answer:To determine the value of ( x ) for which the given complex number ( z ) is a real number, we need to ensure that the imaginary part of ( z ) is zero. The given complex number is:[ z = log_{2}(x^{2}-3x-3) + i log_{2}(x-3) ]For ( z ) to be a real number, the imaginary part ( log_{2}(x-3) ) must be zero. Let's break this down:1. Imaginary Part Must Be Zero: [ log_{2}(x-3) = 0 ]2. Solve for ( x ): [ x - 3 = 2^0 ] [ x - 3 = 1 ] [ x = 4 ]3. Verify the Real Part: After finding ( x = 4 ), substitute it back into the real part of ( z ): [ log_{2}(4^2 - 3 cdot 4 - 3) ] [ log_{2}(16 - 12 - 3) ] [ log_{2}(1) = 0 ]Since both parts of the complex number ( z ) will be zero when ( x = 4 ), ( z ) will indeed be a real number.Let's implement this step-by-step verification in Python using SymPy to ensure accuracy:```pythonimport sympy as sp# Define the variablex = sp.symbols('x')# Define the imaginary part of the complex numberimaginary_part = sp.log(x - 3, 2)# Solve for when the imaginary part is zerosolution = sp.solve(imaginary_part, x)# Verify the value of x by substituting into the real partreal_part = sp.log(x2 - 3*x - 3, 2)real_value_at_solution = real_part.subs(x, solution[0])print(solution[0])print(real_value_at_solution)``````output40```The output confirms our calculations. The value of ( x ) that makes the complex number ( z = log_{2}(x^{2}-3x-3) + i log_{2}(x-3) ) a real number is (boxed{4}).

answer:To solve for the number of ordered triples (a, b, c) of positive integers such that mathrm{lcm}(a, b) = 1000, mathrm{lcm}(b, c) = 2000, and mathrm{lcm}(c, a) = 2000, we can approach the problem step by step. Let's break it down and then use Python for accurate enumeration.First, we need to express the given conditions in terms of prime factorization:- (1000 = 2^3 times 5^3)- (2000 = 2^4 times 5^3)Knowing these, we understand that a, b, c must be of the form 2^x times 5^y, where x and y are bounded by these exponents:- (a = 2^{x_a} times 5^{y_a})- (b = 2^{x_b} times 5^{y_b})- (c = 2^{x_c} times 5^{y_c})Given:1. mathrm{lcm}(a, b) = max(2^{x_a}, 2^{x_b}) times max(5^{y_a}, 5^{y_b}) = 2^3 times 5^32. mathrm{lcm}(b, c) = max(2^{x_b}, 2^{x_c}) times max(5^{y_b}, 5^{y_c}) = 2000 = 2^4 times 5^33. mathrm{lcm}(c, a) = max(2^{x_c}, 2^{x_a}) times max(5^{y_c}, 5^{y_a}) = 2000 = 2^4 times 5^3We need to iterate over all combinations of (x_a, y_a), (x_b, y_b), and (x_c, y_c) within the valid ranges, and count the number of valid triples that satisfy all these conditions.Let's write a Python script to exhaustively search for all valid triples.```pythonfrom itertools import product# Exponents bounds for 2 and 5bounds_two = {'a': (0, 4), 'b': (0, 4), 'c': (0, 4)}bounds_five = {'a': (0, 3), 'b': (0, 3), 'c': (0, 3)}valid_triples = []# Iterate over all possible combinations of exponentsfor xa, xb, xc in product(range(bounds_two['a'][0], bounds_two['a'][1] + 1), range(bounds_two['b'][0], bounds_two['b'][1] + 1), range(bounds_two['c'][0], bounds_two['c'][1] + 1)): for ya, yb, yc in product(range(bounds_five['a'][0], bounds_five['a'][1] + 1), range(bounds_five['b'][0], bounds_five['b'][1] + 1), range(bounds_five['c'][0], bounds_five['c'][1] + 1)): if max(xa, xb) == 3 and max(ya, yb) == 3 and max(xb, xc) == 4 and max(yb, yc) == 3 and max(xc, xa) == 4 and max(yc, ya) == 3: valid_triples.append(((2xa * 5ya), (2xb * 5yb), (2xc * 5yc)))print(len(valid_triples))``````output70```The number of ordered triples ((a, b, c)) of positive integers for which (mathrm{lcm}(a, b) = 1000), (mathrm{lcm}(b, c) = 2000), and (mathrm{lcm}(c, a) = 2000) is (boxed{70}).

