answer:To solve the given problem, we need to find the remainder when the sum of the first 28 terms of the sequence ( {a_n} ) is divided by 1000. We'll start by writing code to compute the values of ( a_n ) for ( n = 1 ) to ( 30 ) based on the recursive relation ( a_{n+3} = a_{n+2} + a_{n+1} + a_n ). Given the initial values:[ a_1 = a_2 = a_3 = 1 ]We also know:[ a_{28} = 6090307 ][ a_{29} = 11201821 ][ a_{30} = 20603361 ]Let's implement this sequence in Python and sum the first 28 terms. Finally, we'll take the sum modulo 1000 to find the result.```python# Define the initial values of the sequencea = [1, 1, 1] # This represents a1, a2, a3# Compute the values up to a30 using the given recurrence relationfor n in range(3, 30): a.append(a[n-1] + a[n-2] + a[n-3])# Given values for a28, a29, and a30 match the problem statementassert a[27] == 6090307assert a[28] == 11201821assert a[29] == 20603361# Calculate the sum of the first 28 termssum_first_28 = sum(a[:28])# Find the remainder when the sum is divided by 1000remainder = sum_first_28 % 1000print(remainder)``````output834```The remainder when the sum of the first 28 terms of the sequence ( {a_n} ) is divided by 1000 is (boxed{834}).

answer:To solve the problem, we need to find the intersection of sets ( A ) and ( B ), and then determine the number of subsets of the resulting set ( C ).Steps:1. Compute the intersection ( C = A cap B ).2. Determine the number of elements in set ( C ).3. Calculate the number of subsets of set ( C ). The number of subsets of a set with ( n ) elements is ( 2^n ).Let's implement this in Python using SymPy.```pythonimport sympy as sp# Define the sets A and BA = {1, 2, 3, 4, 5}B = {1, 3, 5, 7, 9}# Compute the intersection of A and BC = A.intersection(B)# Calculate the number of subsets of set Cnum_elements_C = len(C)num_subsets_C = 2 num_elements_C# Print the number of subsets of set Cprint(num_subsets_C)``````output8```The number of subsets of the set (C = A cap B) is (boxed{8}).

answer:To solve this problem, we need to address the conditions for the roots of the quadratic polynomial (2x^2 + bx + c) to be real and differ by 30. Let's denote the roots as ( alpha ) and ( beta ) such that ( alpha - beta = 30 ).Given a quadratic equation (Ax^2 + Bx + C = 0), the sum of the roots ( alpha + beta ) and the product of the roots ( alpha beta ) are given by:[alpha + beta = -frac{B}{A}][alpha beta = frac{C}{A}]For our specific polynomial (2x^2 + bx + c), we have (A = 2), (B = b), and (C = c). This gives us:[alpha + beta = -frac{b}{2}][alpha beta = frac{c}{2}]Additionally, we know:[alpha - beta = 30]Starting from the equations ( alpha + beta = -frac{b}{2} ) and (alpha - beta = 30), we can solve for (alpha) and (beta).[2alpha = (alpha + beta) + (alpha - beta) = -frac{b}{2} + 30 ][2beta = (alpha + beta) - (alpha - beta) = -frac{b}{2} - 30 ]Thus, the expressions simplify to:[alpha = -frac{b}{4} + 15][beta = -frac{b}{4} - 15]Next, we find ( c ) using (alpha beta = frac{c}{2}):[left( -frac{b}{4} + 15 right) left( -frac{b}{4} - 15 right) = frac{c}{2}]We compute:[left( -frac{b}{4} right)^2 - 15^2 = frac{c}{2}][frac{b^2}{16} - 225 = frac{c}{2}]Multiplying through by 2:[frac{b^2}{8} - 450 = c]Now, we need to minimize ( b + c ) subject to these conditions. We have:[b + c = b + left( frac{b^2}{8} - 450 right)][b + c = b + frac{b^2}{8} - 450]To find the minimum value, we can evaluate this for the smallest ( b ) that results in real roots, meaning (b) gives feasible ( alpha ) and ( beta ). We ensure (c) remains a positive integer. Using code, we check iteratively: 1. Iterate over potential ( b ) values.2. Skip to those that yield positive ( c ).Let's perform the calculations programmatically.```pythonimport mathdef find_least_b_c_sum(): min_sum = float('inf') min_b = 0 min_c = 0 # Iterate over potential values for b for b in range(1, 1000): # upper bound is high to ensure finding valid b # Calculate c c = (b2) / 8 - 450 # Check if c is a positive integer if c.is_integer() and c > 0: c = int(c) current_sum = b + c if current_sum < min_sum: min_sum = current_sum min_b = b min_c = c return min_b, min_c, min_summin_b, min_c, min_sum = find_least_b_c_sum()print((min_b, min_c, min_sum))``````output(64, 62, 126)```The least possible value of ( b + c ) given that the polynomial ( 2x^2 + bx + c ) has two real roots differing by 30 is:[boxed{126}]

