answer:1. **Assumption:** Suppose it is possible that for each pair of numbers (a) and (b) placed in adjacent cells, at least one of the quadratic equations (x^2 - ax + b = 0) or (x^2 - bx + a = 0) has two integer roots. 2. **Prime Number Consideration:** Let (a) be a prime number such that (77 < a leq 150), and let (b) be a number standing in a cell adjacent to (a). We will begin analyzing the quadratic equations given. 3. **Quadratic Equation Analysis:** - Consider the quadratic equation (x^2 - bx + a = 0). This has two integer roots if and only if the discriminant is a perfect square: [ b^2 - 4a = k^2 quad text{for some integer} , k ] - If this equation has two integer roots, then their product is (a) and their sum is (b>0). The possible integer roots are 1 and (a), provided (b = 1 + a). 4. **Other Quadratic Equation:** - Now, consider the quadratic equation (x^2 - ax + b = 0). This has two integer roots (x_1) and (x_2) if: [ x_1 + x_2 = a quad text{and} quad x_1 x_2 = b ] - Assume (x_2 geq x_1 geq 2). Since the function (t(a-t)) increases for (t leq frac{a}{2}): [ b = x_1 (a - x_1) geq 2(a - 2) > 150 ] - The inequality (2(a-2) > 150) results from (a > 76). Since (77 < a leq 150), the values of (x_1) and (x_2) should comply with (b leq 150), which is impossible unless we reconsider root values. 5. **Possible values and contradiction:** - For quadratic equations to hold, possible neighboring numbers for primes (a) in the range must be: - (b = a + 1) - (b = a - 1) 6. **Independent Reasoning:** - **First Method:** Only edge primes could have two possible neighbors fitting prime neighbor placement (b = a pm 1): - Between 77 and 150, multiple primes (such as 79, 83, 89, 97, 101, etc.) exist in non-corner positions, more than four positioned alone contradicting initial assumptions. - **Second Method:** Primes 101 and 103 specific can only align at specific edges given neighbors 100, 102, and 102, 104. The cell 102 cannot effectively align contradicting neighbor layout twice. # Conclusion: Given the contradictions identified using two approaches, the assumption that the described quadratic properties hold for all numbers in the table between adjacent cells is not valid, thus concluding: [ boxed{text{It is not possible.}} ]

answer:Solution: lim_{n to infty} frac{3^n - 2^n}{3^{n+1} + 2^{n+1}} = lim_{n to infty} frac{1 - left(frac{2}{3}right)^n}{3 + 2left(frac{2}{3}right)^n} = frac{1 - 0}{3 + 0} = frac{1}{3}. Therefore, the answer is: boxed{frac{1}{3}}. Directly use the operation rules of sequence limits, divide both the numerator and the denominator by 3^n, and then solve the limit. This question examines the operation rules of sequence limits and the examination of basic knowledge.

answer:Let's denote the third number as X. According to the given information: The first number is 5% of X, so we can write it as: First number = (5/100) * X The second number is 25% of X, so we can write it as: Second number = (25/100) * X Now, we want to find out what percentage the first number is of the second number. To do this, we divide the first number by the second number and multiply by 100 to get the percentage: Percentage = (First number / Second number) * 100 Substituting the values we have: Percentage = [(5/100) * X] / [(25/100) * X] * 100 Since X is common in both the numerator and the denominator, it cancels out: Percentage = (5/25) * 100 Now, simplify the fraction: Percentage = (1/5) * 100 Percentage = 20 Therefore, the first number is boxed{20%} of the second number.

answer:To solve the problem of constructing a point ( P ) inside the right-angled triangle ( ABC ) such that ( angle PAB = angle PBC = angle PCA ), we can follow these steps: 1. **Identify the key properties of ( ABC ):** - Let ( angle C = 90^circ ), meaning ( triangle ABC ) is a right-angled triangle with the right angle at ( C ). 2. **Investigate the angles ( angle APB ) and ( angle BPC ):** - For the construction to be feasible, we need to find a point ( P ) such that each of the angles ( angle PAB, angle PBC, ) and ( angle PCA ) are equal. This means denoting these angles as ( theta ), one for each of the equal angles. - Remembering that the sum of angles in a triangle is ( 180^circ ), we know: [ angle APB = 180^circ - angle PAB - angle PBA ] 3. **Use the property of cyclic quadrilaterals:** - To ensure ( angle APB = 90^circ ), point ( P ) must lie on the circle with diameter ( AB ). This is because an angle subtended by a semicircle is always a right angle (Thales' theorem). 4. **Construct point ( P ) using geometrical properties:** - Now, consider the circle with diameter ( AB ). For ( P ) to be inside ( triangle ABC ) and fulfill ( angle PBC = theta ) and ( angle PCA = theta ), one way to proceed is to use intersections on the perpendicular dropped from ( C ) onto ( AB ). 5. **Determine the intersection:** - Let ( X ) be the intersection of the perpendicular dropped from ( C ) onto ( AB ). - Given ( angle BXC = angle C = 90^circ ), if ( P ) is chosen on the minor arc of ( BC ) of the circumcircle of ( triangle ABC ), it assures ( angle BPC = 180^circ - angle C ). By ensuring that the point ( P ) lies on the minor arc ( BC ), such that it also touches the criteria of right-angle properties, we successfully construct ( P ) inside ( triangle ABC ), where ( PAB = PBC = PCA ). # Conclusion: The point ( P ) that meets the given condition of equal angles ( theta ) inside ( triangle ABC ) can be constructed using the described geometric strategy. Hence, the solution is verified by the geometric properties of the structures involved. (boxed{})