answer:# Problem: On a 5 times 5 grid, Petya marks several cells. Vasya wins if he can cover all the marked cells with non-overlapping and non-out-of-bounds L-shaped pieces of three cells each (the "L" shapes must align with the gridlines). What is the smallest number of cells Petya must mark to ensure that Vasya cannot win? 1. **Initial Example**: Petya can mark cells so that Vasya can cover at most one of the marked cells with any three-cell "L" shape. 2. **Observing the Arrangement**: Let’s consider a specific arrangement on a 5 times 5 grid. If Petya marks cells in a strategic way, he can ensure that any "L" piece placed by Vasya will only cover one marked cell or fewer.  3. **Calculation of Coverage**: - The grid has 25 cells in total. - You can potentially place up to 9 L-shaped pieces each covering 3 cells, but this would cover 27 cells which is greater than 25. - Therefore, it is impossible to fit 9 L-shaped pieces without any overlap or going out-of-bounds. 4. **Strategic Marking**: If Petya marks fewer than 9 cells: - Let's say Petya marks 8 cells. In that case, there would still be 25 - 8 = 17 empty cells left. - Vasya can try to place 8 L-pieces (covering 24 cells in total), but the crucial point here is overlapping. - One marked cell must coincide within some L-shape's region making some other regions unable to form perfect Ls. 5. **Conclusion**: - The minimum effective number of cells Petya must mark to prevent Vasya from covering them all is 9. - This ensures that there are insufficient spots to perfectly place 9 L-shaped pieces without overlap or extending out of the grid. # Conclusion: The smallest number of cells Petya needs to mark to guarantee that Vasya cannot win is: [ boxed{9} ]

answer:The problem requires simplifying frac{sqrt{3}cdotsqrt{5}cdotsqrt{6}}{sqrt{7}cdotsqrt{8}cdotsqrt{9}}. - Firstly, simplify each square root where possible. sqrt{8} = sqrt{4 cdot 2} = 2sqrt{2}, and sqrt{9} = 3. - After substitution, the expression becomes frac{sqrt{3}cdotsqrt{5}cdotsqrt{6}}{sqrt{7}cdot 2sqrt{2}cdot 3}. - Simplify further where possible. sqrt{6} = sqrt{2}cdotsqrt{3}, allowing sqrt{3} from the numerator and denominator to cancel, and sqrt{2} to partly cancel. This leaves us with frac{sqrt{5}}{2cdotsqrt{7}cdot3}. - The resulting expression is frac{sqrt{5}}{6sqrt{7}}. To rationalize the denominator, multiply the numerator and denominator by sqrt{7}, giving frac{sqrt{5}cdotsqrt{7}}{6cdot7}. - Simplify to frac{sqrt{35}}{42}. Conclusion: The solution to the problem is boxed{frac{sqrt{35}}{42}}.

answer:First, let's find out how much work A and B can do together in one day. A can complete the work in 30 days, so A's work rate is ( frac{1}{30} ) of the work per day. B can complete the work in 20 days, so B's work rate is ( frac{1}{20} ) of the work per day. Together, their combined work rate per day is: ( frac{1}{30} + frac{1}{20} = frac{2}{60} + frac{3}{60} = frac{5}{60} = frac{1}{12} ) This means that together, A and B can complete the work in 12 days. Now, let's find out the ratio of their work rates to divide the payment accordingly: A's work rate : B's work rate = ( frac{1}{30} : frac{1}{20} ) To compare these, we need to have a common denominator, which is 60 in this case: A's work rate : B's work rate = ( frac{2}{60} : frac{3}{60} ) This simplifies to: A's work rate : B's work rate = 2 : 3 The total payment is Rs. 1000, and we need to divide it in the ratio of 2:3 for A and B, respectively. The sum of the ratio parts is 2 + 3 = 5 parts. The value of each part is ( frac{1000}{5} = Rs. 200 ) So, B's share (which is 3 parts) is: 3 parts * Rs. 200 per part = Rs. 600 Therefore, B's share of the payment is Rs. boxed{600} .

answer:Given the complex numbers z_1 = -1 + ai and z_2 = b + 2i, First, we find z_1 - z_2 = -1 + ai - (b + 2i) = -1 - b + (a - 2)i. Next, we calculate z_1 cdot z_2 = (-1 + ai)(b + 2i) = -b - 2i + abi - 2a = -b - 2a + (ab - 2)i. Since both z_1 - z_2 and z_1 cdot z_2 are real numbers, It follows that a - 2 = 0 and ab - 2 = 0. Therefore, a = 2 and b = 1. Thus, a + b = 3. Hence, the answer is boxed{3}.