answer:To find all pairs of integers whose sum is equal to their product, let's denote the integers by (x) and (y). We can write the given condition as follows: [ x + y = xy. ] 1. First, manipulate the given equation to bring all terms to one side: [ x + y - xy = 0. ] 2. Now, add 1 to both sides of the equation: [ xy - x - y + 1 = 1. ] 3. Next, we can factorize the left-hand side: [ (x-1)(y-1) = 1. ] 4. We recognize that the integer solutions to the equation ((x-1)(y-1) = 1) are limited to pairs of integers whose products equal 1. The possible pairs are: [ (x-1, y-1) = (1, 1) quad text{or} quad (x-1, y-1) = (-1, -1). ] 5. Solve these pairs for (x) and (y): - For ((x-1, y-1) = (1, 1)): [ x - 1 = 1 quad text{and} quad y - 1 = 1, ] solving these, we get: [ x = 2 quad text{and} quad y = 2. ] - For ((x-1, y-1) = (-1, -1)): [ x - 1 = -1 quad text{and} quad y - 1 = -1, ] solving these, we get: [ x = 0 quad text{and} quad y = 0. ] # Conclusion: The integer pairs ((x, y)) that satisfy the condition (x + y = xy) are: [ boxed{(2, 2) text{ and } (0, 0)}. ]

answer:Given that i is the imaginary unit, we know that i^2 = -1. Therefore, i^{2018} = (i^2)^{1009} = (-1)^{1009} = -1. The conjugate of z=3+4i is overline{z}=3-4i. Thus, i^{2018}overline{z} = -1(3-4i) = -3+4i. Therefore, the correct answer is boxed{text{C}}.

answer:To determine which regular polygons can be used to tile the floor of a laboratory with no gaps, we need to investigate the properties of regular polygons and how they fit together. 1. **Formula for Interior Angle**: The interior angle theta of a regular n-sided polygon is given by: [ theta = frac{(n-2) cdot 180^circ}{n} ] 2. **Condition for Tiling**: If x polygons meet at one vertex, then the sum of internal angles at that vertex must be 360^circ: [ x cdot theta = 360^circ ] 3. **Substitute theta**: Substitute the formula for theta from step 1 into the tiling condition: [ x cdot left( frac{(n-2) cdot 180^circ}{n} right) = 360^circ ] 4. **Solve for x**: Simplifying the equation: [ x cdot left( frac{(n-2) cdot 180^circ}{n} right) = 360^circ ] [ x cdot frac{(n-2)}{n} cdot 180^circ = 360^circ ] [ x cdot frac{(n-2)}{n} = 2 ] [ x = frac{2n}{n-2} ] 5. **Condition for Whole Numbers**: For x to be an integer, frac{4}{n-2} must be an integer: [ frac{2n}{n-2} = 2 + frac{4}{n-2} ] This implies that 4/(n-2) must be an integer. 6. **Constraints for n**: - n must be greater than 2 since a polygon must have at least 3 sides. - Checking possible values of (n-2) that make 4/(n-2) an integer, we have: [ n-2 in { 1, 2, 4 } ] 7. **Possible Values for n**: Adding 2 back to each of these values, we get: - ( n - 2 = 1 Rightarrow n = 3 ) - ( n - 2 = 2 Rightarrow n = 4 ) - ( n - 2 = 4 Rightarrow n = 6 ) 8. **Conclusion**: Therefore, the regular polygons that can tile a plane without gaps are the ones with n = 3, 4, 6 sides. So the valid polygons are: - Triangles (n=3) - Squares (n=4) - Hexagons (n=6) boxed{3, 4, 6}

answer:1. **Define r with new base and exponent changes**: Given the changes, r is computed as: [ r = (4a)^{2b} ] 2. **Set up the equation with r and the product squared**: According to the problem, r equals the square of the product of a^b and x^b: [ (4a)^{2b} = (a^b x^b)^2 ] 3. **Simplify and solve for x**: Expanding the right-hand side: [ 16^b a^{2b} = a^{2b} x^{2b} ] Divide both sides by a^{2b} (assuming a neq 0 and b neq 0): [ 16^b = x^{2b} ] Take the b-th root of both sides: [ x^2 = 16 ] Therefore: [ x = 4 ] Conclusion: The new x value that satisfies the given conditions is 4. Therefore, the answer is: [ 4 ] The final answer is boxed{4}.