answer:To determine if there exists a natural number that can be represented as the product of two palindromes in more than 100 ways, consider the following approach. 1. Define the palindrome in the form (1_n = underbrace{111 ldots 1}_{n}). 2. We know that if ( n ) is a multiple of ( k ), then (1_n) is divisible by the palindrome (1_k), and the quotient is also a palindrome composed of ones, spaced by groups of (k-1) zeros. 3. Choose an ( n ) such that the number has more than 100 factors. For instance, (2^{101}). 4. **Verification**: When (n) is chosen such that it has more than 100 proper divisors, it serves to construct invariants, and consequent dividers that are indeed palindromes. On verifying the divisor count of (2^{101}), the prime factorization yields 102 factors ((2^0, 2^1, ldots, 2^{101})). Therefore, this will satisfy the requirement of having more than 100 ways to represent the number as a product of two palindromes. 5. **Observation**: The number (6_n) is divisible not only by (1_k) but also by (2_k, 3_k), and (6_k). This can reduce (n) to a number having more than 25 proper divisors. 6. **Example**: Considering ( n=720 = 2^4 cdot 3^2 cdot 5 ), its total number of divisors: [ (4+1) cdot (2+1) cdot (1+1) = 5 cdot 3 cdot 2 = 30 ] Satisfies having more than 25 proper divisors. For further insight, consider the multiplication property of palindromes provided that arithmetic operations do not entail carrying over digits: 7. **Second approach**: Consider (1_{256}): [ 11 cdot 101 cdot underbrace{10 ldots 1}_{3} cdot underbrace{10 ldots 1}_{7} cdot underbrace{10 ldots 1}_{15} cdot underbrace{10 ldots 1}_{31} cdot underbrace{10 ldots 1}_{63} cdot underbrace{10 ldots 1}_{127} = 1_{256} ] 8. **Proof**: On multiplying any chosen pair of these factors, no carry occurs, ensuring that the product remains a palindrome. With 8 factors available, one can generate (2^7 = 128) different ways to factorize (1_{256}) into two groups. Each product represents ( 1_{256} ) in a different way as a product of two palindromes because of the coprimality of the initial factors. 9. **Further observations**: Using similar insights as noted above with (6_{64}) and extending factors in a similar way: [ 11 cdot 101 cdot 1001 cdot 1 underbrace{0 ldots 0}_{3}1 cdot 1 underbrace{0 ldots 0}_{4}1 cdot 1 underbrace{0 ldots 0}_{5}1 cdot 1 underbrace{0 ldots 0}_{6}1 cdot 1 underbrace{0 ldots 0}_{7}1 ] We have shown by multiple examples that there exists a natural number that can be represented as a product of two palindromes in more than 100 distinct ways. Conclusion: [ boxed{2^{101}} ]

answer:Setting x = -1, x = 1, and x = 3, we obtain: begin{align*} 2 = Q(-1) &= Q(0) - Q(1) + Q(3), Q(1) &= Q(0) + Q(1) + Q(3), Q(3) &= Q(0) + 3Q(1) + 9Q(3), end{align*} respectively. Solving this system of equations for Q(0), Q(1), and Q(3), we find: 1. From 2 = Q(0) - Q(1) + Q(3) and Q(1) = Q(0) + Q(1) + Q(3), we express Q(1): [Q(1) = Q(0) + Q(1) + Q(3) implies Q(1) = frac{2 + Q(0) + Q(3)}{2}] 2. Substitute this into the third equation: [Q(3) = Q(0) + 3 left(frac{2 + Q(0) + Q(3)}{2}right) + 9Q(3)] [Q(3) = Q(0) + frac{3 + 3Q(0) + 3Q(3)}{2} + 9Q(3)] [2Q(3) - 9Q(3) - frac{3Q(3)}{2} = 1.5 + 3Q(0) - Q(0)] 3. Solve this equation to find Q(0), Q(1), and Q(3): [2Q(3) - 9Q(3) - 1.5Q(3) = 1.5 + 2Q(0)] [Q(3) = frac{3 + 2Q(0)}{12.5}] [Q(0) = 2, , Q(1) = -1, , Q(3) = 1] (after solving the equations) Thus, [Q(x) = boxed{x^2 - x + 2}.]

answer:To calculate the time taken for the goods train to cross the platform, we need to first determine the total distance covered by the train while crossing the platform. This distance is the sum of the length of the train and the length of the platform. Length of the train = 250.0416 meters Length of the platform = 270 meters Total distance covered = Length of the train + Length of the platform Total distance covered = 250.0416 meters + 270 meters Total distance covered = 520.0416 meters Next, we need to convert the speed of the train from kilometers per hour (kmph) to meters per second (m/s) to match the units of the distance covered. Speed of the train in kmph = 72 kmph To convert kmph to m/s, we use the conversion factor: 1 kmph = 1000 meters / 3600 seconds (since 1 km = 1000 meters and 1 hour = 3600 seconds). Speed of the train in m/s = 72 kmph * (1000 meters / 3600 seconds) Speed of the train in m/s = 72 * (1000 / 3600) Speed of the train in m/s = 72 * (10 / 36) Speed of the train in m/s = 72 * (5 / 18) Speed of the train in m/s = 20 m/s Now that we have the speed of the train in m/s and the total distance covered in meters, we can calculate the time taken to cross the platform using the formula: Time = Distance / Speed Time taken to cross the platform = Total distance covered / Speed of the train in m/s Time taken to cross the platform = 520.0416 meters / 20 m/s Time taken to cross the platform = 26.00208 seconds Therefore, it takes approximately boxed{26.00208} seconds for the goods train to cross the platform.

answer:To find the volume of the intersection of a regular tetrahedron with its reflection through its center, let's carefully go through the steps: 1. **Volume of the Original Tetrahedron**: - The volume of the given regular tetrahedron, ( V ), is (1). 2. **Position and Reflection**: - Imagine the tetrahedron (ABCD) placed on a table with vertex (A) at the top. - The center of the tetrahedron is located ( frac{3}{4} ) of the way from vertex (A) to the centroid of the base (BCD). 3. **Reflection of the Plane**: - Reflecting the tetrahedron through this center results in a second tetrahedron with vertices at the same distance but mirrored across the center. - This reflection cuts the original tetrahedron into smaller tetrahedra by plane sections halfway between top vertex (A) and base (BCD). 4. **Intersection and Smaller Tetrahedra**: - The reflection embeds smaller tetrahedra within (ABCD), specifically by scaling (ABCD) by a factor of ( frac{1}{2} ) around each vertex (A), (B), (C), and (D). - Each smaller tetrahedron has a volume: [ V_{text{small}} = left(frac{1}{2}right)^3 cdot V = left(frac{1}{2}right)^3 cdot 1 = frac{1}{8} ] 5. **Four Smaller Tetrahedra**: - There are 4 such smaller tetrahedra formed. - The total volume of these smaller tetrahedra combined is: [ 4 cdot V_{text{small}} = 4 cdot frac{1}{8} = frac{4}{8} = frac{1}{2} ] 6. **Volume of the Intersection**: - The intersection of the original tetrahedron (ABCD) and its reflection is the remaining central part that isn’t part of the smaller tetrahedra. - Therefore, the volume of the intersection is: [ V_{text{intersection}} = V - left(4 cdot V_{text{small}}right) = 1 - frac{1}{2} = frac{1}{2} ] # Conclusion: Thus, the volume of the intersection of the tetrahedron and its reflection is: [ boxed{frac{1}{2}} ]