Divisibility Rules for 3, 6 and 9: Identifying and defining elements

In order to determine if a number divisible by 6 is also divisible by 2, we first review the divisibility rules:

• A number is divisible by 6 if it is divisible by both 2 and 3.
• A number is divisible by 2 if its last digit is even (one of 0, 2, 4, 6, or 8).

Consider a number $n$ that is divisible by 6. By definition, since 6 itself factors into 2 multiplied by 3, any number divisible by 6 must be divisible by 2 and 3. This means that any number divisible by 6 is automatically divisible by 2 because 2 is a part of its factorization.

Therefore, yes, any number divisible by 6 will necessarily be divisible by 2 as per the rule of divisibility.

Thus, the correct choice is:

• Choice 1: Yes

This conclusion adheres strictly to divisibility rules and confirms the assertion that being divisible by 6 includes being divisible by 2.

To determine whether a number divisible by 6 is necessarily divisible by 3, we need to understand the properties of divisibility for the numbers involved.

Let's analyze the problem step by step:

• Step 1: Restate the Problem
  We need to find out if any number that is divisible by 6 is also divisible by 3.
• Step 2: Identify Key Information and Variables
  - A number is divisible by 6 if it can be expressed as $k \times 6$ for any integer $k$.
  - We want to check if such a number is also divisible by 3, meaning it can also be expressed as $m \times 3$ for some integer $m$.
• Step 3: Relevant Theorems
  - A number is divisible by 6 if it is divisible by both 2 and 3.
• Step 4: Choose Approach
  We'll use the divisibility rules for numbers to deduce if a number divisible by 6 must be divisible by 3.
• Step 5: Steps for Solution
  1. Given a number is divisible by 6, it is expressed as a multiple of 6: $n = k \times 6$.
  2. Since 6 can be factored into $2 \times 3$, a number divisible by 6 is also divisible by 3.
  3. Therefore, $n = k \times 6 = k \times (2 \times 3) = (k \times 2) \times 3$, making it divisible by 3.
• Step 6: Assumptions
  We assume the integer $k$ is any integer and does not affect the general proof.
• Step 7: Conclusion
  Every number divisible by 6 is necessarily divisible by both 2 and 3, due to the factorization properties of numbers. Thus, by the rules of divisibility, a number divisible by 6 is necessarily divisible by 3.

Therefore, the answer to the problem is Yes.

To determine if a number divisible by 2 is also divisible by 6, we need to understand the rules of divisibility:

• A number is divisible by 2 if it is an even number, i.e., its last digit is 0, 2, 4, 6, or 8.
• A number is divisible by 6 if it meets two conditions: it is divisible by 2 (i.e., it is even) and divisible by 3.
• A number is divisible by 3 if the sum of its digits is divisible by 3.

Now, let's analyze the implication of these rules:

Since a number divisible by 2 is even, it satisfies the first condition for divisibility by 6. However, it still needs to meet the second condition—divisibility by 3—to be divisible by 6. This implies that not all even numbers (divisible by 2) are multiples of 3.

For example, consider the number 4:

• 4 is divisible by 2 because it is even.
• However, 4 is not divisible by 3 because the sum of its digits (4) is not divisible by 3.
• Therefore, 4 is not divisible by 6 because it doesn't meet both required conditions.

Therefore, understanding the definitions, we can see that a number divisible by 2 is not necessarily divisible by 6. These two criteria must both be met for a number to be divisible by 6. Consequently, the correct answer is "No".

In conclusion, a number being divisible by 2 does not guarantee that it is divisible by 6.

To solve this problem, we need to understand the divisibility rules for 3 and 9:

• A number is divisible by 3 if the sum of its digits is divisible by 3.
• A number is divisible by 9 if the sum of its digits is divisible by 9.

Let's evaluate whether a number divisible by 3 is necessarily divisible by 9:

Consider the number 12. The sum of its digits is $1 + 2 = 3$, which is divisible by 3, so 12 is divisible by 3. However, when we check divisibility by 9, 12 is not divisible by 9 because 3 is not divisible by 9.

Now consider another number, like 18. The sum of its digits is $1 + 8 = 9$, which is divisible by both 3 and 9. Thus, 18 is divisible by both.

These examples demonstrate that while some numbers divisible by 3 are also divisible by 9 (e.g., 18), not all are (e.g., 12).

Therefore, a number being divisible by 3 does not necessarily mean it is divisible by 9.

The correct answer is No.

To solve this problem, we need to apply the divisibility rules for both 9 and 3.

• Step 1: Understand divisibility by 9.
  A number is divisible by 9 when the sum of its digits is a multiple of 9.
• Step 2: Understand divisibility by 3.
  A number is divisible by 3 when the sum of its digits is a multiple of 3.
• Step 3: Relate the rules.
  Since any multiple of 9 (given by the sum of digits) is also a multiple of 3, a number divisible by 9 is necessarily divisible by 3.

Therefore, it follows that if a number is divisible by 9, it must be divisible by 3, because the divisibility by 9 inherently satisfies the divisibility condition for 3.

Thus, the correct answer is Yes.

To determine if a number divisible by 3 is necessarily divisible by 6, we must apply the divisibility rules for both 3 and 6:

• A number is divisible by 3 if the sum of its digits is divisible by 3.
• A number is divisible by 6 if it is divisible by both 2 and 3. Thus, it must also be an even number.

To explore this question, let's consider a counterexample:

Take the number $9$. The sum of its digits is $9$, which is divisible by 3, so 9 is divisible by 3.

However, 9 is not even, so it is not divisible by 2. As a result, 9 is not divisible by 6 (because it does not satisfy the requirement to be divisible by both 2 and 3).

This counterexample demonstrates that a number divisible by 3 is not necessarily divisible by 6.

Therefore, the statement is incorrect, and the answer is No.

To determine if a number divisible by 9 is necessarily divisible by 6, let's explore the divisibility rules.
A number is divisible by 9 if the sum of its digits is divisible by 9. Consequently, such a number is also divisible by 3 since divisibility by 9 implies divisibility by 3.
For divisibility by 6, a number must be divisible by both 2 and 3. We've established that a number divisible by 9 is also divisible by 3, so we now need to check whether it is necessarily divisible by 2.

Consider an example: the number 27 is divisible by 9 since $2 + 7 = 9$, which is divisible by 9. However, 27 is odd (since $27 \div 2 = 13.5$), and thus, not divisible by 2.
Since 27 is not divisible by both 2 and 3, this number is not divisible by 6.

Therefore, a number divisible by 9 is not necessarily divisible by 6. The correct answer is No.

To solve this problem, identify the least common multiple (LCM) of the group sizes required: 4, 6, and 9.

• Calculate the LCM of 4, 6, and 9:
  • Prime factorize each number:
    • 4 = $2^2$
    • 6 = $2 \times 3$
    • 9 = $3^2$
  • The LCM is found by taking the highest power of each prime number present in any factorization: $(2^2 \times 3^2 = 36)$.
• Within the range 29 to 39, check which numbers are divisible by 36.
• The only number within this range that is exactly divisible by 36 is 36 itself.

Therefore, the number of students in the class is $36$.