Correct: C. 42

The greatest common divisor of 84 and 126 is 42 because 42 divides both numbers exactly. The prime factorizations are 84 = 2² × 3 × 7 and 126 = 2 × 3² × 7, so the common prime factors with the smallest exponents give 2 × 3 × 7 = 42. Finding common factors through prime factorization is a standard method for determining the greatest common divisor.

Correct: B. 12

The number 72 has 12 positive factors. Its prime factorization is 72 = 2³ × 3², so the number of positive factors is found by multiplying one more than each exponent: (3 + 1)(2 + 1) = 12. This factor-counting method follows from choosing independently how many times each prime appears in a factor.

Correct: C. 72

The least common multiple of 18 and 24 is 72 because 72 is the smallest positive number divisible by both. Their prime factorizations are 18 = 2 × 3² and 24 = 2³ × 3, so the LCM uses the highest exponent of each prime, giving 2³ × 3² = 72. LCM calculations are useful when combining repeating cycles or finding a common denominator for fractions.

Correct: C. 7

The units digit of 7^45 is 7 because powers of 7 repeat their units digits in the cycle 7, 9, 3, 1. The cycle has length 4, and 45 leaves a remainder of 1 when divided by 4, so the units digit matches the first value in the cycle. This cyclic pattern allows large powers to be handled without calculating the entire number.

Correct: B. 16-May

The fraction 5/16 has a terminating decimal expansion because 16 is 2? and therefore its denominator contains only the prime factor 2 after simplification. A rational fraction in lowest terms has a terminating decimal expansion when its denominator has no prime factors other than 2 and 5. The other denominators contain factors such as 3, so their decimal expansions repeat rather than terminate.

Correct: C. 60

The smallest positive integer divisible by 4, 6, and 15 is 60. Their prime factorizations are 4 = 2², 6 = 2 × 3, and 15 = 3 × 5, so the LCM is 2² × 3 × 5 = 60. Any common multiple must contain enough factors to include each of these prime powers, making 60 the least possible positive value.

Correct: C. 74

The number is 74 because its digits satisfy both conditions: 7 + 4 = 11 and 7 is 3 greater than 4. Let the units digit be x, making the tens digit x + 3, so x + x + 3 = 11 gives x = 4 and the tens digit becomes 7. This digit-based algebraic method is useful for reconstructing numbers from constraints on their place values.

Correct: A. 2² × 3² × 5

The prime factorization of 180 is 2² × 3² × 5. Dividing 180 by 2 twice gives 45, and 45 factors as 3² × 5, so all factors are prime and the product is 180. Prime factorization expresses a composite number uniquely as a product of prime numbers.

Correct: B. 992

The greatest three-digit number divisible by 8 is 992 because 992 ÷ 8 = 124. The next three-digit number after 992 is 993, so no larger three-digit multiple of 8 exists. Divisibility by 8 can also be checked using the last three digits of a whole number.

Correct: A. 7

The smallest positive number is 7 because 7 ÷ 5 leaves remainder 2 and 7 ÷ 4 leaves remainder 3. The number satisfies both remainder conditions simultaneously, so it is the first positive solution. Problems of this type can be solved by listing multiples of one divisor and checking the required remainder for the other.

Correct: A. 1

The units digit of 3²? is 1 because powers of 3 have the repeating units-digit cycle 3, 9, 7, 1. The cycle has length 4, and 20 is divisible by 4, so the fourth digit in the cycle, 1, is the result. Cyclic patterns make it possible to determine the last digit of very large powers without calculating the entire value.

Correct: B. 5

The value of a is 6 because multiplying powers with the same base means adding their exponents, so 2^a × 2³ = 2^(a+3). Equating the exponent to 9 gives a + 3 = 9, so a = 6. This follows the product property of exponents.

Correct: B. 2

The remainder when 2¹? is divided by 7 is 2 because 2¹? = 1024 and 1024 = 7 × 146 + 2. The powers of 2 modulo 7 repeat in the cycle 2, 4, 1, which can also be used to solve the problem efficiently. Since 10 leaves remainder 1 when divided by the cycle length 3, the resulting remainder matches 2¹ modulo 7.

Correct: A. 4

The smallest positive integer with exactly three positive factors is 4 because its factors are 1, 2, and 4. A number has exactly three positive factors precisely when it is the square of a prime number, and 4 = 2². The factor-counting rule for 2² gives (2 + 1) = 3 factors.

Correct: A. 30

The smallest positive integer divisible by every integer from 1 through 5 is 60? Actually, the least common multiple of 1, 2, 3, 4, and 5 is 60, so the correct option should be D. The prime powers needed are 2², 3, and 5, whose product is 60.

Correct: C. ?50

?50 is irrational because ?50 = 5?2, and ?2 is irrational, so multiplying it by the nonzero rational number 5 remains irrational. By contrast, 0.125 and 3.5 are terminating decimals and therefore rational, while ?49 = 7 is an integer. Simplifying radicals before classifying them can make the distinction easier to see.

Correct: C. 48

The greatest common factor of 96 and 144 is 48 because 48 divides both numbers exactly. Their prime factorizations are 96 = 2^5 × 3 and 144 = 2^4 × 3^2, so the common factors with the smaller exponents give 2^4 × 3 = 48. Prime factorization provides a systematic way to determine the greatest common factor of two integers.

