answer:**Given:** We need to show that there exist infinitely many positive integers n such that the numbers from 1 to 3n can be arranged into a 3 times n table satisfying the following conditions: 1. The sums of numbers in each column are equal and are multiples of 6. 2. The sums of numbers in each row are equal and are multiples of 6. **Proof:** 1. Start by expressing the total sum of the numbers from 1 to 3n: [ frac{3n(3n+1)}{2}. ] 2. If the sums of the numbers in each column are equal and multiples of 6, assume the sum of any column is 6s. For such an arrangement to be possible, the sum of all numbers must be divisible by 6: [ 3n cdot 6s = 6t quad text{(where t is an integer)}. ] Thus we need: [ frac{3n(3n+1)}{2} = 6t. ] 3. Rewriting the above equation we get: [ 3n(3n+1) = 12s, ] [ n(3n+1) = 4s. ] 4. Analyzing this, we see it's possible to find values of ( n ) that satisfy this equation. For simplicity, observe the case when n = 12k + 9 for some non-negative integer k. To validate, let’s solve n: [ n(3n + 1) = 12s. ] For n = 12k + 9: [ (12k + 9)(36k + 10) = 432k^2 + 108k + 12s. ] This implies the structured form leads to feasible values. 5. First, show a specific n, say 9, can satisfy the arrangement: Constructing a 3 times 3 table for n = 3: [ begin{array}{lll} 1 & 2 & 3 2 & 3 & 1 3 & 1 & 2 end{array} + begin{array}{lll} 0 & 6 & 3 3 & 0 & 6 6 & 3 & 0 end{array} = begin{array}{lll} 1 & 8 & 6 5 & 3 & 7 9 & 4 & 2 end{array}. ] Here, each row and column adds to 15, thus: [ alpha(3) = (1, 8, 6), quad beta(3) = (5, 3, 7), quad gamma(3) = (9, 4, 2). ] 6. Construct a 3 times 9 table (A_9): Using alpha(3), beta(3), gamma(3): [ A_{9} = left{ begin{array}{ccccccc} alpha(3) & beta(3)+18 & gamma(3)+9 beta(3)+9 & gamma(3) & alpha(3)+18 gamma(3)+18 & alpha(3)+9 & beta(3) end{array} right}. ] Thus: [ begin{array}{ccccccccc} 1 & 8 & 6 & 23 & 21 & 25 & 18 & 13 & 11 14 & 12 & 16 & 9 & 4 & 2 & 19 & 26 & 24 27 & 22 & 20 & 10 & 17 & 15 & 5 & 3 & 7 end{array}. ] Here each column sums to 42, and row sums can be calculated accordingly. 7. Assume ( m in S ), now show 9m in S: Assume a 3 times m table exists such that columns sum to 6u and rows to 6v. Construct a 3 times 3m table (A_{3m}): [ A_{3m} = left{ begin{array}{ccc} alpha(m) & beta(m) + 6m & gamma(m) + 3m beta(m) + 3m & gamma(m) & alpha(m) + 6m gamma(m) + 6m & alpha(m) + 3m & beta(m) end{array} right}. ] Each column sum is: [ 6u + 9m, ] Each row sum is: [ 18v + 9m^2. ] Construct 3 times 9m (A_{9m}): [ A_{9m} = left{ begin{array}{lll} alpha(3m) & beta(3m) + 18m & gamma(3m) + 9m beta(3m) + 9m & gamma(3m) & alpha(3m) + 18m gamma(3m) + 18m & alpha(3m) + 9m & beta(3m) end{array} right}. ] Satisfying: [ 6(u+6m) ] and [ 6(9v + 18m^2). ] Thus, 9^k in S, so S is infinite. Conclusion: boxed{text{The set S is infinite}}.

answer:1. Start by analyzing the given equation: [ 4 - frac{9}{y} + frac{9}{y^2} = 0 ] 2. Multiply each term by (y^2) to eliminate the fractions: [ y^2 cdot 4 - 9y + 9 = 4y^2 - 9y + 9 = 0 ] 3. Factor the quadratic equation: [ 4y^2 - 9y + 9 = (2y - 3)^2 = 0 ] 4. Solve for (y) by setting the factored form equal to zero: [ (2y - 3)^2 = 0 implies 2y - 3 = 0 implies 2y = 3 implies y = frac{3}{2} ] 5. Substitute (y = frac{3}{2}) into (frac{3}{y}) to find the value: [ frac{3}{y} = frac{3}{frac{3}{2}} = 2 ] 6. Conclude that the value of frac{3}{y} is 2. [ 2 ] The final answer is boxed{2}

answer:Given that the lines ax+2y+1=0 and x+y-2=0 are parallel, Their slopes are equal. The slope of the line ax+2y+1=0 is -frac{a}{2}, and the slope of the line x+y-2=0 is -1. So, we have -frac{a}{2} = -1. Solving for a, we get a = 2. Therefore, the answer is boxed{2}. This problem tests our understanding of the properties of parallel lines, which have equal slopes.

answer:Let mathbf{u} = begin{pmatrix} x y end{pmatrix}. First, use the projection formula: [ operatorname{proj}_{begin{pmatrix} 3 4 end{pmatrix}} mathbf{u} = frac{mathbf{u} cdot begin{pmatrix} 3 4 end{pmatrix}}{left| begin{pmatrix} 3 4 end{pmatrix} right|^2} begin{pmatrix} 3 4 end{pmatrix} ] Compute the dot product and the magnitude squared: [ begin{pmatrix} x y end{pmatrix} cdot begin{pmatrix} 3 4 end{pmatrix} = 3x + 4y,quad left| begin{pmatrix} 3 4 end{pmatrix} right|^2 = 3^2 + 4^2 = 25 ] Then, [ operatorname{proj}_{begin{pmatrix} 3 4 end{pmatrix}} mathbf{u} = frac{3x + 4y}{25} begin{pmatrix} 3 4 end{pmatrix} = begin{pmatrix} -frac{9}{2} -6 end{pmatrix} ] This implies: [ frac{3x + 4y}{25} = -frac{9}{25} ] Therefore, [ 3x + 4y = -9 ] Solving for y in terms of x: [ 4y = -9 - 3x quad Rightarrow quad y = -frac{3}{4} x - frac{9}{4} ] Thus, the equation of the line is: [ boxed{y = -frac{3}{4} x - frac{9}{4}} ]