answer:To solve this problem, we can use the Pigeonhole Principle. The Pigeonhole Principle states that if ( n ) items are put into ( m ) containers, with ( n > m ), then at least one container must contain more than one item.Here's the step-by-step approach:1. Identify pairs of numbers whose sum is fixed: The pairs of numbers from 1 to 20 that sum to a fixed value can be identified. For example, pairs that sum to 21 are: (1, 20), (2, 19), (3, 18), ..., (10, 11).2. Count the pairs: There are 10 such pairs that sum to 21.3. Apply the Pigeonhole Principle: If we select 11 numbers from the 20 numbers, and there are only 10 pairs that sum to 21, by the Pigeonhole Principle, we are forced to pick at least one pair that sums to 21. This is because we have one more number than the number of pairs, ensuring at least one pair will be present in any selection of 11 numbers.Let's verify this using Python and sympy.```pythonfrom itertools import combinations# Define the set of numbers from 1 to 20numbers = list(range(1, 21))# Define the pairs that sum to 21pairs = [(i, 21 - i) for i in range(1, 11)]# Check if there exists any selection of 11 numbers without any pair summing to 21found_pair = Falsefor selection in combinations(numbers, 11): found_pair = False for pair in pairs: if pair[0] in selection and pair[1] in selection: found_pair = True break if not found_pair: breakprint(found_pair)``````outputTrue```The output `True` indicates that in any selection of 11 numbers from the set {1, 2, ..., 20}, there will always be at least one pair whose sum is 21. Therefore, the correct answer to the problem is ( boxed{21} ).

answer:To solve this problem, let's carefully analyze the movements and areas involved:1. Initial Contact Area: - The square ink pad initially touches the paper and covers an area of 1 cm² (since the side length of the square is 1 cm).2. Rotation of the Pad: - When the square ink pad rotates 180° about one of its corners, it sweeps out an area that forms a shape which is the union of two sectors of a quarter circle.3. Calculating the Swept Area: - Each sector has a radius equal to the diagonal of the square. The diagonal of a square with side length 1 cm is ( sqrt{2} ) cm. - The area of each quarter circle is ( frac{1}{4} pi (sqrt{2})^2 = frac{1}{4} pi cdot 2 = frac{pi}{2} ).4. Total Swept Area: - Since there are two such sectors, the total swept area is ( 2 times frac{pi}{2} = pi ) cm². 5. Union of the Initial and Swept Areas: - The initial area of 1 cm² and the swept area of π cm² together form the total coloured area on the paper.Let's use Python and SymPy to verify this reasoning and calculate the exact area.```pythonimport sympy as sp# Calculate the side length and the diagonal of the squareside_length = 1diagonal = sp.sqrt(2) # Diagonal of the square# Calculate the area of each quarter circle sectorsector_area = (1/4) * sp.pi * (diagonal2)# Since there are two such sectorstotal_swept_area = 2 * sector_area# Initial area of the square ink padinitial_area = side_length2# Total colored area on the papertotal_colored_area = initial_area + total_swept_area# Simplify the expression to get the final areasimplified_total_area = sp.simplify(total_colored_area)print(simplified_total_area)``````output1 + 1.0*pi```The output of the Python code indicates that the total colored area on the paper is (1 + pi) cm². This corresponds to option E: (pi + 1).Thus, the area of paper, in cm², that is colored black is (boxed{pi + 1}).

answer:To find the number of subsets of an (n)-element set (A = {a_{1}, a_{2}, cdots, a_{n}}), we can use the combinatorial property that each element of the set can either be included or not included in a subset. This gives us (2) choices (include or exclude) per element.Therefore, the total number of subsets of an (n)-element set is (2^n).Let's use Python and sympy to compute this for a general (n).```pythonimport sympy as sp# Define the variable nn = sp.symbols('n')# Calculate the number of subsetsnumber_of_subsets = 2n# Display the resultprint(number_of_subsets)``````output2n```The total number of subsets of an (n)-element set (A = {a_{1}, a_{2}, cdots, a_{n}}) is (2^n).To illustrate this more concretely, let's evaluate this for a few values of (n):1. For (n = 0): [ 2^0 = 1 ] There is 1 subset, which is the empty set.2. For (n = 1): [ 2^1 = 2 ] There are 2 subsets: ({}, {a_1}).3. For (n = 2): [ 2^2 = 4 ] There are 4 subsets: ({}, {a_1}, {a_2}, {a_1, a_2}).To generalize, the formula (2^n) correctly gives us the number of subsets for any integer (n).So, the number of subsets of an (n)-element set is (boxed{2^n}).