=5, C=1, F=-2. 5+1-2=4. Correct.C=1: B=4, D=2, G=-5. 4+2-5=1. Correct.D=2: A=5, C=1, H=-4. 5+1-4=2. Correct.E=-1: A=5, F=-2, H=-4. 5-2-4=-1. Correct.F=-2: B=4, E=-1, G=-5. 4-1-5=-2. Correct.G=-5: C=1, F=-2, H=-4. 1-2-4=-5. Correct.H=-4: D=2, E=-1, G=-5. 2-1-5=-4. Correct.Again, all numbers are distinct and happy. So this seems to hold. Therefore, the answer is indeed 8. So the maximum number of happy numbers is 8.But why does this work? It seems that by choosing B, C, D appropriately, we can generate such a cube where all numbers are happy and distinct. The key was allowing negative numbers, which the problem permits.Therefore, the answer should be boxed{8}. But to confirm, maybe check another resource or think if there's a reason this wouldn't be possible.Alternatively, maybe the problem is from a competition where the answer is 4, but in that case, the example above contradicts it. Maybe the original problem had a condition that numbers must be positive, but here it's stated they can be negative. Therefore, with negatives allowed, 8 is possible.Hence, the maximum number is 8.Final Answerboxed{8}</think>To solve the problem of finding the maximum number of happy numbers on the vertices of a cube, we need to consider that a number is happy if it equals the sum of the three numbers at its adjacent vertices. All eight numbers must be distinct, but they can be positive or negative.1. Understanding the Cube Structure: - A cube has 8 vertices, each connected to three others. - We name the vertices as follows: top face vertices (A, B, C, D) and bottom face vertices (E, F, G, H).2. Setting Up Equations: - Each happy number at a vertex must be the sum of its three adjacent vertices. - We derive a system of equations for each vertex if it is happy.3. Example Solution: - Assign values to some vertices and solve the system of equations to ensure all numbers are distinct and satisfy the happy condition. - Example assignment: - (A = 4), (B = 2), (C = 1), (D = 3) - (E = -1), (F = -3), (G = -4), (H = -2)4. Verification: - Check each vertex to ensure it is the sum of its adjacent vertices: - (A = B + D + E = 2 + 3 + (-1) = 4) - (B = A + C + F = 4 + 1 + (-3) = 2) - (C = B + D + G = 2 + 3 + (-4) = 1) - (D = A + C + H = 4 + 1 + (-2) = 3) - (E = A + F + H = 4 + (-3) + (-2) = -1) - (F = B + E + G = 2 + (-1) + (-4) = -3) - (G = C + F + H = 1 + (-3) + (-2) = -4) - (H = D + E + G = 3 + (-1) + (-4) = -2)5. Conclusion: - All eight numbers are distinct and each is the sum of its adjacent vertices, making them all happy.Thus, the maximum number of happy numbers that can live at the vertices of the cube is (boxed{8}).,<think>Okay, so I need to figure out the maximum number of "happy" numbers on a cube where each vertex has a distinct number. A happy number is one that's equal to the sum of its three adjacent vertices. All eight numbers are different. Hmm, interesting.First, let me visualize a cube. A cube has 8 vertices and each vertex is connected to three others. So each vertex has three neighbors. If a number at a vertex is the sum of its three neighbors, it's happy. The question is, how many such happy numbers can we have at most?Since all numbers are distinct, there might be some constraints. If multiple numbers are sums of their neighbors, those sums might have dependencies that could lead to contradictions, especially with all numbers being distinct. Let me start by considering simpler cases.Suppose there's one happy number. That's trivial; set one number as the sum of its three neighbors. But since all numbers are distinct, maybe the other numbers can be arranged accordingly. But we need to maximize the number of happy numbers. Let's think about two happy numbers.If two vertices are happy, they each are the sum of their three neighbors. But if those two vertices are adjacent, then their equations might be related. Let me try to formalize this.Let's denote the vertices of the cube. Maybe label them as binary numbers from 000 to 111, each bit representing a coordinate. Then each vertex is connected to three others that differ by one bit. For example, 000 is connected to 001, 010, and 100. Maybe assigning variables to each vertex. Let me assign variables A, B, C, D, E, F, G, H to the vertices. Wait, maybe labeling them with coordinates would be better. Let me try:Let the vertices be:(0,0,0) = V1(0,0,1) = V2(0,1,0) = V3(0,1,1) = V4(1,0,0) = V5(1,0,1) = V6(1,1,0) = V7(1,1,1) = V8So each V_i is a vertex. Now, each vertex is connected to three others. For example, V1 (0,0,0) is connected to V2, V3, V5. V2 (0,0,1) is connected to V1, V4, V6. And so on.Suppose I want V1 to be happy. Then the number at V1 must equal the sum of numbers at V2, V3, V5. Similarly, if V2 is happy, it must equal V1 + V4 + V6, and so on.But if both V1 and V2 are happy, then:V1 = V2 + V3 + V5V2 = V1 + V4 + V6So substituting V1 from the first equation into the second:V2 = (V2 + V3 + V5) + V4 + V6Which simplifies to:V2 = V2 + V3 + V5 + V4 + V6Subtracting V2 from both sides:0 = V3 + V4 + V5 + V6So this gives a condition that the sum of V3, V4, V5, V6 must be zero. Hmm, interesting. But all numbers are distinct, so is it possible? Maybe, but I need to check if such a system is possible.But this is just two happy numbers. Let's see if more can be added. Suppose I have three happy numbers. Let's pick three vertices that are not adjacent. For example, V1, V4, V8. Let me check if that's possible.If V1 is happy: V1 = V2 + V3 + V5If V4 is happy: V4 = V2 + V3 + V7 (Wait, V4 is (0,1,1), so its neighbors are V2 (0,0,1), V3 (0,1,0), and V7 (1,1,0). So V4 = V2 + V3 + V7.If V8 is happy: V8 = V5 + V6 + V7 (since V8 is (1,1,1), neighbors are V5, V6, V7).So now, we have three equations:1. V1 = V2 + V3 + V52. V4 = V2 + V3 + V73. V8 = V5 + V6 + V7Now, let's see if we can solve this system with all variables being distinct.From equations 1 and 2, if we subtract equation 1 from equation 2:V4 - V1 = (V2 + V3 + V7) - (V2 + V3 + V5) => V4 - V1 = V7 - V5 => V4 - V1 = V7 - V5 => V4 + V5 = V1 + V7So that's a relation. Similarly, from equation 3, V8 is expressed in terms of V5, V6, V7.But this is getting a bit complicated. Maybe I need to assign some values and see if it works. Let's suppose some numbers.Suppose I set V2, V3, V5, V6, V7 to some numbers such that V1, V4, V8 are sums of their neighbors. Let's assign small integers for simplicity.Let me try:Let V2 = 1, V3 = 2, V5 = 3. Then V1 = 1 + 2 + 3 = 6.Then for V4: V4 = V2 + V3 + V7 => V4 = 1 + 2 + V7. Let's choose V7 = 4. Then V4 = 1 + 2 + 4 = 7.Then for V8: V8 = V5 + V6 + V7 => V8 = 3 + V6 + 4. So V8 = 7 + V6.Now, we need to assign V6 such that all numbers are distinct. So far, numbers are: V1=6, V2=1, V3=2, V4=7, V5=3, V7=4. Remaining numbers: V6 and V8. Since V8 = 7 + V6, and we need all numbers to be distinct, let's choose V6=5. Then V8=12. But 12 is distinct from others. So numbers would be:V1=6, V2=1, V3=2, V4=7, V5=3, V6=5, V7=4, V8=12. All distinct. So here, V1, V4, V8 are happy. That's three happy numbers. But the question is, can we get more?Let's see if we can add another happy number. Suppose we want V6 to be happy. V6 is connected to V2, V5, V8. So V6 should equal V2 + V5 + V8. In our current setup, V6=5, V2=1, V5=3, V8=12. So 1 + 3 + 12 = 16. But V6 is 5, which is not 16. So that's not happy. Alternatively, maybe adjusting the numbers.Alternatively, maybe making another vertex happy. Let's check V7. V7 is connected to V4, V6, V8. So V7 should equal V4 + V6 + V8. In our current setup, V7=4, V4=7, V6=5, V8=12. 7 + 5 + 12 = 24. Not equal to 4. Not happy.What about V5? V5 is connected to V1, V6, V7. So V5 should equal V1 + V6 + V7. 6 + 5 + 4 = 15. V5 is 3. Not equal.So in this case, only three happy numbers. Maybe we can adjust the numbers to get more. Let me try a different approach.Suppose I want four happy numbers. Let's pick four vertices that are not adjacent. For example, V1, V4, V6, V7. Wait, are these non-adjacent? V1 is connected to V2, V3, V5. V4 is connected to V2, V3, V7. V6 is connected to V2, V5, V8. V7 is connected to V3, V4, V8. Hmm, there might be overlaps. Maybe V1, V4, V6, V8. Let's check if they are adjacent. V1 is adjacent to V2, V3, V5; V4 is adjacent to V2, V3, V7; V6 is adjacent to V2, V5, V8; V8 is adjacent to V5, V6, V7. So none of V1, V4, V6, V8 are adjacent to each other. So maybe they can be happy without conflicting dependencies.Let me set up the equations:1. V1 = V2 + V3 + V52. V4 = V2 + V3 + V73. V6 = V2 + V5 + V84. V8 = V5 + V6 + V7Wait, but equation 3 and 4 are related. From equation 3: V6 = V2 + V5 + V8From equation 4: V8 = V5 + V6 + V7Substitute V6 from equation 3 into equation 4:V8 = V5 + (V2 + V5 + V8) + V7Simplify:V8 = 2V5 + V2 + V8 + V7Subtract V8 from both sides:0 = 2V5 + V2 + V7So 2V5 + V2 + V7 = 0That's a constraint. Also, equations 1 and 2:From equation 1: V1 = V2 + V3 + V5From equation 2: V4 = V2 + V3 + V7So subtract equation 1 from equation 2:V4 - V1 = (V2 + V3 + V7) - (V2 + V3 + V5) => V4 - V1 = V7 - V5 => V4 + V5 = V1 + V7So that's another relation.Now, let's try to assign values.Let me suppose some variables. Let's assume V2, V3, V5, V7 are variables we can set.From 2V5 + V2 + V7 = 0 => V7 = -2V5 - V2From equation 2: V4 = V2 + V3 + V7 = V2 + V3 -2V5 - V2 = V3 - 2V5From equation 1: V1 = V2 + V3 + V5From equation 3: V6 = V2 + V5 + V8From equation 4: V8 = V5 + V6 + V7. Substitute V7 = -2V5 - V2 and V6 from equation 3:V8 = V5 + (V2 + V5 + V8) + (-2V5 - V2)Simplify:V8 = V5 + V2 + V5 + V8 -2V5 - V2Simplify the right-hand side:V5 + V2 + V5 + V8 -2V5 - V2 = (V5 + V5 -2V5) + (V2 - V2) + V8 = 0 + 0 + V8 = V8So equation 4 becomes V8 = V8, which is always true. So equation 4 doesn't give us new information; it's dependent on equation 3 and the expression for V7.Therefore, our variables are:V7 = -2V5 - V2V4 = V3 - 2V5V1 = V2 + V3 + V5V6 = V2 + V5 + V8But V8 is also defined by equation 4. Wait, but equation 4 is redundant. So actually, V8 can be expressed in terms of V5, V6, V7, but since V7 is already expressed in terms of V2 and V5, and V6 is expressed in terms of V2, V5, V8, but we need to resolve this.Wait, maybe let's express V8 in terms of V2 and V5.From equation 4: V8 = V5 + V6 + V7. But V7 = -2V5 - V2, so:V8 = V5 + V6 -2V5 - V2 => V8 = -V5 + V6 - V2But from equation 3: V6 = V2 + V5 + V8. Substitute V8 from above into this:V6 = V2 + V5 + (-V5 + V6 - V2) => V6 = V2 + V5 - V5 + V6 - V2 => V6 = V6. Again, an identity. So no new info.This suggests that once we set V2, V3, V5, then V1, V4, V7, V6, and V8 are determined. But V6 and V8 depend on each other. Let me try assigning values.Let me pick V2 = 1, V3 = 2, V5 = 3.Then V7 = -2*3 -1 = -7V4 = 2 - 2*3 = 2 -6 = -4V1 = 1 + 2 + 3 = 6Now, from equation 3: V6 = V2 + V5 + V8 => V6 = 1 + 3 + V8 = 4 + V8From equation 4: V8 = V5 + V6 + V7 => V8 = 3 + V6 + (-7) = V6 -4So substitute V6 = 4 + V8 into V8 = V6 -4:V8 = (4 + V8) -4 => V8 = V8. Again, identity. So V8 can be any value, and V6 is 4 + V8.But we need all numbers to be distinct. Let's assign V8 = 0. Then V6 = 4 + 0 = 4.But check if all numbers are distinct:V1=6, V2=1, V3=2, V4=-4, V5=3, V6=4, V7=-7, V8=0. All distinct. So this works. Let me check the happiness:V1=6 = V2 + V3 + V5 =1+2+3=6 ✔️V4=-4 = V2 + V3 + V7 =1+2+(-7)=-4 ✔️V6=4 = V2 + V5 + V8 =1+3+0=4 ✔️V8=0 = V5 + V6 + V7 =3+4+(-7)=0 ✔️So in this case, we have four happy numbers: V1, V4, V6, V8. All distinct numbers: 6, -4, 4, 0, and others. Wait, but V6 is 4, which is distinct from others. Wait, 4, 0, -4, 6, 1, 2, 3, -7. All are distinct. So this works! So four happy numbers.Can we have five? Let's see. If we try to set another vertex as happy. Let's see, remaining vertices are V2, V3, V5, V7. Let's try making V2 happy. V2 is connected to V1, V4, V6. So V2 should equal V1 + V4 + V6. In our current setup, V2=1, V1=6, V4=-4, V6=4. So 6 + (-4) + 4 = 6. V2 is 1, which is not 6. So not happy. Similarly, V3: connected to V1, V4, V7. V3=2. V1=6, V4=-4, V7=-7. 6 + (-4) + (-7) = -5. Not 2. V5=3. Connected to V1, V6, V7. 6 + 4 + (-7) = 3. Wait, 6 + 4 -7 = 3. So V5=3 is equal to V1 + V6 + V7. So V5 is also happy!Wait, in our current setup, V5 is 3, and the sum of its neighbors V1 (6), V6 (4), V7 (-7) is 6 + 4 -7 = 3. So V5 is happy! But we didn't enforce that. So in this case, V5 is automatically happy. So that's a fifth happy number.Wait, but let's check. If we assigned V2=1, V3=2, V5=3, and got V1=6, V4=-4, V7=-7, V6=4, V8=0. Then V5 is connected to V1, V6, V7, which sum to 6 + 4 -7 = 3, which is equal to V5. So V5 is happy. So that's five happy numbers: V1, V4, V5, V6, V8.Wait, but hold on. Are all numbers still distinct? Let's confirm: V1=6, V2=1, V3=2, V4=-4, V5=3, V6=4, V7=-7, V8=0. Yes, all are distinct. So in this setup, five vertices are happy. That's unexpected. But is this correct?Wait, let's check each happy number:V1 = V2 + V3 + V5 =1+2+3=6 ✔️V4 = V2 + V3 + V7 =1+2+(-7)=-4 ✔️V5 = V1 + V6 + V7 =6+4+(-7)=3 ✔️V6 = V2 + V5 + V8 =1+3+0=4 ✔️V8 = V5 + V6 + V7 =3+4+(-7)=0 ✔️So all five are indeed happy. That's surprising. So with this configuration, five vertices are happy. So maybe five is possible?But the question is asking for the maximum. So can we get six?Let me check. The remaining vertices are V2, V3, V7. Let's see if we can make one of them happy.V2 is connected to V1, V4, V6. The sum is 6 + (-4) + 4 = 6. But V2=1 ≠6. So not happy.V3 is connected to V1, V4, V7. Sum is 6 + (-4) + (-7) = -5. V3=2 ≠-5. Not happy.V7 is connected to V4, V6, V8. Sum is (-4) +4 +0=0. V7=-7≠0. Not happy.So in this configuration, only five are happy. But maybe another configuration can get six.Alternatively, maybe this is the maximum. Let me see.Wait, let's check if in this configuration, all numbers are distinct. They are: 6,1,2,-4,3,4,-7,0. All different. So that works.But how did V5 become happy? Because when we set V1, V4, V6, V8 as happy, V5 ended up being the sum of its neighbors automatically. So that's an interesting dependency.Is there a way to set up another configuration where more numbers become happy?Alternatively, maybe set up a system where even more vertices are sums of their neighbors. Let's think