answer:To solve this problem, we need to determine the value of (a) such that the points (A(4, 3)), (B(5, a)), and (C(6, 5)) are collinear. Three points are collinear if the area of the triangle formed by them is zero. The area of a triangle formed by points ((x_1, y_1)), ((x_2, y_2)), and ((x_3, y_3)) can be calculated using the determinant formula:[text{Area} = frac{1}{2} left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) right|]For the points to be collinear, this area should be zero. Thus, we can set up the equation:[0 = x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]Substituting the given points (A(4, 3)), (B(5, a)), and (C(6, 5)), we get:[0 = 4(a - 5) + 5(5 - 3) + 6(3 - a)]Let's solve this equation step-by-step using Python and sympy:```pythonimport sympy as sp# Define variablea = sp.symbols('a')# Define the pointsx1, y1 = 4, 3x2, y2 = 5, ax3, y3 = 6, 5# Define the collinearity equationcollinearity_eq = x1*(y2 - y3) + x2*(y3 - y1) + x3*(y1 - y2)# Solve for aa_value = sp.solve(collinearity_eq, a)print(a_value)``````output[4]```The value of (a) that makes the points (A(4, 3)), (B(5, a)), and (C(6, 5)) collinear is (boxed{4}).

answer:To solve the problem, we need to follow these steps:1. Determine the General Form of the Arithmetic Sequence: Given the sum of the first ( n ) terms of the arithmetic sequence ( a_n ) is ( S_n = 6n - n^2 ), we can find the general form of ( a_n ).2. Calculate the First 10 Terms of the Arithmetic Sequence ( a_n ): We will find the explicit values of ( a_n ) for ( n = 1, 2, ldots, 10 ).3. Compute the Absolute Values of the First 10 Terms: We will then compute the absolute values of these terms to get ( |a_n| ).4. Sum the Absolute Values: Finally, we sum the absolute values of the first 10 terms.Let's implement this step-by-step in Python using SymPy.```pythonimport sympy as sp# Define the variablesn = sp.symbols('n')# Given sum of the first n termsS_n = 6*n - n2# Find the first term a_1a_1 = S_n.subs(n, 1)# Find the general term a_na_n = S_n - S_n.subs(n, n-1)a_n = sp.simplify(a_n)# Calculate the first 10 terms of a_nterms = [a_n.subs(n, i) for i in range(1, 11)]# Calculate the absolute values of the first 10 termsabs_terms = [abs(term) for term in terms]# Sum the absolute valuessum_abs_terms = sum(abs_terms)print(sum_abs_terms)``````output58```The sum of the first 10 terms of the sequence ( |a_n| ) is (boxed{58}).

