answer:To solve the problem, we need to find all numbers ( n ) that, when divided by ( q^2 ) (for any integer ( q )), have a remainder less than ( q^2 / 2 ). This means we are looking for all ( n ) such that:[ n % q^2 < frac{q^2}{2} ]We will perform the following steps:1. Define the conditions for the property.2. Iterate over possible values of ( n ) and check if they satisfy the condition for a range of ( q ).3. List all numbers that have this property.Let's implement this in Python using Sympy to check the conditions.```pythonimport sympy as sp# Define the range for n and qmax_n = 100 # You can increase this limit to search for more valuesmax_q = 10 # Number of q values to test for each n# List to store numbers that satisfy the propertysatisfying_numbers = []# Check each number n in the rangefor n in range(1, max_n + 1): satisfies_all = True for q in range(1, max_q + 1): q_squared = q2 remainder = n % q_squared if remainder >= q_squared / 2: satisfies_all = False break if satisfies_all: satisfying_numbers.append(n)print(satisfying_numbers)``````output[1, 4]```The numbers that have the property that when divided by ( q^2 ), the remainder is less than ( q^2 / 2 ) for any ( q ) are (boxed{1}) and (boxed{4}).In summary, the list of numbers with the given property is:[ boxed{1, 4} ]

answer:To solve this problem, we need to calculate the total number of small cubes obtained from splitting a cube with an edge of 1 meter into small cubes with an edge of 1 millimeter. Then, we need to determine the height of a column formed by stacking all these small cubes.Steps to solve the problem:1. Calculate the volume of the large cube.2. Calculate the volume of a small cube.3. Determine the number of small cubes that can be formed from the large cube.4. Calculate the height of the column formed by stacking all the small cubes.Let's implement this step-by-step in Python using sympy:```pythonimport sympy as sp# Define the edge length of the large cube and the small cubeedge_large_cube = 1 # in metersedge_small_cube = 1 / 1000 # converting 1mm to meters# Calculate the volume of the large cube and the small cubevolume_large_cube = edge_large_cube 3volume_small_cube = edge_small_cube 3# Calculate the number of small cubes that can be formed from the large cubenumber_of_small_cubes = volume_large_cube / volume_small_cube# Calculate the height of the column formed by stacking all the small cubesheight_of_column = number_of_small_cubes * edge_small_cube# Print the height of the column in metersheight_of_column_in_km = height_of_column / 1000 # converting meters to kilometersprint(height_of_column_in_km)``````output999.9999999999999```The height of the column formed by stacking all the small cubes one on top of another is approximately (1000 , mathrm{km}).Thus, the correct answer is:[boxed{1000 , mathrm{km}}]

answer:To solve the given problem, we can use combinatorial mathematics. The key idea is to calculate the total number of ways to select 3 parts from 10 parts, and then subtract the number of ways to select 3 parts such that all parts are second-class (since we need at least one first-class part).Here's the step-by-step breakdown:1. Calculate the total number of ways to select 3 parts out of 10: This is given by the combination formula (binom{10}{3}).2. Calculate the number of ways to select 3 parts such that all parts are second-class: This is given by the combination formula (binom{4}{3}).3. Subtract the second-class-only selection from the total selection: The result will give us the number of ways to select at least one first-class part.Let's implement this in Python using SymPy.```pythonimport sympy as sp# Define the combination functiondef combination(n, r): return sp.binomial(n, r)# Total number of ways to select 3 parts out of 10total_ways = combination(10, 3)# Number of ways to select 3 parts such that all are second-classsecond_class_ways = combination(4, 3)# Number of ways to select at least one first-class partat_least_one_first_class = total_ways - second_class_ways# Print the resultprint(at_least_one_first_class)``````output116```The number of different ways to select at least one first-class part from the 10 parts is (boxed{116}).

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