Divisibility Rules for 3, 6 and 9: Worded problems

To solve this problem, we'll follow these steps:

• Calculate the LCM of 6 and 4.
• Find the number between 21 and 31 that is a multiple of this LCM.

Step 1: Calculate the LCM of 6 and 4.
- The prime factorization of 6 is $2 \times 3$.
- The prime factorization of 4 is $2^2$.
- The LCM is derived by taking the highest power of each prime number involved: $LCM(6, 4) = 2^2 \times 3 = 12$.

Step 2: Identify the multiples of 12 within the range (21, 31).
- Multiples of 12 are: $12, 24, 36, \ldots$
- In the range between 21 and 31, the only multiple of 12 is 24.

Therefore, the number of students in the class is $24$.

To determine the number of students, let's find all multiples of 8 within the given range of 29 to 39.

• Calculating multiples of 8: $8 \times 1 = 8$, $8 \times 2 = 16$, $8 \times 3 = 24$, $8 \times 4 = 32$, $8 \times 5 = 40$.

The relevant multiple of 8 that falls within the 29 to 39 range is $32$.

Thus, the number of students in the class is $32$.

This means that on any given day, all students can indeed be grouped as specified without any remainder.

Therefore, the solution to the problem is $32$.

To solve the problem, we first determine the LCM of the numbers 2, 4, and 5:

• The prime factors of 2 are $2$.
• The prime factors of 4 are $2^2$.
• The prime factors of 5 are $5$.

The LCM is determined by taking the highest power of each prime number: $2^2$ and $5$, so $\text{LCM}(2, 4, 5) = 2^2 \times 5 = 20$.

Next, we consider numbers between 19 and 29: 20, 21, 22, 23, 24, 25, 26, 27, 28, 29.
Among these, only 20 is divisible by 20.

Therefore, the number of flowers is $20$.

So, the solution to the problem is: $\text{Number of flowers} = 20$.