answer:To solve this problem, we need to set up a system of congruences and solve for the number of candies within the specified range. Specifically, we need to find an integer (n) such that:1. (n equiv 6 pmod{10})2. (n equiv -4 pmod{15})3. (200 leq n leq 250)Let's break this down step-by-step:1. Understand the Congruences: - (n equiv 6 pmod{10}) means that when (n) is divided by 10, the remainder is 6. - (n equiv -4 pmod{15}) is equivalent to (n equiv 11 pmod{15}) because (-4 + 15 = 11).2. Solve the System of Congruences: We can solve the system using the Chinese Remainder Theorem (CRT) which provides a solution for simultaneous congruences when the moduli are coprime.3. Find (n) within the specified range: 200 ≤ (n) ≤ 250Let's write Python code using `sympy` to find the solution:```pythonimport sympy as sp# Variables for the congruencesn = sp.symbols('n')# Congruencescongruence1 = sp.Mod(n, 10) - 6congruence2 = sp.Mod(n, 15) - 11# Solve the system of linear congruencessolution = sp.solve((congruence1, congruence2), n)# General solution formgeneral_solution = solution[n]# Find values within the specified rangelower_bound = 200upper_bound = 250# Generate a list of potential solutions within the rangepotential_solutions = [general_solution.subs(n, k) for k in range(lower_bound // general_solution, upper_bound // general_solution + 1) if lower_bound <= general_solution.subs(n, k) <= upper_bound]# Print the valid solution(s)valid_solutions = [sol.evalf() for sol in potential_solutions]print(valid_solutions)``````outputTraceback (most recent call last): File " solution = sp.solve((congruence1, congruence2), n)NotImplementedError: could not solve Mod(n, 10) - 6```The error encountered indicates that the symbolic solver couldn't handle the modulo operations directly. We can use a different approach by manually solving the system of congruences using properties of the Chinese Remainder Theorem (CRT). Let's implement this step by step.1. Restate the Congruences: - (n equiv 6 pmod{10}) - (n equiv 11 pmod{15}) 2. Use the Chinese Remainder Theorem: Since 10 and 15 are not coprime, let's manually solve by finding a common solution.3. Iterate Over the Potential Solutions: We'll generate potential solutions for (n) based on the first congruence and check if they satisfy the second congruence within the given range.Here's the code to achieve this:```pythonimport sympy as sp# Define the rangelower_bound = 200upper_bound = 250# Iterate over possible solutions for the first congruencesolutions = []for n in range(lower_bound, upper_bound + 1): if n % 10 == 6 and n % 15 == 11: solutions.append(n)print(solutions)``````output[206, 236]```The solutions to the problem are ( n = 206 ) and ( n = 236 ). This means that there are two possible values for the total number of candies within the specified range:[boxed{206 text{ or } 236}]Both numbers satisfy the given conditions:1. When packed into boxes with a capacity of 10 candies, 6 candies remain.2. When packed into boxes with a capacity of 15 candies, 4 candies are missing (equivalently, 11 candies remain since (-4 equiv 11 pmod{15})).Thus, the total number of candies is either 206 or 236.

answer:To solve the problem, we need to count the number of 5-element subsets (S_1 = {a_1, a_2, a_3, a_4, a_5}) of the set (S = {1, 2, 3, ldots, 18}) such that the difference between any two elements of (S_1) is not 1.We can use a combinatorial transformation approach to solve this problem. Let's transform the elements of (S_1) to another set of elements where the differences between any two elements are always at least 2.1. Transformation Approach: If (S_1 = {a_1, a_2, a_3, a_4, a_5}), where (a_1 < a_2 < a_3 < a_4 < a_5), we can define a new set (T_1 = {b_1, b_2, b_3, b_4, b_5}) where (b_i = a_i - (i-1)). This transformation ensures that the difference between any two elements of (T_1) is at least 1.2. Size of the Transformed Set: The set (S = {1, 2, 3, ldots, 18}) is transformed into the set (T = {b_1, b_2, b_3, b_4, b_5}) where (1 leq b_1 < b_2 < b_3 < b_4 < b_5 leq 18 - 4 = 14). So, we need to choose 5 elements from the set ({1, 2, 3, ldots, 14}).3. Combinatorial Calculation: The number of ways to choose 5 elements from a set of 14 elements is given by the binomial coefficient (binom{14}{5}).The correct Python code to compute this using sympy is as follows:```pythonimport sympy as sp# Define the binomial coefficient functionn = 14k = 5# Calculate the binomial coefficientresult = sp.binomial(n, k)print(result)``````output2002```The number of 5-element subsets ( S_1 = {a_1, a_2, a_3, a_4, a_5} ) of the set ( S = {1, 2, 3, ldots, 18} ) such that the difference between any two elements of ( S_1 ) is not 1 is given by the binomial coefficient (binom{14}{5}), which is 2002.Thus, the correct answer from the provided options is:[boxed{C_{14}^{5}}]This corresponds to option D.