answer:We are tasked with finding the number of regions a circle can be divided into by 10 parabolas. Let us consider the general case of n parabolas, for which we aim to demonstrate that the answer is 2n^2 + 1. 1. **Intuition:** - If the parabolas are made steep enough, they resemble pairs of parallel lines. - Each crossing of two parabolas within the circle creates a new region. - Each pair of parabolas intersects in four points, and we can always arrange them so that all these intersection points lie inside the circle. 2. **General Approach:** - When adding the k-th set of parallel lines, there are already 2k - 2 lines. - Each of these lines can intersect the new lines in 2k-1 points, adding new regions. 3. **Detailed Steps:** - Adding the k-th set of lines breaks into: [ 4k-2 ] - The formula for the total number of regions is: [ R(n) = 1 + 2 + 6 + 10 + 14 + cdots + (4n-2) ] - Summarizing the arithmetic series: [ R(n) = 1 + 2(1 + 3 + 5 + 7 + cdots + (2n-1)) ] - This observation is the sum of the first n odd numbers, which is known to be n^2. Therefore: [ 1 + 2n^2 = 2n^2 + 1 ] 4. **Proof By Induction:** - **Base case:** For n=0 parabolas, the circle is one region (1 region). - **Inductive Step:** - Assume for n parabolas the regions are given by 2n^2 + 1. - Adding the (n+1)-th parabola intersects existing parabolas in p points. - Each intersection adds regions, so new regions either p+1 or p+2 depending on circle intersects four points. - This adding confirm maximum constraint: [ 2(n+1)^2 = 2n^2 + 4n + 2 + 1 = 2n^2 + 1 + 4n + 1 ] 5. **Concluding Calculation and Formula Justification:** - Number of intersections K follows: [ K leq 4 binom{n}{2} = 2n^2 - 2n ] [ r leq n ] [ K + n + r + 1 leq 2n^2 + 1 ] - Maximum is attained when parabolas steep act parallel lines. Therefore, the number of regions that 10 parabolas can divide a circle into is calculated as follows: [ n = 10 ] [ 2 times 10^2 + 1 = 201 ] # Conclusion: [ boxed{201} ]

answer:The statement |x - 1| leq 1 implies that x is within 1 unit of 1, or in other words, x is in the interval [0, 2]. Similarly, the statement |y - 2| leq 1 implies that y is within 1 unit of 2, or y is in the interval [1, 3]. To find the maximum of |x - 2y + 1|, we can rewrite this expression as |(x - 1) - 2(y - 1)| by adding and subtracting 1 to x and 2 to y respectively. Now we take advantage of the triangle inequality, which states that |a + b| leq |a| + |b| for any real numbers a and b. Applying the triangle inequality: begin{aligned} |x - 2y + 1| &= |(x - 1) - 2(y - 1)| &leq |x - 1| + |(-2)(y - 1)| &= |x - 1| + 2|y - 1| &leq 1 + 2(|y - 2| + 1) quad text{(since } |y - 1| leq |y - 2| + 1text{)} &leq 1 + 2(1 + 1) &= 1 + 2 cdot 2 &= 5. end{aligned} Thus, the maximum value of |x - 2y + 1| is boxed{5}.

answer:As stated, every four-digit palindrome is of the form ABBA. For it to be divisible by 7, the entire number ABBA, or simplified as 1100A + 11B, must be divisible by 7. Further, A needs to be odd. 1. **Explore A values (odd)**: Since A is odd and cannot be 0 (four-digit number), the possible values are A = 1, 3, 5, 7, 9. 2. **Checking divisibility by 7**: For each A, find the smallest B such that 11(100A + B) is divisible by 7. - For A = 1: 11(100 + B) must be divisible by 7. Start checking from smallest B = 0, 1, 2, ldots until it divides: - When B = 6, 11(100 + 6) = 1166 is divisible by 7 (since 1166 = 7 times 166.5714 which is integer). - Therefore, A = 1 and B = 6 work. Thus, the smallest palindrome satisfying the condition is boxed{1661}.

answer:Let (a, b, c, d) be four consecutive integers. Without loss of generality, we can represent these integers as: [ a = n, quad b = n+1, quad c = n+2, quad d = n+3 ] where (n) is some integer. We need to show that the expression: [ ab + ac + ad + bc + bd + cd + 1 ] is divisible by 12. First, substitute the values of (a, b, c, d) into the expression: [ (n)(n+1) + (n)(n+2) + (n)(n+3) + (n+1)(n+2) + (n+1)(n+3) + (n+2)(n+3) + 1 ] Now we will expand and simplify this expression step by step: [ n(n+1) + n(n+2) + n(n+3) + (n+1)(n+2) + (n+1)(n+3) + (n+2)(n+3) + 1 ] [= n^2 + n + n^2 + 2n + n^2 + 3n + (n^2 + 2n + 1) + (n^2 + 3n + 1) + (n^2 + 5n + 6) + 1 ] Combine all like terms: [= n^2 + n + n^2 + 2n + n^2 + 3n + n^2 + 2n + 1 + n^2 + 3n + 1 + n^2 + 5n + 6 + 1 ] [= 6n^2 + 16n + 9 ] Next, let's group and factor where possible: [= 6(n^2 + 3n + frac{3}{2}) ] Now, consider modulo 12: Calculate the expression under modulo 12: Since (a, b, c, d) are four consecutive numbers, one of them is divisible by 4, and at least one will be odd and one even. In these four numbers, there is guaranteed to be a multiple of 3 as well as a multiple of 4. Testing these individual properties: Since one of these terms is (4k): [k(4k) + k (4k +1) + k(4k+2) + (k+1)(4k) + (k+1)(4k+1) +(k+2)(4k+1) +1 ] Summing parts of the expressions yields properties multiple of 12. Thus the expression: [ab + ac + ad + bc + bd + cd + 1] is divisible by 12. Conclusion: (boxed{12}