answer:To solve for x in this problem, set up the equation: [ frac{x}{5} = 30 + frac{x}{6} ] Subtract frac{x}{6} from both sides to simplify: [ frac{x}{5} - frac{x}{6} = 30 ] Find a common denominator for the fractions, which is 30: [ frac{6x}{30} - frac{5x}{30} = 30 ] [ frac{x}{30} = 30 ] Now, solve for x: [ x = 30 times 30 = 900 ] Thus, the number is boxed{900}.

answer:1. Let's denote the unknown costs of the drum, wife, and leopard skin by ( x ), ( y ), and ( z ) respectively. 2. According to the problem, the following system of equations can be written: [ left{ begin{aligned} 2x + 3y + z &= 111, 3x + 4y - 2z &= -8. end{aligned} right. ] 3. First, we multiply the first equation by 2 to help eliminate ( z ): [ 2(2x + 3y + z) = 2 times 111 implies 4x + 6y + 2z = 222. ] 4. Now, add this new equation to the second original equation: [ (4x + 6y + 2z) + (3x + 4y - 2z) = 222 + (-8) implies 4x + 6y + 2z + 3x + 4y - 2z = 214. ] 5. Simplify the resulting equation: [ 7x + 10y = 214. ] 6. From the equation above, we need to determine ( x ) and ( y ). Notice that: [ 7x + 10y = 230 implies 7x = 230 - 10y. ] 7. For integers ( x ) and ( y ), we need ( 7x ) to be divisible by 10. This implies: [ 230 - 10y equiv 0 pmod{10} implies 230 equiv 10y pmod{10} implies y equiv 0 pmod{1}. ] 8. Thus, ( y ) must be integers, and let's check possible values for ( x ). Firstly, ( x ) must make ( z ) even, so ( x ) must be divisible by 2: - If ( x = 30 ): [ 7(30) + 10y = 230 implies 210 + 10y = 230 implies 10y = 20 implies y = 2. ] - If ( x = 20 ): [ 7(20) + 10y = 230 implies 140 + 10y = 230 implies 10y = 90 implies y = 9. ] - If ( x = 10 ): [ 7(10) + 10y = 230 implies 70 + 10y = 230 implies 10y = 160 implies y = 16. ] 9. We verify possible solutions for ( z ) to be even (as per problem). - For ( y = 2 ), substitute into the system and notice inconsistency in the modulus constraint. - For ( y = 9 ): [ 2x + 3y + z = 111 implies 2(20) + 3(9) + z = 111 implies 40 + 27 + z = 111 implies 67 + z = 111 implies z = 44. ] Thus, ( z ) is even, meeting requirements. # Conclusion: The required costs are: [ boxed{20, 9, 44} ]

answer:1. **Define the sides of the squares:** Let the sides of the squares be denoted by (a_i) for (i = 1, 2, ldots, N), where (N) is the number of squares. The total area of these squares is given by: [ sum_{i=1}^N a_i^2 = 4 ] 2. **Case when a square is larger than the unit square:** If there exists any (a_k > 1), then the (k)-th square alone can cover the unit square of side 1. This is because the area of the (k)-th square, (a_k^2), would be greater than 1. 3. **Case when all squares are smaller than the unit square:** Assume (a_i < 1) for all (i). Each (a_i) must belong to some interval ((2^{-k_i}, 2^{-k_i+1})), i.e., (2^{-k_i} le a_i < 2^{-k_i+1}), where (k_i) are positive integers for all (i). 4. **Decrease the size of each square:** Define (b_i = frac{1}{2^{k_i}}). Then, the area of each square with side (b_i) is: [ b_i^2 = left(frac{1}{2^{k_i}}right)^2 = frac{1}{4^{k_i}} ] Since (1 le frac{a_i}{b_i} < 2), the area of each square with side (a_i) is at most 4 times the area of the square with side (b_i): [ a_i^2 le 4 b_i^2 ] Therefore, the total area of the squares with sides (b_i) is: [ sum_{i=1}^N b_i^2 ge frac{1}{4} sum_{i=1}^N a_i^2 = frac{1}{4} cdot 4 = 1 ] 5. **Tiling the unit square:** We need to show that the squares with sides (b_i) can tile the unit square. Divide the unit square into 4 smaller squares of side (frac{1}{2}). Place the squares with side (frac{1}{2}) (if they exist) in these smaller squares. For the remaining untiled squares of side (frac{1}{2}), divide each into 4 smaller squares of side (frac{1}{4}) and place the squares with side (frac{1}{4}) (if they exist). Continue this process for (k = 3, 4, ldots), placing squares with side (frac{1}{2^k}) in the untiled squares of side (frac{1}{2^{k-1}}), dividing them into 4 equal smaller squares. 6. **Conclusion:** Since the sum of the areas of the squares with sides (b_i) is at least 1, at some step, the unit square will be fully covered. Increasing the side of each square from (b_i) to (a_i) will still cover the unit square, as the total area remains the same. (blacksquare)

answer:Given that iz=4+3i, we want to find z. We can start by expressing z in terms of the given equation: [z = frac{4+3i}{i}] To simplify this expression, we multiply both the numerator and the denominator by -i (the complex conjugate of i) to get rid of the imaginary unit in the denominator: [z = frac{(4+3i)(-i)}{-i^2}] Since i^2 = -1, we can simplify the denominator to 1: [z = frac{-4i - 3i^2}{1}] Substituting i^2 with -1: [z = -4i - 3(-1)] [z = 3 - 4i] Therefore, the correct answer is boxed{C}.