Sets and their representations

First list the multiples of 5 between 1 and 50: 5, 10, 15, 20, 25, 30, 35, 40, 45, and 50. The phrase “but not by 10” removes 10, 20, 30, 40, and 50, because these are multiples of 10. The remaining values are 5, 15, 25, 35, and 45. Thus the roster form is option A.

The positive natural-number divisors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. We now apply x² > 24. For x = 1, 2, 3, and 4, the squares are not greater than 24; in particular, 4² = 16. For 6, 8, 12, and 24, the squares exceed 24. Therefore X = {6, 8, 12, 24}, so option A is correct.

Because n is a natural number satisfying 1 ≤ n ≤ 5, substitute n = 1, 2, 3, 4, and 5 into x = n² − 1. The resulting values are 1² − 1 = 0, 2² − 1 = 3, 3² − 1 = 8, 4² − 1 = 15, and 5² − 1 = 24. Therefore the roster form is A = {0, 3, 8, 15, 24}, making option A correct.

The positive multiples of 4 that do not exceed 25 are 4, 8, 12, 16, 20, and 24. Among these, 8, 16, and 24 are divisible by 8, so they must be excluded. The remaining numbers, 4, 12, and 20, are divisible by 4 but not by 8 and are all at most 25. Therefore option A is correct.

For n = 1, 2, 3, 4, and 5, the expression n² + 1 gives 2, 5, 10, 17, and 26 respectively. Therefore, the rule generating every element is x = n² + 1. The restriction 1 ≤ n ≤ 5 is essential because it produces exactly the five listed elements; without this restriction, option D would generate infinitely many additional values.

Using the usual school convention N = {1, 2, 3, ...}, every natural number is at least 1. Therefore, no natural number satisfies x < 1. A set whose defining condition is satisfied by no element has zero elements and is called the empty set, commonly denoted by ∅. Hence option A is correct.

The governing concept is roster representation using the integer solutions of the defining equation. Factor the quadratic: x² + x − 6 = (x + 3)(x − 2). Therefore (x + 3)(x − 2) = 0 gives x = −3 or x = 2. Both values belong to the integers, so D = {−3, 2}. The other options arise from incorrect signs or from using roots that do not satisfy the equation. Hence A is correct.

Because 2 is positive, multiplying the compound inequality by 2 preserves both inequality signs: −6 ≤ x < 8. Since x must be an integer, the values are −6, −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, 6, and 7. Counting them gives 8 − (−6) = 14 integers, with the upper endpoint 8 excluded. Therefore option A is correct.

The set is formed by filtering the positive divisors of 18 according to a second condition. The divisors are 1, 2, 3, 6, 9, and 18. Adding 1 gives 2, 3, 4, 7, 10, and 19. The prime results are 2, 3, 7, and 19, corresponding to x = 1, 2, 6, and 18. Thus F = {1, 2, 6, 18}; option A is correct.

To convert the condition into roster form, list the positive divisors of 18: 1, 2, 3, 6, 9, and 18. Check x + 1 for each: 2, 3, 4, 7, 10, and 19. Only 2, 3, 7, and 19 are prime, so the corresponding x-values are 1, 2, 6, and 18. Therefore G = {1, 2, 6, 18}, making option A correct; the other options omit or include unsuitable divisors.

For integer x, the inequality x² ≤ 9 is equivalent to |x| ≤ 3, or −3 ≤ x ≤ 3. Every integer in this closed interval satisfies the condition: −3, −2, −1, 0, 1, 2, and 3. Values outside the interval have square greater than 9. Hence option A is the complete roster form; B omits negative integers, C omits the endpoints, and D includes only two values.

A two-digit natural number has a nonzero tens digit. If both digits are equal, the repeated digit can be 1 through 9, producing 11, 22, 33, 44, 55, 66, 77, 88, and 99. The form 00 is not a two-digit natural number because its tens digit is zero. Option C contains unequal digits, while D contains one-digit numbers. Therefore A is correct.

Use the defining formula for each allowed value of n. When n = 1, 2, 3, 4, and 5, x = 3n − 1 gives 2, 5, 8, 11, and 14 respectively. The inequality includes both endpoints, so n = 5 must be used and 14 cannot be omitted. Option B lists multiples of 3, C follows 3n − 2, and D misses the final value. Hence A is correct.

A positive integer has exactly three positive divisors precisely when it is the square of a prime: if x = p², its divisors are 1, p, and p². The prime squares less than 40 are 2² = 4, 3² = 9, and 5² = 25. The next prime square, 7² = 49, is not less than 40. Hence K = {4, 9, 25}.

The condition |x + 1| < 3 is equivalent to −3 < x + 1 < 3. Subtracting 1 throughout gives −4 < x < 2. Since x must be an integer, the possible values strictly between −4 and 2 are −3, −2, −1, 0, and 1. The endpoints −4 and 2 are excluded because the inequality is strict, so option A is correct.

A number divisible by both 3 and 5 must be divisible by their least common multiple, LCM(3, 5) = 15. The positive multiples of 15 that are at most 50 are 15, 30, and 45. Therefore, in roster form, M = {15, 30, 45}. The numbers 3 and 5 separately do not satisfy both conditions, and 60 is excluded because it is greater than 50.

The listed elements 5, 10, 15, 20, and so on are precisely the positive multiples of 5. Every positive multiple of 5 can be written as x = 5n, where n belongs to the natural numbers. The expression 5 + n gives consecutive numbers after 5, while 5ⁿ gives powers of 5, not all multiples. Hence option A is correct.

Solve the defining equation x² = 2x by bringing all terms to one side: x² − 2x = 0. Factoring gives x(x − 2) = 0. By the zero-product property, either x = 0 or x − 2 = 0, so x = 2. Both values are integers, as required. Hence the roster form is O = {0, 2}. Dividing by x at the beginning would incorrectly discard the valid solution x = 0.

A number whose last digit is 2 appears in the sequence 2, 12, 22, 32, and so on, increasing by 10 each time. Restricting the numbers to the interval from 1 through 100 gives 2, 12, 22, 32, 42, 52, 62, 72, 82, and 92. The number 102 is not allowed because it exceeds 100. Therefore, option A is the complete roster form of P.

Perfect cubes in the relevant positive range are obtained from 1³, 2³, 3³, and so on. We have 1³ = 1 and 2³ = 8, both of which lie between 1 and 20. The next cube is 3³ = 27, which is outside the upper limit. Thus the members satisfying every condition are 1 and 8, so Q = {1, 8}. The numbers in option D are perfect squares, not perfect cubes.