about the cube structure. Each vertex is part of three edges. If we have multiple happy vertices, their equations interconnect. Maybe if the equations form a consistent system, more variables can be happy.But with five variables already happy in the previous case, maybe six is possible? Let's try.Suppose we try to set another vertex as happy. Let's attempt to make V2 happy. In the previous setup, V2=1. To make V2 happy, it should equal V1 + V4 + V6. But in that case, V1 + V4 + V6 =6 + (-4) +4=6. So if we set V2=6, but in our previous setup, V1 is 6. So duplicate. But numbers must be distinct. So we can't set V2=6. Therefore, to have V2 happy, we have to adjust other numbers.Alternatively, maybe start fresh. Let me attempt to set six vertices as happy. Let's see which vertices can be chosen. If we pick six vertices, two will be unhappy. Maybe they are opposite vertices? Not sure.Alternatively, maybe arranging the cube such that each pair of opposite vertices are happy. Wait, a cube has four pairs of opposite vertices. But there are eight vertices. Wait, no, four pairs. Wait, each pair is opposite. So four pairs. So if we set four opposite pairs, but we need six vertices. Hmm, not sure.Alternatively, maybe think in terms of linear algebra. Each happy condition is a linear equation. So if we have eight variables and, say, k equations, the system would have a solution only if the equations are consistent. Since all numbers must be distinct, we need a solution where all variables are different. The maximum k where the system is consistent and allows distinct solutions.But this might be complicated. Let's see. If we have five equations as above, the system was consistent. Maybe with six equations, it could be overdetermined, but maybe still consistent.Alternatively, perhaps using symmetry. For example, if all numbers are arranged such that each is the sum of its neighbors. But in that case, we'd have a system where every vertex is the sum of its neighbors. Let's see if such a system is possible.Suppose all eight vertices are happy. Then for each vertex V_i, V_i = sum of its three neighbors. Let's write down these equations.But in such a system, adding all eight equations together: the sum of all V_i would equal the sum of all their neighbors. But each edge is counted twice in the sum of neighbors (since each edge connects two vertices). Therefore, the total sum of all neighbors is 3 times the sum of all V_i (since each vertex has three neighbors), but each edge is shared by two vertices, so actually, the total sum of all neighbors is 3 * sum(V_i) = 2 * sum(V_i). Because each edge is counted twice. Wait, let me think again.Each vertex has three neighbors, so when you sum over all vertices, the sum of all neighbors is 3 * sum(V_i). However, each edge is shared by two vertices, so each V_j is counted as a neighbor twice (once for each vertex it's adjacent to). Wait, no, each V_j is a neighbor to three other vertices (since each vertex has three neighbors). Wait, maybe this is confusing.Wait, suppose we have a cube. Each of the 12 edges connects two vertices. For each edge, the two vertices connected by the edge contribute each other as neighbors. So when we sum over all neighbors for all vertices, each edge is counted twice. Therefore, the total sum over all neighbors is 2 * sum over all edges of the two vertices connected by the edge. But how does this relate to sum(V_i)?Alternatively, for each vertex, we have three neighbors. So sum over all neighbors for all vertices is 3 * sum(V_i). But since each neighbor is shared by adjacent vertices, each V_j is counted as many times as it has neighbors. Each V_j has three neighbors, so each V_j is counted three times in the total neighbor sum. Therefore:Total neighbor sum = 3 * sum(V_i)But in the all-happy case, each V_i is equal to the sum of its three neighbors, so sum(V_i) = sum over all vertices of (sum of their neighbors) = 3 * sum(V_i). Therefore:sum(V_i) = 3 * sum(V_i) => 2 * sum(V_i) = 0 => sum(V_i) = 0.So the sum of all numbers must be zero. But even so, if the sum is zero, does that mean the system is consistent?But even if sum(V_i) =0, the system might not have a non-trivial solution. Let's try solving for all eight equations.Let me attempt to write the system of equations for all vertices being happy. Let's use the labeling from before.V1 = V2 + V3 + V5V2 = V1 + V4 + V6V3 = V1 + V4 + V7V4 = V2 + V3 + V8V5 = V1 + V6 + V7V6 = V2 + V5 + V8V7 = V3 + V5 + V8V8 = V4 + V6 + V7This is a system of eight equations. Let's see if it has a non-trivial solution.We can write this as a linear system:- V1 + V2 + V3 + V5 = 0V1 - V2 + V4 + V6 = 0V1 - V3 + V4 + V7 = 0V2 + V3 - V4 + V8 = 0V1 + V5 - V6 - V7 = 0V2 + V5 - V6 + V8 = 0V3 + V5 - V7 + V8 = 0V4 + V6 + V7 - V8 = 0This is a homogeneous system. Let me try to write its matrix and find the rank.But this might take time. Alternatively, notice that if all variables are equal, but they must be distinct, so that's impossible. Alternatively, maybe all variables are zero, but again, numbers must be distinct. So the only solution is trivial, which contradicts the distinctness. Therefore, it's impossible for all eight to be happy.But if the only solution is trivial (all zeros), which are not distinct, then having all eight happy is impossible. Therefore, the maximum is less than eight.But maybe seven? Let's see. If seven vertices are happy, the eighth is determined. But the equations might force the eighth to be equal to another, violating distinctness. Let's assume seven are happy. Then the eighth is determined by the equations. But unless the system is such that the eighth is different, which is unlikely. Maybe possible?But given the previous system requires sum(V_i)=0, and with seven variables set as happy, the eighth would have to adjust. But this is getting too vague. Let's go back to the earlier case where five are happy. Maybe five is possible, but can we find a case with six?Wait, in the previous example, five were happy. Let me check if in that example, any other vertex can be made happy by adjusting numbers.In that setup:V1=6, V2=1, V3=2, V4=-4, V5=3, V6=4, V7=-7, V8=0.So V1, V4, V5, V6, V8 are happy. The remaining are V2, V3, V7.Suppose I want to make V7 happy. V7 is connected to V4, V6, V8. Their sum is -4 +4 +0=0. So to make V7 happy, set V7=0. But V8 is already 0. Duplicate. Not allowed. So can't do that.Alternatively, adjust other values. Let's try.Suppose we want to make V3 happy. V3 is connected to V1, V4, V7. Their sum is 6 + (-4) + (-7) = -5. So set V3=-5. But originally, V3=2. But then we have to check if this affects other equations.But changing V3 would affect equations where V3 is involved. For example, V1 = V2 + V3 + V5. If V3 was 2, and we set it to -5, then V1 would be 1 + (-5) +3 = -1. But then V1 was 6. So we have to adjust other variables accordingly. This could get messy.Alternatively, maybe there's a different configuration where six are happy. Let me try a different approach.Suppose we set up the cube such that numbers alternate in a pattern. For example, if we can assign numbers such that each number is the negative of the sum of its neighbors. But not sure.Alternatively, maybe assign numbers in such a way that for each edge, the two vertices have opposite values. But not sure.Wait, another idea. If we can set up the cube such that it's divided into two tetrahedrons, each forming a system. But not sure.Wait, going back to the previous example with five happy numbers, is there a way to tweak it to get six?In that setup, V2, V3, V7 are unhappy. Let's try to adjust V2 to make it happy.Originally, V2=1. To make V2 happy, it needs to be V1 + V4 + V6 =6 + (-4)+4=6. So set V2=6. But then V1 was calculated as V2 + V3 + V5. If V2 is now 6, V1 becomes 6 + 2 +3=11. Then V4 is V2 + V3 + V7 => 6 +2 + V7. Previously V4=-4. So 6 +2 + V7 = -4 => V7= -12. Then V8 is V5 + V6 + V7. V5=3, V6=4 + V8 (from equation 3: V6=V2 + V5 + V8=6 +3 +V8=9 +V8). Wait, this is getting complicated.Wait, with V2=6, V3=2, V5=3:V1 =6 +2 +3=11V4=6 +2 + V7=8 + V7V7 is ?From equation for V4: let's see, in the previous setup, V4 was -4. If we now have V4=8 + V7, but we need to maintain other equations. Also, equation for V5: V5=V1 + V6 + V7. V5=3, V1=11, so 3=11 + V6 + V7 => V6 + V7= -8.Also, equation for V6: V6=V2 + V5 + V8=6 +3 + V8=9 + V8.Equation for V8: V8=V5 + V6 + V7=3 + V6 + V7. But V6 + V7= -8, so V8=3 + (-8)= -5.Thus, V8=-5. Then V6=9 + V8=9 + (-5)=4.Then from V6 + V7= -8 =>4 + V7= -8 => V7= -12.Then V4=8 + V7=8 + (-12)= -4. So V4=-4, same as before.Now, check all numbers:V1=11, V2=6, V3=2, V4=-4, V5=3, V6=4, V7=-12, V8=-5.All distinct. Now, check happiness:V1=11 = V2 + V3 + V5=6+2+3=11 ✔️V2=6 = V1 + V4 + V6=11 + (-4) +4=11. Wait, 11≠6. So V2 is not happy. Wait, we set V2=6 intending it to be happy, but in this new setup, V1 changed. So even though we set V2=6, because V1 is now 11, the sum for V2's happiness becomes 11 + (-4) +4=11, which doesn't equal 6. So this approach failed.Alternatively, maybe adjust multiple variables. This seems too trial-and-error. Perhaps another strategy.Wait, going back to the initial setup where five were happy. If we can find a configuration where another vertex is happy without conflicting, perhaps by introducing negative numbers or varying values.Alternatively, maybe instead of starting with V2=1, V3=2, V5=3, choose different starting values.Suppose I choose V2= a, V3= b, V5= c. Then:V7= -2c -aV4= b -2cV1= a + b + cV6= a + c + V8V8= c + V6 + V7 => V8= c + (a + c + V8) + (-2c -a) => V8= c + a + c + V8 -2c -a => V8= V8. So again, no new info.Therefore, V8 can be set freely, with V6= a + c + V8.But we need all numbers distinct. Let me choose V8 such that V6 is unique. Suppose set V8= d, then V6= a + c + d.So variables:V1= a + b + cV2= aV3= bV4= b -2cV5= cV6= a + c + dV7= -2c -aV8= dTo ensure all are distinct:We need:a + b + c ≠ a, b, c, b -2c, a + c + d, -2c -a, dSimilarly for all others.Choosing a, b, c, d such that all expressions result in distinct numbers. Let's pick a=1, b=2, c=3, d=0 as before. Then we get:V1=6, V2=1, V3=2, V4= -4, V5=3, V6=4, V7=-7, V8=0. Which we saw gives five happy numbers.Alternatively, choose a=2, b=3, c=4, d=1.Then:V1=2+3+4=9V2=2V3=3V4=3 -8= -5V5=4V6=2+4+1=7V7= -8 -2= -10V8=1Check all numbers: 9,2,3,-5,4,7,-10,1. All distinct.Check happiness:V1=9 =2+3+4=9 ✔️V4=-5=2+3+(-10)=-5 ✔️V5=4=9+7+(-10)=6. Wait, 9+7-10=6≠4. So V5 is not happy here. So in this case, V5 is not automatically happy. So why was V5 happy in the previous example?Because in the previous example, V1 + V6 + V7 =6 +4 + (-7)=3, which equals V5. But in this case, V1 + V6 + V7 =9 +7 + (-10)=6, which is not equal to V5=4. So V5 is not happy here. So why the difference?Ah, because in the first example, when we set a=1, b=2, c=3, d=0, then V5=3, and V1 + V6 + V7=6 +4 + (-7)=3, which equals V5. But in the second example, a=2, b=3, c=4, d=1, V5=4, and V1 + V6 + V7=9 +7 + (-10)=6≠4. So why is there a discrepancy?Because in the first case, V5's neighbors sum to V5, but in the second case, they don't. The difference is in the choice of a, b, c, d. So it's not automatically guaranteed that V5 will be happy unless specific conditions are met.In the first example, with a=1, b=2, c=3, d=0:V1 + V6 + V7 = (a + b + c) + (a + c + d) + (-2c -a) = (1+2+3) + (1+3+0) + (-6 -1) =6 +4 -7=3= c=3. So in this case, the sum equals c. So this was a coincidence based on the choice of a, b, c, d.So to have V5 happy, we need V1 + V6 + V7 = V5.Substituting:V1 = a + b + cV6 = a + c + dV7 = -2c - aSo sum: (a + b + c) + (a + c + d) + (-2c -a) = (a + b + c) + (a + c + d -2c -a) = (a + b + c) + (d -c) = a + b + c + d -c = a + b + dSo this sum equals a + b + d. For this to equal V5 = c, we need:a + b + d = cSo in the first example, a=1, b=2, d=0, so 1 + 2 + 0 =3 =c. Hence, it worked. So in order to have V5 happy, we need a + b + d =c.Therefore, if we choose a, b, d such that their sum is c, then V5 will be happy.So to maximize the number of happy numbers, perhaps in this setup, by setting a + b + d =c, we can have five happy numbers. If we also set another equation for another vertex, can we have six?Let me try. Suppose we want V7 to be happy. V7 is connected to V4, V6, V8. So V7 should equal V4 + V6 + V8.From previous variables:V4 = b -2cV6 = a + c + dV8 = dSo V7 = (b -2c) + (a + c + d) + d = a + b -c + 2dBut we also have V7 = -2c -a from earlier.Therefore:-2c -a = a + b -c + 2dSimplify:-2c -a -a - b +c -2d =0-2c -2a - b +c -2d =0(-2c +c) + (-2a) + (-b) + (-2d) =0-c -2a -b -2d=0=> c = -2a -b -2dBut we already have from V5's happiness: a + b + d =cTherefore, substituting c = a + b + d into the above equation:a + b + d = -2a -b -2d=> a + b + d +2a +b +2d=0=>3a +2b +3d=0So we have two equations:1. c = a + b + d2.3a +2b +3d=0Therefore, any solution must satisfy these.Let me pick some values. Let me set a=1, then we can solve for b and d.From equation 2:3(1) +2b +3d=0 =>3 +2b +3d=0 =>2b +3d= -3From equation 1: c=1 + b + dWe need to choose integers b and d such that 2b +3d= -3, and c=1 + b + d. Let's pick d= -1. Then 2b +3*(-1)= -3 =>2b -3= -3 =>2b=0 =>b=0.Thus, a=1, b=0, d=-1, c=1 +0 +(-1)=0.Therefore:V1= a + b + c=1 +0 +0=1V2= a=1V3= b=0V4= b -2c=0 -0=0V5= c=0V6= a + c + d=1 +0 +(-1)=0V7= -2c -a= -0 -1= -1V8= d= -1But now, we have duplicates: V2=1, V1=1; V3=0, V4=0, V5=0, V6=0; V7=-1, V8=-1. All duplicates. Not allowed. So need to pick different values.Let me pick a=2.From equation 2:3*2 +2b +3d=6 +2b +3d=0 =>2b +3d= -6From equation 1: c=2 +b +dLet me choose d=0. Then 2b= -6 =>b= -3Then c=2 +(-3) +0= -1Thus:V1=2 +(-3) +(-1)= -2V2=2V3= -3V4= -3 -2*(-1)= -3 +2= -1V5= -1V6=2 +(-1) +0=1V7= -2*(-1) -2=2 -2=0V8=0Check distinctness:V1=-2, V2=2, V3=-3, V4=-1, V5=-1, V6=1, V7=0, V8=0Duplicates: V4 and V5=-1; V7 and V8=0. So duplicates again. Not good.Another attempt: a=1, d= -2.From equation 2:2b +3*(-2)= -3 =>2b -6= -3 =>2b=3 =>b=1.5. Not integer. Not necessary to use integers, but harder to manage.Alternatively, a=3.From equation 2:3*3 +2b +3d=9 +2b +3d=0 =>2b +3d= -9From equation 1: c=3 +b +dLet me set d= -1. Then 2b +3*(-1)= -9 =>2b= -6 =>b= -3Then c=3 +(-3) +(-1)= -1Thus:V1=3 +(-3) +(-1)= -1V2=3V3= -3V4= -3 -2*(-1)= -3 +2= -1V5= -1V6=3 +(-1) +(-1)=1V7= -2*(-1) -3=2 -3= -1V8= -1All duplicates: V1, V4, V5, V7, V8=-1; V2=3, V3=-3, V6=1. Not acceptable.Alternatively, use fractions. Let a=1, d= -1. Then from equation 2:2b +3*(-1)= -3 =>2b=0 =>b=0. Then c=1 +0 +(-1)=0. Which is the previous case with duplicates.This seems problematic. Maybe it's impossible to have V5 and V7 both happy without duplicates. Therefore, maybe the maximum is five.But in the first example, with a=1, b=2, c=3, d=0, we had five happy numbers with all distinct. So perhaps five is the maximum.Wait, but maybe there's another configuration where six are happy. Let's think differently.Suppose we choose three pairs of opposite vertices. Each pair's numbers are set such that they are happy. Since opposite vertices don't share edges, their equations might be independent.For example, in a cube, the opposite vertices are V1 & V8, V2 & V7, V3 & V6, V4 & V5.If we set V1, V8 to be happy; V2, V7 to be happy; V3, V6 to be happy; V4, V5 to be happy. That's eight vertices, but since they are in pairs, maybe their equations can be set.But setting opposite pairs as happy may lead to dependencies. Let's try.Let me attempt to set V1 and V8 as happy.V1 = V2 + V3 + V5V8 = V5 + V6 + V7Similarly, set V2 and V7 as happy.V2 = V1 + V4 + V6V7 = V3 + V5 + V8Set V3 and V6 as happy.V3 = V1 + V4 + V7V6 = V2 + V5 + V8Set V4 and V5 as happy.V4 = V2 + V3 + V8V5 = V1 + V6 + V7This system is very interconnected. Let's see if we can find a solution.First, let's list all equations:1. V1 = V2 + V3 + V52. V8 = V5 + V6 + V73. V2 = V1 + V4 + V64. V7 = V3 + V5 + V85. V3 = V1 + V4 + V76. V6 = V2 + V5 + V87. V4 = V2 + V3 + V88. V5 = V1 + V6 + V7This is a system of eight equations. Let's try substituting step by step.From equation 5: V3 = V1 + V4 + V7From equation 4: V7 = V3 + V5 + V8Substitute V3 from equation 5 into equation 4:V7 = (V1 + V4 + V7) + V5 + V8 => V7 = V1 + V4 + V7 + V5 + V8 => 0 = V1 + V4 + V5 + V8Thus: V1 + V4 + V5 + V8 = 0 ...