Statement: If a number is divisible by 2 and by 3, then it is divisible by 6 and vice versa. Is this statement true? Give an example or counter example.
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To determine whether the statement is true or not, we can use the concept of divisibility. A number \(x\) is divisible by another number \(y\) if the remainder of \(x\) divided by \(y\) is zero.
In this case, the statement claims that if a number is divisible by both 2 and 3, then it must also be divisible by 6. Additionally, if a number is divisible by 6, it should be divisible by both 2 and 3.
To verify if the statement is true, let's consider an example. We can start by selecting a number that is divisible by both 2 and 3, such as 12. When we divide 12 by 2, we get a remainder of 0, and when we divide it by 3, we also get a remainder of 0. This confirms that 12 is divisible by 2 and 3.
Now, let's check if 12 is divisible by 6. Dividing 12 by 6 also gives us a remainder of 0, confirming that 12 is indeed divisible by 6. Hence, in this case, the statement holds true.
On the other hand, to check whether the vice versa part of the statement is true, we need to find a number that is divisible by 6 but not by either 2 or 3.
Consider the number 9. When we divide 9 by 2, we get a remainder of 1, indicating that it is not divisible by 2. Similarly, when we divide 9 by 3, we get a remainder of 0, indicating that it is divisible by 3. However, when we divide 9 by 6, we get a remainder of 3, meaning that it is not divisible by 6.
Therefore, the vice versa part of the statement is not true, as we found a counterexample (9) that is divisible by 3 but not by 6.
In conclusion, the first part of the statement (if a number is divisible by 2 and by 3, then it is divisible by 6) is true, but the vice versa part (if a number is divisible by 6, it is divisible by 2 and by 3) is false.