14. Read the following conjecture. Any number that is divisible by 3 is also divisible by 6. Find a counterexample to show that the conjecture is false. 21 24 12 18
A counterexample to the conjecture is the number 21. 21 is divisible by 3, but it is not divisible by 6 since 21 divided by 6 equals 3 with a remainder of 3.
To find a counterexample, we need to find a number that is divisible by 3 but not by 6. Let's look at the given numbers:
21 is divisible by 3, but it is not divisible by 6 (since 21 รท 6 = 3 remainder 3).
24 is divisible by both 3 and 6.
12 is divisible by both 3 and 6.
18 is divisible by both 3 and 6.
Therefore, the counterexample to the conjecture is 21 since it is divisible by 3 but not by 6.
To understand the given conjecture, let's first define what it means for a number to be divisible by another number. A number "a" is divisible by a number "b" if the remainder of the division a/b is zero. In other words, when you divide a by b, there is no remainder left.
According to the given conjecture, any number divisible by 3 should also be divisible by 6. So, we need to find a counterexample, which is a number divisible by 3 but not divisible by 6, to show that the conjecture is false.
Let's examine the numbers provided: 21, 24, 12, and 18. We need to determine if they are divisible by 3 and 6 or not.
Starting with 21, when we divide it by 3, we get 7 with no remainder. This means 21 is divisible by 3. However, when we divide 21 by 6, we get 3 and a remainder of 3. Since there is a remainder, 21 is not divisible by 6.
Moving on to 24, when we divide it by 3, we get 8 with no remainder. This shows that 24 is divisible by 3. And when we divide 24 by 6, we also get 4 with no remainder. Therefore, 24 is divisible by 3 and 6.
Next, considering 12, when we divide it by 3, we get 4 with no remainder. Thus, 12 is divisible by 3. Also, when we divide 12 by 6, we get 2 with no remainder, indicating that 12 is divisible by 6.
Lastly, looking at 18, when we divide it by 3, we get 6 with no remainder. Similar to the previous numbers, this demonstrates that 18 is divisible by 3 as well. Moreover, when we divide 18 by 6, we get 3 with no remainder, meaning 18 is divisible by 6.
Out of the numbers provided (21, 24, 12, and 18), none of them are counterexamples to the conjecture. All of them are divisible by 3, and the ones divisible by 3 are also divisible by 6. Therefore, the given conjecture that any number divisible by 3 is also divisible by 6 is actually true based on these examples.