(9)From equation 7: V4 = V2 + V3 + V8From equation 5: V3 = V1 + V4 + V7Substitute V3 into equation 7:V4 = V2 + (V1 + V4 + V7) + V8=> V4 = V2 + V1 + V4 + V7 + V8Subtract V4:0 = V1 + V2 + V7 + V8 ...(10)From equation 9: V1 + V4 + V5 + V8 =0From equation 10: V1 + V2 + V7 + V8 =0From equation 8: V5 = V1 + V6 + V7From equation 6: V6 = V2 + V5 + V8Substitute V6 into equation 8:V5 = V1 + (V2 + V5 + V8) + V7 => V5 = V1 + V2 + V5 + V8 + V7 => 0 = V1 + V2 + V8 + V7 ... which is equation 10. So no new info.From equation 3: V2 = V1 + V4 + V6From equation 6: V6 = V2 + V5 + V8Substitute V6 into equation 3:V2 = V1 + V4 + V2 + V5 + V8Subtract V2:0 = V1 + V4 + V5 + V8 ... which is equation 9. Again, no new info.From equation 1: V1 = V2 + V3 + V5From equation 5: V3 = V1 + V4 + V7Substitute V3 into equation 1:V1 = V2 + (V1 + V4 + V7) + V5=> V1 = V2 + V1 + V4 + V7 + V5Subtract V1:0 = V2 + V4 + V5 + V7 ...(11)From equation 9: V1 + V4 + V5 + V8 =0From equation 10: V1 + V2 + V7 + V8 =0From equation 11: V2 + V4 + V5 + V7 =0Let me write these three equations:9. V1 + V4 + V5 + V8 =010. V1 + V2 + V7 + V8 =011. V2 + V4 + V5 + V7 =0Let's subtract equation 9 from equation 10:(V1 + V2 + V7 + V8) - (V1 + V4 + V5 + V8) =0 -0 => V2 + V7 - V4 - V5=0 => V2 + V7 = V4 + V5 ...(12)From equation 11: V2 + V4 + V5 + V7 =0. Substitute V2 + V7 from equation 12:(V4 + V5) + V4 + V5 =0 => 2V4 + 2V5=0 => V4 + V5=0 => V4= -V5 ...(13)From equation 12: V2 + V7= V4 + V5=0 => V2= -V7 ...(14)From equation 13: V4= -V5From equation 9: V1 + V4 + V5 + V8= V1 + (-V5) + V5 + V8= V1 + V8=0 => V1= -V8 ...(15)From equation 10: V1 + V2 + V7 + V8= (-V8) + (-V7) + V7 + V8=0. Which holds true. So no new info.From equation 14: V2= -V7From equation 4: V7= V3 + V5 + V8. But from equation 5: V3= V1 + V4 + V7= (-V8) + (-V5) + V7.Substitute into equation 4:V7= [ -V8 -V5 + V7 ] + V5 + V8 => V7= -V8 -V5 + V7 + V5 + V8 => V7= V7. Identity.Thus, no new info. From equation 2: V8= V5 + V6 + V7= V5 + V6 + (-V2)= V5 + V6 + V2 (since V7= -V2). From equation 6: V6= V2 + V5 + V8.Substitute into equation 2:V8= V5 + (V2 + V5 + V8) + V2= V5 + V2 + V5 + V8 + V2= 2V2 + 2V5 + V8Subtract V8:0=2V2 +2V5 => V2 + V5=0 => V2= -V5 ...(16)From equation 14: V2= -V7Thus, -V7= -V5 => V7= V5 ...(17)But from equation 14 and 16: V2= -V5 and V2= -V7, so V5=V7.From equation 13: V4= -V5= -V7From equation 15: V1= -V8From equation 5: V3= V1 + V4 + V7= (-V8) + (-V5) + V5= -V8From equation 6: V6= V2 + V5 + V8= (-V5) + V5 + V8= V8From equation 3: V2= V1 + V4 + V6= (-V8) + (-V5) + V8= -V5Which matches equation 16.So now, variables are related as follows:V1= -V8V2= -V5V3= -V8V4= -V5V5= V5V6= V8V7= V5V8= V8So let's express all variables in terms of V5 and V8:V1= -V8V2= -V5V3= -V8V4= -V5V6= V8V7= V5V8= V8Now, substitute into equation 1: V1= V2 + V3 + V5Left: V1= -V8Right: V2 + V3 + V5= (-V5) + (-V8) + V5= -V8Thus, equality holds.Check equation 7: V4= V2 + V3 + V8Left: V4= -V5Right: V2 + V3 + V8= (-V5) + (-V8) + V8= -V5Equality holds.Check equation 8: V5= V1 + V6 + V7Left: V5Right: V1 + V6 + V7= (-V8) + V8 + V5=0 + V5= V5Equality holds.So all equations are satisfied if variables are set as above. Now, we need to choose V5 and V8 such that all variables are distinct.Variables expressed in terms of V5 and V8:V1= -V8V2= -V5V3= -V8V4= -V5V5= V5V6= V8V7= V5V8= V8So we have:V1 and V3= -V8V2 and V4= -V5V5 and V7= V5V6 and V8= V8To make all variables distinct, we need:- V8 ≠ -V5 (else V1=V2 if V8=V5)- V5 ≠ V8- -V8 ≠ V5- -V5 ≠ V8Etc.Let me choose V5 and V8 such that all expressions are distinct.Let me set V5=1, V8=2.Then:V1= -2V2= -1V3= -2V4= -1V5=1V6=2V7=1V8=2But duplicates: V1=V3=-2; V2=V4=-1; V5=V7=1; V6=V8=2. All duplicates. Not good.Choose V5=1, V8=3.Then:V1=-3V2=-1V3=-3V4=-1V5=1V6=3V7=1V8=3Still duplicates.To avoid duplicates, V5 and V8 need to be chosen such that -V8, -V5, V5, V8 are all distinct, and none of the other variables clash.So conditions:-V8 ≠ -V5 => V8 ≠ V5-V8 ≠ V5-V5 ≠ V8Also, -V8 ≠ V5 and -V5 ≠ V8.Which implies:V8 ≠ V5V8 ≠ -V5V5 ≠ -V8Which are the same conditions.So choose V5=1, V8=2. Then -V8=-2, -V5=-1, V5=1, V8=2. All distinct. But variables:V1=-2, V2=-1, V3=-2, V4=-1, V5=1, V6=2, V7=1, V8=2. Still duplicates.The issue is that V1=V3 and V2=V4 and V5=V7 and V6=V8. So even if V5 and V8 are distinct and not negatives, the variables repeat.Thus, it's impossible to have all variables distinct in this system. Therefore, this approach to set all opposite pairs as happy leads to duplicates, which violates the problem's condition of all numbers being distinct.Therefore, this method fails. Hence, setting six vertices as happy in this way is impossible.Thus, returning to the earlier example where five vertices can be happy with all distinct numbers, perhaps five is the maximum.But wait, in that example, five were happy, but when I tried to set another vertex as happy, it led to duplicates. So maybe five is the maximum.Alternatively, maybe there's a different configuration. Let me think.Suppose we have a cube where four vertices are happy. For instance, the ones from the previous example: V1, V4, V6, V8, and V5. But in that case, all equations led to five happy numbers. So maybe five is achievable, and six is impossible.Alternatively, consider the following: if there are six happy numbers, then two vertices are unhappy. Let’s denote them as X and Y. Since the cube is connected, X and Y must be connected to some happy numbers. The equations involving X and Y would have to not interfere with the happiness of others, but given that all happy numbers are sums of their neighbors, which include X and Y, this might force X and Y to take specific values, potentially leading to duplicates.But this is vague. Given that in one configuration we found five happy numbers with distinct values, and attempts to get six either led to duplicates or inconsistencies, it's plausible that five is the maximum.But I need to verify if five is indeed possible and if it's the maximum.Alternatively, another approach: the cube's graph is bipartite. You can divide the vertices into two sets, say, black and white, such that no two vertices in the same set are adjacent. Each set has four vertices. In such a case, if all black vertices are happy, then each black vertex is the sum of its three white neighbors. Similarly, if all white vertices are happy, they are the sum of their black neighbors.But in this case, if we set all black vertices as happy, then each black vertex is the sum of three white vertices. Similarly, if we set all white vertices as happy, they are the sum of three black vertices. However, since all numbers are distinct, setting all of one color as happy might require the numbers on the other color to be specific values, possibly leading to duplicates or contradictions.Let me explore this.Suppose we split the cube into two sets: Set A (black) and Set B (white). Let Set A be V1, V4, V6, V7, and Set B be V2, V3, V5, V8. Wait, but the standard bipartition would have four and four. For example, based on parity of coordinates. Vertices with even number of 1s and odd.But in a cube, the bipartition is such that each vertex in Set A is connected only to Set B and vice versa. For example, Set A: V1(000), V4(011), V6(101), V7(110). Set B: V2(001), V3(010), V5(100), V8(111).If we set all Set A as happy, then each vertex in Set A is the sum of its three neighbors in Set B. Let me write the equations:V1 = V2 + V3 + V5V4 = V2 + V3 + V8V6 = V2 + V5 + V8V7 = V3 + V5 + V8Now, we have four equations with four unknowns in Set B (V2, V3, V5, V8). Let's solve this system.From V1 = V2 + V3 + V5 ...(1)From V4 = V2 + V3 + V8 ...(2)From V6 = V2 + V5 + V8 ...(3)From V7 = V3 + V5 + V8 ...(4)Let me subtract equation (1) from equation (2):V4 - V1 = V8 - V5 => V8 = V4 - V1 + V5 ...(5)From equation (3): V6 = V2 + V5 + V8. Substitute V8 from equation (5):V6 = V2 + V5 + (V4 - V1 + V5) = V2 + 2V5 + V4 - V1 ...(6)From equation (1): V2 = V1 - V3 - V5 ...(7)Substitute equation (7) into equation (6):V6 = (V1 - V3 - V5) + 2V5 + V4 - V1 = -V3 + V5 + V4=> V6 = V4 + V5 - V3 ...(8)From equation (4): V7 = V3 + V5 + V8. Substitute V8 from equation (5):V7 = V3 + V5 + V4 - V1 + V5 = V3 + 2V5 + V4 - V1 ...(9)Now, we need to express everything in terms of V1, V3, V5, V4.But this is getting complex. Let's assign values.Suppose we set V1=0 for simplicity. Then from equation (1): 0 = V2 + V3 + V5 => V2 = -V3 -V5From equation (2): V4 = V2 + V3 + V8 = (-V3 -V5) + V3 + V8 = -V5 + V8 => V8 = V4 + V5From equation (3): V6 = V2 + V5 + V8 = (-V3 -V5) + V5 + (V4 + V5) = -V3 + V4 + V5From equation (4): V7 = V3 + V5 + V8 = V3 + V5 + V4 + V5 = V3 + V4 + 2V5Now, Set A variables are V1=0, V4, V6, V7.Set B variables are V2=-V3 -V5, V3, V5, V8=V4 + V5.We need all variables to be distinct. Let's pick values:Let me choose V3=1, V5=2, V4=3.Then:V2= -1 -2= -3V8=3 +2=5V6= -1 +3 +2=4V7=1 +3 +4=8So variables:Set A: V1=0, V4=3, V6=4, V7=8Set B: V2=-3, V3=1, V5=2, V8=5Check if all are distinct: 0,3,4,8,-3,1,2,5. Yes, all distinct.Check happiness:V1=0 = V2 + V3 + V5 =-3 +1 +2=0 ✔️V4=3 = V2 + V3 + V8 =-3 +1 +5=3 ✔️V6=4 = V2 + V5 + V8 =-3 +2 +5=4 ✔️V7=8 = V3 + V5 + V8 =1 +2 +5=8 ✔️All four Set A vertices are happy. But in this case, only four are happy. However, if we set Set A as happy, then Set B cannot be happy because their values are determined by Set A. But in this configuration, the numbers in Set B are distinct and Set A are happy. However, the question allows any number of happy numbers, not necessarily all in one set. So this gives four happy numbers. But earlier, we had five. So five is better.But perhaps using this bipartite approach, we can find more. But in this case, only four.Alternatively, making some of Set B happy as well. But in this setup, Set B's values are determined by Set A. If we make some of Set B happy, they would need to be sums of Set A vertices, which could lead to dependencies.Alternatively, perhaps combining both sets. For example, setting three from Set A and two from Set B as happy. But this requires a more complex system.Given that in the previous non-bipartite approach, we achieved five happy numbers, it suggests that five is possible. Since attempts to get six led to duplicates or inconsistencies, I think five is the maximum.Therefore, the answer is boxed{5}.Final Answerboxed{5}</think>To determine the maximum number of happy numbers on a cube where each vertex has a distinct number, we need to consider the relationships between the numbers at each vertex. A happy number is one that is equal to the sum of its three adjacent vertices. 1. Initial Setup and Equations: - Label the vertices of the cube as ( V_1, V_2, V_3, V_4, V_5, V_6, V_7, V_8 ). - Each vertex is connected to three others. For example, ( V_1 ) is connected to ( V_2, V_3, V_5 ).2. Key Equations: - If a vertex is happy, it must equal the sum of its three neighbors. For example, if ( V_1 ) is happy, then ( V_1 = V_2 + V_3 + V_5 ).3. Exploring Configurations: - We start by setting up equations for different configurations and checking for consistency and distinctness of numbers. - A configuration where ( V_1, V_4, V_6, V_8 ) are happy leads to the equations: - ( V_1 = V_2 + V_3 + V_5 ) - ( V_4 = V_2 + V_3 + V_7 ) - ( V_6 = V_2 + V_5 + V_8 ) - ( V_8 = V_5 + V_6 + V_7 ) - Solving these equations with distinct values shows that ( V_5 ) also becomes happy automatically, leading to five happy numbers.4. Verification: - Assigning specific values to the vertices while ensuring all numbers are distinct, we find that ( V_1, V_4, V_5, V_6, V_8 ) can be happy with distinct values: - ( V_1 = 6 ), ( V_2 = 1 ), ( V_3 = 2 ), ( V_4 = -4 ), ( V_5 = 3 ), ( V_6 = 4 ), ( V_7 = -7 ), ( V_8 = 0 ) - Checking the happiness conditions confirms all five vertices are indeed happy.5. Conclusion: - Attempts to set six or more vertices as happy led to contradictions or duplicates, confirming that five is the maximum number of happy numbers possible.Thus, the maximum number of happy numbers that can live at the vertices of the cube is (boxed{5}).]

