Divisibility Rules: Is Every Multiple of 6 Also a Multiple of 3?

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.