This question tests a property of the greatest common factor (GCF). The GCF of two numbers that share a common factor is that common factor multiplied by the GCF of the remaining parts. Let G = GCF(m, n). Then m = Gx and n = Gy, where x and y are coprime. Column A is GCF(12Gx, 12Gy) = 12G × GCF(x, y) = 12G × 1 = 12G. Column B is 12 × GCF(m, n) = 12G. The two quantities are always equal.

This question tests middle school quantitative reasoning skills: understanding and applying divisibility and factors. Divisibility means a number can be divided by another number without leaving a remainder. Factors are numbers you can multiply to get another number. In this problem, the scenario involves finding the missing factor in 9 × __ = 72, requiring understanding of inverse operations. The correct answer, choice B (8), is correct because 9 × 8 = 72, making 8 the missing factor. This can be found by dividing: 72 ÷ 9 = 8. Choice C (9) would give 9 × 9 = 81, not 72. To help students: Encourage using division to find missing factors (72 ÷ 9 = ?). Teach checking answers by multiplying back. Emphasize the relationship between multiplication and division as inverse operations.

This question tests middle school quantitative reasoning skills: understanding and applying divisibility and factors. Divisibility means a number can be divided by another number without leaving a remainder. Factors are numbers you can multiply to get another number. In this problem, the scenario involves finding the greatest common factor (GCF) of 60 cookies and 36, requiring understanding of common factors. The correct answer, choice B (12), is correct because the factors of 60 are {1,2,3,4,5,6,10,12,15,20,30,60} and the factors of 36 are {1,2,3,4,6,9,12,18,36}, with 12 being the largest number in both lists. Choice C (18) is incorrect because while 18 is a factor of 36, it is not a factor of 60 (60 ÷ 18 = 3.33...). To help students: Encourage listing all factors of each number systematically. Teach identifying common factors by finding numbers in both lists. Emphasize that the GCF is the largest of the common factors.

This question tests middle school quantitative reasoning skills: understanding and applying divisibility and factors. Divisibility means a number can be divided by another number without leaving a remainder. Factors are numbers you can multiply to get another number. In this problem, the scenario involves identifying the complete set of factors of 60 cookies, requiring systematic factor finding. The correct answer, choice A, is correct because it lists all factors of 60: 1×60, 2×30, 3×20, 4×15, 5×12, and 6×10 are all the factor pairs. Choice B incorrectly includes 7, which is not a factor (60 ÷ 7 = 8.57...). To help students: Encourage finding factors in pairs starting from 1. Teach checking completeness by verifying each number divides 60 evenly. Emphasize that the list should include all numbers from 1 up to the square root, plus their pairs.

This question tests middle school quantitative reasoning skills: understanding and applying divisibility and factors. Divisibility means a number can be divided by another number without leaving a remainder. Factors are numbers you can multiply to get another number. In this problem, the scenario involves identifying which number is a factor of 36 chairs, requiring understanding of what makes a number a factor. The correct answer, choice B (6), is correct because 6 × 6 = 36, making 6 a factor of 36. Choice A (7) is incorrect because 36 ÷ 7 = 5.14..., which is not a whole number, so 7 is not a factor of 36. To help students: Encourage practice with factor pairs (1×36, 2×18, 3×12, 4×9, 6×6). Teach systematic checking by dividing to see if you get whole numbers. Emphasize that factors always divide evenly into the original number.

This question tests middle school quantitative reasoning skills: understanding and applying divisibility and factors. Divisibility means a number can be divided by another number without leaving a remainder. Factors are numbers you can multiply to get another number. In this problem, the scenario involves identifying which number is NOT a factor of 36 chairs, requiring understanding of factor relationships. The correct answer, choice D (8), is correct because 36 ÷ 8 = 4.5, which is not a whole number, meaning 8 is not a factor of 36. Choices A (9), B (3), and C (4) are all factors of 36 because 36 ÷ 9 = 4, 36 ÷ 3 = 12, and 36 ÷ 4 = 9, all giving whole number results. To help students: Encourage listing all factor pairs systematically. Teach checking each option by division. Emphasize that non-whole quotients indicate non-factors.

When you see a question about divisibility and remainders, think about what the given information tells you about the number's properties. If a positive integer $$N$$ is divisible by 6, this means $$N = 6k$$ for some positive integer $$k$$.

Since $$6 = 2 \times 3$$, any number divisible by 6 must also be divisible by both 2 and 3. This is a fundamental property of divisibility: if a number is divisible by a product of factors, it's divisible by each individual factor.

For Column A: When $$N$$ is divided by 2, since $$N$$ is divisible by 2, the remainder is 0.

For Column B: When $$N$$ is divided by 3, since $$N$$ is divisible by 3, the remainder is also 0.

Therefore, both quantities equal 0, making them equal.

Looking at the wrong answers: Choice B suggests the remainder when dividing by 3 is greater, but both remainders are 0. Choice C suggests the remainder when dividing by 2 is greater, which is also incorrect for the same reason. Choice D claims the relationship cannot be determined, but the divisibility by 6 gives us complete information about both remainders.

Study tip: Remember that divisibility by a composite number automatically means divisibility by all its prime factors. When a number is divisible by another, the remainder is always 0. Questions testing this concept often try to confuse you by asking about remainders when the given information already tells you the remainder must be zero.

When you see questions about distinct prime factors, you need to find the prime factorization of each number and count how many different prime numbers appear in each factorization.

Let's find the prime factorization of 120 first. Start by dividing by the smallest prime, 2: $$120 = 2^3 \times 15 = 2^3 \times 3 \times 5$$. So 120 has the distinct prime factors 2, 3, and 5 - that's 3 distinct prime factors.

Now for 90: $$90 = 2 \times 45 = 2 \times 9 \times 5 = 2 \times 3^2 \times 5$$. So 90 also has the distinct prime factors 2, 3, and 5 - that's also 3 distinct prime factors.

Both quantities equal 3, making them equal.

Choice A is correct because both numbers have exactly 3 distinct prime factors. Choice B is wrong because Column B (3) is not greater than Column A (3). Choice C is wrong because Column A (3) is not greater than Column B (3). Choice D is wrong because we have sufficient information to determine that both quantities are equal - there's no ambiguity here.

Remember that "distinct prime factors" means you count each prime number only once, regardless of how many times it appears in the factorization. The exponent doesn't matter for this count - whether it's $$2^1$$ or $$2^3$$, the number 2 contributes just one distinct prime factor. Practice prime factorization systematically by always starting with the smallest prime and working your way up.

This problem requires finding the least common multiple (LCM) of 12 and 10 to determine the smallest number that is a multiple of both. The prime factorization of 12 is 2² × 3. The prime factorization of 10 is 2 × 5. The LCM is the product of the highest powers of all prime factors involved: LCM(12, 10) = 2² × 3 × 5 = 4 × 3 × 5 = 60. Therefore, the minimum number of pencils (and erasers) the store must sell to have an equal amount of each is 60.

Let the three consecutive even integers be represented by \(2n\), \(2n+2\), and \(2n+4\), where \(n\) is an integer. Their sum is \(2n + (2n+2) + (2n+4) = 6n + 6\). This sum can be factored as \(6(n+1)\). Since the sum can always be expressed as 6 times an integer, it is always divisible by 6. To confirm 6 is the greatest such integer, we can test a few cases. For 2, 4, 6, the sum is 12 (divisible by 6). For 4, 6, 8, the sum is 18 (divisible by 6). Since the sums are not always divisible by 12 (e.g., 18), the greatest integer that must be a factor is 6.