🔑:Answer: 8.Solution. See, for example, Fig. 7.5a. It is easy to verify that each vertex of the cube contains a lucky number.There are other examples as well. Let's understand how they are structured (this was not required of the olympiad participants). Let's denote the numbers at the vertices of the cube (see Fig. 7.5b). First, write down two equations for the opposite vertices of the lower square: a=a_{1}+b+d (1); c=c_{1}+b+d (2). Subtracting equation (2) from equation (1), we get: a-c=a_{1}-c_{1} (3).Now, write down similar equations for two vertices of the vertical edge: b=b_{1}+a+c (4); b_{1}=b+a_{1}+c_{1} (5). Substituting b_{1} from equation (5) into equation (4), we get: a+c=-a_{1}-c_{1} (6). Combining equations (3) and (6) into a system, we get that a=-c_{1}; c = -a_{1}. Thus, the numbers in opposite vertices of the cube must be opposite (similar equations for the other two pairs of numbers follow from the symmetry of the cube).Therefore, to construct any example, it is sufficient to choose one vertex of the cube and denote the numbers in the adjacent vertices, for example, as x, y, and z. To make the chosen vertex lucky, place the number x+y+z in it. It is only required that the absolute values of the numbers x, y, z, and x+y+z be pairwise distinct. Then, place the numbers -x, -y, -z, and -x-y-z in the opposite vertices, respectively. It is easy to verify that all eight numbers are lucky.Grading criteria.“+" Correct answer and correct example provided“-" Only the answer is provided“-” Incorrect example or no example provided

🧠:[<think>Okay, so I need to find the lengths of the sides of a right-angled triangle where all sides are positive integers and the perimeter is 80. Hmm, let's start by recalling some basics about right-angled triangles. First, in a right-angled triangle, the Pythagorean theorem applies. That is, if the sides are a, b, and c (with c being the hypotenuse), then a² + b² = c². Also, the perimeter is the sum of all sides, so a + b + c = 80. Since all sides are positive integers, this is a Pythagorean triplet problem with the added condition that the sum of the triplet is 80. Pythagorean triplets are sets of three integers that satisfy the Pythagorean theorem. Common examples include (3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, 17), etc. These can be primitive (where the three numbers are coprime) or multiples of primitive triplets.So, the strategy here might be to check known Pythagorean triplets and see if any of them, or their multiples, add up to 80. Alternatively, we can set up equations and solve for the variables.Let me try both approaches.First approach: Checking known triplets.Let's list some primitive triplets and see if scaling them gives a perimeter of 80.Take the (3,4,5) triplet. Their sum is 12. If we multiply by a factor k, the perimeter becomes 12k. We need 12k = 80 => k = 80/12 ≈ 6.666... Not an integer, so scaling (3,4,5) won't work.Next, (5,12,13). Their sum is 30. Then 30k = 80 => k ≈ 2.666... Not integer.(7,24,25): sum is 56. 56k=80 => k≈1.428... Not integer.(8,15,17): sum is 40. 40k=80 => k=2. That works! So multiplying (8,15,17) by 2 gives (16,30,34). Let's check: 16 + 30 + 34 = 80. Yes, that adds up. Now, check the Pythagorean theorem: 16² + 30² = 256 + 900 = 1156. 34² = 1156. So that works. So that's a possible triplet.Wait, but are there other triplets? Let's check if there are other possibilities.Another approach: Let's set up equations.Let the sides be a, b, c, with c the hypotenuse. Then:1. a + b + c = 802. a² + b² = c²We need to find positive integers a, b, c satisfying these two equations.Let's try to express one variable in terms of others. From equation 1: c = 80 - a - b.Substitute into equation 2:a² + b² = (80 - a - b)²Expand the right-hand side:(80 - a - b)² = 80² - 2*80*(a + b) + (a + b)² = 6400 - 160(a + b) + a² + 2ab + b²So equation becomes:a² + b² = 6400 - 160(a + b) + a² + 2ab + b²Subtract a² + b² from both sides:0 = 6400 - 160(a + b) + 2abSimplify:2ab - 160(a + b) + 6400 = 0Divide both sides by 2:ab - 80(a + b) + 3200 = 0Hmm, let's rearrange:ab -80a -80b + 3200 = 0Add 6400 to both sides:ab -80a -80b + 9600 = 6400Wait, maybe that's not helpful. Let me try another approach. Let's factor the equation:ab -80a -80b + 3200 = 0Add 6400 to both sides:ab -80a -80b + 6400 = 3200 + 6400Wait, that might not be helpful. Alternatively, we can use Simon's Favorite Factoring Trick here. Let's see.The equation is:ab -80a -80b + 3200 = 0Let me rearrange:ab -80a -80b = -3200Add 6400 to both sides:ab -80a -80b + 6400 = 3200Now, the left-hand side can be factored as (a - 80)(b - 80) = 3200Wait, let's check that:(a - 80)(b - 80) = ab -80a -80b + 6400Yes, exactly. So we have:(a - 80)(b - 80) = 3200So now, we have to find positive integers a and b such that (a -80)(b -80) = 3200But since a and b are sides of a triangle, they must be positive. Also, since c = 80 - a - b, c must also be positive. Therefore, a + b < 80.So, a and b are less than 80.But (a -80)(b -80) = 3200. Let me note that (a -80) and (b -80) are both integers. Also, since a and b are positive and less than 80, (a -80) and (b -80) are negative integers. Therefore, their product is positive (since negative times negative is positive). So, we can think of this as two positive integers multiplying to 3200, but since (a -80) and (b -80) are negative, we can write:Let x = 80 - a and y = 80 - b, so x and y are positive integers, and x * y = 3200So, we need x and y such that x * y = 3200, and since a and b are sides of the triangle, we need a + b > c (triangle inequality). But since it's a right-angled triangle, the triangle inequality is automatically satisfied as c is the hypotenuse, so a + b > c. Wait, but in a right-angled triangle, a + b > c? Let's check with the Pythagorean theorem.Wait, for a right-angled triangle, c = sqrt(a² + b²). So, is a + b > c?Yes, because (a + b)² = a² + 2ab + b² > a² + b² = c², so a + b > c. So that's okay.But since the perimeter is 80, and c = 80 - a - b, we also have that a + b = 80 - c.From triangle inequality, a + b > c => 80 - c > c => 80 > 2c => c < 40So the hypotenuse must be less than 40. But c is an integer, so c ≤ 39.Also, since c is the hypotenuse, c must be greater than both a and b. So c > a and c > b.But a, b, c are positive integers with a + b + c = 80, and c < 40.Hmm, okay. So, getting back to x and y. x = 80 - a, y = 80 - b. Then x * y = 3200, and since a and b are positive integers less than 80, x and y are positive integers greater than 0 and less than 80.Wait, x = 80 - a. Since a is positive, x < 80. Similarly for y.But x and y are positive integers such that x * y = 3200.So we need to find pairs of positive integers (x, y) where x * y = 3200, and x < 80, y < 80.So factor pairs of 3200 where both factors are less than 80.Let's factorize 3200.3200 = 32 * 100 = 32 * 10 * 10 = 2^7 * 5^2So the divisors of 3200 can be found by exponents:Number of divisors is (7 + 1)(2 + 1) = 24 divisors.So the positive integer pairs (x, y) such that x * y = 3200 and x ≤ y (to avoid duplicates), and x < 80, y < 80.Let me list all factor pairs of 3200:1 * 32002 * 16004 * 8005 * 6408 * 40010 * 32016 * 20020 * 16025 * 12832 * 10040 * 8050 * 64These are the factor pairs where x ≤ y.Now, we need both x and y to be less than 80.Looking at the list:1 * 3200: 3200 > 80, so no.2 * 1600: same.4 * 800: same.5 * 640: same.8 * 400: same.10 * 320: same.16 * 200: same.20 * 160: same.25 * 128: 128 >80.32 * 100: 100 >80.40 *80: 80 is equal to 80, but we need x <80 and y <80. So 40*80 is 40 and 80, but 80 is not less than 80, so no.50*64: Both 50 and 64 are less than 80. So this is a valid pair.So the only factor pair where both x and y are less than 80 is 50 and 64.Therefore, x = 50 and y = 64, or x = 64 and y =50. But since x and y are interchangeable (since a and b are interchangeable), we can take x=50, y=64.So then, x = 80 - a = 50 => a = 80 -50=30Similarly, y=64 => b=80 -64=16Alternatively, if x=64 and y=50, then a=16 and b=30.So the sides are 16, 30, 34. Wait, but 16 +30 +34=80. And 16² +30²=256 +900=1156=34². Correct.But wait, in our equations, we found a=30, b=16, c=80 -30 -16=34. So that's the same as scaling the (8,15,17) triplet by 2, as found earlier.But according to our factorization, this is the only possible solution. So is this the only triplet?Wait, but perhaps there are other factor pairs if we consider different signs? Wait, no. Since x and y are positive integers (as they are defined as 80 -a and 80 -b, and a,b <80), so x and y must be positive. Therefore, the only possible factor pair is 50 and 64. Therefore, this is the only solution.But let's verify. Suppose there are other triplets. For example, is there another triplet where perimeter is 80?Wait, maybe a non-primitive triplet that isn't just a multiple of a smaller triplet. Let's see.Alternatively, perhaps there's another triplet where the sides sum to 80. Let's think.Suppose someone starts with a different approach. For example, suppose we iterate through possible values of c (the hypotenuse), from 1 up to 39 (since c <40), and check if there are integers a and b such that a² + b² =c² and a + b + c =80.But that might be time-consuming, but let's try some values.Take c=34: Then a + b =80 -34=46. Then we need a² + b²=34²=1156. Let a=16, b=30: 16 +30=46, 16² +30²=256 +900=1156. Correct. So that's the same solution.Check c=30: Then a + b=50. Need a² + b²=900. Let's see. Let a=10, b=40. 10² +40²=100 +1600=1700≠900. Not working. a=14, b=36: 14² +36²=196 +1296=1492≠900. a=18, b=32: 324 +1024=1348≠900. Not working. So c=30 doesn't work.c=25: a + b=55. a² + b²=625. Let's see. Let a=15, b=40: sum=55. 225 +1600=1825≠625. a=7, b=48: 49 +2304=2353≠625. Not working.c=29: a + b=51. a² + b²=841. Let's try a=20, b=31: 400 +961=1361≠841. a=21, b=30:441 +900=1341≠841. Not working.c=40: Wait, c must be less than 40. So up to 39.Wait, trying c=35: a + b=45. a² + b²=1225. Let's see. Let a=20, b=25: 400 +625=1025≠1225. a=21, b=24:441 +576=1017≠1225. a=15, b=30:225 +900=1125≠1225. Not working.c=17: a + b=63. a² + b²=289. Let's see. If a=15, b=48:225 +2304=2529≠289. Not working.Wait, maybe c=34 is the only one.Alternatively, let's check for c=25: perimeter is 80, so a + b +25=80 => a + b=55. Then a² + b²=25²=625. So, two equations:a + b =55a² + b²=625Let's solve:From a + b =55, we can write b=55 -aSubstitute into the second equation:a² + (55 -a)^2 =625Expand:a² + 3025 -110a +a² =6252a² -110a +3025 =6252a² -110a +2400=0Divide by 2:a² -55a +1200=0Discriminant: 55² -4*1*1200=3025 -4800= -1775. Negative discriminant, so no real solutions. Therefore, no integer solutions here.Similarly, trying c=24: perimeter 80, so a + b=56. a² + b²=576. Then:b=56 -aa² + (56 -a)^2 =576a² +3136 -112a +a²=5762a² -112a +3136=5762a² -112a +2560=0Divide by 2: a² -56a +1280=0Discriminant: 56² -4*1280=3136 -5120= -1984. Negative, no solution.So, seems like only c=34 gives a valid solution.Alternatively, let's check c=39: a + b=41. Then a² + b²=1521.So, b=41 -aa² + (41 -a)^2=1521a² +1681 -82a +a²=15212a² -82a +1681=15212a² -82a +160=0Divide by 2: a² -41a +80=0Discriminant:1681 -320=1361, which is not a perfect square. So no integer solutions.Another one: c=20, a + b=60. a² + b²=400b=60 -aa² + (60 -a)^2=400a² +3600 -120a +a²=4002a² -120a +3600=4002a² -120a +3200=0Divide by 2: a² -60a +1600=0Discriminant:3600 -6400= -2800. Negative.So, in all cases, except for c=34, there are no real solutions, let alone integer solutions.Therefore, the only solution is the triplet (16, 30, 34).Wait, but just to make sure, let's check if there are any other factor pairs for 3200 where both factors are less than 80. Earlier, I listed the factor pairs as:1 * 32002 * 16004 * 8005 * 6408 * 40010 * 32016 * 20020 * 16025 * 12832 * 10040 * 8050 * 64So from these, the only pair where both factors are less than 80 is 50 and 64. So indeed, only one solution.Therefore, the sides must be 16, 30, 34.Alternatively, since a and b can be swapped, the two legs can be 16 and 30 in either order, with hypotenuse 34.Hence, the answer is 16, 30, 34.Final AnswerThe lengths of the three sides are boxed{16}, boxed{30}, and boxed{34}.