To find the maximum number of students, we need the GCF of 48 and 72. Using prime factorization: 48 = 2⁴ × 3 and 72 = 2³ × 3². The GCF is 2³ × 3 = 24. So 24 students can receive items, and each gets 48 ÷ 24 = 2 stickers. Choice B uses 12 (a factor but not the greatest), Choice C uses 8 (a factor but not the greatest), and Choice D uses 6 (a factor but not the greatest).

To use the distributive property correctly, we need the GCF of 45 and 75. Since 45 = 3² × 5 and 75 = 3 × 5², the GCF is 3 × 5 = 15. So 45 + 75 = 15(3 + 5). Choice A uses 5 (a common factor but not the greatest), Choice C incorrectly calculates 9(5 + 15) = 9(20) = 180 ≠ 120, and Choice D uses 25 (not a factor of 45).

We need the LCM of 8 and 12. Since 8 = 2³ and 12 = 2² × 3, the LCM is 2³ × 3 = 24. The buses meet at the station together at 0 minutes (start), 24 minutes (first time together again), and 48 minutes (second time together again). Choice A gives the first time they meet again, Choice C is 1.5 × 24 (not a multiple), and Choice D is 4 × 24 (fourth time together).

When you see problems about events happening on regular schedules, you need to find the Least Common Multiple (LCM) to determine when they'll coincide.

Item X restocks every 9 days and Item Y restocks every 12 days. To find when they'll both restock together, you need the LCM of 9 and 12. Start by finding the prime factorization: $$9 = 3^2$$ and $$12 = 2^2 \times 3$$. The LCM uses the highest power of each prime factor: $$2^2 \times 3^2 = 4 \times 9 = 36$$. This means both items will restock together every 36 days.

Since they both restocked today (day 0), the pattern will be:

• First coincidence: 36 days from now
• Second coincidence: 72 days from now
• Third coincidence: 108 days from now

The answer is C) 108 days from now.

Looking at the wrong answers: A) 72 days represents the second coincidence, not the third. B) 144 days would be the fourth coincidence ($$36 \times 4 = 144$$). D) 36 days is the first coincidence when both items restock together again.

The key strategy here is recognizing that "coinciding schedules" problems always require finding the LCM. Once you have that base interval, multiply it by the position number you need (first, second, third, etc.). Don't get confused by thinking the LCM itself is the final answer—it's just the repeating interval between coincidences.

We need the GCF of 36 and 60. Since 36 = 2² × 3² and 60 = 2² × 3 × 5, the GCF is 2² × 3 = 12. This means 12 chairs per row is the maximum possible. Choice A (6 chairs) is a common factor but not the greatest, Choice C (18 chairs) doesn't divide 60 evenly (60 ÷ 18 = 3.33...), and Choice D (4 chairs) is a common factor but not the greatest.

First find when they meet: LCM(8, 10) = 40 minutes. They meet at the start at 0, 40, 80, and 120 minutes. The third meeting is at 120 minutes. In 120 minutes, Sarah completes 120 ÷ 8 = 15 laps. Choice B (12 laps) is what she completes at the second meeting (80 minutes), Choice C (30 laps) doubles the correct answer, and Choice D (10 laps) is David's lap count at the third meeting.

When you see a problem asking for the maximum number of identical groups using all items, you're looking for the Greatest Common Factor (GCF). Emma needs to divide both her stickers and erasers equally among the bags, so the number of bags must be a factor of both 40 and 64.

To find the GCF of 40 and 64, you can list the factors or use prime factorization. Let's find the factors: 40 has factors 1, 2, 4, 5, 8, 10, 20, 40, and 64 has factors 1, 2, 4, 8, 16, 32, 64. The common factors are 1, 2, 4, and 8, making 8 the greatest common factor.

This means Emma can make 8 bags maximum. To find erasers per bag, divide the total erasers by the number of bags: $$64 ÷ 8 = 8$$ erasers per bag. You can verify this works for stickers too: $$40 ÷ 8 = 5$$ stickers per bag.

Answer A (4 erasers) would require 16 bags, but $$40 ÷ 16 = 2.5$$ stickers per bag, which isn't whole. Answer B (16 erasers) would need only 4 bags, but this isn't the maximum possible. Answer D (2 erasers) would require 32 bags, but $$40 ÷ 32 = 1.25$$ stickers per bag, again not whole.

The correct answer is C.

Remember: when a problem asks for the maximum number of equal groups using all items, find the GCF of the quantities. This ensures you can divide everything evenly with no items left over.

When you see a problem asking to factor out a common factor so that the remaining terms have no common factors, you're working with the greatest common factor (GCF) and the concept of relatively prime numbers.

To solve $$56 + 84$$, you need to find the GCF of 56 and 84, then factor it out completely. Let's find the GCF by listing the factors: 56 = 2³ × 7 and 84 = 2² × 3 × 7. The GCF is 2² × 7 = 28.

Factoring out 28: $$56 + 84 = 28(2 + 3)$$. Now check if 2 and 3 have any common factors greater than 1. Since 2 and 3 are both prime numbers, their only common factor is 1, making them relatively prime. This confirms that answer choice D is correct.

Let's examine why the other choices fail. Choice A factors out only 4, leaving $$4(14 + 21)$$. While 14 and 21 might seem relatively prime at first glance, they actually share a common factor of 7 (since 14 = 2 × 7 and 21 = 3 × 7). Choice B factors out 14, giving $$14(4 + 6)$$, but 4 and 6 share a common factor of 2. Choice C factors out 7, resulting in $$7(8 + 12)$$, but 8 and 12 share a common factor of 4.

The key strategy is to always factor out the complete GCF first, then verify that the remaining terms are truly relatively prime by checking if they share any common factors greater than 1.

This question tests correcting a misconception about LCM, not always the product. LCM is smallest common multiple, like LCM(6,8)=24, not 48. Student said 48, a multiple but not least; correct LCM=24 from multiples. Statement B is correct: incorrect, LCM is 24. Errors: thinking product always LCM (only if coprime), or LCM smaller than numbers. Find LCM by multiples list or primes: 6=2*3, $8=2^3$, $LCM=2^3$*3=24. Applications: timing like bus schedules.

When you see a problem about events happening at regular intervals and need to find when they'll occur together again, you're looking for the Least Common Multiple (LCM). This tells you the smallest amount of time that passes before all cycles repeat simultaneously.

To find when all three clocks ring together again, you need the LCM of their intervals: 6 minutes, 9 minutes, and 4 minutes. Start by finding the prime factorization of each number:

• 6 = 2 × 3
• 9 = 3²
• 4 = 2²

For the LCM, take the highest power of each prime factor that appears: 2² × 3² = 4 × 9 = 36. So all three clocks will ring together every 36 minutes.

Since they all rang at 3:00 PM, add 36 minutes: 3:00 PM + 36 minutes = 3:36 PM.

Looking at the wrong answers: Choice A (3:18 PM) represents 18 minutes after 3:00, which is the LCM of just clocks A and C (6 and 4). Choice B (4:12 PM) is 72 minutes later, which would be when they ring together for the second time after 3:00. Choice C (3:54 PM) doesn't correspond to any meaningful LCM calculation and may result from adding the intervals incorrectly.

Study tip: When solving "meeting again" problems, always find the LCM of all the given intervals. Don't confuse this with finding when just two of the items will coincide - you need all of them together.