Divisibility Rules for 3, 6 and 9: Practice Problems

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.

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 if the number $132$ is divisible by $3$, we can apply the rule for divisibility by $3$, which involves summing the digits of the number.

Step-by-step solution:

• Step 1: Sum the digits of the number $132$.
  $1 + 3 + 2 = 6$
• Step 2: Check if the sum is divisible by $3$.
  $6 \div 3 = 2$, which is an integer.

Since the sum of the digits is $6$ and $6$ is divisible by $3$, the number $132$ is also divisible by $3$.

Therefore, the number $132$ is divisible by $3$, and the correct choice is:

Yes

To determine if 352 is divisible by 3, we need to follow these steps:

• Calculate the sum of its digits.
• Check if this sum is divisible by 3.

Let's work through the procedure:

The number $352$ consists of the digits 3, 5, and 2.

Step 1: Calculate the sum of the digits.

The sum is $3 + 5 + 2 = 10$.

Step 2: Check if 10 is divisible by 3.

Since 10 divided by 3 gives a remainder, 10 is not divisible by 3.

Therefore, the number 352 is not divisible by 3.

The correct answer is No.

To determine if 673 is divisible by 3, we must use the divisibility rule for 3, which states that a number is divisible by 3 if the sum of its digits is divisible by 3.

First, we'll calculate the sum of the digits: $6 + 7 + 3$.

Calculating this, we get: $6 + 7 + 3 = 16$.

Next, we check if 16 is divisible by 3. Dividing 16 by 3 gives a quotient of 5 and a remainder of 1.

Since 16 is not divisible by 3 (as it leaves a remainder), we conclude that 673 is not divisible by 3.

Thus, the correct answer is No.