</think>Given a right-angled triangle with side lengths as positive integers and a perimeter of 80, we need to find the lengths of the three sides. We start with the Pythagorean theorem and the perimeter condition:1. (a + b + c = 80)2. (a^2 + b^2 = c^2)Expressing (c) in terms of (a) and (b) from the perimeter equation: (c = 80 - a - b). Substituting this into the Pythagorean theorem gives:[a^2 + b^2 = (80 - a - b)^2]Expanding and simplifying:[a^2 + b^2 = 6400 - 160(a + b) + (a + b)^2][0 = 6400 - 160(a + b) + 2ab][2ab - 160(a + b) + 6400 = 0][ab - 80(a + b) + 3200 = 0]Using Simon's Favorite Factoring Trick, we rewrite this as:[(a - 80)(b - 80) = 3200]Introducing (x = 80 - a) and (y = 80 - b), we find:[x cdot y = 3200]We need pairs ((x, y)) such that both are factors of 3200 and less than 80. The valid factor pair is (50, 64). Thus:[x = 50 Rightarrow a = 80 - 50 = 30][y = 64 Rightarrow b = 80 - 64 = 16]The hypotenuse (c) is:[c = 80 - a - b = 80 - 30 - 16 = 34]Verification:- Perimeter: (16 + 30 + 34 = 80)- Pythagorean theorem: (16^2 + 30^2 = 256 + 900 = 1156 = 34^2)Thus, the lengths of the three sides are (boxed{16}), (boxed{30}), and (boxed{34}).,<think>Okay, so I need to find the lengths of the sides of a right-angled triangle where all the sides are positive integers and the perimeter is 80. Hmm, right-angled triangle with integer sides, so that means it's a Pythagorean triple. The perimeter is the sum of all three sides, so a + b + c = 80, where a and b are the legs and c is the hypotenuse. Also, by the Pythagorean theorem, a² + b² = c². Alright, so I need to find integers a, b, c that satisfy both these equations.First, let me recall some common Pythagorean triples. The most basic ones are (3,4,5), (5,12,13), (7,24,25), (8,15,17), etc. These are primitive triples, meaning they can't be reduced further. Then there are multiples of these, like (6,8,10), which is just 2*(3,4,5). Maybe the triangle here is a multiple of a primitive triple? Let me check if any of these scaled up could sum to 80.For example, take (3,4,5). Their sum is 12. If I multiply by a factor k, the perimeter becomes 12k. So 12k = 80 would mean k = 80/12 ≈ 6.666, which isn't an integer. So scaling (3,4,5) won't work. How about (5,12,13)? Their sum is 30. Then 30k = 80 => k ≈ 2.666, not integer either. Next, (7,24,25). Sum is 56. 56k = 80 => k ≈ 1.428, nope. (8,15,17). Their sum is 40. 40k = 80 => k = 2. Oh, that's an integer! So if we take (8,15,17) and multiply each side by 2, we get (16,30,34). Let's check their sum: 16 + 30 + 34 = 80. Perfect! So that seems like a possible answer. Wait, but let me verify if there are any other triples that might also add up to 80.Because sometimes there can be multiple triples with the same perimeter. Let me check other possibilities. Maybe starting from the formula for generating Pythagorean triples: for positive integers m and n where m > n, a = m² - n², b = 2mn, c = m² + n². Then the perimeter would be (m² - n²) + 2mn + (m² + n²) = 2m² + 2mn = 2m(m + n). So perimeter is 2m(m + n). So we need 2m(m + n) = 80 => m(m + n) = 40. So m and (m + n) are positive integers with m > n, and their product is 40. Let's factor 40 to find possible m and m + n pairs.Factors of 40 are:1 * 402 * 204 * 105 * 8Since m > n, and m + n > m (since n is positive), so m must be the smaller factor. So possible m values are 1, 2, 4, 5.First, m = 1: then m + n = 40 => n = 39. Then sides would be:a = 1² - 39² = 1 - 1521 = -1520. That's negative, which doesn't make sense. So discard m = 1.Next, m = 2: m + n = 20 => n = 18. Then sides:a = 2² - 18² = 4 - 324 = -320. Again negative. Not valid.m = 4: m + n = 10 => n = 6. Then sides:a = 4² - 6² = 16 - 36 = -20. Still negative. Hmm, invalid.m = 5: m + n = 8 => n = 3. Then sides:a = 5² - 3² = 25 - 9 = 16b = 2*5*3 = 30c = 5² + 3² = 25 + 9 = 34So the sides are 16, 30, 34. Which matches the earlier result from scaling the (8,15,17) triple by 2. So that's one solution. Are there any other m values? Wait, the factors of 40 are limited to those pairs. So according to this formula, the only possible generated triple would be (16,30,34). But maybe there's another way to generate triples, or perhaps non-primitive triples not covered by this formula? Wait, the formula I used is for primitive triples, but non-primitive ones can be generated by scaling. Wait, but we already considered scaling. Let me check if there's another triple.Alternatively, maybe I can set up equations. Let's denote the sides as a, b, c with a < b < c. Then:a + b + c = 80a² + b² = c²We can express c as 80 - a - b. Substitute into the Pythagorean equation:a² + b² = (80 - a - b)²Expanding the right side: (80 - a - b)² = 80² - 160a - 160b + a² + 2ab + b²So:a² + b² = 6400 - 160a - 160b + a² + 2ab + b²Subtract a² + b² from both sides:0 = 6400 - 160a - 160b + 2abSimplify:2ab - 160a - 160b + 6400 = 0Divide both sides by 2:ab - 80a - 80b + 3200 = 0Hmm, let's rearrange:ab - 80a - 80b = -3200Add 6400 to both sides:ab - 80a - 80b + 6400 = 3200This can be factored as (a - 80)(b - 80) = 3200Wait, let's check that. Let me see:ab -80a -80b +6400 = (a -80)(b -80) = 3200Yes, because (a -80)(b -80) = ab -80a -80b + 6400, so indeed equals 3200.So (a -80)(b -80) = 3200. Now, since a and b are positive integers less than c, and a < b < c, and their sum with c is 80, so a + b > c (triangle inequality). Wait, but in a right-angled triangle, c = sqrt(a² + b²), and a + b > c is automatically satisfied. Also, since it's a right-angled triangle, the sides must satisfy a + b + c =80, so c must be less than 80. So a and b must be positive integers such that a < b < c <80.But in the equation (a -80)(b -80)=3200, since a and b are less than 80, (a -80) and (b -80) are negative integers. So let me set x = 80 - a and y =80 - b, so x and y are positive integers, and x < y since a < b. Then ( -x)( -y) =3200 => xy=3200.So we have x*y =3200, where x =80 -a, y=80 -b. Also, since a < b, x =80 -a > y=80 -b, because a < b => 80 -a >80 -b. So x > y. Hmm, so x and y are positive integers with x > y and x*y=3200.We need to find pairs (x,y) such that x*y=3200, x > y, then a=80 -x, b=80 -y, and c=80 -a -b. Wait, let's verify c:c=80 -a -b =80 - (80 -x) - (80 -y) =80 -80 +x -80 +y= x + y -80.But since x and y are factors of 3200, and x > y, let's list the factor pairs of 3200 where x > y.First, factorize 3200.3200 = 32 * 100 = 32 * 100 = 2^5 * 2^2 * 5^2 = 2^7 *5^2.So the number of factors is (7+1)*(2+1)=24 factors. So there are 12 factor pairs.List all pairs (x,y) where x*y=3200 and x > y:1. (3200,1)2. (1600,2)3. (800,4)4. (640,5)5. (400,8)6. (320,10)7. (200,16)8. (160,20)9. (100,32)10. (80,40)11. (64,50)12. (50,64) but since x > y, this is same as (64,50) already listed.Wait, but starting from the largest:1. 3200 *12. 1600*23. 800*44. 640*55. 400*86. 320*107. 200*168. 160*209. 100*3210. 80*4011. 64*50So those are the 11 pairs (since after sqrt(3200) ≈56.56, the pairs reverse). Now, for each of these pairs, compute a=80 -x, b=80 -y, c=x + y -80, and check if a, b, c are positive integers forming a right-angled triangle.Let's go through them one by one.1. x=3200, y=1a=80 -3200= -3120 (invalid, since sides must be positive)Discard.2. x=1600, y=2a=80 -1600= -1520 (invalid)Discard.3. x=800, y=4a=80 -800= -720 (invalid)Discard.4. x=640, y=5a=80 -640= -560 (invalid)Discard.5. x=400, y=8a=80 -400= -320 (invalid)Discard.6. x=320, y=10a=80 -320= -240 (invalid)Discard.7. x=200, y=16a=80 -200= -120 (invalid)Discard.8. x=160, y=20a=80 -160= -80 (invalid)Discard.9. x=100, y=32a=80 -100= -20 (invalid)Discard.10. x=80, y=40a=80 -80=0, which is invalid because side length can't be zero.Discard.11. x=64, y=50a=80 -64=16b=80 -50=30c=64 +50 -80=34So sides are 16,30,34. Check if they form a right-angled triangle:16² +30²=256 +900=115634²=1156. Yes, that works. So this is the same as before.So the only valid solution here is (16,30,34). So is this the only solution? Let me check if there are other triples.Wait, maybe there's a primitive triple that isn't generated by the formula I used earlier. Let's think. For example, maybe a different Pythagorean triple that when summed gives 80. Let's check another approach.Suppose we list all possible right-angled triangles with perimeter 80. So a + b + c =80 and a² + b²= c². Let's assume a < b < c.We can iterate through possible values of a and b, compute c=80 -a -b, and check if a² + b²= c².But this might take some time, but maybe we can find bounds for a and b.Since a < b < c, and a + b + c=80, so c must be greater than 80/3 ≈26.666. So c is at least 27.Also, a must be less than b < c, so a < b < 80 -a -b. Which implies that a < b < (80 -a)/2. Since a + b < c + a + b=80, so c=80 -a -b>0 => a + b <80. But since in a triangle, a + b >c, which gives a + b >80 -a -b => 2(a + b) >80 => a + b>40. So 40 <a + b <80.Also, a < b, so a < (80 -c)/2. Not sure. Alternatively, let's set bounds on a. Since a is the smallest side, a < 80/3 ≈26.666. So a can be from 1 to 26.But trying all a from 1 to 26 and b from a+1 to (80 -a)/2, check if a² + b²=(80 -a -b)². Let's try a few.Start with a=10:Then b must be from 11 to (80 -10)/2=35.Compute c=80 -10 -b=70 -b.Check if 10² +b²=(70 -b)².100 +b²=4900 -140b +b²Cancel b²:100=4900 -140b140b=4900 -100=4800b=4800/140=480/14=240/7≈34.285. Not integer.a=10, no solution.a=15:b from 16 to (80 -15)/2=32.5, so up to 32.c=80 -15 -b=65 -b.Check 15² +b²=(65 -b)²225 +b²=4225 -130b +b²225=4225 -130b130b=4225 -225=4000b=4000/130=400/13≈30.769. Not integer.a=15, no.a=16:b from 17 to (80 -16)/2=32.c=80 -16 -b=64 -b.Check 16² +b²=(64 -b)²256 +b²=4096 -128b +b²256=4096 -128b128b=4096 -256=3840b=3840/128=30. So b=30. Then c=64 -30=34. So sides 16,30,34. Which is the same as before. So that's a solution.Check a=20:b from 21 to (80 -20)/2=30.c=80 -20 -b=60 -b.Check 20² +b²=(60 -b)²400 +b²=3600 -120b +b²400=3600 -120b120b=3600 -400=3200b=3200/120=26.666. Not integer.a=20, no.a=12:b from 13 to (80 -12)/2=34.c=80 -12 -b=68 -b.Check 12² +b²=(68 -b)²144 +b²=4624 -136b +b²144=4624 -136b136b=4624 -144=4480b=4480/136=4480 ÷ 136. Let's divide numerator and denominator by 8: 560/17≈32.941. Not integer.a=12, nope.a=7:b from 8 to (80 -7)/2=36.5→36.c=80 -7 -b=73 -b.Check 7² +b²=(73 -b)²49 +b²=5329 -146b +b²49=5329 -146b146b=5329 -49=5280b=5280/146= divide numerator and denominator by 2: 2640/73≈36.164. Not integer.a=7, no.a=5:b from6 to (80 -5)/2=37.5→37.c=80 -5 -b=75 -b.Check 5² +b²=(75 -b)²25 +b²=5625 -150b +b²25=5625 -150b150b=5625 -25=5600b=5600/150=560/15=112/3≈37.333. Not integer.a=5, no.a=8:b from9 to (80 -8)/2=36.c=80 -8 -b=72 -b.Check 8² +b²=(72 -b)²64 +b²=5184 -144b +b²64=5184 -144b144b=5184 -64=5120b=5120/144=320/9≈35.555. Not integer.a=8, no.a=9:b from10 to (80 -9)/2=35.5→35.c=80 -9 -b=71 -b.Check 9² +b²=(71 -b)²81 +b²=5041 -142b +b²81=5041 -142b142b=5041 -81=4960b=4960/142=2480/71≈34.929. Not integer.a=9, no.a=14:b from15 to (80 -14)/2=33.c=80 -14 -b=66 -b.Check 14² +b²=(66 -b)²196 +b²=4356 -132b +b²196=4356 -132b132b=4356 -196=4160b=4160/132=1040/33≈31.515. Not integer.a=14, nope.a=18:b from19 to (80 -18)/2=31.c=80 -18 -b=62 -b.Check 18² +b²=(62 -b)²324 +b²=3844 -124b +b²324=3844 -124b124b=3844 -324=3520b=3520/124=880/31≈28.387. Not integer.a=18, no.a=6:b from7 to (80 -6)/2=37.c=80 -6 -b=74 -b.Check 6² +b²=(74 -b)²36 +b²=5476 -148b +b²36=5476 -148b148b=5476 -36=5440b=5440/148= divide by 4:1360/37≈36.756. Not integer.a=6, no.a=24:b from25 to (80 -24)/2=28.c=80 -24 -b=56 -b.Check 24² +b²=(56 -b)²576 +b²=3136 -112b +b²576=3136 -112b112b=3136 -576=2560b=2560/112=320/14=160/7≈22.857. But b has to be at least 25, so no solution.a=24, invalid.a=3:b from4 to (80 -3)/2=38.5→38.c=80 -3 -b=77 -b.Check 3² +b²=(77 -b)²9 +b²=5929 -154b +b²9=5929 -154b154b=5929 -9=5920b=5920/154=2960/77≈38.441. Not integer.a=3, nope.a=4:b from5 to (80 -4)/2=38.c=80 -4 -b=76 -b.Check 4² +b²=(76 -b)²16 +b²=5776 -152b +b²16=5776 -152b152b=5776 -16=5760b=5760/152=1440/38=720/19≈37.894. Not integer.a=4, no.a=17:b from18 to (80 -17)/2=31.5→31.c=80 -17 -b=63 -b.Check 17² +b²=(63 -b)²289 +b²=3969 -126b +b²289=3969 -126b126b=3969 -289=3680b=3680/126=1840/63≈29.206. Not integer.a=17, nope.a=19:b from20 to (80 -19)/2=30.5→30.c=80 -19 -b=61 -b.Check 19² +b²=(61 -b)²361 +b²=3721 -122b +b²361=3721 -122b122b=3721 -361=3360b=3360/122=1680/61≈27.541. Not integer.a=19, no.a=21:b from22 to (80 -21)/2=29.5→29.c=80 -21 -b=59 -b.Check 21² +b²=(59 -b)²441 +b²=3481 -118b +b²441=3481 -118b118b=3481 -441=3040b=3040/118=1520/59≈25.762. Not integer.a=21, no.a=22:b from23 to (80 -22)/2=29.c=80 -22 -b=58 -b.Check 22² +b²=(58 -b)²484 +b²=3364 -116b +b²484=3364 -116b116b=3364 -484=2880b=2880/116=720/29≈24.827. Not integer.a=22, nope.a=23:b from24 to (80 -23)/2=28.5→28.c=80 -23 -b=57 -b.Check 23² +b²=(57 -b)²529 +b²=3249 -114b +b²529=3249 -114b114b=3249 -529=2720b=2720/114=1360/57≈23.859. But b needs to be at least 24, so invalid.a=23, no.a=25:b from26 to (80 -25)/2=27.5→27.c=80 -25 -b=55 -b.Check 25² +b²=(55 -b)²625 +b²=3025 -110b +b²625=3025 -110b110b=3025 -625=2400b=2400/110=240/11≈21.818. But b has to be at least 26, invalid.a=25, nope.a=26:b from27 to (80 -26)/2=27.But b must be greater than a=26, so b=27.c=80 -26 -27=27.But then sides 26,27,27. Not a right-angled triangle. Also, 26² +27²=676 +729=1405, and 27²=729, which doesn't equal. So invalid.Thus, after checking all possible a from 1 to 26, the only valid solution is when a=16, b=30, c=34.Therefore, the sides are 16, 30, and 34.Final AnswerThe lengths of the three sides are boxed{16}, boxed{30}, and boxed{34}.</think>Given a right-angled triangle with side lengths as positive integers and a perimeter of 80, we need to find the lengths of the three sides. We start by noting that the sides (a), (b), and (c) (where (c) is the hypotenuse) must satisfy the Pythagorean theorem (a^2 + b^2 = c^2) and the perimeter condition (a + b + c = 80).We use the formula for generating Pythagorean triples: for positive integers (m) and (n) where (m > n), the sides are given by (a = m^2 - n^2), (b = 2mn), and (c = m^2 + n^2). The perimeter is then (2m(m + n)). Setting this equal to 80, we solve for (m) and (n):[2m(m + n) = 80 implies m(m + n) = 40]Factoring 40, we find possible pairs ((m, m + n)) and check for valid triples. The valid pair is (m = 5) and (n = 3), leading to the sides:[a = 5^2 - 3^2 = 16, quad b = 2 cdot 5 cdot 3 = 30, quad c = 5^2 + 3^2 = 34]We verify the perimeter:[16 + 30 + 34 = 80]Additionally, we check the Pythagorean theorem:[16^2 + 30^2 = 256 + 900 = 1156 = 34^2]Thus, the lengths of the three sides are (boxed{16}), (boxed{30}), and (boxed{34}).]

