answer:First, identify the points of intersection of the lines y=x, y=-x, and y=8. - The intersection of y = x and y = 8 is at (8, 8). - The intersection of y = -x and y = 8 is at (-8, 8). - The base of the triangle, AB, is the distance between these two points. Calculate this as AB = 8 - (-8) = 16. - The height from the origin (0, 0) to the line y = 8 is 8. Now, calculate the area of triangle OAB: [ text{Area} = frac{1}{2} times text{base} times text{height} = frac{1}{2} times 16 times 8 = 64. ] Thus, the area of the triangle is boxed{64}.

answer:To evaluate the given expression, we follow the steps closely related to the standard solution: 1. **Evaluate the inner ceiling functions:** - For leftlceilfrac{27}{17}rightrceil, we find the smallest integer greater than frac{27}{17}. Since frac{27}{17} is slightly more than 1 but less than 2, the ceiling of frac{27}{17} is 2. - For leftlceilfrac{7cdot17}{27}rightrceil, we calculate frac{7cdot17}{27} = frac{119}{27}. This fraction is more than 4 but less than 5, so the ceiling of frac{119}{27} is 5. 2. **Substitute the values from step 1 into the original expression:** The expression becomes frac{leftlceilfrac{17}{7}-2rightrceil}{leftlceilfrac{27}{7}+5rightrceil}. Simplifying the arithmetic inside the ceiling functions gives us frac{leftlceilfrac{17}{7}-frac{14}{7}rightrceil}{leftlceilfrac{27}{7}+frac{35}{7}rightrceil} = frac{leftlceilfrac{3}{7}rightrceil}{leftlceilfrac{62}{7}rightrceil}. 3. **Evaluate the ceiling functions:** - The smallest integer greater than frac{3}{7} is 1. - The smallest integer greater than frac{62}{7} is 9, since frac{62}{7} is slightly less than 9. 4. **Simplify the fraction:** The expression simplifies to frac{1}{9}. Therefore, the final answer is boxed{frac{1}{9}}.

answer:1. **Introduction and Definitions**: Given sequence a_1, a_2, ldots, a_n where each a_k in {0, 1} and sequence b_1, b_2, ldots, b_{n-1} defined as: - b_k = 0 if a_k = a_{k+1} - b_k = 1 if a_k neq a_{k+1}. 2. **Examples and Initial Values**: - Calculate for small values of n to identify a pattern. - For n = 1, x_1 = 1. - For n = 2, x_2 = 2. - For n = 3, considering the 8 possible combinations of a_1, a_2, a_3, only three configurations reach a maximum of 4 ones, thus x_3 = 4. 3. **Identifying Recursion**: - Generalize x_n with sequences: [ begin{array}{llllllll} a_1 & a_2 & a_3 & cdots & a_n & a_{n+1} & a_{n+2} & a_{n+3} end{array} ] [ begin{array}{lllllll} b_1 & b_2 & b_3 & cdots & b_n & b_{n+1} & b_{n+2} end{array} ] [ begin{array}{llllll} c_1 & c_2 & c_3 & cdots & c_n & c_{n+1} end{array} ] - Group sequences into triplets (a_i, b_i, c_i). 4. **Analyzing Triplets**: - Breakdown of triplets: Identify zeros within triplets for minimizing the number of zeros. - There exist at least n+2 zeros, hence a maximum of 2n+4 ones are achievable: [ x_{n+3} leq x_n + (2n + 4). ] 5. **Recursive Inequality**: - Develop the inequalities: [ x_{3k} leq x_{3(k-1)} + 6(k-1) + 4, ] [ ldots ] [ x_6 leq x_3 + 6 cdot 1 + 4. ] - Summing these results to find: [ x_{3k} leq 6(1 + 2 + cdots + (k - 1)) + 4k = k(3k + 1). ] 6. **Special Cases**: - For n = 3k + 1: [ x_{3k+1} leq 3k(k+1) + 1. ] - For n = 3k + 2: [ x_{3k+2} leq (k + 1)(3k + 2). ] 7. **General Pattern**: - Combine results and formulate: [ x_n leq leftlfloorfrac{n(n+1)+1}{3}rightrfloor. ] 8. **Verification and Conclusion**: - Verify correctness by constructing sequences manually. - Confirm that the sequence pattern meets the specified relationships. # Conclusion The maximum value of the sum of all 1s in the triangular table formed by repeating the operation is: [ boxed{leftlfloorfrac{n(n+1)+1}{3}rightrfloor} ]

