16. Read the following conjecture. Any number that is divisible by 3 is also divisible by 9. Find a counterexample to show that the conjecture is false. 27 48 45 18
The conjecture states that any number divisible by 3 is also divisible by 9. A counterexample is a number that is divisible by 3 but not by 9.
In this case, the number 48 is divisible by 3 (48 ÷ 3 = 16), but it is not divisible by 9 (48 ÷ 9 = 5 remainder 3). Therefore, 48 is a counterexample to the conjecture, showing that it is false.
To find a counterexample to show that the conjecture is false, we need to find a number that is divisible by 3 but not by 9. Let's analyze the numbers provided:
27: This number is divisible by 3, and it is also divisible by 9 since 27 divided by 9 is 3.
48: This number is divisible by 3 since the sum of its digits (4 + 8 = 12) is divisible by 3. However, 48 is also divisible by 9 since 48 divided by 9 is 5 remainder 3.
45: This number is divisible by 3 since the sum of its digits (4 + 5 = 9) is divisible by 3. However, 45 is not divisible by 9 since 45 divided by 9 is 5 remainder 0.
18: This number is divisible by 3 and it is also divisible by 9 since 18 divided by 9 is 2.
So, out of the given numbers, the counterexample to show that the conjecture is false is 45, as it is divisible by 3 but not by 9.
To determine if the conjecture is false, we need to find a number that is divisible by 3 but not divisible by 9.
First, let's understand what it means for a number to be divisible by 3. A number is divisible by 3 if the sum of its digits is divisible by 3.
Now let's check if each number provided is divisible by 3:
27: The sum of the digits 2 and 7 is 9, which is divisible by 3. So, 27 is divisible by 3.
48: The sum of the digits 4 and 8 is 12, which is divisible by 3. So, 48 is divisible by 3.
45: The sum of the digits 4 and 5 is 9, which is divisible by 3. So, 45 is divisible by 3.
18: The sum of the digits 1 and 8 is 9, which is divisible by 3. So, 18 is divisible by 3.
From the given numbers, it seems that all of them are divisible by 3. Therefore, we have not found any counterexamples to disprove the conjecture.
To show that the conjecture is false, we need a number that is divisible by 3 but not divisible by 9. Let's try a counterexample:
Counterexample: 33
To check if 33 is divisible by 3, we need to determine if the sum of its digits is divisible by 3. The sum of the digits 3 and 3 is 6, which is divisible by 3. So, 33 is divisible by 3.
However, to check if 33 is divisible by 9, we need to determine if the sum of its digits is divisible by 9. The sum of the digits 3 and 3 is 6, which is not divisible by 9. So, 33 is not divisible by 9.
Therefore, the number 33 serves as a counterexample to disprove the given conjecture that any number divisible by 3 is also divisible by 9.