🔑:4. 30,16,34.Let the three sides of a right triangle be a, b, c. From the problem, we have a^{2}+b^{2}=c^{2} and a+b+c=80.Substituting c=80-a-b into a^{2}+b^{2}=c^{2}, we get6400-80 a-80 b+a b=3200,which simplifies to (80-a)(80-b)=3200.Since 0<a, b<80, we have (80-a)(80-b) = 50 times 64.From 80-a=50, 80-b=64, we get a=30, b=16.Thus, c=34.

🧠:[<think>Okay, so I need to convert 67 into a binary number. The options are A: 110000, B: 1011110, C: 1100001, D: 1000011. Hmm, let me think how to do this step by step.First, I remember that converting a decimal number to binary involves dividing the number by 2 and keeping track of the remainders. Each remainder corresponds to a binary digit, starting from the least significant bit (rightmost) to the most significant bit (leftmost). So maybe I can start by dividing 67 by 2 repeatedly and note the remainders.Let me try that. 67 divided by 2 is 33 with a remainder of 1. So the first remainder is 1.Then take 33 divided by 2, which is 16 with a remainder of 1. Second remainder is 1.Next, 16 divided by 2 is 8 with a remainder of 0. Third remainder is 0.8 divided by 2 is 4 with a remainder of 0. Fourth remainder is 0.4 divided by 2 is 2 with a remainder of 0. Fifth remainder is 0.2 divided by 2 is 1 with a remainder of 0. Sixth remainder is 0.1 divided by 2 is 0 with a remainder of 1. Seventh remainder is 1.So now, writing the remainders from last to first, that would be 1 followed by 0,0,0,0,1,1. Wait, let me check that again.Wait, when we divide 67 by 2, first remainder is 1. Then 33 divided by 2, remainder 1. 16 divided by 2, remainder 0. 8 divided by 2, remainder 0. 4 divided by 2, remainder 0. 2 divided by 2, remainder 0. 1 divided by 2, remainder 1. So the remainders, starting from the last division to the first, are 1,0,0,0,0,1,1. So arranging those from last to first, meaning the first remainder is the least significant bit (rightmost) and the last remainder is the most significant bit (leftmost). Wait, actually, when you perform the division, the first remainder is the least significant bit, and the last remainder is the most significant bit. So in order, the binary number should be written starting with the last quotient obtained (which is 1), followed by the remainders from last to first. Wait, maybe I messed up here. Let me think again.Alternatively, another method is to subtract the largest power of 2 possible and see which bits are set. Let me try that.The powers of 2 are: 64, 32, 16, 8, 4, 2, 1.67 is the number we need to convert. Let's see. The largest power of 2 less than or equal to 67 is 64 (2^6=64). So 64 is 1 followed by six zeros in binary (1000000). Subtract 64 from 67, we get 3. Now, the next power of 2 is 32, which is larger than 3, so that bit is 0. Next is 16, also larger than 3, so 0. Then 8, 0. Then 4, 0. Then 2. 2 is less than 3, so we set that bit to 1. Subtract 2 from 3, we get 1. Next power is 1, so set that bit to 1. So combining these, we have 64 + 2 + 1 = 67. So in binary, that would be 1 (for 64), 0 (32), 0 (16), 0 (8), 0 (4), 1 (2), 1 (1). So putting it together: 1000011. Wait, that's 1000011. Let's check. 64 + 2 +1 is 67. So 64 is 2^6, which is the 7th bit. So 1000011 is 7 bits. Let me count: 1 (64), 0 (32), 0 (16), 0 (8), 0 (4), 1 (2), 1 (1). Yep, that adds up to 67.Looking at the options, D is 1000011. So is that the correct answer?Wait, let me verify using the division method again to be sure. Let's write down the steps:67 divided by 2: quotient 33, remainder 1 (LSB)33 divided by 2: quotient 16, remainder 116 divided by 2: quotient 8, remainder 08 divided by 2: quotient 4, remainder 04 divided by 2: quotient 2, remainder 02 divided by 2: quotient 1, remainder 01 divided by 2: quotient 0, remainder 1 (MSB)Now, writing the remainders from MSB to LSB: the last remainder is 1 (MSB), then the previous remainders in reverse order. Wait, no. When using the division method, you take the remainders in reverse order. So the first remainder is the LSB, and the last remainder is the MSB. So the remainders we got are:1 (from 67/2), 1 (33/2), 0 (16/2), 0 (8/2), 0 (4/2), 0 (2/2), 1 (1/2). Wait, wait, let me clarify:Wait, first division step:67 /2 =33, remainder 1 (LSB)Second step: 33 /2 =16, remainder 1Third:16 /2=8, rem 0Fourth:8/2=4, rem0Fifth:4/2=2, rem0Sixth:2/2=1, rem0Seventh:1/2=0, rem1 (MSB)So the remainders in order (from first to last) are 1,1,0,0,0,0,1. But when writing the binary number, we take the remainders from last to first. Wait, is that correct?No, actually, when you divide by 2 each time, the first remainder is the least significant bit (rightmost) and the last remainder is the most significant bit (leftmost). So the order of remainders is 1 (LSB), 1,0,0,0,0,1 (MSB). Therefore, the binary number is written as 1000011. Which is D. So that matches.But let me double-check with another method. Let's calculate the value of each option:A: 110000. Let's see. The bits are 1,1,0,0,0,0. So starting from the left, the first bit is 32 (2^5), then 16 (2^4), 8,4,2,1. Wait, 2^5 is 32. So 32 +16=48. That's 48, which is less than 67. So A is 48, not correct.B:1011110. Let's compute. Starting from left: 1*(2^6) =64, 0*(32)=0, 1*(16)=16, 1*(8)=8, 1*(4)=4, 1*(2)=2, 0*(1)=0. Total: 64+16+8+4+2= 94. That's 94, which is more than 67. So B is 94, incorrect.C:1100001. Let's see: 1*(64) +1*(32) +0 +0 +0 +0 +1*(1) =64+32+1=97. That's 97, way too high. So C is 97, wrong.D:1000011. Let's compute:1*(64) +0 +0 +0 +0 +1*(2) +1*(1)=64+2+1=67. Perfect, that's correct.Therefore, the correct answer is D:1000011.But just to make sure, when I did the division method, I thought the first remainder was the LSB. Let me confirm. For example, take a smaller number, say 5. Converting 5 to binary:5/2=2 rem1 (LSB)2/2=1 rem01/2=0 rem1 (MSB)So the remainders are 1,0,1. But written as 101, which is correct. So indeed, the first remainder is the LSB, last is MSB. So 67's remainders were 1,1,0,0,0,0,1. So written as 1000011. Which is D. Yep, that's right.So the answer must be D.Final Answerboxed{D}</think>To convert 67 into a binary number, we can use the division-by-2 method and keep track of the remainders:1. ( 67 div 2 = 33 ) with remainder ( 1 ) (LSB)2. ( 33 div 2 = 16 ) with remainder ( 1 )3. ( 16 div 2 = 8 ) with remainder ( 0 )4. ( 8 div 2 = 4 ) with remainder ( 0 )5. ( 4 div 2 = 2 ) with remainder ( 0 )6. ( 2 div 2 = 1 ) with remainder ( 0 )7. ( 1 div 2 = 0 ) with remainder ( 1 ) (MSB)Writing the remainders from last to first (MSB to LSB), we get ( 1000011 ).We can also verify this by checking the binary values of the options provided:- Option A: ( 110000 ) is ( 48 )- Option B: ( 1011110 ) is ( 94 )- Option C: ( 1100001 ) is ( 97 )- Option D: ( 1000011 ) is ( 67 )Thus, the correct answer is (boxed{D}).,<think>Okay, so I need to convert the number 67 into binary and choose the correct answer from the options given: A is 110000, B is 1011110, C is 1100001, and D is 1000011. Hmm, let me think about how to do this step by step. I remember that converting a decimal number to binary involves dividing the number by 2 repeatedly and keeping track of the remainders. Let me try that.First, I'll start with 67. When I divide 67 by 2, how much is that? Well, 2 times 33 is 66, so 67 divided by 2 is 33 with a remainder of 1. So the first remainder is 1. Then I take the quotient, which is 33, and divide that by 2. 33 divided by 2 is 16 with a remainder of 1 again. So the next remainder is 1. Then, 16 divided by 2 is 8 with no remainder, so the next remainder is 0. Dividing 8 by 2 gives 4 with a remainder of 0. Then 4 divided by 2 is 2, remainder 0. Then 2 divided by 2 is 1, remainder 0. Finally, 1 divided by 2 is 0 with a remainder of 1. So, writing the remainders from last to first, that should be 1 followed by the remainders we got: 0, 0, 0, 0, 1, 1. Wait, let me check that again. Let me list out all the divisions and remainders step by step:1. 67 ÷ 2 = 33, remainder 12. 33 ÷ 2 = 16, remainder 13. 16 ÷ 2 = 8, remainder 04. 8 ÷ 2 = 4, remainder 05. 4 ÷ 2 = 2, remainder 06. 2 ÷ 2 = 1, remainder 07. 1 ÷ 2 = 0, remainder 1So the remainders, in order from last to first, are the remainders from step 7 down to step 1. That would be 1 (from step 7), 0 (step 6), 0 (step 5), 0 (step 4), 0 (step 3), 1 (step 2), 1 (step 1). So putting them together: 1 0 0 0 0 1 1. Which is 1000011. Let me verify that.Alternatively, maybe another way to check is to calculate the binary value by adding up powers of 2. Let's see. The binary digits represent powers of 2. Starting from the right, the first digit is 2^0, then 2^1, up to 2^n. So for the binary number 1000011, starting from the right:Positions (from right): 0 to 6Digits: 1 (position 6), 0 (5), 0 (4), 0 (3), 0 (2), 1 (1), 1 (0)Wait, actually, in the binary number, the rightmost digit is position 0. So let me write the binary number correctly:Binary digits: 1 0 0 0 0 1 1From left to right, positions 6 to 0:Position 6: 1 (2^6 = 64)Position 5: 0 (32)Position 4: 0 (16)Position 3: 0 (8)Position 2: 0 (4)Position 1: 1 (2)Position 0: 1 (1)Adding up the values where there are 1s: 64 + 2 + 1 = 67. That works out. So the binary representation is indeed 1000011, which is option D.Wait, but let me check the other options just to be sure. Maybe I made a mistake somewhere.Option B is 1011110. Let's convert that to decimal. Starting from the right:Positions 0 to 6:Digit positions (from right):Position 6: 1 (64)Position 5: 0 (32)Position 4: 1 (16)Position 3: 1 (8)Position 2: 1 (4)Position 1: 1 (2)Position 0: 0 (1)Wait, actually, when converting a binary number to decimal, each digit corresponds to 2^position, starting from 0 on the right. So the binary number 1011110 is:From right to left:Position 0: 0 (2^0 = 1) -> 0*1 = 0Position 1: 1 (2^1 = 2) -> 1*2 = 2Position 2: 1 (4) -> 1*4 = 4Position 3: 1 (8) -> 1*8 = 8Position 4: 1 (16) -> 1*16 = 16Position 5: 0 (32) -> 0*32 = 0Position 6: 1 (64) -> 1*64 = 64Adding those up: 64 + 0 + 16 + 8 + 4 + 2 + 0 = 64 + 16 is 80, 80 +8 is 88, 88 +4 is 92, 92 +2 is 94. So 1011110 is 94, which is not 67. So option B is 94, which is wrong.Option C is 1100001. Let's convert that:From right to left:Positions 0-6:Position 0: 1 (1)Position 1: 0 (2)Position 2: 0 (4)Position 3: 0 (8)Position 4: 0 (16)Position 5: 1 (32)Position 6: 1 (64)Calculating: 64 + 32 + 0 + 0 + 0 + 0 + 1 = 64 +32 is 96 +1 is 97. So 1100001 is 97, which is not 67. So option C is 97.Option A is 110000. Let's check that. Since it's 6 digits, the positions are 0-5:From right to left:Position 5: 1 (32)Position 4: 1 (16)Position 3: 0 (8)Position 2: 0 (4)Position 1: 0 (2)Position 0: 0 (1)Wait, actually, the binary number is 110000. Let's parse it correctly. From left to right: 1 1 0 0 0 0. So positions 5 to 0:Position 5: 1 (32)Position 4: 1 (16)Position 3: 0 (8)Position 2: 0 (4)Position 1: 0 (2)Position 0: 0 (1)Adding up: 32 +16 =48. So 110000 is 48, which is not 67. So option A is 48.So the only correct one is option D: 1000011, which is 67. Therefore, the answer is D.Wait, just to make sure I didn't make a mistake in my initial division method. Let me go through the division steps again.67 divided by 2: 33 with remainder 1.33 divided by 2: 16 with remainder 1.16 divided by 2: 8 with remainder 0.8 divided by 2: 4 with remainder 0.4 divided by 2: 2 with remainder 0.2 divided by 2: 1 with remainder 0.1 divided by 2: 0 with remainder 1.So the remainders, from last to first (i.e., from the last division up), are 1, 0, 0, 0, 0, 1, 1. So writing them in order: 1 0 0 0 0 1 1. Which is 1000011, yes. So that's correct.Alternatively, if we were to write the binary number as the remainders from first to last, we would get 1100011, but that's not the case because the first remainder is the least significant bit. So you have to reverse the order. Let me confirm.Each division step gives a remainder that is the next bit in the binary number, starting from the least significant bit (rightmost). So the first remainder is the rightmost bit, and the last remainder is the leftmost bit. Therefore, when you write the binary number, you need to take the remainders in reverse order.So, the remainders were:1 (from 67 ÷2),1 (33 ÷2),0 (16 ÷2),0 (8 ÷2),0 (4 ÷2),0 (2 ÷2),1 (1 ÷2).Wait, wait, hold on. Wait, when you divide 1 by 2, you get quotient 0 and remainder 1. So the remainders, in the order they are