Correct: A. 2728

2728 is divisible by 11 because 2728 ÷ 11 = 248 exactly. The divisibility test for 11 involves finding the difference between alternating sums of the digits, and for 2728 this gives (2 + 2) ? (7 + 8) = -11, a multiple of 11. This test can identify divisibility without carrying out the full division.

Correct: D. 12

The remainder is 11 because 13 × 18 = 234 and 245 ? 234 = 11. Therefore, 245 can be written as 13 × 18 + 11, which follows the division algorithm. The remainder must always be a nonnegative integer smaller than the divisor.

Correct: C. 12

The number 48 has 10 positive factors. Its prime factorization is 48 = 2^4 × 3, so the number of positive factors is (4 + 1)(1 + 1) = 10. Each factor is formed by independently choosing an exponent for 2 from 0 through 4 and for 3 from 0 through 1.

Correct: C. 60

The least common multiple of 12, 15, and 20 is 60 because 60 is divisible by all three numbers. Their prime factorizations are 12 = 2^2 × 3, 15 = 3 × 5, and 20 = 2^2 × 5, so the required highest prime powers give 2^2 × 3 × 5 = 60. The LCM is the smallest positive integer containing enough prime factors to be a multiple of every given number.

Correct: D. 9

The units digit of 9^37 is 9 because powers of 9 alternate between units digits 9 and 1. Odd powers of 9 end in 9, while even powers end in 1, and 37 is odd. This repeating pattern makes it unnecessary to calculate the full value of 9^37.

Correct: C. 6

The smallest number that must be subtracted from 1000 is 6 because 1000 leaves a remainder of 6 when divided by 7. Subtracting 6 gives 994, and 994 ÷ 7 = 142 exactly. The required subtraction is therefore the remainder itself when the goal is to reach the nearest lower multiple.

Correct: C. 36

The smallest positive integer divisible by 9, 12, and 18 is 36. Their prime factorizations are 9 = 3^2, 12 = 2^2 × 3, and 18 = 2 × 3^2, so the LCM is 2^2 × 3^2 = 36. The least common multiple gives the smallest number that contains the necessary prime factors for all three divisors.

Correct: C. 42

The smallest positive integer that is a multiple of both 14 and 21 is 42. Since 14 = 2 × 7 and 21 = 3 × 7, their least common multiple is 2 × 3 × 7 = 42. Any common multiple must contain the factors needed by both numbers, making 42 the smallest possible positive common multiple.

Correct: A. 6

The number 6 has exactly four positive factors: 1, 2, 3, and 6. Its prime factorization is 2 × 3, so the number of factors is (1 + 1)(1 + 1) = 4. Numbers with a prime factorization containing two distinct primes to the first power therefore have four positive factors.

Correct: C. n + 3

n + 3 must be even because adding an odd number to an even integer produces an odd result, so this option actually does not satisfy the condition. The correct expression must be checked carefully: n² is even when n is even, making n² + 1 odd as well. Therefore none of the listed expressions is necessarily even, so this question is invalid as written.

Correct: A. 21

21 is divisible by 3 because its digit sum is 3, but it is not divisible by 9 because 3 is not a multiple of 9. The divisibility rule for 3 uses the sum of the digits, while divisibility by 9 uses the same sum but requires it to be a multiple of 9. For 21, the digit sum provides an immediate distinction between the two tests.

Correct: B. 108

The smallest positive integer greater than 100 divisible by both 6 and 8 is 120. The least common multiple of 6 and 8 is 24, and the multiples around 100 are 96, 120, and 144, so 120 is the first one greater than 100. Using the LCM is the efficient way to combine the two divisibility requirements.

Correct: B. 2^3 × 3^2 × 5

The prime factorization of 360 is 2^3 × 3^2 × 5. Dividing 360 repeatedly by 2 gives 180, 90, 45, and then 45 factors as 3 × 3 × 5. Prime factorization expresses a composite number as a product of prime numbers and is unique apart from the order of the factors.

Correct: C. 24

The greatest common factor of 72 and 120 is 24 because 24 divides both numbers exactly. Their prime factorizations are 72 = 2^3 × 3^2 and 120 = 2^3 × 3 × 5, so the common prime factors give 2^3 × 3 = 24. Prime factorization provides a systematic method for finding the greatest common factor.

Correct: C. 48

The least common multiple of 16 and 24 is 48 because 48 is the smallest positive number divisible by both. Their prime factorizations are 16 = 2^4 and 24 = 2^3 × 3, so the LCM is 2^4 × 3 = 48. The LCM method is useful when two or more quantities must align at their first common multiple.

Correct: B. 12

The number 12 has exactly six positive factors: 1, 2, 3, 4, 6, and 12. Its prime factorization is 2^2 × 3, so the factor-count formula gives (2 + 1)(1 + 1) = 6. This formula counts every possible choice of prime exponents in a factor.

Correct: B. 2

The units digit of 2^31 is 2 because the units digits of powers of 2 repeat in the cycle 2, 4, 8, 6. Since 31 leaves a remainder of 3 when divided by 4, the third value in the cycle is selected. Cyclic patterns allow the final digit of very large powers to be found without calculating the entire power.

Correct: C. 60

The smallest positive integer greater than 50 divisible by both 4 and 9 is 72. The least common multiple of 4 and 9 is 36, and the multiples around 50 are 36 and 72, making 72 the first common multiple greater than 50. Because 4 and 9 have no common prime factor, their LCM is their product, 36.

Correct: A. 30

The smallest positive integer divisible by 3, 4, and 10 is 60, so the correct option should be C; therefore this question is invalid as written.