Sets and their representations

List the positive factors of 72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72. Now retain only those that are multiples of 6. The qualifying numbers are 6, 12, 24, 36, and 72. Although 18 is a multiple of 6, it is not a factor of 72 because 72 ÷ 18 = 4 exactly—actually it is a factor; therefore check carefully: 18 should also be included. This reveals that option B is incomplete and option A is correct.

The elements can be written as 1 = 1³, 8 = 2³, 27 = 3³, and 64 = 4³. Therefore every element has the form n³, where n is a natural number from 1 through 4. Option A gives squares, option C gives powers of 2, and option D gives an arithmetic sequence, so none of those represents the given set.

A natural number has exactly two distinct positive factors precisely when it is prime: the number itself and 1. The prime numbers less than 15 are 2, 3, 5, 7, 11, and 13. The number 1 is not prime because it has only one positive factor, namely 1. Hence the required set is option B.

Since x/2 is an integer, x must be divisible by 2; therefore x must be an even integer. The strict inequality −5 < x < 5 allows the integers −4 through 4, but among them the even integers are −4, −2, 0, 2, and 4. Zero is included because 0/2 = 0, which is an integer. Thus option A is correct.

Factor the quadratic: x² − 5x + 6 = (x − 2)(x − 3). This product is negative strictly between its roots, so 2 < x < 3. There is no natural number strictly between 2 and 3. The roots themselves do not qualify because the inequality is strict and gives zero at x = 2 or x = 3. Therefore, J₁ is the empty set, option A.

Factor the quadratic: x² − 5x + 6 = (x − 2)(x − 3). Since the parabola opens upward, the expression is less than or equal to zero between the roots, including the endpoints. Thus 2 ≤ x ≤ 3. The natural numbers in this interval are only 2 and 3, so the roster form is K₁ = {2, 3}.

For an absolute-value equation |x + 1| = 3, there are two possible cases: x + 1 = 3 or x + 1 = −3. These give x = 2 and x = −4, respectively. Both values are integers and satisfy the original equation, so both must be included. Therefore, the roster form is L₁ = {−4, 2}.

The governing concept is finding the intersection of the divisors of 100 with the perfect squares. The positive divisors are 1, 2, 4, 5, 10, 20, 25, 50, and 100. Among them, 1 = 1², 4 = 2², 25 = 5², and 100 = 10² are squares. Thus M₁ = {1, 4, 25, 100}, so option A is correct. Option C lists all divisors rather than only squares, and 16 in D is square but does not divide 100.

A number divisible by both 4 and 6 must be divisible by their least common multiple. Since lcm(4, 6) = 12, the required numbers are multiples of 12. The positive multiples of 12 that are at most 50 are 12, 24, 36 and 48. Thus, the roster form is O₁ = {12, 24, 36, 48}. Numbers such as 4 or 6 satisfy only one of the two divisibility conditions.

The positive multiples of 5 less than 30 are 5, 10, 15, 20, and 25. The positive multiples of 7 less than 30 are 7, 14, 21, and 28. Since “or” means that either condition may hold, combine both lists and remove repetitions. The result is option A; 30 is excluded because the inequality is strict.

The governing concept is combining an inequality with an integer and parity condition. From x² ≤ 9, we get |x| ≤ 3, so the possible integers are −3, −2, −1, 0, 1, 2, and 3. Filtering these for odd values leaves −3, −1, 1, and 3. Therefore R₁ = {-3, -1, 1, 3}, making option A correct. B includes even integers, C omits valid endpoints, and D incorrectly includes zero.

The natural-number factors of 45 are 1, 3, 5, 9, 15 and 45. Adding 2 to these values gives 3, 5, 7, 11, 17 and 47, respectively. Every one of these results is prime. Hence every factor of 45 satisfies the second condition as well, so the complete roster form is S₁ = {1, 3, 5, 9, 15, 45}. Option C is incorrect because it omits valid factors.

Factor the quadratic equation: x² − 2x − 8 = (x − 4)(x + 2) = 0. Hence x = 4 or x = −2. Both values belong to the integers and satisfy the defining equation. A set has no order requirement, so {−2, 4} and {4, −2} would describe the same set; among the given choices, option A is correct.

The roster form lists every natural number from 10 through 40 that contains the digit 3 in at least one position. The numbers 13 and 23 contain 3 in the units place, while 30 through 39 contain it in the tens place. Number 40 does not contain 3, and 43 is outside the interval. Therefore option A is complete and correct.

A natural number that is both a perfect square and a perfect cube must be a perfect sixth power, because the least common multiple of the exponents 2 and 3 is 6. The sixth powers below 100 are 1⁶ = 1 and 2⁶ = 64; 3⁶ = 729 is already greater than 100. Therefore, the required set in roster form is {1, 64}, so option A is correct. Option B lists ordinary cubes, while option C lists squares without applying both conditions.

To convert the set-builder form into roster form, substitute every permitted natural-number value of n into x = n/(n + 1). For n = 1, 2, 3, and 4, the values are respectively 1/2, 2/3, 3/4, and 4/5. Listing these distinct values gives A = {1/2, 2/3, 3/4, 4/5}. Option B reverses the fractions, while C and D do not use the stated formula correctly.

We need the natural numbers less than 10 that have exactly two positive divisors: 1 and the number itself. These are 2, 3, 5, and 7. The number 1 is not prime because it has only one positive divisor, and 9 is composite because it has divisors 1, 3, and 9. Hence A = {2, 3, 5, 7}.

In the usual elementary classification of the English alphabet, the vowels are a, e, i, o, and u. The letters b, c, and d are consonants, so option B is incorrect. The letter y is not included in the standard vowel list used in this question. Hence the complete roster form is option A.

Since x is an integer, we list the integers between the boundary values. The strict inequality -3 < x excludes -3, while x ≤ 2 includes 2. The integers satisfying both conditions are -2, -1, 0, 1, and 2. Therefore, the roster form is C = {-2, -1, 0, 1, 2}.

A mathematical set must be well-defined, meaning that membership can be decided objectively and consistently. The even natural numbers less than 10 are exactly 2, 4, 6, and 8, so there is no ambiguity. In contrast, good, beautiful, and interesting depend on personal opinion. Therefore option C describes a set.