Sets and their representations

A factor of 20 is a natural number that divides 20 exactly, leaving no remainder. The factor pairs are 1 × 20, 2 × 10, and 4 × 5. Thus the complete list of positive natural factors is 1, 2, 4, 5, 10, and 20. Both 1 and 20 must be included, because 1 divides every natural number and every number divides itself. Hence option A is correct.

Two sets are equal when they contain exactly the same distinct elements; the order of listing and repetition do not matter. In option A, A contains 1, 2, and 3, while B also contains the same distinct elements because the repeated 2 is counted only once. Thus A = B. Every other option has at least one different or additional element.

Let the tens digit be a and the units digit be b. For a two-digit number, a is at least 1, and a + b = 3. The possible digit pairs are (1,2), (2,1), and (3,0), producing 12, 21, and 30. The number 3 is not two-digit, and 11 has digit sum 2. Therefore option A is correct.

Rewrite the equation as x³ − x = 0 and factor it: x(x² − 1) = 0, so x(x − 1)(x + 1) = 0. The possible solutions are x = 0, x = 1, and x = −1. All three are integers and therefore satisfy the restriction x ∈ ℤ. Hence the set is U = {−1, 0, 1}, making option A correct.

The listed elements are exactly the first five positive multiples of 4. Substituting n = 1, 2, 3, 4, and 5 into 4n gives 4, 8, 12, 16, and 20, with no extra elements. Option B produces every positive even number up to 20, option C describes infinitely many multiples of 4, and option D includes every natural number from 4 through 20. Therefore, option A is the only exact set-builder representation.

Because x is an integer and x² < 25, x must lie between −4 and 4 inclusive. Thus the possible values are −4, −3, −2, −1, 0, 1, 2, 3, and 4. The additional condition 1 ≤ x² excludes only x = 0, because 0² = 0. The remaining eight integers are −4, −3, −2, −1, 1, 2, 3, and 4. Hence option A is correct.

A set records distinct objects, so a repeated letter is written only once. The word “mathematics” contains the letters m, a, t, h, e, m, a, t, i, c, and s. Removing repetitions leaves {m, a, t, h, e, i, c, s}. Option B repeats letters and therefore is not a proper roster form of a set, option C omits letters, and option D lists vowels rather than all distinct letters. Thus A is correct.

The positive divisors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36. We must retain only those divisors that are also perfect squares. Here, 1 = 1², 4 = 2², 9 = 3², and 36 = 6². The numbers 2, 3, 6, 12, and 18 are not perfect squares. Therefore, the required set is R = {1, 4, 9, 36}, so option A is correct.

Examine the integers 1 through 10. The numbers divisible by 2 are 2, 4, 6, 8, and 10; the numbers divisible by 3 are 3, 6, and 9. Since the phrase “neither 2 nor 3” excludes every number in either group, remove their union. The numbers left are 1, 5, and 7. Hence the required members are {1, 5, 7}, making option A correct.

A product is zero when at least one of its factors is zero. Therefore, solve x − 1 = 0 or x + 2 = 0. The first equation gives x = 1, while the second gives x = −2. Both values are integers and satisfy the original equation: (1−1)(1+2)=0 and (−2−1)(−2+2)=0. Thus the roster form is {−2, 1}; order does not matter in a set, so A is correct.

The elements 7, 14, 21, 28, and 35 are the first five positive multiples of 7. Writing an element as 7n and allowing n to take the values 1 through 5 produces exactly those five numbers. Option B includes every natural number up to 35, option C gives multiples of 5, and option D gives the first five square numbers. Therefore, the correct description is option A.

First list the integers satisfying −5 ≤ x < 2: they are −5, −4, −3, −2, −1, 0, and 1. Now select the even integers from this list. The even numbers are −4, −2, and 0; negative integers can also be even because divisibility by 2 is what matters. The endpoints −5 and 2 are not included as even members: −5 is odd and 2 is outside the strict upper bound. Hence A = {−4, −2, 0}.

Since x is a natural number and x < 10, the possible values are 1 through 9. A two-digit square must be at least 10 and at most 99. The first natural number whose square is at least 10 is 4, because 3² = 9 but 4² = 16. Every value from 4 through 9 has a square between 16 and 81, so each qualifies. Therefore B = {4, 5, 6, 7, 8, 9}, making option A correct.

The positive factors of 15 are 1, 3, 5, and 15, while the positive factors of 25 are 1, 5, and 25. The numbers occurring in both lists are 1 and 5. Therefore, the set of common factors is C = {1, 5}. A number must divide both 15 and 25 exactly to belong to C.

Every element of D is twice an integer, and the required values of n are 0, 1, 2, 3, and 4. Thus 2n produces 0, 2, 4, 6, and 8 exactly when n ∈ Z and 0 ≤ n ≤ 4. Option B can depend on whether a convention includes zero in N, so option A is the unambiguous form.

The positive divisors of 10 are 1, 2, 5, and 10. Set x + 2 equal to each divisor: x = −1, 0, 3, and 8, respectively. With the usual school convention N = {1, 2, 3, ...}, only 3 and 8 belong to the natural numbers. Hence E = {3, 8}, making option A correct. Option C incorrectly includes 0, and option D lists divisors rather than the corresponding values of x.

Rearrange the inequality: x² − 2x − 8 < 0. Factoring gives (x − 4)(x + 2) < 0, which holds strictly when −2 < x < 4. Since x must be an integer, the possible values are −1, 0, 1, 2, and 3. The endpoints −2 and 4 give equality, not a strict inequality, so they are excluded. Thus option A is correct.

The inequalities are strict, so neither 50 nor 70 can be included. The multiples of 5 between these endpoints are 55, 60, and 65. Each is a natural number and is divisible by 5, so all three satisfy the definition. Therefore, the roster form is I = {55, 60, 65}, which is option A. The other options incorrectly include one or both boundary values.

The positive divisors of 30 are 1, 2, 3, 5, 6, 10, 15, and 30. Applying x > 5 removes 1, 2, 3, and 5. The remaining divisors are 6, 10, 15, and 30, so J = {6, 10, 15, 30}. Option A is correct. Option B wrongly includes 5, while C omits 6 and D ignores the inequality condition.

The absolute-value condition |x| ≤ 2 means that x is at a distance of at most 2 from zero. Among the integers, the values satisfying this are −2, −1, 0, 1, and 2. Both endpoints are included because the inequality is less than or equal to, and zero is also an integer satisfying the condition. Therefore, option A gives the correct roster form.