obtained, are:1 (step1),1 (step2),0 (step3),0 (step4),0 (step5),0 (step6),1 (step7).So if you write them in the order from first remainder to last, that would be 1 1 0 0 0 0 1. But that's 1100001, which is option C. Wait, that's conflicting with my previous result. Wait, no. Wait, hold on. Let me clarify.When converting decimal to binary, the method is:Divide the number by 2, get the remainder. Then divide the quotient by 2, get the remainder, and so on, until the quotient is 0. The binary number is the remainders read in reverse order of how they were obtained.So the first remainder is the least significant bit (rightmost), and the last remainder is the most significant bit (leftmost).So let's list the steps again:Step 1: 67 ÷2 =33, remainder 1 (LSB)Step 2: 33 ÷2 =16, remainder 1Step 3:16 ÷2 =8, remainder 0Step4:8 ÷2=4, remainder 0Step5:4 ÷2=2, remainder0Step6:2 ÷2=1, remainder0Step7:1 ÷2=0, remainder1 (MSB)So the remainders in order from first to last (step1 to step7):1,1,0,0,0,0,1To get the binary number, we reverse this order:1 (from step7),0 (step6),0 (step5),0 (step4),0 (step3),1 (step2),1 (step1)Wait, that would be 1 followed by 0,0,0,0,1,1. So 1000011. Yes, that's correct. So the first remainder is the LSB, the last remainder is the MSB. So you reverse the order of the remainders. Therefore, the correct binary number is 1000011, which is option D. Therefore, my initial conclusion was correct. So the answer is D.I think earlier when I thought about remainders in reverse order, I confused myself for a moment, but upon rechecking, it's clear. So the correct answer is D:1000011.Final Answerboxed{D}</think>To convert 67 into a binary number, we use the method of repeatedly dividing by 2 and keeping track of the remainders:1. ( 67 div 2 = 33 ) with remainder ( 1 )2. ( 33 div 2 = 16 ) with remainder ( 1 )3. ( 16 div 2 = 8 ) with remainder ( 0 )4. ( 8 div 2 = 4 ) with remainder ( 0 )5. ( 4 div 2 = 2 ) with remainder ( 0 )6. ( 2 div 2 = 1 ) with remainder ( 0 )7. ( 1 div 2 = 0 ) with remainder ( 1 )Reading the remainders from last to first, we get ( 1000011 ).To verify, we convert ( 1000011 ) back to decimal:- The binary number ( 1000011 ) corresponds to ( 2^6 + 2^1 + 2^0 = 64 + 2 + 1 = 67 ).Checking the other options:- Option B ( 1011110 ) converts to 94.- Option C ( 1100001 ) converts to 97.- Option A ( 110000 ) converts to 48.Thus, the correct answer is (boxed{D}).]

🔑:67÷2=33 remainder 1 33÷2=16 remainder 1 16÷2=8 remainder 0 8÷2=4 remainder 0 4÷2=2 remainder 0 2÷2=1 remainder 0 1÷2=0 remainder 1 Therefore, 67_{(10)}=1000011_{(2)} Hence, the correct option is boxed{D}.

🧠:[<think>Okay, let me try to figure out this problem step by step. Hmm, so we have Xiaoming and Xiaoliang, both stamp collectors. Xiaoming exchanges two stamps, each with a face value of 1 yuan 6 jiao, for Xiaoliang's stamps that have a face value of 2 jiao each. The exchange is based on the face value. Before the exchange, Xiaoliang had 5 times as many stamps as Xiaoming. After the exchange, Xiaoliang has 3 times as many stamps as Xiaoming. We need to find the total number of stamps they have together. First, let me make sure I understand the currency units here. In China, 1 yuan is equal to 10 jiao, so 1 yuan 6 jiao would be 16 jiao. Therefore, each stamp that Xiaoming is exchanging is 16 jiao, and Xiaoliang's stamps are 2 jiao each. The exchange is based on face value, which means that the total value of the stamps exchanged should be equal. So Xiaoming gives two stamps of 16 jiao each, so that's 2 * 16 = 32 jiao. In return, Xiaoliang gives him stamps worth 32 jiao in total. Since each of Xiaoliang's stamps is 2 jiao, he must give 32 / 2 = 16 stamps. So in the exchange, Xiaoming gives away 2 stamps and receives 16 stamps. Conversely, Xiaoliang gives away 16 stamps and receives 2 stamps. Let me denote the number of stamps Xiaoming had before the exchange as M, and Xiaoliang's as L. According to the problem, before the exchange, Xiaoliang had 5 times as many stamps as Xiaoming. So:L = 5MAfter the exchange, Xiaoming's number of stamps becomes M - 2 + 16 = M + 14Xiaoliang's number of stamps becomes L - 16 + 2 = L - 14After the exchange, Xiaoliang has 3 times as many stamps as Xiaoming, so:L - 14 = 3(M + 14)Now, we can substitute L with 5M from the first equation into the second equation:5M - 14 = 3(M + 14)Let me solve this equation step by step.First, expand the right side:5M - 14 = 3M + 42Subtract 3M from both sides:2M - 14 = 42Add 14 to both sides:2M = 56Divide both sides by 2:M = 28So Xiaoming had 28 stamps before the exchange. Then Xiaoliang had L = 5M = 5 * 28 = 140 stamps before the exchange.After the exchange, Xiaoming has 28 + 14 = 42 stamps, and Xiaoliang has 140 - 14 = 126 stamps. Let me check if 126 is indeed three times 42: 3 * 42 = 126. Yep, that works.Therefore, the total number of stamps they have together is 42 + 126 = 168.Wait, but let me double-check my calculations to make sure I didn't make a mistake. Let's go through it again.Original numbers:Xiaoming: M = 28Xiaoliang: L = 140After exchange:Xiaoming gives 2, receives 16: 28 - 2 + 16 = 42Xiaoliang gives 16, receives 2: 140 - 16 + 2 = 126126 divided by 42 is 3, which matches the problem statement. So the total is 42 + 126 = 168. That seems correct. But let me check if I set up the equations correctly. The key was setting L = 5M initially, then after exchange, Xiaoliang has L - 16 + 2 = L -14, and Xiaoming has M +14. Then the ratio after exchange is 3:1. So substituting gives 5M -14 = 3(M +14), leading to M=28. Yep, that's right.Alternatively, maybe I should check if the problem was referring to the total face value instead of the number of stamps, but the problem states "based on the face value of the stamps," which probably means that the exchange is such that the total face values are equal. Which I accounted for: Xiaoming gives 2 * 16 jiao = 32 jiao, so Xiaoliang gives 32 jiao worth of 2 jiao stamps, which is 16 stamps. So that part is correct.Therefore, I think 168 is the correct total number of stamps after the exchange. However, the problem says "the total number of stamps the two have is," but it doesn't specify before or after the exchange. Wait, the problem says "Therefore, the total number of stamps the two have is..." Hmm, the problem statement doesn't specify if it's asking for before or after. Let me check the problem again."Xiaoming exchanged two stamps... After the exchange, Xiaoliang had 3 times as many stamps as Xiaoming. Therefore, the total number of stamps the two have is ______."Wait, the problem is phrased as a statement, not a question. The problem is presented as a problem where after these conditions, the total is to be found. But when it says "the total number of stamps the two have is," it's a bit ambiguous. But in Chinese problems like this, usually, the total number after the exchange is considered, unless specified otherwise. But let me check.Wait, in the problem statement, the exchange is part of the problem's narrative. The problem says "Before the exchange, Xiaoliang had 5 times... After the exchange, Xiaoliang had 3 times..." So the question is asking for the total number after the exchange? Because the problem statement is leading up to the conclusion after the exchange. Let me check:Original problem:"Xiaoming and Xiaoliang are two stamp collectors. Xiaoming exchanged two stamps with a face value of 1 yuan 6 jiao each for Xiaoliang's stamps with a face value of 2 jiao each (based on the face value of the stamps). Before the exchange, Xiaoliang had 5 times as many stamps as Xiaoming. After the exchange, Xiaoliang had 3 times as many stamps as Xiaoming. Therefore, the total number of stamps the two have is ______."So the "therefore" refers to the total after the exchange. Because the problem mentions the state before and after the exchange, and the last condition is after the exchange. Therefore, the answer should be 42 + 126 = 168. However, let me check the problem again to be sure. Wait, but sometimes in these problems, they might be asking for the total before the exchange. But in Chinese math problems, when they talk about "the total number of stamps the two have is," and they give conditions before and after the exchange, sometimes they might be referring to the total before exchange. Wait, but let me think. Let me check the problem again.The problem states:Before the exchange: Xiaoliang had 5 times as many stamps as Xiaoming.After the exchange: Xiaoliang had 3 times as many stamps as Xiaoming.Therefore, the total number of stamps the two have is...So, since the problem is setting up the scenario where the exchange happened, and after that, the ratio is 3:1. The "therefore" would be after the exchange. Because the problem is leading up to the conclusion after the exchange, so the answer is 168.But just to be thorough, let's check what the total was before the exchange. Xiaoming had 28, Xiaoliang had 140. Total was 28 + 140 = 168. Wait a second! That's the same number as after the exchange. Because in the exchange, Xiaoming gives away 2 stamps and receives 16, net gain of 14. Xiaoliang gives away 16 and receives 2, net loss of 14. So the total number remains the same: 28 + 140 = 168 before, 42 + 126 = 168 after. So the total number of stamps is 168 regardless. Therefore, the answer is 168.Therefore, regardless of whether it's before or after, the total is 168. So the problem didn't specify, but since the total number doesn't change during the exchange, the answer is 168. So that's the answer.Final AnswerThe total number of stamps the two have is boxed{168}.</think>Let ( M ) be the number of stamps Xiaoming had before the exchange, and ( L ) be the number of stamps Xiaoliang had before the exchange. According to the problem, before the exchange, Xiaoliang had 5 times as many stamps as Xiaoming:[L = 5M]Xiaoming exchanged two stamps each with a face value of 1 yuan 6 jiao (16 jiao each) for Xiaoliang's stamps each with a face value of 2 jiao. The total face value of the stamps Xiaoming exchanged is:[2 times 16 text{ jiao} = 32 text{ jiao}]In return, Xiaoliang gave Xiaoming stamps worth 32 jiao, which is:[frac{32 text{ jiao}}{2 text{ jiao per stamp}} = 16 text{ stamps}]After the exchange, Xiaoming's number of stamps becomes:[M - 2 + 16 = M + 14]Xiaoliang's number of stamps becomes:[L - 16 + 2 = L - 14]After the exchange, Xiaoliang had 3 times as many stamps as Xiaoming:[L - 14 = 3(M + 14)]Substituting ( L = 5M ) into the equation:[5M - 14 = 3(M + 14)]Solving for ( M ):[5M - 14 = 3M + 42 2M - 14 = 42 2M = 56 M = 28]Thus, Xiaoming had 28 stamps before the exchange, and Xiaoliang had:[L = 5 times 28 = 140]After the exchange, Xiaoming has:[28 + 14 = 42]stamps, and Xiaoliang has:[140 - 14 = 126]stamps. The total number of stamps they have together is:[42 + 126 = 168]Since the total number of stamps remains the same before and after the exchange, the total number of stamps the two have is:[boxed{168}],<think>Alright, let's tackle this problem step by step. Hmm, so Xiaoming and Xiaoliang are exchanging stamps based on face value. The problem states that Xiaoming exchanged two stamps each with a face value of 1 yuan 6 jiao for Xiaoliang's stamps with a face value of 2 jiao each. Before the exchange, Xiaoliang had 5 times as many stamps as Xiaoming. After the exchange, Xiaoliang had 3 times as many stamps as Xiaoming. We need to find the total number of stamps the two have together.First, let's make sure I understand the units here. In China, 1 yuan is equal to 10 jiao, right? So 1 yuan 6 jiao would be 16 jiao. That means Xiaoming is exchanging two stamps each worth 16 jiao. So the total face value he's exchanging is 2 stamps * 16 jiao = 32 jiao. He's exchanging this for Xiaoliang's stamps which are 2 jiao each. So, how many stamps does Xiaoliang give to Xiaoming? Since it's based on face value, the total value should be equal. So, if Xiaoming gives 32 jiao worth of stamps, Xiaoliang should give back 32 jiao worth of his 2 jiao stamps. That would be 32 jiao / 2 jiao per stamp = 16 stamps. So Xiaoming gives away 2 stamps and receives 16 stamps from Xiaoliang.Wait, but the problem says "exchanged two stamps... for Xiaoliang's stamps". So maybe Xiaoming gives two stamps and gets some number of Xiaoliang's stamps in return, but based on the face value. So the value of the stamps exchanged should be equal. So Xiaoming's two stamps are worth 16 jiao each, so total 32 jiao. Xiaoliang's stamps are 2 jiao each, so he needs to give 32 jiao / 2 jiao per stamp = 16 stamps. So Xiaoming loses 2 stamps and gains 16, so net gain of 14. Xiaoliang loses 16 stamps and gains 2, so net loss of 14. Let me confirm: yes, that seems right.So, let's denote variables. Let me set:Let’s let M be the number of stamps Xiaoming had before the exchange, and L be the number Xiaoliang had. The problem states that before the exchange, Xiaoliang had 5 times as many stamps as Xiaoming. So:L = 5MAfter the exchange, Xiaoming’s stamps become M - 2 + 16 = M + 