answer:To solve this problem, we need to determine the probability that a randomly chosen 3-digit positive integer is in the set mathcal{S}, which consists of numbers that are multiples of 3 and have at least one digit that is 1. We will use the Principle of Inclusion-Exclusion (PIE) to count the numbers in mathcal{S}. 1. **Total number of 3-digit integers:** The range of 3-digit integers is from 100 to 999. Therefore, there are: [ 999 - 100 + 1 = 900 text{ three-digit integers.} ] 2. **Counting multiples of 3:** A number is a multiple of 3 if the sum of its digits is divisible by 3. We need to count the 3-digit numbers that are multiples of 3. - The smallest 3-digit multiple of 3 is 102. - The largest 3-digit multiple of 3 is 999. - These numbers form an arithmetic sequence with the first term (a = 102) and common difference (d = 3). The number of terms in this sequence is given by: [ n = frac{999 - 102}{3} + 1 = 300. ] So, there are 300 multiples of 3 among the 3-digit numbers. 3. **Counting multiples of 3 with at least one digit being 1:** We will use PIE to count the numbers that are multiples of 3 and have at least one digit equal to 1. - **Case 1: 1 is the first digit (hundreds place):** If the first digit is 1, the number is of the form (1xy), where (x) and (y) are digits. The sum (1 + x + y) must be divisible by 3. There are 10 choices for (x) and 10 choices for (y), but we need to count only those combinations where (1 + x + y) is divisible by 3. - For each (x), there are 3 values of (y) that make (1 + x + y) divisible by 3. - Therefore, there are (10 times 3 = 30) such numbers. - **Case 2: 1 is the second digit (tens place):** If the second digit is 1, the number is of the form (x1y). Similar to the previous case, we need (x + 1 + y) to be divisible by 3. - For each (x), there are 3 values of (y) that make (x + 1 + y) divisible by 3. - Therefore, there are (10 times 3 = 30) such numbers. - **Case 3: 1 is the third digit (units place):** If the third digit is 1, the number is of the form (xy1). We need (x + y + 1) to be divisible by 3. - For each (x), there are 3 values of (y) that make (x + y + 1) divisible by 3. - Therefore, there are (10 times 3 = 30) such numbers. - **Subtracting overcounted cases:** We have overcounted numbers where two digits are 1. We need to subtract these cases. - If two digits are 1, the number is of the form (11x), (1x1), or (x11). In each case, the remaining digit (x) must be such that the sum of the digits is divisible by 3. - For each of these forms, there are 3 values of (x) that make the sum divisible by 3. - Therefore, there are (3 + 3 + 3 = 9) such numbers. - **Adding back the case where all three digits are 1:** The number 111 is a multiple of 3 and has all digits as 1. We have subtracted this case three times, so we need to add it back once. Combining all these, we get: [ 30 + 30 + 30 - 9 + 1 = 82. ] 4. **Calculating the probability:** The probability that a randomly chosen 3-digit number is in mathcal{S} is: [ frac{82}{900}. ] Simplifying this fraction, we get: [ frac{82}{900} = frac{41}{450}. ] 5. **Finding (m + n):** The fraction (frac{41}{450}) is in simplest form, where (m = 41) and (n = 450). Therefore: [ m + n = 41 + 450 = 491. ] The final answer is (boxed{491}).