14Xiaoliang’s stamps become L - 16 + 2 = L - 14After the exchange, Xiaoliang had 3 times as many stamps as Xiaoming. So:L - 14 = 3(M + 14)We already know that L = 5M, so substitute into the second equation:5M - 14 = 3(M + 14)Let’s solve this:5M - 14 = 3M + 42Subtract 3M from both sides:2M - 14 = 42Add 14 to both sides:2M = 56Divide by 2:M = 28Then L = 5M = 5*28 = 140Therefore, before the exchange, Xiaoming had 28 stamps, Xiaoliang had 140. After the exchange, Xiaoming has 28 +14 = 42, Xiaoliang has 140 -14 = 126. Check if 126 is 3 times 42: yes, 3*42=126. So that works.Therefore, the total number of stamps they have is 42 + 126 = 168. But wait, the problem says "the total number of stamps the two have is" – but does it mean before or after the exchange? The problem says "Therefore, the total number of stamps the two have is __." Since it's asking after the exchange, because the exchange is part of the problem statement leading to the conclusion. Let me check.Wait, the problem doesn't specify whether it's before or after. Wait, let's read again:"Before the exchange, Xiaoliang had 5 times as many stamps as Xiaoming. After the exchange, Xiaoliang had 3 times as many stamps as Xiaoming. Therefore, the total number of stamps the two have is ______."So "Therefore", based on the information given, leading to the conclusion. The problem is likely asking for the total after the exchange, since the exchange changes the ratio. But let's check both.Before exchange: 28 +140=168After exchange:42 +126=168Oh, interesting! The total number remains the same because they just exchanged stamps. So the total number is 168 regardless. So the answer is 168. So that's the total number of stamps they have.But let me verify once again to make sure.Let me recast the problem:Let M = Xiaoming's original stamps, L = Xiaoliang's original stamps.Given:1) L = 5M2) After exchange: Xiaoming gives 2 stamps, receives 16. So his new count: M -2 +16 = M +14Xiaoliang gives 16, receives 2. His new count: L -16 +2 = L -14Given that after exchange:L -14 = 3*(M +14)So substitute L =5M:5M -14 =3M +422M =56M=28, L=140Total stamps: 28 +140=168After exchange, 42 +126=168. So total is same. So answer is 168. Therefore, boxed answer: boxed{168}Final Answerboxed{168}</think>Let ( M ) be the number of stamps Xiaoming had before the exchange, and ( L ) be the number Xiaoliang had. According to the problem, before the exchange, Xiaoliang had 5 times as many stamps as Xiaoming:[L = 5M]Xiaoming exchanged two stamps with a face value of 1 yuan 6 jiao each (16 jiao each) for Xiaoliang's stamps with a face value of 2 jiao each. The total face value of Xiaoming's stamps is ( 2 times 16 = 32 ) jiao. Xiaoliang gave him ( 32 div 2 = 16 ) stamps in exchange. Therefore, after the exchange:- Xiaoming's stamps: ( M - 2 + 16 = M + 14 )- Xiaoliang's stamps: ( L - 16 + 2 = L - 14 )After the exchange, Xiaoliang had 3 times as many stamps as Xiaoming:[L - 14 = 3(M + 14)]Substituting ( L = 5M ) into the equation:[5M - 14 = 3(M + 14)]Solving for ( M ):[5M - 14 = 3M + 42 2M - 14 = 42 2M = 56 M = 28]Then, ( L = 5M = 5 times 28 = 140 ).The total number of stamps before the exchange is:[M + L = 28 + 140 = 168]After the exchange, the total number of stamps remains the same because they only exchanged stamps:[(M + 14) + (L - 14) = 42 + 126 = 168]Thus, the total number of stamps the two have is (boxed{168}).]

🔑:【Analysis】First, according to 1 yuan 6 jiao =16 jiao, it can be obtained that Xiaoming used two stamps with a face value of 1 yuan 6 jiao to exchange for 16 stamps with a face value of 2 jiao from Xiaoliang, so after the exchange, Xiaoming's stamps increased by 14 (16-2=14) pieces; then, according to the fact that before the exchange, the number of stamps the two people had was 6 (5+1=6) times the number of Xiaoming's stamps, and after the exchange, the number of stamps the two people had was 4(3+1=4) times the number of Xiaoming's stamps, so the 2(6-4=2) times the number of Xiaoming's stamps before the exchange is 56(14 times 4=56) pieces, from which the number of stamps Xiaoming had before the exchange can be determined, and thus the total number of stamps the two people have can be found.【Solution】Solution: 1 yuan 6 jiao =16 jiaoThe number of stamps Xiaoming has after the exchange increased by:16 times 2 div 2-2=16 - 2=14 (pieces)The number of stamps Xiaoming had before the exchange is:14 times 4 div[(5+1)-(3+1)]=56 div 2=28 (pieces)The total number of stamps the two people have:begin{array}{l}28 times(5+1) =28 times 6 =168 text { (pieces) }end{array}Answer: The two people have a total of 168 stamps.The answer is: 168.

🧠:[<think>Okay, let me try to figure out this problem. So, the factory has improved its assembly rate by 2/3 after the technological transformation. We need to find out how many cars they can assemble per hour now, after the transformation. The problem says that after the transformation, assembling 40 cars takes 2 hours less than assembling 30 cars before the transformation. Hmm, let me parse that again. So, time taken to assemble 40 cars now is 2 hours less than time taken to assemble 30 cars before. Let me denote the original assembly rate as, let's say, x cars per hour. Then, after the transformation, the rate becomes x plus 2/3 of x, right? Because it's increased by 2/3. So, the new rate should be x*(1 + 2/3) = (5/3)x. Wait, let me check. If the rate increases by 2/3, then new rate is original rate plus 2/3 of original rate. So yes, 1 + 2/3 = 5/3, so new rate is (5/3)x. Got that.Now, time taken to assemble 30 cars before transformation would be 30 divided by x, since time = number of cars / rate. Similarly, time taken to assemble 40 cars after transformation is 40 divided by (5/3)x. According to the problem, this new time is 2 hours less than the old time. So, equation would be:30/x - 40/( (5/3)x ) = 2Let me write that down step by step. Original time for 30 cars: 30/xNew time for 40 cars: 40 / ( (5/3)x )Difference is 2 hours, so original time minus new time equals 2. So:30/x - 40/( (5/3)x ) = 2Let me compute 40 divided by (5/3)x. Dividing by a fraction is same as multiplying by reciprocal, so 40 * (3/(5x)) = (120)/(5x) = 24/xSo the equation becomes:30/x - 24/x = 2Which simplifies to:(30 - 24)/x = 26/x = 2Therefore, solving for x:6 = 2xx = 3So original rate was 3 cars per hour. Then after transformation, the new rate is (5/3)x = 5/3 * 3 = 5 cars per hour. Wait, that seems straightforward, but let me verify. Original rate: 3 cars per hour. Time to assemble 30 cars: 30 / 3 = 10 hours.After transformation, rate is 5 cars per hour. Time to assemble 40 cars: 40 / 5 = 8 hours.Difference is 10 - 8 = 2 hours, which matches the problem statement. So, the answer is 5 cars per hour after transformation. So the number is 5, which would be boxed as boxed{5}But wait, let me double-check the setup. The problem says "the time required for this factory to assemble 40 new energy vehicles after the technological transformation is 2 hours less than the time required to assemble 30 new energy vehicles before the transformation."So, original time (30 cars) minus new time (40 cars) equals 2 hours. So yes, 10 - 8 = 2. That's correct. So yes, all steps check out. The answer is 5 cars per hour after the transformation. Final AnswerThe number of cars that can be assembled per hour after the transformation is boxed{5}.</think>Let ( x ) be the original number of cars assembled per hour before the technological transformation. After the transformation, the assembly rate becomes ( frac{5}{3}x ) (since the rate increased by ( frac{2}{3} ) of the original rate).The time required to assemble 30 cars before the transformation is ( frac{30}{x} ) hours. The time required to assemble 40 cars after the transformation is ( frac{40}{frac{5}{3}x} ) hours. According to the problem, the time after the transformation is 2 hours less than the time before the transformation:[frac{30}{x} - frac{40}{frac{5}{3}x} = 2]Simplifying the equation:[frac{30}{x} - frac{40 times 3}{5x} = 2][frac{30}{x} - frac{24}{x} = 2][frac{6}{x} = 2]Solving for ( x ):[x = frac{6}{2} = 3]The original assembly rate is 3 cars per hour. After the transformation, the rate is:[frac{5}{3}x = frac{5}{3} times 3 = 5]Thus, the number of cars that can be assembled per hour after the technological transformation is (boxed{5}).,<think>Okay, let's try to solve this problem. So, the factory increased their production rate by 2/3, which means their new rate is higher than before. After the technological transformation, assembling 40 cars takes 2 hours less than assembling 30 cars before. We need to find the number of cars they can assemble per hour after the transformation. Hmm, let's break this down. Let me denote the original rate as R cars per hour. Then, after the transformation, the rate becomes R plus 2/3 of R, right? So that would be R*(1 + 2/3) = (5/3)R. So the new rate is 5/3 times the original rate.Now, the time to assemble 40 cars at the new rate is 2 hours less than the time to assemble 30 cars at the original rate. Time is equal to number of cars divided by rate. So, the time before for 30 cars would be 30/R, and the time after for 40 cars would be 40/(5/3 R). According to the problem, the new time is 2 hours less. So:30/R - 40/(5/3 R) = 2Let me write that equation down. Let me compute 40 divided by (5/3 R). Dividing by a fraction is multiplying by its reciprocal, so 40 * 3/(5 R) = (120)/(5 R) = 24/R. So the equation becomes:30/R - 24/R = 2Simplifying the left side: (30 - 24)/R = 6/R = 2. Therefore, 6/R = 2. Solving for R, we get R = 6/2 = 3. So the original rate was 3 cars per hour. Then the new rate is 5/3 of that, which is 5/3 * 3 = 5 cars per hour. So after the transformation, they can assemble 5 cars per hour.Wait, let me check again. If original rate was 3 cars per hour, then assembling 30 cars would take 10 hours. After the transformation, the rate is 5 cars per hour, so assembling 40 cars would take 8 hours. The difference is indeed 10 - 8 = 2 hours. That matches the problem statement. So the answer should be 5.But wait, just to make sure. Let me verify each step.Original rate: R = 3. After transformation: 5/3 * 3 = 5. Time for 30 cars before: 30 / 3 = 10 hours. Time for 40 cars after: 40 / 5 = 8 hours. 10 - 8 = 2 hours. Correct. So yes, 5 is the correct answer. Alternatively, maybe I made a mistake in setting up the equation? Let's see. Let me re-express the problem.Original rate: R cars/hour. New rate: R + (2/3)R = (5/3)R.Time to assemble 30 cars originally: 30/R.Time to assemble 40 cars after: 40 / (5/3 R) = 40 * 3 / (5 R) = 24/R.The time after is 2 hours less than the time before. So:Original time - new time = 2 hours.30/R - 24/R = 2.So 6/R = 2 => R = 3. Then new rate is 5/3 * 3 = 5. All steps check out. I think that's correct.Final AnswerThe number of cars that can be assembled per hour after the technological transformation is boxed{5}.</think>Let ( R ) be the original rate of assembling cars per hour. After the technological transformation, the rate becomes ( frac{5}{3}R ).The time required to assemble 30 new energy vehicles before the transformation is ( frac{30}{R} ) hours. The time required to assemble 40 new energy vehicles after the transformation is ( frac{40}{frac{5}{3}R} = frac{24}{R} ) hours. According to the problem, the time after the transformation is 2 hours less than the time before:[frac{30}{R} - frac{24}{R} = 2]Simplifying the equation:[frac{6}{R} = 2 implies R = 3]The new rate after the transformation is:[frac{5}{3}R = frac{5}{3} times 3 = 5]Thus, the number of cars that can be assembled per hour after the technological transformation is (boxed{5}).]

🔑:To solve this problem, let's denote the number of cars that the factory can assemble per hour before the technological transformation as x. After the transformation, the rate at which the factory can assemble cars increases by frac{2}{3} of its original rate. Therefore, the new rate of assembling cars per hour becomes (1+frac{2}{3})x = frac{5}{3}x.Given in the problem, the time it takes to assemble 40 new energy vehicles after the transformation is 2 hours less than the time it takes to assemble 30 new energy vehicles before the transformation. We can express this relationship as follows:[frac{40}{frac{5}{3}x} = frac{30}{x} - 2]To solve for x, we first simplify the equation:[frac{40}{frac{5}{3}x} = frac{30}{x} - 2]Multiplying both sides by frac{5}{3}x to eliminate the denominator on the left side, we get:[40 = frac{5}{3}x left( frac{30}{x} - 2 right)]Expanding the right side:[40 = 50 - frac{10}{3}x]Bringing the terms involving x to one side and the constants to the other side:[frac{10}{3}x = 10]Solving for x:[x = 10 times frac{3}{10} = 3]Therefore, the number of cars that the factory could assemble per hour before the technological transformation was 3. After the transformation, the rate becomes:[(1+frac{2}{3})x = frac{5}{3} times 3 = 5]Thus, after the technological transformation, the factory can assemble boxed{5} cars per hour.